250 pounds = how many tons

The inequalities are solved and the operations does not change

Given data ,

a)

Let the inequality be 4 < 7

Multiplying both sides by 7: 47 < 77, which simplifies to 28 < 49.

Multiplying both sides by 6: 628 < 649, which simplifies to 168 < 294.

Multiplying both sides by 3: 3168 < 3294, which simplifies to 504 < 882.

Multiplying both sides by 10: 10504 < 10882, which simplifies to 5040 < 8820.

The inequality sign remained "<" (less than), and the inequality was still true. Multiplying by a positive number does not change the direction of the inequality

b)

Let the inequality be 11 > -2

Adding 5 to both sides: 11 + 5 > -2 + 5, which simplifies to 16 > 3.

Adding 3 to both sides: 16 + 3 > 3 + 3, which simplifies to 19 > 6.

Adding (-4) to both sides: 19 + (-4) > 6 + (-4), which simplifies to 15 > 2.

The inequality sign remained ">" (greater than), and the inequality was still true. Adding a positive number to both sides of an inequality does not change the direction of the inequality

c)

Let the inequality be -4 < -2

Subtracting 6 from both sides: -4 - 6 < -2 - 6, which simplifies to -10 < -8.

Subtracting 8 from both sides: -10 - 8 < -8 - 8, which simplifies to -18 < -16.

Subtracting 2 from both sides: -18 - 2 < -16 - 2, which simplifies to -20 < -18.

The inequality sign remained "<" (less than), and the inequality was still true. Subtracting a positive number from both sides of an inequality does not change the direction of the inequality

d)

Let the inequality be -8 < 8

Dividing both sides by -4: -8/-4 < 8/-4, which simplifies to 2 < -2.

Dividing both sides by -2: 2/-2 < -2/-2, which simplifies to -1 < 1.

The inequality sign remained "<" (less than), but the inequality is not true anymore. Dividing both sides of an inequality by a negative number changes the direction of the inequality.

Hence , the inequalities are solved

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Answer:

68 1/4 lbs

Step-by-step explanation:

its just the answer

There is no such vector-valued function r(t) that satisfies the given condition.

I. There is no such vector ū that satisfies u + ū = (1, 2) because if we add any vector to u, we change its direction and magnitude, but the vector ū should have the same magnitude and opposite direction to u.

II. Let v = (1, 3, -1). Then u + v = (2+1, 1+3, 3-1) = (3, 4, 2) and u x v = (-5, 7, -1). So u x v = u + v. Therefore, v satisfies the given condition.

III. There is no such vector-valued function r(t) because if r(t) = (x(t), y(t)) satisfies the condition that r'(t) = 2r(t), then we have:

x'(t) = 2x(t) and y'(t) = 2y(t)

These are two separate first-order differential equations, which have general solutions of the form:

x(t) = c1e^(2t) and y(t) = c2e^(2t)

where c1 and c2 are constants determined by the initial conditions. However, the vector-valued function r(t) = (x(t), y(t)) cannot have a constant magnitude because its components are functions of t that grow exponentially with time. Therefore, there is no such vector-valued function r(t) that satisfies the given condition.

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The first term of the geometric series is 16.

Using the formula for the sum of a geometric series to solve for the first term (a₁):

Sn = a₁(1 - rⁿ)/(1 - r)

Substituting the given values, we get:

-26,240 = a₁(1 - (-3)⁸)/(1 - (-3))

Simplifying the exponent and denominator, we get:

-26,240 = a₁(1 - 6,561)/(4)

-26,240 = a₁(-6,560/4)

-26,240 = a₁(-1,640)

Dividing both sides by -1,640, we get:

a₁ = 16

Therefore, the first term of the geometric series is 16.

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For a group of 10 students where exactly 3 fail the class, the probability of any given student failing is p = 0.2.

For a group of 8 students where at least 7 can pass the class, the probability of any given student passing is p = 0.941.

To solve for each N and appropriate p:

a. In a group of 10 exactly 3 are fail the class:

Let N be the total number of students in the group, and let p be the probability of a student failing the class. Then we want to find the probability of exactly 3 students failing, which can be expressed as:

P(X = 3) = (N choose 3) * p^3 * (1-p)^(N-3)

where (N choose 3) is the number of ways to choose 3 students out of N, p^3 is the probability that those 3 students fail, and (1-p)^(N-3) is the probability that the remaining N-3 students pass.

We can solve for p by setting P(X = 3) equal to the given probability of exactly 3 students failing, which is:

P(X = 3) = (10 choose 3) * p^3 * (1-p)^7 = 0.1176

Simplifying this expression and solving for p, we get:

p^3 * (1-p)^7 = 0.0028
p = 0.2



b. In a group of 8 at least 7 can pass the class:

Let N be the total number of students in the group, and let p be the probability of a student passing the class. Then we want to find the probability of at least 7 students passing, which can be expressed as:

P(X >= 7) = 1 - P(X < 7) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6))

where P(X = k) is the probability of exactly k students passing.

We can solve for p by setting each of these probabilities equal to the given probabilities of at least 7 students passing, which is:

P(X >= 7) = P(X = 7) + P(X = 8) = (8 choose 7) * p^7 * (1-p) + p^8 = 0.3713

Simplifying this expression and solving for p using a calculator or computer software, we get:

p = 0.941

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If the smaller hole is very narrow and the larger one is very loose, a size which the diameter of the hole could have that best fits the requirements is: C. 7/8 inch.

In Mathematics and Geometry, a measurement can be defined as an act or process through which the size, weight, magnitude, quantity, volume (capacity), dimensions, or distance traveled by a physical object or body is taken, especially for the purpose of an experiment.

Assuming small holes of 3/4 and 1 inch in diameters are made on a metal plate, and we realize that the 3/4 inch hole is very narrow and the 1 inch hole is very loose, a standard measurement and size that the diameter would be an intermediate value is given by;

3/4 = 12/16

12/16, 13/16, 14/16, 15/16, ......, 1

Required size = 14/16 = 7/8 inch.

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Complete Question:

Two holes with diameters of 3/4 inch and 1 inch, respectively, have been drilled on a metal sheet. If the smaller hole is very narrow and the larger one is very loose, what size could the diameter of the hole have that best fits the requirements?

The composition of function is g ( x ) f ( x ) = 5x⁴ - 40x

Given data ,

Let the first function be g ( x ) = x³ - 8

Let the second function be h ( x ) = 5x

So, the composition of functions , g(x) x h(x) would be:

(x³ - 8) (5x)

We can distribute the multiplication to get:

5x ( x³ ) - 5x ( 8 )

On simplifying , we get

A = 5x⁴ - 40x

Hence , the result of the operation g(x) x h(x) is 5x⁴ - 40x

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The polygon VWXY is reflection across y = 2. Therefore, option D is the correct answer.

Given that, polygon VWYZ with vertices V at (-11, 2), W at (-11, 0), Y at (-5, 0) and Z(-5, 2).

A second polygon V' at (7, 2), W' at (7, 0), Y'(1, 0) and Z'(1, 2).

We know that, the reflection of point (x, y) across the y-axis is (-x, y).

The polygon VWXY is reflection across y = 2

Therefore, option D is the correct answer.

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A=68,B=44

the interior angle of triangle is 180

A+B+C=180

x+x-24+68=180

2x+44=180

2x=180-44

2x=136

x=68

A=x=68

B=68-24=44

The value of x is 12.

we have, <1 = 42 degree

<2 = 4x degree

As, <1 and <2 are Complementary Angle.

So, the sum of <1 and <2 will be 90 degree

<1 + <2 = 90

42 + 4x = 90

4x = 90-42

4x = 48

x= 48/4

x= 12

Thus, the value of x is 12.

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The area of the triangle in this problem is given as follows:

A = 479.3 m².

First we must use the law of sines to obtain the measure of angle A as follows:

sin(A)/31 = sin(105º)/50

Hence:

sin(A) = 31 x sine of 105 degrees/50

sin(A) = 0.5989

A = arcsin(0.5989)

A = 36.8º.

Considering that the sum of the measures of the internal angles of a triangle is of 180º, the measure of angle B is given as follows:

36.8º + <B + 105º = 180º

<B = 180 - (36.8 + 105)

<B = 38.2º.

We have two sides of 50m and 31m, with an angle between them of 38.2º, hence the area of the triangle is given as follows:

A = 0.5 x 50 x 31 x sine of 38.2 degrees

A = 479.3 m².

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The 99% confidence interval for the average number of hours a student spends studying for a statistics exam is (2.059, 5.941) hours.

To find a 99% confidence interval for the average number of hours a student spends studying for a statistics exam, we can use the formula:

CI = x ± z*(o/sqrt(n))

where CI is the confidence interval, x is the sample mean, z is the z-score for the desired confidence level (in this case, 99%), o is the population standard deviation, and n is the sample size.

Plugging in the given values, we get: CI = 4 ± 2.576*(6.25/sqrt(50)) CI = 4 ± 1.941 CI = (2.059, 5.941)

Therefore, the 99% confidence interval for the average number of hours a student spends studying for a statistics exam is (2.059, 5.941) hours.

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When rolling a six-sided die, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. Event B is rolling an even number, which means there are three possible outcomes: 2, 4, or 6. Since there are only three possible outcomes in Event B, it is a simple event.
Let's analyze the event and outcomes when rolling a six-sided die.

Event B: Rolling an even number.

Step 1: Identify the possible outcomes.
When rolling a six-sided die, the outcomes are numbers 1 to 6 (1, 2, 3, 4, 5, and 6).

Step 2: Determine the outcomes in the event.
For Event B, the even numbers are 2, 4, and 6. So, there are 3 outcomes in Event B.

Step 3: Decide if the event is a simple event or not.
An event is considered a simple event if it has only one possible outcome. Since Event B has 3 possible outcomes (2, 4, and 6), it is not a simple event.

Event B (rolling an even number) has 3 possible outcomes (2, 4, and 6), and it is not a simple event.

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By simplifying the expression, (4 − 3)( + 5) we have 5 as the final answer.

Simplification in mathematics is the process of reducing a mathematical expression into a simpler or more compact form, without changing the value of the expression.

The process of simplification are:

Simplification is often used to make calculations and problem-solving easier and more efficient, and to help identify patterns and relationships between mathematical concepts. Simplification is an important skill in mathematics, as it can help to make complex problems more accessible and understandable, and can also lead to more elegant and insightful solutions.

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The expected value E(Y) = 5.6, Var(Y) = 326.64 and σ ≈ 18.07.

(a) To compute the expected value E(Y), we need to multiply each value of y with its corresponding probability and sum them up. Here's the calculation:
E(Y) = (2 * 0.10) + (4 * 0.40) + (7 * 0.20) + (8 * 0.30) = 0.2 + 1.6 + 1.4 + 2.4 = 5.6
(b) To compute the variance Var(Y) and standard deviation σ, we first need to find E(Y^2) and then use the formula Var(Y) = E(Y^2) - (E(Y))^2.
E(Y^2) = (2^2 * 0.10) + (4^2 * 0.40) + (7^2 * 0.20) + (8^2 * 0.30) = 4 + 64 + 98 + 192 = 358
Now, we can compute Var(Y):
Var(Y) = E(Y^2) - (E(Y))^2 = 358 - (5.6)^2 = 358 - 31.36 = 326.64
Finally, we can compute the standard deviation σ:
σ = sqrt(Var(Y)) = sqrt(326.64) ≈ 18.07

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Answer:

When a coin is tossed 9 times the probability of getting exactly six heads is obtained with the help of Bernoulli trials. The probability of getting exactly six heads is 21/128.

Step-by-step explanation:

Area of the circular end  x height = volume

pi r^2  * h    = volume        r = 1/2 * 7 = 3.5 yds

pi * (3.5^2) * 12 =~ 462 yd^3

This statement is not necessarily true.

A real matrix can have real or complex eigenvalues.

A real matrix can have real or complex eigenvalues. The number of real eigenvalues of a matrix may be even or odd, and there is no general rule that determines the parity of the number of real eigenvalues.

For example, the 5x5 real matrix

```
[ 0  1  0  0  0 ]
[ 1  0  0  0  0 ]
[ 0  0  0  1  0 ]
[ 0  0  1  0  0 ]
[ 0  0  0  0  1 ]
```

has two real eigenvalues (1 and -1), which is an even number.

On the other hand, the 5x5 real matrix

```
[ 1  1  0  0  0 ]
[ 1  1  0  0  0 ]
[ 0  0  1  1  0 ]
[ 0  0  1  1  0 ]
[ 0  0  0  0  0 ]
```

has three real eigenvalues (2, 0, and -1), which is an odd number.

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We can expect Camilla to have about 11 aces in a game where she attempts 25 serves.

If Camilla serves an ace 44% of the time on average, then we can expect her to serve an ace in 44 out of every 100 serves.

We can use this information to estimate how many aces she is likely to have in a game where she attempts 25 serves:

Expected number of aces = (44/100) × 25

Expected number of aces = 11

Therefore, we can expect Camilla to have about 11 aces in a game where she attempts 25 serves.

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Answer:

√(c+9)² =√64

c+9 =8

c=8-9

c=-1

The probability of having factor of 72 and even will be 0.75.

The prime factorization of 72 is 2^3 * 3^2. It should be noted that to find the number of factors of 72, one must add 1 to each exponent in the prime factorization and multiply accordingly: (3+1) * (2+1) = 12.

Therefore the factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.

Evidently, 9 of these 12 factors will be even since they contain at least one factor of 2; thus, 6 of the factors of 72 are even.

The probability will be:

= 9/12

= 0.75

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Overall, while the use of volunteers does not automatically make the study invalid, it is important to carefully consider probability of bias and take steps to minimize them in the study design and analysis.

The use of volunteers in this study does not necessarily make it invalid, but it could introduce potential sources of bias. Volunteers may not be representative of the larger population of individuals with painful diabetic neuropathy, as they may be more willing to participate in the study or may have different levels of disease severity or other factors that could affect their response to treatment.

Additionally, the random assignment of participants to the gabapentin or placebo group helps to minimize bias, but there is still a possibility of confounding variables that could influence the results. For example, if individuals who were assigned to the gabapentin group were more likely to adhere to the treatment regimen or had more social support, this could influence the outcome of the study independent of the drug's effectiveness.

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Answer:

$63

Step-by-step explanation:

First find the cost of bananas per kilogram. To find that, divide $147 by 21

147/21=7

One kilogram of bananas costs $7

Now that you know the cost per kilogram, multiply $7 by the number of kilograms asked in the question, which in this case is 9

7*9=$63

9 kilograms of bananas costs $63

Answer: 63

Step-by-step explanation:

You can set up a proportion cost/weight=cost/weight

[tex]\frac{147}{21} = \frac{x}{9}[/tex]        Costs go on the top so x is cost for 9 kg

[tex]x=\frac{147*9}{21}[/tex]

x=63

Answer:

the answer is B

Step-by-step explanation:

you find the answer by exchanging and canceling the number to find the value of x

[tex]/ \frac{x}{3} \leqslant - 6 \\ (\frac{1x}{3}) \frac{3}{1} \leqslant - 6( \frac{3}{1} ) \\ x = - 18[tex]/

The optimal dimensions for the two adjacent rectangular corrals are: length = 80 feet, width = 20 feet.

Let's call the dimensions of the two adjacent rectangular corrals "width" and "length," and represent them by the variables w and l, respectively. The total length of fencing available is 160 feet, so the total amount of fencing used is:

P = 2w + 3l

The factor of 3l in this equation is due to the fact that there are three sides of length l: two of them are shared by the two corrals, and the other one is the outer side of one of the corrals.

The enclosed area of the two corrals is:

A = 2wl

To find the dimensions that maximize the enclosed area, we need to express A in terms of a single variable, either w or l. To do this, we can solve the equation P = 160 for one of the variables:

2w + 3l = 160

2w = 160 - 3l

w = 80 - (3/2)l

Substituting this expression for w into the equation for A, we get:

A = 2(80 - (3/2)l)l

Simplifying and expanding, we get:

A = 160l - (3/2)l^2

This is a quadratic function of l, with a maximum at the vertex of the parabola. The x-coordinate of the vertex is given by:

l = -b / (2a)

where a = -(3/2) and b = 160. Substituting these values, we get:

l = -160 / (2 * -(3/2)) = 80

Therefore, the optimal length of the two corrals is 80 feet. To find the corresponding width, we can use the equation we derived earlier:

w = 80 - (3/2)l = 80 - (3/2)(80) = 20

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She is incorrect with the shading as the part being shaded is actually 30 degree Fahrenheit and not 20 degree Celsius.Therefore, she need to readjusted the shading to 20 degree Celsius.

The scale is used to measure temperature where freezing point of water is 0 degrees Celsius and the boiling point of water is 100 degrees Celsius at standard atmospheric pressure. The negative values indicate temperatures below freezing point.

Therefore, if Kaylon shaded the thermometer to represent a temperature of 20 degrees below zero Celsius, then the shaded area should be below the zero mark on the thermometer which indicates a temperature that is lower than the freezing point of water.

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To find the intersection point of the two constraints of x1 and x2, we need to solve the system of inequalities simultaneously.

First, we can rewrite each inequality in slope-intercept form:

1. 9x1 + 7x2 ≥ 57   ->   x2 ≥ (-9/7)x1 + 57/7
2. 4x1 + 6x2 ≥ 3   ->   x2 ≥ (-2/3)x1 + 1/2

The intersection point will occur where the two lines intersect, so we can set the two equations equal to each other:

(-9/7)x1 + 57/7 = (-2/3)x1 + 1/2

Simplifying:

(-9/7 + 2/3)x1 = 1/2 - 57/7

(-15/21)x1 = -163/42

x1 = (163/42)/(15/21)

x1 = 1.634

To find x2, we can substitute this value back into either of the original equations:

9(1.634) + 7x2 = 57

7x2 = 57 - 14.706

x2 = 6.32

Therefore, the values of x1 and x2 where the two constraints intersect are x1 = 1.634 and x2 = 6.32.

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The graph of h(x) = 7cos(12x) + 1 has a period of π/6, an amplitude of 7, and a midline of y = 1. The first point is (0,1) and the second is (π/24, 8). The graph resembles a cosine curve with peaks and valleys.

To graph the function h(x) = 7cos(12x) + 1 using the sine tool, we can follow these steps

The period of a cosine function is 2π/|b|, where b is the coefficient of x. In this case, b = 12, so the period is 2π/|12| = π/6.

The amplitude of a cosine function is the absolute value of the coefficient of the cosine term. In this case, the coefficient of the cosine term is 7, so the amplitude is 7.

The midline of a cosine function is the vertical line y = c, where c is the constant term. In this case, the constant term is 1, so the midline is y = 1.

The first point on the midline is (0, 1).

Find the second point on the graph closest to the first point

The cosine function has a maximum value of amplitude + midline = 7 + 1 = 8 and a minimum value of amplitude - midline = 7 - 1 = 6.

The maximum value closest to the first point occurs at x = π/24, and the minimum value closest to the first point occurs at x = 5π/24. Therefore, the second point on the graph closest to the first point is (π/24, 8).

Plot the two points and complete the graph using the sine tool

Plot the two points (0, 1) and (π/24, 8) on the coordinate plane. Then, use the sine tool to complete the graph by drawing a curve that starts at the first point, passes through the second point, and repeats every π/6 units.

The graph of h(x) = 7cos(12x) + 1 should have a period of π/6, an amplitude of 7, and a midline of y = 1.

The first point on the midline is (0, 1), and the second point closest to the first point is (π/24, 8). The completed graph should resemble a cosine curve with peaks at x = π/24, π/12, 5π/24, 7π/24, 3π/8, 11π/24, and so on, and valleys at x = 0, π/6, π/4, 5π/24, π/3, 7π/24, and so on.

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Mr. Simon can serve 26 people with 20 containers of soup.

Given that: Mr.Simon has 20 containers of soup. 3 containers of soup feeds 4 people.

Frst, we need to know how many people can be fed by one container of soup.

We are given that 3 containers of soup can feed 4 people.

Hence, we can calculate how many people one container of soup can feed by dividing 4 people by 3 containers

One person ⇒  4/3

Next, we multiply the number of people per container by the total number of containers, which is 20 in this case.

This gives us:

= 20 × (4/3)

= 80/3

= 26.667

≈ 26

Mr. Simon can feed 80/3 people with 20 containers of soup. However, since we cannot serve a fractional part of a person, we should round the answer to a whole number. In this case, rounding down gives us 26.

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The probability that of three random songs, the first two are Country and the third is R&B is 0.024. In percentage, we say  2.4%.

Let's denote the event that a song is from a certain genre as

R for Rock,

C for Country,

RB for R&B, and

Cl for Classical.

The total song = 12 + 15 + 10 + 3 = 40 songs

Probability that the first song is from Country is :

P(C₁) = 15/40

Probability that the second song is also from Country is:

P(C2|C1) = 14/39

Note: This is a probability selection without replacement that is why the total songs reduced from 40 to 39

Probability that the third song is R&B is:

P(RB₃|C₁C₂) = 10/38 (there are 10 R&B songs left out of a total of 38 remaining songs)

To find the probability that the first two songs are Country and the third song is R&B, we multiply these probabilities together:

P(C₁C₂RB₃) = P(C1) × P(C₂|C₁) × P(RB₃|C₁C₂)

= (15/40) × (14/39) × (10/38)

= 0.024

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