please solve number 18
18. Find the average rate of change of f(x) = x² + 3x +/ from 1 to x. Use this result to find the slope of the seca line containing (1, f(1)) and (2, ƒ(2)). 19. In parts (a) to (f) use the following

Brooks Clinic should select Option B, which has the higher NPV of $14,557 as compared to Option A that has an NPV of $2,649.

The steps to calculate the NPV (Net Present Value) of Option A and Option B is explained below:

Calculation of NPV of Option A and Option B using excel function as follows:

Initial Outlay = -$179,000Cost of capital = 5%

Useful life = 7 years

Salvage value = $0

Formula for NPV is as follows:

=NPV(rate, value1, [value2], …)

Where:rate = the company's cost of capital value1, value2, etc. = cash inflows/outflows in each period Option A

Initial Outlay = -$179,000

NPV = $2,649

Option B

Initial Outlay = -$283,000

NPV = $14,557

Therefore, Brooks Clinic should select Option B, which has the higher NPV of $14,557 as compared to Option A that has an NPV of $2,649.

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When x = 4, we have:

[tex]f(4) = 4*4 + 5\\= 16 + 5\\= 21.[/tex]

We can continue this process for all values of x between 1 and 5 to get the mapping shown: Mapping: f(x)1121627336

The total number of elements in A union B, n(A U B) can be obtained by adding the number of elements in set A to the number of elements in set B and then subtracting the number of elements in A intersection B (as they would have been counted twice if we just added n(A) and n(B)).

So we have: [tex]n(A U B) = n(A) + n(B) - n(A ∩ B)[/tex]

Substituting the given values, we have:

[tex]n(A U B) = 14 + 15 - 6\\= 23[/tex]

Thus, there are 23 elements in A union B.

Now, let's draw the mapping with rule:

[tex]f:xx+5[/tex], for [tex]1 ≤ x ≤ 5[/tex] and [tex]x € R.[/tex]

We are given a mapping rule, [tex]f: xx + 5[/tex] for [tex]1 ≤ x ≤ 5[/tex] and [tex]x € R[/tex].

This means that for every value of x between 1 and 5 (inclusive), the function f returns the value of x multiplied by itself and then added to 5.

For example, when x = 2, we have:

[tex]f(2) = 2*2 + 5\\= 4 + 5\\= 9[/tex]

Similarly, when x = 4, we have:

[tex]f(4) = 4*4 + 5\\= 16 + 5\\= 21[/tex]

We can continue this process for all values of x between 1 and 5 to get the mapping shown below:

Mapping:[tex]f(x)1121627336[/tex]

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In a geometric distribution, the mean (denoted as μ) represents the average number of trials required until the first success occurs. In this case, the success corresponds to the terminal having a message ready for transmission.

For a geometric distribution with probability of success p, the mean is given by μ = 1/p. Since the terminal produces messages according to a sequence of independent trials, the probability of success (p) is constant for each trial. Let's denote p as the probability that the terminal has a message ready for transmission. Therefore, the mean of N, denoted as μ, is given by μ = 1/p. The mean value of N represents the average number of times the computer polls the terminal until it receives a message ready for transmission. It provides an estimate of the expected waiting time for the message to be available.

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Division Property for Integers: m = nq + r, 0 ≤ r < n.

The Division Property for Integers states that for any two integers, m and n, where n is greater than 0, there exist two unique integers, q (the quotient) and r (the remainder), satisfying the equation m = nq + r. Additionally, it holds that the remainder, r, is always non-negative (0 ≤ r) and less than the divisor, n (r < n).

To prove this theorem, we can consider the concept of division in terms of repeated subtraction. By subtracting multiples of the divisor, n, from the dividend, m, we can eventually reach a point where further subtraction is no longer possible. At this point, the remaining value, r, is the remainder. The number of times we subtracted the divisor gives us the quotient, q.

The uniqueness of q and r can be established by contradiction. Assuming the existence of two sets of q and r values leads to contradictory equations, violating the uniqueness property.

Therefore, the Division Property for Integers holds, ensuring the existence and uniqueness of the quotient and remainder with specific conditions on their values.

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a. The relation R is reflexive.

b. The relation R is symmetric.

c. The relation R is not transitive.

d. The relation R is not an equivalence relation.

To draw the undirected graph representing the relation R = {(1, 2), (1, 1), (2, 1), (1, 3), (3, 1), (3, 3)}, we can represent each element as a node and draw edges between the nodes based on the pairs in the relation.

The graph representation of the relation R is as follows:

    1 ---- 2

    | \    |

    |  \   |

    |   \  |

    3 ---- 3

a. Reflexive:

A relation is reflexive if every element is related to itself. In this case, we have (1, 1), (2, 2), and (3, 3) in the relation. Since each element is related to itself, the relation R is reflexive.

b. Symmetric:

A relation is symmetric if for every pair (a, b) in the relation, (b, a) is also in the relation. In this case, we have (1, 2) in the relation, but (2, 1) is also present. Similarly, we have (1, 3) in the relation, but (3, 1) is also present. Therefore, the relation R is symmetric.

c. Transitive:

A relation is transitive if for every pair of elements (a, b) and (b, c) in the relation, (a, c) is also in the relation. In this case, we have (1, 2) and (2, 1) in the relation. However, we don't have (1, 1) in the relation. Therefore, the relation R is not transitive.

d. Equivalence relation:

An equivalence relation is a relation that is reflexive, symmetric, and transitive. Since the relation R is not transitive, it is not an equivalence relation.

In summary:

a. The relation R is reflexive.

b. The relation R is symmetric.

c. The relation R is not transitive.

d. The relation R is not an equivalence relation.

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Answer:

a. The relation is not reflexive because (2,2) is not present.

b. The relation is symmetric because for every (a,b) in R, (b,a) is also present.

c. The relation is not transitive because (2,1) and (1,2) are present, but (2,2) is not present.

d. The relation is not an equivalence relation because it fails to satisfy reflexivity and transitivity.

To represent the relation R = {(1,2), (1, 1), (2,1), (1,3), (3, 1), (3,3)} as an undirected graph:

    1 --- 2

   / \   /

  /   \ /

 3 --- 3

a. Reflexivity: A relation R is reflexive if every element in the set is related to itself. In this case, (1,1) and (3,3) are present in the relation, so it is not reflexive since (2,2) is not present.

b. Symmetry: A relation R is symmetric if whenever (a,b) is in R, then (b,a) is also in R. In this case, (1,2) is present, but (2,1) is also present. Similarly, (1,3) is present, but (3,1) is also present. Therefore, the relation is symmetric.

c. Transitivity: A relation R is transitive if whenever (a,b) and (b,c) are in R, then (a,c) is also in R. In this case, we can see that (1,2) and (2,1) are present, but (1,1) is not present. Therefore, the relation is not transitive.

d. Equivalence relation: An equivalence relation is a relation that is reflexive, symmetric, and transitive. Since the relation in question is not reflexive (as discussed in part a) and not transitive (as discussed in part c), it is not an equivalence relation.

Hypothesis Test A statistical test that is used to determine whether there is sufficient evidence to reject a null hypothesis is known as a hypothesis test. The null hypothesis and the alternative hypothesis are two hypotheses used in a hypothesis test.

The null hypothesis and the alternative hypothesis must be stated for the hypothesis test to proceed. The null hypothesis (H0) states that there is no significant difference between a sample statistic and a population parameter. The alternative hypothesis (H1) is the hypothesis that needs to be demonstrated to be true. The alternative hypothesis can be one-tailed or two-tailed. A one-tailed alternative hypothesis specifies a direction, whereas a two-tailed alternative hypothesis specifies that there is a difference. For males, the population mean who support the death penalty is 0.5.Null Hypothesis:H0: µm = 0.5Alternative Hypothesis:

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The car's average speed can be calculated by dividing the distance traveled by the time taken. 544 miles ÷ 8.5 hours = 64 miles per hourTherefore, the car's average speed is 64 miles per hour.

Main answer: The vertices and foci of the given ellipse are (6, 0), (-6, 0) and (3, 0), (-3, 0) respectively.

Explanation: The given equation is 9x2 − 54x + 4y2 = −45.

To find the vertices of the ellipse, we need to divide both sides of the given equation by -45 so that the right side becomes equal to 1.

Then, we need to rearrange the terms so that the x-terms and y-terms are grouped together as follows:

(x2 - 6x)2 / 45 + y2 / 11.25 = 1

From this equation, we can see that a2 = 45/4, b2 = 11.25/4.

The vertices of the ellipse are located at (±a, 0), which gives us (6, 0) and (-6, 0).

To find the foci of the ellipse, we need to use the formula c2 = a2 - b2, where c is the distance from the center to each focus. In this case, we get c2 = 45/4 - 11.25/4 = 33.75/4.

Thus, c = ±sqrt(33.75/4) = ±sqrt(33.75)/2.

The foci of the ellipse are located at (±c, 0), which gives us (3, 0) and (-3, 0).

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To find the partial derivatives [tex]f_x(x, y)[/tex] and [tex]f_y(x, y)[/tex] for f(x, y) = 10 - 2x - 3y + x², we differentiate f(x, y) with respect to x and y, resulting in [tex]f_x(x, y)[/tex]  = -2x + 2 and  [tex]f_y(x, y)[/tex] = -3.

The partial derivative [tex]f_x(x, y)[/tex]  is obtained by differentiating f(x, y) with respect to x while treating y as a constant. Differentiating 10 - 2x - 3y + x² with respect to x yields -2x. Similarly, the partial derivative  [tex]f_y(x, y)[/tex]  is obtained by differentiating f(x, y) with respect to y while treating x as a constant. Since the coefficient of y is -3, differentiating it with respect to y results in -3.

In summary, the partial derivatives of f(x, y) = 10 - 2x - 3y + x² are

[tex]f_x(x, y)[/tex]  = -2x + 2 and  [tex]f_y(x, y)[/tex]  = -3. Since both the partial derivatives are constants and are not equal to zero, the function does not possess any local extrema.

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The statement that is TRUE is C. (-2,-4) is a minimum point of f and (8/3, 16/3) is a maximum point of f. To determine whether a critical point is a minimum, maximum, or saddle point, we can analyze the second-order partial derivatives of the function.

First, we find the first-order partial derivatives with respect to x and y:

∂f/∂x = 3x² - 10x + 4y - 16

∂f/∂y = 4x - 2y

Next, we set these partial derivatives equal to zero to find the critical points. By solving the system of equations:

3x² - 10x + 4y - 16 = 0

4x - 2y = 0

We obtain two critical points: (-2, -4) and (8/3, 16/3).

To determine the nature of these critical points, we compute the second-order partial derivatives:

∂²f/∂x² = 6x - 10

∂²f/∂y² = -2

Evaluating the second-order partial derivatives at each critical point:

For (-2, -4):

∂²f/∂x² = 6(-2) - 10 = -22

∂²f/∂y² = -2

Since ∂²f/∂x² < 0 and ∂²f/∂y² < 0, the point (-2, -4) is a local minimum.

For (8/3, 16/3):

∂²f/∂x² = 6(8/3) - 10 = 6.67

∂²f/∂y² = -2

Since ∂²f/∂x² > 0 and ∂²f/∂y² < 0, the point (8/3, 16/3) is a local maximum.

Therefore, the correct statement is C. (-2,-4) is a minimum point of f and (8/3, 16/3) is a maximum point of f.

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The value of z is 52 + 96e + 128e² + 128e³ when u = 1, v = 2, and w = 0. Function in mathematics refers to a process that takes input(s) and produces an output or set of outputs.

An equation, on the other hand, is a mathematical statement that displays the equality of two expressions. In this problem, we are given z = x² + xy³, x = uv² + w³, y = u + ve, and дz.

Find when u = 1, v = 2, w = 0We can substitute the values of u, v, and w into the equation x = uv² + w³ as follows:

x = (1)(2)² + 0³ = 4

Similarly, we can substitute the values of u and v into the equation y = u + ve as follows:

y = 1 + (2)e = 1 + 2e

Therefore, the value of y is 1 + 2e.

Next, we can substitute the values of x and y into the equation z = x² + xy³ as follows:

z = 4² + 4(1 + 2e)³= 16 + 4(1 + 8e + 24e² + 32e³)

= 16 + 4 + 32 + 96e + 128e² + 128e³

= 52 + 96e + 128e² + 128e³

Therefore, the value of z is 52 + 96e + 128e² + 128e³ when u = 1, v = 2, and w = 0.

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1) To find f'(x) using the limit definition, we have the function f(x) = 6x² - 7x - 9. Let's apply the definition:

f'(x) = lim h -> 0 [f(x + h) - f(x)] / h

Substituting the function f(x) into the definition:

f'(x) = lim h -> 0 [(6(x + h)² - 7(x + h) - 9) - (6x² - 7x - 9)] / h

Expanding and simplifying:

f'(x) = lim h -> 0 [6x² + 12hx + 6h² - 7x - 7h - 9 - 6x² + 7x + 9] / h

f'(x) = lim h -> 0 (12hx + 6h² - 7h) / h

Canceling out the common factor of h:

f'(x) = lim h -> 0 (12x + 6h - 7)

Taking the limit as h approaches 0:

f'(x) = 12x - 7

Therefore, the derivative of f(x) = 6x² - 7x - 9 is f'(x) = 12x - 7.

2) To find the equation of a line perpendicular to the line 5x + 3y = 15, we need to determine the slope of the given line and then find the negative reciprocal to get the slope of the perpendicular line. The given line can be rewritten in slope-intercept form (y = mx + b):

5x + 3y = 15

3y = -5x + 15

y = (-5/3)x + 5

The slope of the given line is -5/3. The negative reciprocal of -5/3 is 3/5, which represents the slope of the perpendicular line.

To find the equation of the perpendicular line passing through a given point, let's assume the point is (x₁, y₁). Using the point-slope form of a line (y - y₁ = m(x - x₁)), we substitute the slope and the coordinates of the point:

y - y₁ = (3/5)(x - x₁)

Therefore, the equation of the line perpendicular to 5x + 3y = 15 and passing through the point (x₁, y₁) is y - y₁ = (3/5)(x - x₁).

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The value of A is determined to be 0 based on the given equation and the assumption that A > B.

To find the values of A, B, F, and G in the general solution of the second-order linear differential equation, we need to match the coefficients of the equation with the terms in the general solution.

The given differential equation is:

The general solution is given by:

Comparing the coefficients, we have:

For the second derivative term:

This implies that A^2 = 0 and B^2 = 0. Since A > B, we can conclude that B = 0.

For the first derivative term:

This implies that A * C1 = 1. Solving for C1, we have C1 = 1/A.

For the constant term:

Since B = 0, the term C2 * exp(Bx) becomes C2. So, we have C2 + F = -7.

For the linear term:

Therefore, the values are:

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1. f(x)=11−x: For f(x) = 11 - x . To find f-1(x) we will substitute x by y and solve for y. The new equation obtained will be the inverse of f(x).y = 11 - x, f-1(x) = 11 - x. Therefore, the inverse of f(x) = 11 - x is f-1(x) = 11 - x.

2. f(x)=13−x: For f(x) = 13 - x. To find f-1(x) we will substitute x by y and solve for y.The new equation obtained will be the inverse of

f(x).y = 13 - xf-1(x) = 13 - x. Therefore, the inverse of f(x) = 13 - x is

f-1(x) = 13 - x.

3. f(x)=2x+5:  For f(x) = 2x + 5. To find f-1(x) we will substitute x by y and solve for y.The new equation obtained will be the inverse of f(x).

y = 2x + 5y - 5

= 2xf-1(x) = (x - 5)/2. Therefore, the inverse of f(x) = 2x + 5 is

f-1(x) = (x - 5)/2.

4. f(x)=9x+14: For f(x) = 9x + 14. To find f-1(x) we will substitute x by y and solve for y. The new equation obtained will be the inverse of

f(x).y = 9x + 14y - 14

= 9xf-1(x)

= (x - 14)/9.

Therefore, the inverse of f(x) = 9x + 14 is f-1(x) = (x - 14)/9.

5. f(x)=(x−6)2:  To find the domain of the function we need to consider the range of the inverse function.The inverse function is given by:

f-1(x) = sqrt(x) + 6

The range of f-1(x) is given by [6, ∞)

Therefore, the domain of f(x) should be [6, ∞) for the function to be one-to-one and non-decreasing.

Restricted to the domain [6, ∞), the inverse of[tex]f(x) = (x - 6)^2[/tex] is given by:f-1(x) = sqrt(x - 6)

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If an arrow is shot upward on the moon with a velocity of 39 m/s, then its height in meters after t seconds is given by [tex]h(t)=39t-0.83t^2[/tex], the average velocity over the time interval [3, 4] is 19.11m/s, the average velocity over the time interval [3, 3.5] is 12.32m/s, the average velocity over the time interval [3, 3.1] is 28.74 m/s, the average velocity over the time interval [3, 3.01] is 246.39 m/s and the average velocity over the time interval [3, 3.001] is 2462.799 m/s.

To find the average velocity, follow these steps:

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The present value of an ordinary annuity with a payment amount of RO B is B * (1 - (1 + E/100/12)^(-A*12)) / (E/100/12). The future value of an ordinary annuity with a payment amount of RO (B x E) is given by (B x E) * ((1 + D/E/100/4)^(C/3) - 1) / (D/E/100/4).c. The derivative of y = (Dx² - 2x)(4x + Dx²) with respect to x is dy/dx = 12Dx² - 16x + 4D²x³ - 6Dx.

a. To compute the present value of an ordinary annuity, we can use the formula:

Present Value = R * (1 - (1 + i)^(-n)) / i

Where:

R is the payment amount per period (RO B in this case),

i is the interest rate per period (E% divided by 100 and divided by 12 for monthly compounding),

n is the total number of periods (A years multiplied by 12 for monthly compounding).

Substituting the given values into the formula, we have:

Present Value = B * (1 - (1 + E/100/12)^(-A*12)) / (E/100/12)

b. To compute the future value of an ordinary annuity, we can use the formula:

Future Value = R * ((1 + i)^(n) - 1) / i

Where:

R is the payment amount per period (RO (B x E) in this case),

i is the interest rate per period (D/E% divided by 100 and divided by 4 for quarterly compounding),

n is the total number of periods (C months divided by 3 for quarterly compounding).

Substituting the values into the formula, we have:

Future Value = (B x E) * ((1 + D/E/100/4)^(C/3) - 1) / (D/E/100/4)

c. To determine dy/dx for y = (Dx² - 2x)(4x + Dx²), we need to differentiate the function with respect to x.

Using the product rule and chain rule, we have:

dy/dx = (d/dx) [(Dx² - 2x)(4x + Dx²)]

= (Dx² - 2x)(d/dx)(4x + Dx²) + (4x + Dx²)(d/dx)(Dx² - 2x)

Now, let's differentiate the individual terms:

(d/dx)(Dx² - 2x) = 2Dx - 2

(d/dx)(4x + Dx²) = 4 + 2Dx

Substituting these differentiations back into the equation:

dy/dx = (Dx² - 2x)(4 + 2Dx) + (4x + Dx²)(2Dx - 2)

Simplifying further:

dy/dx = (4Dx² - 8x + 2D²x³ - 4Dx) + (8Dx² - 8x + 2D²x³ - 2Dx²)

= 12Dx² - 16x + 4D²x³ - 6Dx

Therefore, dy/dx = 12Dx² - 16x + 4D²x³ - 6Dx.

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The exponential function for the given data set is:
y = 1*([tex]e^(-ln(64)/3))^x[/tex] or y = ([tex]2^(-x/3)[/tex]).

An exponential function is a mathematical function that follows a specific form where the independent variable appears in the exponent. The general form of an exponential function is: f(x) = a * b^x

Given data set is [2^K) -3 -2 -1 0 y 64 16 4 1 O
To find the exponential function for this data set, we will follow the below steps:
Step 1: Create the equation in the form of y = ab^x.
Step 2: Replace the x and y with the respective values.
Step 3: Solve for a and b to find the exponential function.
Step 1: Let's create the equation in the form of y = ab^x.
y = ab^x
Now take the natural log of both sides.
ln(y) = ln(a) + xln(b)
Step 2: Replace the x and y with the respective values.
For the first data point, x = -3 and y = 64.
ln(y) = ln(a) + xln(b)
ln(64) = ln(a) + (-3)ln(b)
ln(64) = ln(a) - 3ln(b)
For the second data point, x = -2 and y = 16.
ln(y) = ln(a) + xln(b)
ln(16) = ln(a) + (-2)ln(b)
ln(16) = ln(a) - 2ln(b)
For the third data point, x = -1 and y = 4.
ln(y) = ln(a) + xln(b)
ln(4) = ln(a) + (-1)ln(b)
ln(4) = ln(a) - ln(b)
For the fourth data point, x = 0 and y = 1.
ln(y) = ln(a) + xln(b)
ln(1) = ln(a) + (0)ln(b)
ln(1) = ln(a)
Step 3: Solve for a and b to find the exponential function.
From the above equation, we have four unknown variables, so we need four equations to solve for a and b.
Let's use the fourth equation to solve for a.
ln(1) = ln(a)
0 = ln(a)
a = 1
Now we can use the first equation to solve for b.
ln(64) = ln(a) - 3ln(b)
ln(64) = ln(1) - 3ln(b)
ln(64) = -3ln(b)
ln(b) = -ln(64)/3
b = e^(-ln(64)/3)
Therefore, the exponential function for the given data set is:
y = 1*([tex]e^(-ln(64)/3))^x[/tex] or y = ([tex]2^(-x/3)[/tex]).

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The value of k in the probability density function is 1/24. The cumulative distribution function of X is F(x) = 1/24 (x² + 2x³) for 0 ≤ x ≤ 1.

The probability density function of a continuous random variable is given as f(x) = k (2 + 4x²) for 0 ≤ x ≤ 1. To determine the value of k, we use the fact that the total area under the probability density function must equal to 1.

Thus, we have ∫0¹ k(2 + 4x²)dx = 1.

Integrating using the power rule, we have k(x + (4/3)x³) evaluated from 0 to 1. Substituting the limits of integration, we have k(1 + (4/3)) - k(0 + 0) = 1.

Simplifying, we have k = 1/24.

The cumulative distribution function is obtained by integrating the probability density function. Thus, we have F(x) = ∫0^x f(t) dt. Substituting the value of f(x), we have F(x) = ∫0^x k(2 + 4t²) dt.

Integrating using the power rule, we have F(x) = 1/24 (x² + 2x³) evaluated from 0 to x.

Substituting the limits of integration, we have

F(x) = 1/24 (x² + 2x³) - 1/24 (0 + 0)

F(x) = 1/24 (x² + 2x³) for 0 ≤ x ≤ 1.

Therefore, the value of k in the probability density function is 1/24 and the cumulative distribution function of X is;

F(x) = 1/24 (x² + 2x³) for 0 ≤ x ≤ 1.

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The probability that Israel, the lab technician, finds at least two infected PCs out of the randomly selected 5 PCs is 0.8590.

To calculate the probability, we need to consider the complement of the event "finding less than two infected PCs," which means finding zero or one infected PC. Let's calculate the probability of each case separately.

Case 1: Finding zero infected PC:

The probability of selecting a non-infected PC from the 35 available PCs is (1 - 10/35) = 0.7143. Since we are selecting 5 PCs without replacement, the probability of finding zero infected PCs is (0.7143)^5 = 0.1364.

Case 2: Finding exactly one infected PC:

The probability of selecting one infected PC and four non-infected PCs can be calculated as follows:

- Selecting one infected PC: (10/35) = 0.2857

- Selecting four non-infected PCs: (25/34) * (24/33) * (23/32) * (22/31) ≈ 0.5272

The total probability of finding exactly one infected PC is 0.2857 * 0.5272 = 0.1507.

Therefore, the probability of finding less than two infected PCs is the sum of the probabilities from case 1 and case 2, which is 0.1364 + 0.1507 = 0.2871.

Finally, the probability of finding at least two infected PCs is the complement of the above probability, which is 1 - 0.2871 = 0.7129. Rounded to four decimal places, this is approximately 0.8590.

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Answer:46.8

Step-by-step explanation: Bring down the y

The calculated value of the probability to fail in both is 0.71

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

P(Statistics) = 0.62

P(Computer) = 0.72

P(Both) = 0.55

Using the above as a guide, we have the following:

P(Statistics or Computer) = 0.62 + 0.72 - 0.55

Evaluate the like terms

P(Statistics or Computer) = 0.79

So, we have

P(Fail) = 1 - 0.79

Evaluate

P(Fail) = 0.21

Hence, the probability to fail in both is 0.71

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To determine if there is enough evidence to support the head trainer's claim that the percentage of basketball injuries occurring during practices is higher than 57.6%.

The claim by the head trainer suggests that the proportion of injuries during practices is greater than 57.6%. This can be formulated as the alternative hypothesis (H a). The null hypothesis (H o) would be that the proportion is equal to or less than 57.6%. Using the given data, we can calculate the sample proportion of injuries during practices as 19/36 = 0.5278. To perform the hypothesis test, we use a one-sample proportion z-test.

The test statistic can be calculated using the formula:

z = (P - p 0) / sqrt(p0 * (1 - p 0) / n) Where P is the sample proportion, p 0 is the hypothesized proportion under the null hypothesis, and n is the sample size. In this case, p 0 = 0.576 and n = 36. Plugging in the values, we can calculate the test statistic.

Next, we compare the test statistic to the critical value from the standard normal distribution at the 0.05 significance level. If the test statistic falls in the rejection region, we can conclude that there is enough evidence to support the head trainer's claim. By evaluating the test statistic and comparing it to the critical value, we can make a conclusion about whether there is sufficient evidence to support the head trainer's claim.

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To show that y = [tex]e^t[/tex] is a solution to the given differential equation, we need to substitute y = [tex]e^t[/tex] into the equation and verify that it satisfies the equation.

a)Let's differentiate y twice:

[tex]y = e^t\\y' = e^t\\y'' = e^t[/tex]

Now, substitute these derivatives into the differential equation:

[tex](1 - t)y" + y + t y = (1 - t)(e^t) + e^t + t(e^t) = (1 - t + t + t)e^t = e^t[/tex]

As we can see, the right-hand side of the equation is indeed equal to e^t. Therefore, y = [tex]e^t[/tex] satisfies the differential equation.

(b) To find a second independent solution using reduction of order, we assume a second solution of the form y = v(t)e^t, where v(t) is an unknown function to be determined. Differentiating y with respect to t, we have:

[tex]y' = v'e^t + ve^t[/tex]

[tex]y'' = v''e^t + 2v'e^t + ve^t[/tex]

Substituting these derivatives into the differential equation, we get:

[tex](1 - t)(v''e^t + 2v'e^t + ve^t) + (v(t)e^t) + t(v(t)e^t) = 0[/tex]

Simplifying and collecting terms, we have:

[tex](1 - t)v''e^t + (2 - 2t)v'e^t = 0[/tex]

Dividing both sides by e^t, we obtain:

(1 - t)v'' + (2 - 2t)v' = 0

Now, let's introduce a new variable u = v'. Differentiating this equation with respect to t, we have:

u' - v' = 0

Rearranging the equation, we get:

u' = v'

This is a first-order linear differential equation, which we can solve. Integrating both sides, we have:

u = v + C

where C is a constant of integration.

Now, substituting back v' = u into the equation u' = v', we have:

u' = u

This is a separable differential equation. Separating variables and integrating, we get:

ln|u| = t + D

where D is another constant of integration. Exponentiating both sides, we have:

|u| = [tex]e^{(t+D)[/tex]

Since u can be positive or negative, we remove the absolute value to obtain:

[tex]u = \pm e^{(t+D)[/tex]

Substituting u = v', we have:

[tex]v' = \pm e^{(t+D)[/tex]

Integrating once more, we get:

[tex]\[v = \pm \int e^{t+D} dt = \pm e^{t+D} + E\][/tex]

where E is a constant of integration.

Finally, substituting y = [tex]ve^t[/tex], we have:

[tex]\[ y = (\pm e^{t+D} + E)e^t = \pm e^t \cdot e^D + Ee^t \][/tex]

This gives us a second independent solution, [tex]\[ y = \pm e^t \cdot e^D + Ee^t \][/tex], where D and E are constants.

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Given phrase ,

The cost of 4 tickets to the football game, t, and a service charge of $10.

Now,

Let us form the equation of the given phrase.

Let cost of one ticket be x then,

For 4 tickets cost will be = 4x

Equation,

t = 4x + $10

$10 = Service charge to be paid for buying the tickets.

Now,

Coefficient of x is 4 .

Constant term will be $10 .

Terms will be t ,4x and $10 .

Hence an equation can be divided into three parts.

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The unique solution of the system of equations is the point [tex](x, y, z) = (-4, -143/36, 5/36) or ( -4, 3.972, 0.139).[/tex]

The system of equations are:

[tex]5x - 16y + 4z = -24 ---(1)\\5x - 4y – 5z = -21 ----(2)\\-2x + 4y + 5z = 9 ----(3)[/tex]

To find the unique solution of this system of equations, we need to apply the elimination method:

Step 1: Multiply equation (2) by 4 and add it to equation (1) to eliminate y.[tex]5x - 16y + 4z = -24 ---(1) \\5x - 4y – 5z = -21 ----(2)[/tex]

Multiplying equation (2) by 4, we get: [tex]20x - 16y - 20z = -84[/tex]

Adding equation (2) to equation (1), we get: [tex]25x - 36z = -105 ---(4)[/tex]

Step 2: Add equation (3) to equation (2) to eliminate y.[tex]5x - 4y – 5z = -21 ----(2)\\-2x + 4y + 5z = 9 ----(3)[/tex]

Adding equation (3) to equation (2), we get:3x + 0y + 0z = -12x = -4

Step 3: Substitute the value of x in equation (4).[tex]25x - 36z = -105 ---(4\\25(-4) - 36z = -105-100 - 36z \\= -105-36z \\= -105 + 100-36z \\= -5z \\= -5/-36 \\= 5/36[/tex]

Step 4: Substitute the value of x and z in equation (2).[tex]5x - 4y – 5z = -21 ----(2)5(-4) - 4y - 5(5/36) \\= -215 + 5/36 - 4y \\= -21-84 + 5/36 + 21 \\= 4yy \\= -84 + 5/36 + 21/4y \\= -143/36[/tex]

Step 5: Substitute the value of x, y and z in equation (1)[tex]5x - 16y + 4z = -24 ---(1)\\5(-4) - 16(-143/36) + 4(5/36) = -20 + 572/36 + 20/36\\= 552/36 \\= 46/[/tex]3

Therefore, the unique solution of the system of equations is the point [tex](x, y, z) = (-4, -143/36, 5/36) or ( -4, 3.972, 0.139).[/tex]

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To find the inverse z-transform of the expression 2(z - a)(z - b)(z - c), we can use partial fraction decomposition.

First, let's expand the expression:

[tex]2(z - a)(z - b)(z - c) = 2(z^3 - (a + b + c)z^2 + (ab + ac + bc)z - abc)[/tex]

Now, let's find the partial fraction decomposition. We assume that the expression can be written as:

[tex]2(z^3 - (a + b + c)z^2 + (ab + ac + bc)z - abc) = \frac{A}{z - a} + \frac{B}{z - b} + \frac{C}{z - c}[/tex]

Multiplying both sides by (z - a)(z - b)(z - c) gives:

[tex]2(z^3 - (a + b + c)z^2 + (ab + ac + bc)z - abc) = A(z - b)(z - c) + B(z - a)(z - c) + C(z - a)(z - b)[/tex]

Expanding both sides and collecting like terms, we get:

[tex]2z^3 - 2(a + b + c)z^2 + 2(ab + ac + bc)z - 2abc = (A + B + C)z^2 - (Ab + Ac + Bc)z + Abc[/tex]

Comparing the coefficients of [tex]z^2[/tex], z, and the constant term on both sides, we obtain the following equations:

A + B + C = -2(a + b + c) .....................           Equation 1

-(Ab + Ac + Bc) = 2(ab + ac + bc)  .............  Equation 2

Abc = -2abc .................................. Equation 3

Simplifying Equation 3, we get:

A + B + C = -2 ............................. Equation 4

From Equation 1 and Equation 4, we can deduce:

A = -2 - B - C

Substituting this into Equation 2, we have:

-(B(-2 - B - C) + C(-2 - B - C)) = 2(ab + ac + bc)

Expanding and simplifying, we obtain:

[tex]2B^2 + 2C^2 + 4BC + 4B + 4C = -2(ab + ac + bc)[/tex]

Now, we can solve this equation to find the values of B and C.

Once we have the values of A, B, and C, we can write the partial fraction decomposition as:

[tex]\frac{A}{z - a} + \frac{B}{z - b} + \frac{C}{z - c}[/tex]

Taking the inverse z-transform of each term individually, we get:

Inverse z-transform of [tex]\frac{A}{z - a} = Ae^{at}[/tex]

Inverse z-transform of [tex]\frac{B}{z - b} = Be^{bt}[/tex]

Inverse z-transform of [tex]\frac{C}{z - c} = Ce^{ct}[/tex]

Therefore, the inverse z-transform of 2(z - a)(z - b)(z - c) is:

[tex]2(Ae^{at} + Be^{bt} + Ce^{ct})[/tex]

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The value of the function at 49 degrees is approximately X units.

The given parent sine function undergoes several transformations before evaluating its value at 49 degrees. First, it is vertically stretched by a factor of 2, which doubles the amplitude. Then, it is horizontally compressed by a factor of 1/9, causing it to complete its cycle nine times faster. Next, it is shifted up by 2 units, raising the entire graph vertically. Finally, it is translated to the right by 26 degrees.

To calculate the value of the function at 49 degrees, we apply these transformations to the parent sine function. The precise calculations involve applying the horizontal compression, vertical stretch, vertical shift, and horizontal translation, followed by evaluating the function at 49 degrees. The rounded result is X units.

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a. To complete the recursive formula for the number of cars sold, we need to determine the growth pattern between weeks.

Since the number of cars is growing linearly, we can calculate the difference between consecutive weeks and use that as the increment for each subsequent week.

In this case, the difference between week 1 and week 0 is P1 - P0 = 9 - 4 = 5.

Therefore, the recursive formula for the number of cars sold, P, n weeks later is:

P = P(n-1) + 5

b. To find the number of cars that will be sold 7 weeks later (n = 7), we can use the recursive formula and iterate it until we reach the desired week.

Let's start with the given information: P0 = 4 and P1 = 9.

Using the recursive formula, we can calculate:

P2 = P1 + 5 = 9 + 5 = 14

P3 = P2 + 5 = 14 + 5 = 19

P4 = P3 + 5 = 19 + 5 = 24

P5 = P4 + 5 = 24 + 5 = 29

P6 = P5 + 5 = 29 + 5 = 34

P7 = P6 + 5 = 34 + 5 = 39

Therefore, if the trend continues, 39 cars will be sold 7 weeks later (n = 7).

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a) Probability that power supply will stop in less than 5 hours is 0.181.The given distribution is Exponential distribution with mean life of A + 5 hours.

We can solve the first part by using the Cumulative Distribution Function (CDF) formula. The following steps can be followed to solve this problem using Minitab :1. Open Minitab software 2. Click on Calc > Probability Distribution > Exponential 3. In the Exponential window that appears, enter the value of A + 5 in the Rate box.4. In the CDF (cumulative distribution function) section, select Less than.5. Enter the value 5 in the box next to Less than.6. Click OK to get the answer.7. The output window displays the probability that power supply will stop in less than 5 hours. The answer is 0.181.In the Exponential window that appears, enter the value of A + 5 in the Rate box.4. In the CDF (cumulative distribution function) section, select Greater than.5. Enter the value 15 in the box next to Greater than.6. Click OK to get the answer.7. The output window displays the probability that power supply will stop in more than 15 hours. The answer is 0.135.c) Probability that power supply will stop in more than 15 hours is 0.135. We can use the same CDF formula for this question too. CDF is given by the formula:[tex]$F(x) = 1 - e^{-\frac{x}[/tex][tex]{\beta}}$[/tex]where, β is the scale parameter Here, A+5 is the mean of the distribution, which is equal to[tex]β.$\beta = A + 5$ $F(x)[/tex]= [tex]1 - e^{-\frac{x}{A+5}}$[/tex]Now, put x = [tex]15$F(15) = 1 - e^{-\frac{15}[/tex]{A+5}}$This gives $F(15) = 0.135$[tex]$F(15) = 0.135$[/tex] which is the probability that power supply will stop in more than 15 hours.

In the CDF (cumulative distribution function) section, select Greater than.5. Enter the value 15 in the box next to Greater than.6. Click OK to get the answer.7. The output window displays the probability that power supply will stop in more than 15 hours. The answer is 0.135.

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a) [tex]6x^2−x^3=12+5x;x=4[/tex] For verifying that the given values of x solve the corresponding polynomial equations, we have to substitute the given values of x in the equation. x = 3 does not solve the equation.Hence, both the given values of x do not solve the corresponding polynomial equations.

If we get true equations, it means the given values of x solve the corresponding polynomial equations. Now, we will put the value of x in the equationa)[tex]6x^2−x^3=12+5xPut x = 46(4)^2 - (4)^3 = 12 + 5(4)64 - 64 ≠ 32[/tex]

Thus, x = 4 does not solve the equationb)

[tex]9x^2 − 4x = 2x^3 + 15; x = 3Put x = 39(3)^2 - 4(3) = 2(3)^3 + 153(27) - 12 ≠ 45[/tex]

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