The statement is false as it **improperly **applies the law of quadratic reciprocity without **providing **the necessary parameters.

(a) False. The law of **quadratic **reciprocity states a relationship between two odd prime numbers p and q. It states that the Legendre symbol (p/q) is equal to (q/p) under certain conditions. In this case, (17) does not represent a valid Legendre symbol because it lacks the second parameter. Therefore, the statement is false.

(b) False. The statement claims that if a is a quadratic residue of an odd prime p, then -a is **also **a quadratic residue of p. However, this is not always true. Quadratic residues are the **values **that satisfy the quadratic congruence x^2 ≡ a (mod p). If a is a quadratic residue, it means there exists an x such that x^2 ≡ a (mod p). However, if we consider -a, it may or may not have a **corresponding **x such that x^2 ≡ -a (mod p). Hence, the statement is false.

(c) True. If ab ≡ r (mod p), where r is a quadratic residue of an odd prime p, then a and b are both **quadratic **residues of p. This statement is valid because the product of two quadratic residues modulo an odd prime will always result in another quadratic residue. Therefore, if r is a quadratic residue and ab is congruent to r modulo p, then both a and b must also be quadratic residues.

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For each probability and percentile problem, draw the picture. A random number generator picks a number from 1 to 8 in a uniform manner. Part (a) Give the distribution of X.

Part (b) Part (c) Enter an exact number as an integer, fraction, or decimal. f(x) = ____, where ____

Part (d) Enter an exact number as an integer, fraction, or decimal. μ = ___

Part (e) Round your answer to two decimal places. σ = ____

Part (f) Enter an exact number as an integer, fraction, or decimal. P(3.75 < x < 7.25) = ____

Part (g) Round your answer to two decimal places. P(x > 4.33) =____ Part (h) Enter an exact number as an integer, fraction, or decimal. P(x > 5 | x > 3) =____ Part (i) Find the 90th percentile. (Round your answer to one decimal place.)

To answer the given **probability **and **percentile **problems, let's go through each part step by step.

(a) The distribution of X is a discrete uniform distribution with values ranging from 1 to 8, inclusive.

(b) The probability mass function (PMF) is given by:

f(x) = 1/8 for x = {1, 2, 3, 4, 5, 6, 7, 8}; 0 otherwise

(c) The PMF is:

f(x) = 1/8, where x = {1, 2, 3, 4, 5, 6, 7, 8}

(d) The mean (μ) is the average of the values in the distribution, which in this case is:

μ = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) / 8

= 4.5

(e)The standard deviation (σ) is a measure of the dispersion of the values in the distribution. For a discrete uniform distribution, it can be calculated using the formula:

σ = [tex]\sqrt{{((n^2 - 1) / 12)\\} }[/tex], where n is the number of values in the distribution.

In this case, n = 8, so:

σ =[tex]\sqrt{ ((8^2 - 1) / 12)\\}[/tex]

= [tex]\sqrt{(63 / 12)}[/tex]

≈ 2.29

(f) To find the probability of a specific range, we need to calculate the cumulative probability for the lower and upper bounds and subtract them.

P(3.75 < x < 7.25) = P(x < 7.25) - P(x < 3.75)

Since the distribution is discrete, we round the bounds to the nearest whole number:

P(x < 7.25) = P(x ≤ 7)

= 7/8

P(x < 3.75) = P(x ≤ 3)

= 3/8

P(3.75 < x < 7.25) = (7/8) - (3/8)

= 4/8

= 1/2

= 0.5

(g) To find the probability of x being greater than a specific value, we need to calculate the cumulative probability for that value and subtract it from 1.

P( > 4.33) = 1 - P(x ≤ 4)

= 1 - 4/8

= 1 - 1/2

= 1/2

= 0.5

(h) To find the **conditional** probability of x being greater than 5 given that x is greater than 3, we calculate:

P(x > 5 | x > 3) = P(x > 5 and x > 3) / P(x > 3)

Since the condition "x > 3" is already satisfied, we only need to consider the probability of x being greater than 5:

P(x > 5 | x > 3) = P(x > 5)

= 1 - P(x ≤ 5)

= 1 - 5/8

= 3/8

= 0.375

(i) The percentile represents the value below which a given percentage of **observations **falls.

To find the 90th percentile, we need to determine the value x such that 90% of the observations fall below it.

For a discrete uniform **distribution**, each value has an equal probability, so the 90th percentile corresponds to the value at the 90th percentile rank.

Since the distribution has 8 values, the 90th percentile rank is:

90th percentile rank = (90/100) * 8

= 7.2

Since the values are discrete, we round up to the nearest whole number:

90th percentile ≈ 8

Therefore, the 90th percentile is 8 (rounded to one decimal place).

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Solve the system. Give answers as (x, y, z)

6x-3y-5z= -21

12x+3y-4z= 12

-24x + 3y + 1z = -9

Therefore, the **solution **of the system is (x, y, z) = (-5/3, -10.067, -2.8).

(x, y, z) = (-5/3, -10.067, -2.8).

The given system of **linear **equations is 6x - 3y - 5z = -21, 12x + 3y - 4z = 12 and -24x + 3y + z = -9.

To solve the system, we'll use elimination method to find the values of x, y, and z:1.

Multiply the first equation by 2:6x - 3y - 5z = -2112x - 6y - 10z = -42

Adding both equations will eliminate y and z:18x = -30x = -30/18x = -5/32.

Substituting the value of x in the first and third equation will eliminate y:-24(-5/3) + 3y + z = -9-40 + 3y + z = -9

→ 3y + z = 31 ... (i)6(-5/3) - 3y - 5z = -21-10 + 3y + 5z = 21

→ 3y + 5z = 31 ... (ii)From (i) and (ii), we have:

3y + z = 31 ... (i)

3y + 5z = 31 ... (ii)

Multiplying (i) by -5 and adding to (ii) will eliminate

y:3y + z = 31 ... (i)-15y - 5z = -155z = -14z = 14/-5z = -2.8

Substituting z = -2.8 and x = -5/3 in the second **equation **will give y:-24(-5/3) + 3y - 2.8 = -9 40 + 3y - 2.8 = -9 3y = -30.2y = -10.067

Therefore, the solution of the system is (x, y, z) = (-5/3, -10.067, -2.8).

(x, y, z) = (-5/3, -10.067, -2.8).

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Let r be a primitive root of the odd prime p. Prove the following:

If p = 3 (mod4), then -r has order (p - 1)/2 modulo p.

Let r be a primitive root of the odd prime p.

Then, r has order (p - 1) modulo p.

This indicates that $r^{p-1} \equiv 1\pmod{p}$.

Therefore, $r^{(p-1)/2} \equiv -1\pmod{p}$.

Also, we can write that $(p-1)/2$ is an odd **integer**.

As p is 3 (mod 4), we can say that $(p-1)/2$ is an odd integer.

For example, when p = 7, (p-1)/2 = 3.

Let's consider $(-r)^{(p-1)/2} \equiv (-1)^{(p-1)/2} \cdot r^{(p-1)/2} \pmod{p}$;

as we know, $(p-1)/2$ is odd, we can say that $(-1)^{(p-1)/2} = -1$.

Therefore, $(-r)^{(p-1)/2} \equiv -1 \cdot r^{(p-1)/2} \equiv -1 \cdot (-1) = 1 \pmod{p}$.

This shows that the order of $(-r)^{(p-1)/2}$ modulo p is (p-1)/2.

As $(-r)^{(p-1)/2}$ has order (p-1)/2 modulo p, then -r has order (p-1)/2 modulo p.

This completes the proof.

The word "**modulus**" has not been used in the solution as it is a technical term in number theory and it was not necessary for this proof.

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For the matrix A shown below, x = (0, 1,-1) is an eigenvector corresponding to a second order eigenvalue X. Use x to find X. Hence determine a vector of the form y = (1, a, b) such that x and y form an orthogonal basis for the subspace spanned by the eigenvectors coresponding to eigenvalue X. 1 2 2 A = 1 2 -1 -1 1 4 Enter your answers as follows: If any of your answers are integers, you must enter them without a decimal point, e.g. 10 • If any of your answers are negative, enter a leading minus sign with no space between the minus sign and the number. You must not enter a plus sign for positive numbers. If any of your answers are not integers, then you must enter them with at most two decimal places, e.g. 12.5 or 12.34, rounding anything greater or equal to 0.005 upwards. Do not enter trailing zeroes after the decimal point, e.g. for 1/2 enter 0.5 not 0.50. These rules are because blackboard does an exact string match on your answers, and you will lose marks for not following the rules. Your answers: a: b:

For the dot product to be zero, a must be equal to b. So, we can choose a = b , a vector y of the form (1, a, a) will form an **orthogonal basis** with x.

To find the** eigenvalue** corresponding to the eigenvector x = (0, 1, -1), we need to solve the equation Ax = Xx, where A is the given matrix. Substituting the values, we have:

A * (0, 1, -1) = X * (0, 1, -1)

Simplifying, we get:

(2, -1, 1) = X * (0, 1, -1)

From the **equation**, we can see that the second component of the vector on the left side is -1, while the second component of the vector on the right side is X. Therefore, we can conclude that X = -1.

To find a **vector** y = (1, a, b) that forms an orthogonal basis with x, we need y to be orthogonal to x. This means their dot product should be zero. The dot product of x and y is given by:

x · y = 0 * 1 + 1 * a + (-1) * b = a - b

For the dot product to be zero, a must be equal to b. So, we can choose a = b. a vector y of the form (1, a, a) will form an orthogonal basis with x.

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Use the information below to find the probability that a flight arrives on time given that it departed on time.

The probability that an airplane flight departs on time is 0.890

The probability that a flight arrives on time is 0.87

The probability that a flight departs and arrives on time is 0.83

The probability that a flight arrives on time given that it departed on time is.......

Therefore, the **probability **that a flight arrives on time given that it departed on time is approximately 0.932.

To find the probability that a flight arrives on **time **given that it departed on time, we can use the formula for conditional probability:

P(Arrival on time | Departure on time) = P(Arrival on time and Departure on time) / P(Departure on time)

From the given information, we have:

P(Arrival on time and **Departure **on time) = 0.83

P(Departure on time) = 0.890

Plugging these values into the formula, we get:

P(Arrival on time | Departure on time) = 0.83 / 0.890 ≈ 0.932

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problem 1: let's calculate the average density of the red supergiant star betelgeuse. betelgeuse has 16 times the mass of our sun and a radius of 500 million km. (the sun has a mass of 2 × 1030 kg.)

The **average density **of the red supergiant star **Betelgeuse** is 1.45 × 10⁻¹¹ kg/m³.

To **calculate** the average density of the red supergiant star Betelgeuse,

we need to use the formula for average density, which is:

Average density = **Mass**/VolumeHere,

Betelgeuse has 16 times the mass of our sun.

Therefore, its mass (M) is given by:

M = 16 × (2 × 10²³) kg

M = 32 × 10²³ kg

M = 3.2 × 10²⁴ kg

Betelgeuse has a radius (r) of 500 million km.

We need to **convert **it to meters:r = 500 million

km = 500 × 10⁹ m

The volume (V) of Betelgeuse can be calculated as:

V = 4/3 × π × r³V = 4/3 × π × (500 × 10⁹)³

V = 4/3 × π × 1.315 × 10³⁵V = 2.205 × 10³⁵ m³

Therefore, the average density (ρ) of Betelgeuse can be calculated as:

ρ = M/Vρ = (3.2 × 10²⁴) / (2.205 × 10³⁵)

ρ = 1.45 × 10⁻¹¹ kg/m³

Thus, the average density of the red supergiant star Betelgeuse is 1.45 × 10⁻¹¹ kg/m³.

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show working out clearly

A. Given the function f(x) = x(3x - x²). Determine: i. The critical value/s; ii. The nature of the critical point/s. (4 marks) (6 marks)

The function f(x) = x(3x - x²) can be written as f(x) = 3x² - x³, and we will find its** critical value/s** and the nature of the critical point/s.i).

To find the critical value/s, we need to find the **derivative **of the function: `f'(x) = 6x - 3x²`. Now we need to solve for x to get the critical values:`f'(x) = 0`Solving for x, we get:`6x - 3x² = 0`Factorizing, we get:`3x(2 - x) = 0`So the critical values are x = 0 and x = 2.ii) To find the nature of the critical points, we can use the second derivative test. We know that `f''(x) = 6 - 6x`.Substituting x = 0, we get:`f''(0) = 6 - 0 = 6`Since `f''(0) > 0`, the function has a local minimum at x = 0.Substituting x = 2, we get:`f''(2) = 6 - 12 = -6`Since `f''(2) < 0`, the function has a local **maximum **at x = 2.Therefore, the critical values are x = 0 and x = 2, and the nature of the critical points is a local minimum at x = 0 and a local maximum at x = 2.

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Solve the equation and in the answer sheet write down the sum of

the roots of the equation.

Solve the equation of the equation. 5x-2 x²+3x-1 3 4 = -1 and in the answer sheet write down the sum of the roots

The given equation is 5x - 2x² + 3x - 1/3 + 4 = -1 . The sum of the roots of the **quadratic equation **ax² + bx + c = 0. The sum of the roots of the equation is 4.

Step by step answer:

Step 1: Rearrange the equation5x - 2x² + 3x + 1/3 + 4 + 1 = 0 Multiplying the whole **equation **by 3, we get,15x - 6x² + 9x + 1 + 12 + 3 = 0

Step 2: **Simplify **the equation-6x² + 24x + 16 = 0 Dividing the whole equation by -2, we get,3x² - 12x - 8 = 0

Step 3: Find the roots of the quadratic equation

3x² - 12x - 8

= 0ax² + bx + c

= 0x

= [-b ± √(b² - 4ac)] / 2a

Here, a = 3,

b = -12,

c = -8x

= [12 ± √(12² - 4(3)(-8))] / 2(3)x

= [12 ± √216] / 6x

= [12 ± 6√6] / 6x

= 2 ± √6

Therefore, the **roots **of the quadratic equation are 2 + √6 and 2 - √6

Step 4: Find the **sum **of the roots The sum of the roots of the quadratic equation ax² + bx + c = 0 is given by the formula, Sum of roots = -b/a Here,

a = 3 and

b = -12

Sum of roots = -b/a= -(-12) / 3

= 4

Hence, the sum of the roots of the equation is 4.

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The graph of a polynomial function is shown, State the interval(s) on which is increasing and the interval(s) on which is decreasing. (Enter your answers using interval notation)

increasing____

decreasing____

In the graph of a polynomial function shown below, it is required to determine the interval(s) on which it is increasing and the **interval**(s) on which it is decreasing. **Polynomial Function **Graph The solution can be found by determining the turning points of the polynomial function.

**Turning points **are points where the polynomial changes direction. This means that if we can determine the x-values of these turning points, we can identify the intervals of increasing and decreasing of the polynomial function.

The turning points of the polynomial function can be found by identifying the **roots **of its derivative. The roots of the derivative indicate the values of x where the function changes from increasing to decreasing or decreasing to increasing.

Thus, we differentiate the polynomial function to obtain its derivative.

f(x) = 2x³ - 3x² - 12x + 20

Differentiating both sides with respect to x gives;

f'(x) = 6x² - 6x - 12

Setting f'(x) equal to zero and solving for x yields: 6x² - 6x - 12 = 0

Factoring out 6 from the expression on the left gives;

6(x² - x - 2) = 0

Factorizing x² - x - 2 gives;

(x - 2)(x + 1) = 0

The roots of the equation are;`

[tex]x - 2 = 0 or x + 1 = 0[/tex]

Thus, the roots of the **derivative **are [tex]`x = 2` and `x = -1`[/tex]. Therefore, the polynomial function has two turning points at [tex]x = 2 and x = -1.[/tex]

The intervals of increasing and decreasing of the polynomial function can now be identified as shown below;*Interval of Decrease: [tex]`(-∞, -1) ∪ (2, ∞)[/tex]`*Interval of Increase:[tex]`(-1, 2)`[/tex]

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find the radius of convergence, r, of the series.[infinity](−9)nnnxnn = 1

The **radius of convergence**, r, of the series is 1/9.

To obtain the radius of convergence, we can use the **ratio test**.

The ratio test states that if we have a **power series** of the form ∑(aₙxⁿ), then the radius of convergence, r, is given by:

r = lim┬(n→∞)|aₙ/aₙ₊₁|

In this case, we have the series ∑((-9)ⁿⁿ/n!)xⁿ.

Let's apply the ratio test to find the radius of convergence.

We start by evaluating the ratio:

|aₙ/aₙ₊₁| = |((-9)ⁿⁿ/n!)xⁿ / ((-9)ⁿ⁺¹⁺¹/(n+1)!)xⁿ⁺¹|

= |-9ⁿ⁺¹⁺¹xⁿ / (-9)ⁿⁿ⁺¹ xⁿ⁺¹(n+1)/n!|

Simplifying the expression:

|aₙ/aₙ₊₁| = |(-9)(n+1)/(n+1)|

= 9

Taking the limit as n approaches infinity:

lim┬(n→∞)|aₙ/aₙ₊₁| = 9

Since the **limit** is a finite positive number (9), the radius of convergence is given by:

r = 1 / lim┬(n→∞)|aₙ/aₙ₊₁| = 1/9

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Table 1 shows scores given to 4 sessions by a network intrusion detection system. The "True Label" column gives the ground truth (i.e., the type each session actually is). Sessions similar to the attack signature are expected to have higher scores while those dissimilar are expected to have lower scores. Draw an ROC curve for the scores in Table 1. Clearly show how you computed the ROC points. Assume "Attack" as the positive ('p') class.

Table 1. Intrusion detector's scores and corresponding "true" labels.

Session No. Score True Label

1

0.1

Normal

2

0.5

Attack

3

0.6

Attack

4

0.7

Normal

The ROC **Curve **can be used to evaluate the performance of the binary classifier that differentiates two classes.

The ROC Curve is generated by plotting the **True Positive Rate** (TPR) against the False Positive Rate (FPR) for a range of threshold settings.

The ROC Curve is a good way to visually evaluate the sensitivity and specificity of the binary classifier.

The ROC Curve is a graphical representation of the binary classifier's true-positive rate (TPR) versus its false-positive rate (FPR) for various classification **thresholds**.

The ROC Curve is often utilized to evaluate the sensitivity and specificity of binary classifiers. Since an ROC Curve can only be produced for binary classifiers, it is not appropriate for classifiers with more than two classes.

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Which of the following statements is true? Los enlaces sencillos se forman compartiendo dos electrones Single bonds are made by sharing two electrons. Un enlace covalente se forma a través de la transferencia de electrones de un átomo a otro. A covalent bond is formed through the transfer of electrons from one atom to another. No es posible que dos átomos compartan más de dos electrones, formando enlaces multiples. It is not possible for two atoms to share more than two electrons, in a multiple bond. Un par de electrones involucrados en un enlace covalente a veces se conocen como "pares solitarios A pair of electrons involved in a covalent bond are sometimes referred to as "lone pairs."

The statement "Single **bonds** are made by sharing two **electrons**" is true.

In a **covalent** bond, atoms share electrons to achieve a stable electron configuration. A single **bond** is formed when two atoms share a pair of electrons. This means that each atom contributes one electron to the shared pair, resulting in a total of two electrons being shared between the atoms.

The statement "A covalent bond is formed through the transfer of electrons from one atom to another" is false. In a covalent bond, there is no transfer of electrons between atoms. Instead, the electrons are shared.

The statement "It is not possible for two atoms to share more than two electrons, in a multiple bond" is also false. In a multiple bond, such as a double or triple bond, atoms can share more than two electrons. In a double bond, two **pairs** of electrons are shared (four electrons in total), and in a triple bond, three pairs of electrons are shared (six electrons in total).

The statement "A pair of electrons involved in a covalent bond are sometimes referred to as 'lone pairs'" is true. In a covalent bond, there are two types of electron pairs: **bonding** pairs, which are involved in the formation of the bond, and lone pairs, which are not involved in bonding and are localized on one atom. These lone pairs play a role in the shape and properties of molecules.

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show that the vectors ⟨1,2,1⟩,⟨1,3,1⟩,⟨1,4,1⟩ do not span r3 by giving a vector not in their span

It is not possible to find a **vector in R3** that cannot be written as a linear combination of ⟨1,2,1⟩,⟨1,3,1⟩, and ⟨1,4,1⟩.

It is required to show that the vectors ⟨1,2,1⟩,⟨1,3,1⟩,⟨1,4,1⟩ do not span R3 by providing a vector that is not in their** span. **Here is a long answer of 200 words:The given vectors are ⟨1,2,1⟩,⟨1,3,1⟩, and ⟨1,4,1⟩, and it is required to prove that they do not span R3.

The span of vectors is the set of all** linear combinations **of these vectors, which can be written as the following:Span {⟨1,2,1⟩, ⟨1,3,1⟩, ⟨1,4,1⟩} = {a ⟨1,2,1⟩ + b ⟨1,3,1⟩ + c ⟨1,4,1⟩ | a, b, c ∈ R}where R represents real numbers.To show that the given vectors do not span R3, we need to find a vector in R3 that cannot be written as a linear combination of ⟨1,2,1⟩,⟨1,3,1⟩, and ⟨1,4,1⟩.Suppose the vector ⟨1,0,0⟩, which is a three-dimensional vector, is not in the span of the given vectors.

Now, we need to prove it.Let the vector ⟨1,0,0⟩ be the linear combination of ⟨1,2,1⟩,⟨1,3,1⟩, and ⟨1,4,1⟩.⟨1,0,0⟩ = a⟨1,2,1⟩ + b⟨1,3,1⟩ + c⟨1,4,1⟩Taking dot products of the above equation with each of the given vectors, we get,⟨⟨1,0,0⟩, ⟨1,2,1⟩⟩ = a⟨⟨1,2,1⟩, ⟨1,2,1⟩⟩ + b⟨⟨1,3,1⟩, ⟨1,2,1⟩⟩ + c⟨⟨1,4,1⟩, ⟨1,2,1⟩⟩⟨⟨1,0,0⟩, ⟨1,2,1⟩⟩ = a(6) + b(8) + c(10)1 = 6a + 8b + 10c

Similarly,⟨⟨1,0,0⟩, ⟨1,3,1⟩⟩ = 7a + 9b + 11c⟨⟨1,0,0⟩, ⟨1,4,1⟩⟩ = 8a + 11b + 14cNow, we have three equations and three unknowns.

Solving these equations simultaneously, we **geta** = 1/2, b = -1/2, and c = 0

The vector ⟨1,0,0⟩ can be expressed as a linear combination of ⟨1,2,1⟩ and ⟨1,3,1⟩, which implies that it is not possible to find a vector in R3 that cannot be written as a linear combination of ⟨1,2,1⟩,⟨1,3,1⟩, and ⟨1,4,1⟩.

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4. (20) In two jars (jar-1, jar-2) containing black and white balls, the probability of drawing a white ball from jar-1 is equal to drawing a black ball from jar-2. The balls are drawn according to the following rules: • The balls are drawn without replacement (i.e. the ball drawn is put back to the jar). • If a black ball is drawn, the next ball is drawn from the other jar. Else the next ball is drawn from the same jar. If an is the probability of having nth draw from jar-1 (a) (10) Prove that an+1 equals drawing a black ball from jar-2 (b) (10) If the first ball is drawn from jar-1, what is the probability of drawing 1000th ball from jar-1?

(a) an+1 = probability of drawing a black ball from jar-2 (b) The **probability** of drawing the 1000th ball from jar-1, given that the first ball was drawn from jar-1, is the same as the probability of drawing a white ball from jar-1.

(a) To prove that an+1 equals **drawing** a black ball from jar-2, we can analyze the different possibilities for the nth draw:

1. If the nth draw is from jar-1 and a white ball is drawn, then an+1 will be equal to an (drawing from jar-1 again).

2. If the nth draw is from jar-1 and a black ball is drawn, then an+1 will be equal to the probability of drawing a black ball from jar-2 (since the next draw will be from jar-2).

3. If the nth draw is from jar-2 and a white ball is drawn, then an+1 will be equal to the probability of drawing a **white ball** from jar-1 (since the next draw will be from jar-1).

4. If the nth draw is from jar-2 and a black ball is drawn, then an+1 will be equal to an (drawing from jar-2 again).

Based on these possibilities, it can be concluded that an+1 equals drawing a black ball from jar-2.

(b) If the first ball is drawn from jar-1, the probability of drawing the 1000th ball from jar-1 can be **calculated** as the product of probabilities for each draw. Since the balls are drawn with replacement (put back after each draw), the probability of drawing a ball from jar-1 remains the same for each draw. Therefore, the probability of drawing the 1000th ball from jar-1 is the same as the probability of drawing the first ball from jar-1, which is given as the probability of drawing a white ball from jar-1.

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All vectors and subspaces are in R". Check the true statements below: A. If W is a subspace of R" and if v is in both W and W, then v must be the zero vector. B. In the Orthogonal Decomposition Theorem, each term y=y.u1/u1.u1 u1 +.... + y.up/up.up up is itself an orthogonal projection of y onto a subspace of W.

C. If y = 21 + 22, where 2₁ is in a subspace W and z2 is in W, then 2₁ must be the orthogonal projection of Y onto W. D. The best approximation to y by elements of a subspace W is given by the vector y – projw(y). E. If an n x p matrix U has orthonormal columns, then UUT x = x for all x in R".

A. The **statement **given is **true**.

This is because if v is in both W and W, then it must be the zero vector.

B. The statement given is also true. In the **Orthogonal Decomposition Theorem**, each term

y=y.u1/u1.u1 u1 +.... + y.up/up.up up is itself an orthogonal projection of y onto a subspace of W. C.

The best approximation to y by elements of a subspace W is given by the vector y – projw(y).E. If an n x p matrix U has orthonormal columns, then UUT x = x for all x in R".The summary of the answers are:A is true.B is true.C is false.D is true.E is true.

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find a power series representation for the function. (give your power series representation centered at x = 0.) f(x)=1/(3 x)

The **power series **representation for the function is [tex]f(x) = \sum\limits^{\infty}_{0} {(-\frac x3)^n}[/tex]

From the question, we have the following parameters that can be used in our computation:

f(x) = 1/(3 + x)

Rewrite the **function **as

[tex]f(x) = \frac{1}{3(1 + \frac x3)}[/tex]

Expand

[tex]f(x) = \frac{1}{3(1 - - \frac x3)}[/tex]

So, we have

[tex]f(x) = \frac{1}{3} * \frac{1}{(1 - (-\frac x3)}[/tex]

The power series centered at x = 0 can be calculated using

[tex]f(x) = \sum\limits^{\infty}_{0} {r^n}[/tex]

In this case

r = -x/3 i.e. the expression in bracket

So, we have

[tex]f(x) = \sum\limits^{\infty}_{0} {(-\frac x3)^n}[/tex]

Hence, the **power series **for the **function **is [tex]f(x) = \sum\limits^{\infty}_{0} {(-\frac x3)^n}[/tex]

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**Question**

Find a power series representation for the function. (give your power series representation centered at x = 0

f(x) = 1/(3 + x)

find the standardized test statistic estimate, z, to test the hypothesis that p1 > p2. use 0.01. the sample statistics listed below are from independent samples.

sample statistics: n1 = 100, x1 = 38, and n2 = 140, x2 = 50 a.0.638 b.0.362 c.2.116 d.1.324 100, 38, and 140, 50

Therefore, the** standardized test statistic** estimate (z) is approximately 0.323. None of the given answer choices (a. 0.638, b. 0.362, c. 2.116, d. 1.324) match the calculated value.

To find the standardized test statistic estimate (z) to test the hypothesis that p₁ > p₂, we can use the following **formula**:

z = (p₁ - p₂) / √(p * (1 - p) * (1/n₁ + 1/n₂))

where:

p₁ = x₁ / n₁ (proportion in sample 1)

p₂= x₂/ n₂(proportion in sample 2)

n₁ = sample size of sample 1

n₂ = sample size of **sample **2

Given:

n₁ = 100, x₁ = 38

n₂ = 140, n₂ = 50

First, we need to calculate p1 and p2:

p₁ = 38 / 100

= 0.38

p₂ = 50 / 140

= 0.3571 (approximately)

Next, we can calculate the standardized test statistic estimate (z):

z = (0.38 - 0.3571) / √( (0.38 * 0.62) * (1/100 + 1/140) )

z = 0.0229 / √(0.2368 * (0.0142 + 0.0071))

z = 0.0229 / √(0.2368 * 0.0213)

z = 0.0229 / √(0.00503504)

z ≈ 0.0229 / 0.07096

z ≈ 0.323

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A rental car company charges $40 plus 15 cents per each mile driven. Part1. Which of the following could be used to model the total cost of the rental where m represents the miles driven. OC=1.5m + 40 OC= 0.15m + 40 OC= 15m + 40 Part 2. The total cost of driving 225 miles is, 10 9 8 7 6 5 4 3 2 Member of People ILI 16-20 21-25 28-30 31-33 A frisbee-golf club recorded the ages of its members and used the results to construct this histogram. Find the number of members 30 years of age or younger

The total cost of driving 225 miles is $73.75. The given **histogram **is as follows: From the histogram, we can see that the number of members 30 years of age or younger is 12. Therefore, the correct answer is 12.

A rental car company charges $40 plus 15 cents per mile driven.

Part 1. Which of the following could be used to model the total cost of the **rental **where m represents the miles driven?OC=0.15m + 40

The given information tells us that a rental car company charges $40 plus 15 cents per mile driven. Here, m represents the miles driven.

Thus, the option that could be used to model the total cost of the rental where m represents the miles driven is:

OC = 0.15m + 40.

Part 2. The total **cost **of driving 225 miles isOC = 0.15m + 40 (given)

Now, we have to find the cost of driving 225 miles.

Thus, we have to put the value of m = 225 in the above equation.OC = 0.15m + 40OC = 0.15 × 225 + 40OC = 33.75 + 40OC = $73.75

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Write the system of linear equations in the form Ax = b and solve this matrix equation for x. -2x1 3x2 -11 6x1 + X2 H -39 CHCE =

The given system of** linear equations** is as follows:-2x1 + 3x2 = -11 (Equation 1)6x1 + x2 = -39 (Equation 2)To write the above system of linear equations in the form Ax = b.

we can represent it as given below:

A = [ -2 3 ; 6 1 ]

x = [ x1 ; x2 ]

b = [ -11 ; -39 ]

Therefore, Ax = b becomes [ -2 3 ; 6 1 ] [ x1 ; x2 ] = [ -11 ; -39 ]Now, to solve this matrix equation, we need to find the **inverse **of matrix A. Let A^-1 be the inverse of matrix A, then we can write x = A^-1 b

So, first we find the determinant of matrix A using the **formula**: Determinant of

A = (ad - bc)

where, a = -2, b = 3, c = 6 and d = 1.So, Determinant of A = (-2)(1) - (3)(6) = -20

As the determinant is not equal to zero, the inverse of matrix A exists. Now, we find the inverse of matrix A using the formula: A^-1 = (1/Determinant of A) [ d -b ; -c a ]where, a = -2, b = 3, c = 6 and d = 1.So, A^-1 = (1/-20) [ 1 -3 ; -6 -2 ]= [ -1/20 3/20 ; 3/10 1/10 ]

Now, we can find the solution to the given **system **of linear equations as follows:

x = A^-1 b= [ -1/20 3/20 ; 3/10 1/10 ] [ -11 ; -39 ]

= [ 2 ; -5 ]

Therefore, the solution to the given system of linear equations isx1 = 2 and x2 = -5.

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Compute the following limit using L'Hospital's rule if appropriate. Use INF to denote oo and MINF to denote -oo.

lim x -> [infinity] (1 - 4/x)^x =

To compute the limit of the function (1 - 4/x)^x as x approaches infinity, we can apply **L'Hôpital's rule**.

Let's rewrite the function as:

f(x) = (1 - 4/x)^x

Taking the** natural logarithm** of both sides:

ln(f(x)) = ln[(1 - 4/x)^x]

Using the property ln(a^b) = b * ln(a):

ln(f(x)) = x * ln(1 - 4/x)

Now, we can find the limit of ln(f(x)) as x approaches infinity:

**lim x -> infinity ln(f(x))** = lim x -> infinity x * ln(1 - 4/x)

This is an indeterminate form of infinity times zero. We can apply L'Hôpital's rule by taking the derivative of the** numerator** and **denominator**:

lim x -> infinity ln(f(x)) = lim x -> infinity [ln(1 - 4/x) - (x * (-4/x^2))] / (-4/x)

Simplifying the expression:

lim x -> infinity ln(f(x)) = lim x -> infinity [ln(1 - 4/x) + 4/x] / (-4/x)

As x approaches infinity, both ln(1 - 4/x) and 4/x approach 0:

lim x -> infinity ln(f(x)) = lim x -> infinity [0 + 0] / 0

This is an indeterminate form of 0/0. We can apply L'Hôpital's rule again by taking the derivative of the numerator and denominator:

lim x -> infinity ln(f(x)) = lim x -> infinity [(d/dx ln(1 - 4/x)) + (d/dx 4/x)] / (d/dx (-4/x))

Differentiating each term:

lim x -> infinity ln(f(x)) = lim x -> infinity [(-4/(x - 4)) * (-1/x^2) + (-4/x^2)] / (4/x^2)

Simplifying the expression:

lim x -> infinity ln(f(x)) = lim x -> infinity [4/(x - 4x) - 4] / (4/x^2)

As x approaches infinity, (x - 4x) becomes -3x:

lim x -> infinity ln(f(x)) = lim x -> infinity [4/(-3x) - 4] / (4/x^2)

Simplifying further:

lim x -> infinity ln(f(x)) = lim x -> infinity [-4/(3x) - 4] / (4/x^2)

Taking the limit as x approaches infinity, the terms with x in the denominator approach 0:

lim x -> infinity ln(f(x)) = [-4/(3 * infinity) - 4] / 0

Simplifying:

lim x -> infinity ln(f(x)) = (-4/INF - 4) / 0 = (-4/INF) / 0 = 0/0

Once again, we have an indeterminate form of 0/0. We can apply **L'Hôpital's rule** one more time:

lim x -> infinity ln(f(x)) = lim x -> infinity [(d/dx (-4/(3x))) + (d/dx -4)] / (d/dx 0).

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\Use the chain rule to find the partial derivatives w = xy + yz + zx, x = rcose, y = rsine, z = r0,- , when r = 2,0 = = aw aw ar' de Q3(c). A rectangular box without a lid to be made from 12m² of cardboard. Find the maximum volume of such a box.

To find the maximum volume of a rectangular box made from 12m² of cardboard, we need to maximize the **volume function **subject to the constraint that the **surface area** is equal to 12m².

Let's denote the length, width, and height of the box as x, y, and z, respectively. The volume of the box is given by V = xyz. According to the given information, the surface area of the box is 12m², which gives us the constraint equation 2xy + 2xz + 2yz = 12. To find the maximum volume, we can use the method of **Lagrange multipliers**. We define the Lagrangian function L(x, y, z, λ) as the volume function V minus the constraint equation multiplied by a Lagrange multiplier λ:

L(x, y, z, λ) = xyz - λ(2xy + 2xz + 2yz - 12)

Next, we need to find the** partial derivatives** of L with respect to x, y, z, and λ, and set them equal to zero to find the critical points.

∂L/∂x = yz - 2λy - 2λz = 0

∂L/∂y = xz - 2λx - 2λz = 0

∂L/∂z = xy - 2λx - 2λy = 0

∂L/∂λ = 2xy + 2xz + 2yz - 12 = 0

Solving this system of equations will give us the **critical points**. From there, we can determine which point(s) correspond to the maximum volume. Once we find the critical points, we substitute their values into the volume function V = xyz to calculate the corresponding volumes. The largest volume among these points will be the maximum volume of the box. By comparing the volumes obtained at the critical points, we can determine the maximum volume of the rectangular box that can be made from 12m² of cardboard.

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n 3n2 + n. 2. For every integer n > 1, prove that Σ(6i – 2) 1=1

**Answer:**

Here the answer

**Step-by-step explanation:**

Hope you get it

Assume you select seven bags from the total number of bags the farmers collected. What is the probability that three of them weigh between 86 and 91 lbs.

4.3.8 For the wheat yield distribution of exercise 4.3.5 find

A. the 65th percentile

B. the 35th percentile

Assuming that the seven bags are selected randomly, we can use the binomial **probability **distribution.

The binomial distribution is used in situations where there are only two possible outcomes of an experiment and the probabilities of success and failure remain constant throughout the experiment.

.Using the standard normal distribution table, we can find that the z-score corresponding to the 65th percentile is approximately 0.385. We can use the **formula **z = (x - μ) / σ to find the value of x corresponding to the z-score. Rearranging the formula, we get:x = zσ + μ= 0.385 * 80 + 1500≈ 1530.8Therefore, the 65th percentile is approximately 1530.8 lbs.B.

To find the 35th percentile, we can follow the same steps as above. Using the standard normal distribution table, we can find that the z-score corresponding to the 35th percentile is approximately -0.385. Using the formula, we get:x = zσ + μ= -0.385 * 80 + 1500≈ 1469.2Therefore, the 35th percentile is approximately 1469.2 lbs.

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let , be vectors in given by a) find a vector with the following properties: for any linear transformation which satisfies we must have . enter the vector in the form

If the result is zero, then we need to choose another **vector** and repeat the process. Therefore, we choose any non-zero vector and apply T to it.

Given, vectors , are given as:

We need to find a vector such that for any linear transformation T satisfying we must have , i.e.,

Here, is the** null space** of the linear transformation T.

Let us first find the basis for the null space of T.

Let be the matrix representing the **linear transformation** T with respect to the standard basis.

Since the columns of A represent the** images** of the standard basis vectors under T, the null space of A is precisely the space of all linear combinations of the vectors that map to zero.

Therefore, we can find a basis for the null space of A by computing the reduced row echelon form of A and looking for the special solutions of the corresponding homogeneous system.

Now, we need to find a vector which is not in the null space of T.

This can be done by taking any **non-zero vector **and applying T to it. If the result is non-zero, then we have found our vector.

If the result is zero, then we need to choose another vector and repeat the process.

Therefore, we choose any non-zero vector and apply T to it.

Let . Then,

Since this is non-zero, we have found our vector. Therefore, we can take as our vector.

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Please take your time and answer both questions. Thank

you!

3. List the possible rational zeros of f. Then determine all the real zeros of f. f(x) = 15x³ - 26x² + 13x - 2 4. Solve for x: log x + log (x + 3)

The possible **rational **zeros of f are ±1/3, ±2/3, ±1/5, ±2/5, ±1/15, and ±2/15. The real zeros of f are x = 1/3 and x = 2/5.

To find the possible rational zeros of f, we use the Rational Root Theorem. According to the **theorem**, the possible rational zeros are of the form p/q, where p is a factor of the constant term (-2) and q is a factor of the leading **coefficient** (15). The factors of -2 are ±1 and ±2, while the factors of 15 are ±1, ±3, ±5, and ±15. Combining these factors, we get the possible rational zeros ±1/3, ±2/3, ±1/5, ±2/5, ±1/15, and ±2/15.

To determine the real zeros of f, we need to solve the equation f(x) = 0. One way to do this is by factoring. However, in this case, factoring the cubic equation may not be straightforward. Alternatively, we can use numerical methods such as graphing or the **Newton-Raphson** method. Using graphing or a graphing calculator, we can observe that the function crosses the x-axis at approximately x = 1/3 and x = 2/5. These are the real zeros of f.

In summary, the possible rational zeros of f are ±1/3, ±2/3, ±1/5, ±2/5, ±1/15, and ±2/15. After evaluating the function or graphing it, we find that the real zeros of f are x = 1/3 and x = 2/5. These values satisfy the equation f(x) = 0. Therefore, the solution to the equation log x + log (x + 3) is x = 1/3 and x = 2/5.

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Suppose x and y are positive real numbers. If x < y, then x^2 < y^2. Prove the statement using the method of direct proof.

Given that x and y are positive real numbers and x < y, we have to prove that x² < y² by direct proof. Method of direct proof Let P and Q are statements. To prove P → Q by the direct proof, we assume that P is true. Then we use only logic and the given information to prove that Q is true. It is also called a proof by deduction. Now, let's begin the proof. Assume that x < y, where x and y are positive real numbers. Squaring both sides, we get$x^2 < y^2$Therefore, it is proved that x² < y² by direct proof.

Hence, we have proved that if x < y, then x² < y² using the **method of direct proof**.

To prove the **statement **"If x < y, then x² < y²" using a direct proof, we will assume the premise that x < y and then show that x² < y².

Let's proceed with the direct proof:

Assumption: x < y

To prove: x² < y²

Proof:

Since x < y, we can multiply both sides of the inequality by x and y, respectively, without changing the inequality direction because both x and y are positive:

x * x < x * y (multiplying both sides by x)

y * x < y * y (multiplying both sides by y)

Simplifying the **inequalities**:

x² < xy

yx < y²

Since x < y, we know that xy < y² because multiplying a smaller number by y will result in a smaller product than multiplying y by itself.

Combining the two inequalities:

x² < xy < y²

Therefore, x² < y²

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Which of the following statements is true about arithmetic sequence?

A. a sequence having a common ratio

C. a sequence having a common difference

B. a sequence which is always finite

D. a sequence which is always infinite

The correct statement about an **arithmetic sequence **is:

C. a sequence having a common difference

What is an arithmetric sequence

An** arithmetic sequence **is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is often referred to as the "common difference." For example, in the arithmetic sequence 2, 5, 8, 11, 14, the common difference is 3, as each term is obtained by adding 3 to the previous term.

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A rectangular pond has a width of 50m and a length of 400m. The area of the pond covered by an alga is denoted by A (in mm²) and is measured at time t (in weeks) after a biologist begins to observe the growth. The rate at which A is changing can be modelled as be modelled as being proportional to √Ā. Initially the algae cover an area of 900m² and three weeks later this has increased to 1296m². How many days after the initial observation will it take for the algae to cover more than 10% of the pond's surface?

To determine the number of days it will take for the algae to cover more than 10% of the pond's **surface**, we need to find the relationship between the **area** covered by the algae and time.

The rate of change of the area is proportional to the square root of the area. By setting up a **differential equation** and solving it, we can find the time required for the algae to exceed 10% of the pond's surface area.

Let A(t) represent the area covered by the algae at time t. According to the problem, the rate of change of A is **proportional** to √A. This can be expressed as dA/dt = k√A, where k is the constant of proportionality.

We know that initially, A(0) = 900 m², and after three weeks, A(3) = 1296 m².

To find the value of k, we can substitute the given values into the differential equation:

dA/dt = k√A

√A dA = k dt

**Integrating** both sides, we have:

(2/3)[tex]A^(3/2)[/tex] = kt + C

Using the initial condition A(0) = 900, we can solve for C:

(2/3)[tex](900)^(3/2)[/tex] = k(0) + C

C = (2/3)[tex](900)^(3/2)[/tex]

Now we can solve for the time when the algae covers more than 10% of the pond's **surface area**, which is 0.10 * (50m * 400m) = 2000 m²:

(2/3)[tex]A^(3/2)[/tex] = kt + (2/3)[tex](900)^(3/2)[/tex]

Solving for t, we find the number of days it will take for the algae to exceed 10% of the pond's surface area.

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Let X be a continuous random variable with the probabilty density function; f(x) = kx 0

To determine the value of the constant k in the probability density **function **(PDF) f(x) = kx^2, we need to integrate the PDF over its entire range and set the result equal to 1, as the total area under the PDF must equal 1 for a valid** probability **distribution.

The given PDF is defined as:

f(x) = kx^2, 0 < x < 1

To find k, we integrate the PDF over its **range**:

∫[0,1] kx^2 dx = 1

Using the **power rule **for integration, we have:

k∫[0,1] x^2 dx = 1

**Integrating **x^2 with respect to x gives:

k * (x^3/3) | [0,1] = 1

Plugging in the limits of** integration,** we have:

k * (1^3/3 - 0^3/3) = 1

Simplifying, we get:

k/3 = 1

Therefore, k = 3.

Hence, the value of the **constant** k in the PDF f(x) = kx^2 is k = 3.

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In Exercises 17-18, use the method of Example 6 to compute the matrix A¹0 0 17. A = 0 3

2 -1

18. A = 1 0

-1 2

The method of Example 6 is the diagonalization of a **matrix**. For diagonalization of a matrix, we need to find the eigenvalues and eigenvectors of the matrix.

Once we have the eigenvalues and eigenvectors, we can construct the diagonal matrix from the eigenvalues and the matrix of eigenvectors. Then, we can write the matrix as the product of the matrix of eigenvectors, diagonal matrix, and the inverse of the matrix of **eigenvectors**. Exercise 17Let A = 0 3 2 -1

To find the eigenvalues of A, we need to solve the characteristic equation

|A - λI| = 0So,

we have |0 - λ 3 2 -1 - λ| = 0 ⇒ λ² + λ - 6 = 0

On solving this quadratic equation,

we get λ₁ = 2 and λ₂ = -3

Now, we need to find the eigenvectors of A corresponding to these eigenvalues.

For λ = 2, we get(A - 2I)X

= 0⇒(0-2 3 2-2)X = 0⇒-2x₁ + 3x₂

= 0 and 2x₁ - 2x₂ = 0Or, x₁ = (3/2)x₂ Let x₂

= 2, then x₁ = 3

Now, the eigenvector corresponding to

λ = 2 is[3 2]TFor

λ = -3, we get(A + 3I)X = 0⇒(0+3 3 2+3)X

= 0⇒3x₁ + 3x₂ = 0 and 3x₁ + 5x₂ = 0Or,

x₁ = -x₂ Let x₂ = 1, then x₁ = -1Now, the eigenvector **corresponding **to λ = -3 is[-1 1]T So, we have D = 2 0 0 -3andP = 3 -1 2 1

Diagonalizing the matrix A, we get A = PDP⁻¹A = 3 -1 2 1 0 3 2 -1 = 1/6 [9 -3] [-2 6] [2 2] [-1 -1] [3 0] [-2 -2]Multiplying A and [1 0 0; 0 0 1; 0 1 0], we getA¹0 0 17 = 1/6 [9 -3] [-2 6] [2 2] [-1 -1] [3 0] [-2 -2] × [1 0 0; 0 0 1; 0 1 0] = 1/6 [9 0 3] [-2 0 2] [2 17 2] [-1 0 -1] [3 0 -2] [-2 0 -2]

Therefore, A¹0 0 17 = 1/6 [9 0 3] [-2 0 2] [2 17 2] [-1 0 -1] [3 0 -2] [-2 0 -2]Exercise 18Let A = 1 0 -1 2To find the eigenvalues of A, we need to solve the characteristic **equation **|A - λI| = 0So, we have |1 - λ 0 -1 2 - λ| = 0 ⇒ (1 - λ)(2 - λ) = 0⇒ λ₁ = 1 and λ₂ = 2.

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Newhard Company assigns overhead cost to jobs on the basis of 117% of direct labor cost. The job cost sheet for Job 313 includes $19,681 in direct materials cost and $10,700 in direct labor cost. A total of 1,650 units were produced in Job 313. Required: a. What is the total manufacturing cost assigned to Job 313? b. What is the unit product cost for Job 313? a. Total manufacturing cost b. Unit product cost
Short-term scheduling is important to efficiency and to cost reduction, but its impact is not of strategic importance. True/False?
If a firm can sell a product, but in order to sell that product the selling price will be less than its variable costs, then the sale should be made because at least the fixed costs will be covered. TRUE OR FALSEAn investment of $1,000 with annual benefits of $150 per year for the first five years of its life and $100 per year the next five years of its life has a payback period of?A. 7 yearsB. 7.5 yearsC. 8 yearsD. none of the aboveOne thousand dollars invested today at 5 percent per year, compounded annually, for five years will be wortha. $1,050b. $1,250c. $1,276d. None of the aboveIf fixed costs are $500,000, the selling price is $10/unit and variable cost is $6/unit, then breakeven in dollars isa. $5,000,000b. $125,000c. $1,250000d. None of the above
Rapid Growth Pty Ltd is presently operating at full capacity. They received a special order that, if accepted, would require refusing some sales to regular customers. Which of the following factors should management consider when making their decision?1) Variable costs of special order2) Fixed costs of equipment3) Opportunity costsSelect one:a. All 3 factorsb. 1 and 2 but NOT factor 3c. 1 and 3 but NOT factor 2d. Only factor 1e. 2 and 3 but NOT factor 1
In the Nowhere Land a "4 out of 16" lottery is very popular. Each ticket costs $2 and contains numbers from 1 through 16. Participants need to choose 4 numbers. If all their numbers are winning, they receive $100; if three out of 4 are winning, they receive $40; if 2 out of 4 are winning, they get $2. Otherwise, they get nothing. Should one play this lottery? In other words, what is the average winning if the cost of the ticket is taken into account?
Consider the well failure data given below. (a) What is the probability of a failure given there are more than 1,000 wells in a geological formation? (b) What is the probability of a failure given there are fewer than 500 wells in a geological formation? Wells Geological Formation Group Gneiss Granite Loch raven schist Total 1685 28 3733 Failed 170 443 14 Marble Prettyboy schist Other schists Serpentine 1403 39
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Need help for belowquestion and explain which criminology theory or theories apply toit.What is yourpersonal theory on why people commit crime?Thank you somuch
Let f(n) = n + 1. Find f(3), f(0), f(-3) Is f a one-to-one function from the set of integers to the set of integers? Is f an onto function from the set of integers to the set of integers? (Explain the reasons behind your answers).
Section 3 1- Purchase & sales Stock (Jan) Capital Bank overdraft Cash Discounts Return inwards Return outwards Carriage outwards Rent & insurance Provision for doubtful debts Fixtures & fittings Delivery van Debtors & creditors Wages & salaries General office expenses Drawings Additional Information 1- The year end stock is RM 429 2- The wages and salaries accrued is RM 210. Prepare the income statement for the year ended 2021 for the above company. (15m) DR (RM) 22860 5160 90 1440 810 2160 1740 1200 2100 11910 8940 450 2880 CR (RM) 41970 7200 4350 930 570 660 6060
which code segment correctly uses the margin shorthand to assign the values to a div? bottom: 15px top: 10px left: 5px right: 20px
ethanoic acid has a pka of 4.75. find the ph of the solution that results from the addition of 40.0 ml of .040 m naoh to 5.0 ml of .0075 m ethanoic acid
Wells, Inc., has identified an investment project with the following cash flows. Cash Flow 865 1,040 1,290 1,385 Year 1 a. If the discount rate is 8 percent, what is the future value of these cash flows in Year 4?
For summer to occur in the Northern Hemisphere, which scenario is correct? a. the south pole points towards b. the Sun the north and south poles receive equal sunlight c. the north pole points towards the Sun d. none of the above e. the north pole points away from the Sun
Which of the following pressure measurements is the equivalent of 2.50 atm? View Available Hint(s) 253 Pa O 14.7 psi 304 mmHq O 1.90 x 103 torr
Discuss some of the reasons companies go global. Whyare the advantages of comparative advantage only temporary
Problem 6 (10 marks) Consider the polynomial 20 (x-1)" p(x) = n! A=0 For parts a) and b) do not include any factorial notation in your final answers. (a) [3 marks] Determine p(1). p(10 (1) and p(20) (1). (b) [3 marks]Determine the tangent line approximation to p about x = 1. (c) [2 marks]Determine the degree 10 Taylor polynomial of p(x) about x = 1. (d) [2 marks]If possible, determine the degree 30 Taylor polynomial of p(x) about x = 1. Hint: this problem requires no computations.
2 What can you say of the skewness in each of the following cases? (09) i) The median is 60 while the two quartiles are 40 and 80. ii) Mean= 140 and Mode = 140. The first three moments about 16 are respectively -0.35, 2.09 and -1.93. Discuss the various measures or quantities by which the characteristics of frequency (06) distributions are measured and compared. (c) Differentiate between descriptive and inferential statistics. (05) (20)
when x= -1. If y=u and u=2x + 5, find dy = dx x= -1 dx (Simplify your answer.)
Your credit card billing period starts on the 11th of the month. Assume that you carry a $1,000 unpaid credit card balance as of July 11. For the first 15 days of the billing period, your balance remains at $1,000. Then, you make a $600 payment. This reduces the balance on your card to $400 for the remainder of the billing period. Interest is charged at a rate of 18%, compounded daily. How much interest will you be charged using the adjusted balance method?$15.29$4.06$6.12$10.55