5.8. Prove that if \( A, B, C \), and \( D \) are finite sets such that \( A \subseteq B \) and \( C \subseteq D \) \( A \times C \subseteq B \times D \).

The probability that the entire shipment will be kept by Dell even though the shipment has 10% defective microprocessors is approximately 0.5905. Hence the correct answer is (a) 0.5905.

To calculate the probability that the entire shipment will be kept by Dell even though the shipment has 10% defective microprocessors, we can use the concept of binomial probability.

Let's denote the probability of a microprocessor being defective as p = 0.10 (10% defective) and the number of microprocessors Dell tests as n = 5.

We want to calculate the probability that all five tested microprocessors are non-defective, which is equivalent to the probability of having zero defective microprocessors in the sample.

Using the binomial probability formula, the probability of getting exactly k successes (non-defective microprocessors) in n trials is:

[tex]\[P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}\][/tex]

For this case, we want to calculate P(X = 0), where X represents the number of defective microprocessors.

[tex]\[P(X = 0) = \binom{5}{0} \cdot 0.10^0 \cdot (1 - 0.10)^{5 - 0} \\= 1 \cdot 1 \cdot 0.9^5 \\\\approx 0.5905\][/tex]

Therefore, the correct answer is (a) 0.5905.

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Cov (⋅⋅) is linear in each of its arguments. Hence proved.

Let X, Y, Z, and W be random variables, and a, b, c, and d be real numbers. We must show that Cov (aX + bY, cZ + dW) = acCov(X, Z) + adCov(X, W) + bcCov(Y, Z) + bdCov(Y, W).The covariance of two random variables is the expected value of the product of their deviations from their respective expected values. Consider the following linearity of expectation: E(aX + bY) = aE(X) + bE(Y) and E(cZ + dW) = cE(Z) + dE(W). Therefore, Cov(aX+bY,cZ+dW) = E((aX + bY) (cZ + dW)) − E(aX + bY) E(cZ + dW)   {definition of covariance}      = E(aXcZ + aX dW + bYcZ + bYdW) − (aE(X) + bE(Y)) (cE(Z) + dE(W))   {linearity of expectation}       = E(aXcZ) + E(aX dW) + E(bYcZ) + E(bYdW) − acE(X)E(Z) − adE(X)E(W) − bcE(Y)E(Z) − bdE(Y)E(W)    {distributivity of expectation}       = acE(XZ) + adE(XW) + bcE(YZ) + bdE(YW) − acE(X)E(Z) − adE(X)E(W) − bcE(Y)E(Z) − bdE(Y)E(W)   {definition of covariance}       = ac(Cov(X,Z)) + ad(Cov(X,W)) + bc(Cov(Y,Z)) + bd(Cov(Y,W)).  Therefore, Cov (⋅⋅) is linear in each of its arguments. Hence proved.

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The derivative of f(x) = (1 + 2x^2)^2 + 2x^5 is f'(x) = 8x(1 + 2x^2) + 10x^4. To find the derivative of the function f(x) = (1 + 2x^2)^2 + 2x^5, we can apply the rules of differentiation.

First, we differentiate each term separately using the power rule and the constant multiple rule:

The derivative of (1 + 2x^2)^2 can be found using the chain rule. Let u = 1 + 2x^2, then (1 + 2x^2)^2 = u^2. Applying the chain rule, we have:

d(u^2)/dx = 2u * du/dx.

Differentiating 2x^5 gives us:

d(2x^5)/dx = 10x^4.

Now, let's differentiate each term:

d((1 + 2x^2)^2)/dx = 2(1 + 2x^2) * d(1 + 2x^2)/dx

                  = 2(1 + 2x^2) * (4x)

                  = 8x(1 + 2x^2).

d(2x^5)/dx = 10x^4.

Putting it all together, the derivative of f(x) is:

f'(x) = d((1 + 2x^2)^2)/dx + d(2x^5)/dx

     = 8x(1 + 2x^2) + 10x^4.

Therefore, the derivative of f(x) = (1 + 2x^2)^2 + 2x^5 is f'(x) = 8x(1 + 2x^2) + 10x^4.

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The general solution for the given differential equation y' = x^2 - x^3 + x^6 is y = (x^3/3) - (x^4/4) + (x^7/7) + C, where C is an arbitrary constant.

To find the general solution for the differential equation y' = x^2 - x^3 + x^6, we can integrate both sides with respect to x.

Integrating the right-hand side term by term, we get:

∫(x^2 - x^3 + x^6) dx = ∫(x^2) dx - ∫(x^3) dx + ∫(x^6) dx

Integrating each term separately, we have:

(x^3/3) - (x^4/4) + (x^7/7) + C

where C is the constant of integration.

Therefore, the general solution for the differential equation y' = x^2 - x^3 + x^6 is:y = (x^3/3) - (x^4/4) + (x^7/7) + C where C is an arbitrary constant.

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The relationship between the standard deviation and variance is that the standard deviation is the square root of the variance.

The correct option is -C

Hence, the correct option is (c) Standard deviation is the square root of the variance. Variance is the arithmetic mean of the squared differences from the mean of a set of data. It is a statistical measure that measures the spread of a dataset. The squared difference from the mean value is used to determine the variance of the given data set.

It is represented by the symbol 'σ²'. Standard deviation is the square root of the variance. It is used to calculate how far the data points are from the mean value. It is used to measure the dispersion of a dataset. The symbol 'σ' represents the standard deviation. The formula for standard deviation is:σ = √(Σ(X-M)²/N) Where X is the data point, M is the mean value, and N is the number of data points.

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The sample standard deviation is approximately 2.73.The 95% confidence interval for the population mean heart rate is approximately (73.833, 76.087).

a. To calculate the sample standard deviation, we first need to find the sample mean. The sample mean is the sum of all observations divided by the sample size:

X = (70 + 74 + 75 + 78 + 74 + 64 + 70 + 78 + 81 + 73 + 82 + 75 + 71 + 79 + 73 + 79 + 85 + 79 + 71 + 65 + 70 + 69 + 76 + 77 + 66) / 25

X= 74.96

Next, we calculate the sum of the squared differences between each observation and the sample mean:

Σ(xᵢ - X)² = (70 - 74.96)² + (74 - 74.96)² + ... + (66 - 74.96)²

Σ(xᵢ - X)² = 407.04

Finally, the sample standard deviation is the square root of the sum of squared differences divided by (n-1), where n is the sample size:

s = √(Σ(xᵢ - X)² / (n-1))

s = √(407.04 / 24)

s ≈ 2.73

Therefore, the sample standard deviation is approximately 2.73.

b. The variance is the square of the standard deviation:

σ² = s² ≈ 2.73²

σ² ≈ 7.46

Therefore, the sample variance is approximately 7.46.

c. To calculate the 95% confidence interval (CI) for the population mean heart rate, we can use the formula:

CI = X ± (tα/2 * (s / √n))

where X is the sample mean, tα/2 is the critical value from the t-distribution for a 95% confidence level with (n-1) degrees of freedom, s is the sample standard deviation, and n is the sample size.

For the given sample, n = 25. The critical value tα/2 can be obtained from the t-distribution table or using a statistical software. For a 95% confidence level with 24 degrees of freedom, tα/2 is approximately 2.064.

Plugging in the values, we have:

CI = 74.96 (2.064 * (2.73 / √25))

CI = 74.96  (2.064 * 0.546)

CI ≈ 74.96  1.127

Therefore, the 95% confidence interval for the population mean heart rate is approximately (73.833, 76.087).

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The actual distance between East York and Oshawa is 80 miles.

The actual distance between East York and Oshawa, we can use the scale on the map. We know that the distance between Brampton and East York is 270 miles and is represented as 4.5 inches on the map. Therefore, the scale is 270 miles/4.5 inches = 60 miles per inch.

Next, we can use the scale to calculate the distance between East York and Oshawa. On the map, this distance is represented as 12 inches. Multiplying the scale (60 miles per inch) by 12 inches gives us the actual distance between East York and Oshawa: 60 miles/inch × 12 inches = 720 miles.

Therefore, the actual distance between East York and Oshawa is 720 miles,  80 miles.

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a. Calculation of payback period for investment A is: Initial Investment = P400,000Incremental cash flow = Year 1: P100,000 Year 2: P120,000 Year 3: P140,000 Year 4: P120,000 Year 5: P100,000 + P20,000

= P120,000Total cash inflows

= Year 1: P100,000 Year 2: P120,000 Year 3: P140,000 Year 4: P120,000 Year 5: P120,000Therefore, the cumulative cash flow for year 4

= P480,000, and the cumulative cash flow for year 5 is P600,000 (P480,000 + P120,000)Payback period

= Year 4 + Unrecovered amount / Cumulative cash flow in year 5

= 4 + (P220,000 / P600,000)

= 4.37 years

Therefore, the payback period for investment A is 4.37 years.

b) Discounted payback period = Year before recovery + (Unrecovered amount / Discounted cash flow ) Present value of cash flow

= Cash flow / (1 + Discount rate)nYear 0: Initial Investment

= P450,000Year 1: P130,000 / (1 + 0.10)1 = P118,182Year 2: P130,000 / (1 + 0.10)2

= P107,439Year 3: P130,000 / (1 + 0.10)3 = P97,672Year 4: P130,000 / (1 + 0.10)4

= P89,000Year 5: (P150,000 + P20,000) / (1 + 0.10)5

= P95,425Therefore, the discounted cash flows are as follows: Year 1: P118,182 Year 2: P107,439 Year 3: P97,672 Year 4: P89,000 Year 5: P95,425 Therefore, the cumulative discounted cash flow for year 4 = P412,293, and the cumulative discounted cash flow for year 5 is P507,718 (P412,293 + P95,425) The discounted payback period is as follows: Discounted payback period = Year before recovery + (Unrecovered amount / Discounted cash flow of the year)Discounted payback period

= 4 + (P42,282 / P95,425)

= 4.44Therefore, the discounted payback period for investment B is 4.44 years.

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The work done by the person is approximately 7,071 ft-lb.

To calculate the work done, we need to consider the weight of the person and the vertical distance they have climbed. The weight of the person is given as 141 lb. Since the person is walking up a circular, spiral staircase, the vertical distance they have climbed after one revolution is 10 ft.

The total distance covered after one and a half revolutions is (2 * π * 5 ft * 1.5) = 47.12 ft. Since work is equal to force multiplied by distance, we can calculate the work done by multiplying the weight (141 lb) by the vertical distance climbed (47.12 ft) to get approximately 7,071 ft-lb.

Therefore, the work done by the person weighing 141 lb walking one and a half revolution(s) up the circular, spiral staircase is approximately 7,071 ft-lb.

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The initial velocity of Ball 2 is 158.69 feet/sec.

Take downside is positive so here θ is negative here.

Initial velocity of Ball 1 is = v₁ = 145 ft./sec = 44.196 m/sec

The balls are to collide at an altitude of 257 ft that is,

H = 257 feet = 78.3336 m

Using Equation of Motion we get,

v² = u² + 2as

Now here v₀ is the final velocity of the Ball 1

u = v₁ = 44.196 m/sec

a = g = 9.8 m/s²

s = H = 78.3336 m

So,

v₀² = v₁² + 2gH

v₀² = (44.196)² + 2 (9.8) (78.3336)

v₀² = 3488.625

v₀ = √3488.625

v₀ = ± 59.06 m/s

Now calculating time for each velocity using equation of motion we get,

v₀ = v₁ + gt

t = (v₀ - v₁)/g

t = (59.06 - 44.196)/(-9.8)

t = - 1.51 second

Time cannot be negative so t = 1.51 second.

When v₀ = - 59.06 m/s

v₀ = v₁ + gt

t = (v₀ - v₁)/g

t = (-59.06 - 44.196)/(-9.8)

t = 10.53 second

Since the second ball throws after 2.7 seconds of ball 1 so we can avoid the case of t = 1.51 second.

So at the time of collision the velocity of ball 1 is decreasing.

Time of fling of ball 2 is given by

= t - Initial time after ball 2 launched

= 10.53 - 2.7

= 7.83 seconds

Height travelled by Ball 2 is, H = 257 feet = 78.3336 m.

Now we need to find the initial velocity of Ball 2 using equation of motion,

S = ut + 1/2 at²

H = v₂t - 1/2 gt² [Since downside is positive so g is negative]

v₂ = H/t + (1/2) gt

Substituting the values H = 78.3336 m; t = 7.83 seconds; g = 9.8 m/s²

v₂ = 48.37 m/s = 158.69 feet/sec.

Hence the initial velocity of Ball 2 is 158.69 feet/sec.

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The probability of getting 8 tails out of 12 tosses is 0.169 or 16.9%..

To find the probability of getting exactly 8 tails out of 12 tosses, we need to use the binomial probability formula:

P(X = k) = (n choose k) * p^k * (1-p)^(n-k)where n is the number of trials (in this case, 12), k is the number of successes (in this case, 8), p is the probability of a success on any one trial (in this case, 0.5 since it's a fair coin toss), and (n choose k) is the binomial coefficient that gives the number of ways to choose k successes out of n trials.(n choose k) = n! / (k! * (n-k)!)

Using this formula, we get:P(X = 8) = (12 choose 8) * 0.5^8 * (1-0.5)^(12-8)P(X = 8) = 495 * 0.0039 * 0.0625P(X = 8) = 0.169 (rounded to three decimal places).

Therefore, the probability of getting exactly 8 tails out of 12 tosses is approximately 0.169 or 16.9%.

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P(X>5) = P(miss on the first five attempts) = (0.6)(0.6)(0.6)(0.6)(0.6) = 0.07776Therefore, P(X>5) is 7.776%.

The possible values that X can take and whether X is discrete or continuous for Maya, who is a basketball player making 40% of her three point field goal attempts, is discussed below.According to the problem statement, the random variable X is the total number of shots Maya attempts in a practice session until she makes one and then stops. Since X can only take integer values, X is a discrete random variable.In principle, conducting a simulation using physical objects (coins, cards, dice, etc) requires tossing a coin, a die, or drawing a card repeatedly until a certain condition is met.

For example, to simulate X for Maya, a spinner could be constructed with three outcomes (miss, hit, and stop), with probabilities of 0.6, 0.4, and 1, respectively. Each spin represents one shot attempt. The simulation could be stopped after a hit is recorded, and the number of attempts recorded to determine X. Repeating this process many times could generate data for estimating probabilities associated with X.P(X=1) represents the probability that Maya makes the first three-point shot attempt.

Given that the probability of making a shot is 0.4, while the probability of missing is 0.6, it follows that:P(X=1) = P(miss on the first two attempts and make on the third attempt)P(X=1) = (0.6)(0.6)(0.4)P(X=1) = 0.144, which means the probability of making the first shot is 14.4%.P(X=2) represents the probability that Maya makes the second three-point shot attempt. This implies that she must miss the first shot, make the second shot, and stop. Therefore:P(X=2) = P(miss on the first attempt and make on the second attempt and stop)P(X=2) = (0.6)(0.4)(1)P(X=2) = 0.24, which means the probability of making the second shot is 24%.P(X=3) represents the probability that Maya makes the third three-point shot attempt. This implies that she must miss the first two shots, make the third shot, and stop.

Therefore:P(X=3) = P(miss on the first two attempts and make on the third attempt and stop)P(X=3) = (0.6)(0.6)(0.4)(1)P(X=3) = 0.096, which means the probability of making the third shot is 9.6%.The probability mass function of X lists all the possible values of X and their corresponding probabilities. Since Maya keeps shooting until she makes one, she could take one, two, three, four, and so on, attempts. The possible values that X can take are X = 1, 2, 3, 4, ..., and the corresponding probabilities are:P(X = 1) = 0.144P(X = 2) = 0.24P(X = 3) = 0.096P(X = 4) = 0.064P(X = 5) = 0.0384...and so on.

To compute P(X>5) without summing, we need to determine the probability that the first five attempts result in a miss, given that X is the total number of shots Maya attempts until she makes one. Thus:P(X>5) = P(miss on the first five attempts) = (0.6)(0.6)(0.6)(0.6)(0.6) = 0.07776Therefore, P(X>5) is 7.776%.

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The recurrence system for this sequence is:

x1 = 0.4483

xn = 2.9 * xn-1 for n ≥ 2

(a) To find the values of the first term (a) and the common ratio (r), we can observe the pattern in the given sequence.

From the first term to the second term, we can see that multiplying by 2.9 (approximately) gives us the second term:

1.3 * 2.9 ≈ 3.77

Similarly, from the second term to the third term, we multiply by approximately 2.9:

3.77 * 2.9 ≈ 10.933

And from the third term to the fourth term, we multiply by approximately 2.9:

10.933 * 2.9 ≈ 31.7057

So, we can determine that the common ratio is approximately 2.9.

To find the first term (a), we can divide the second term by the common ratio:

1.3 / 2.9 ≈ 0.4483

Therefore, the first term (a) is approximately 0.4483 and the common ratio (r) is approximately 2.9.

(b) To write down the closed form for this sequence, we can use the formula for the nth term of a geometric sequence:

xn = a * r^(n-1)

For this sequence, the closed form is:

xn = 0.4483 * 2.9^(n-1)

(c) To calculate the 10th term of the sequence, we substitute n = 10 into the closed form equation:

x10 = 0.4483 * 2.9^(10-1)

x10 ≈ 0.4483 * 2.9^9 ≈ 419.136

Therefore, the 10th term of the sequence is approximately 419.136.

(d) To determine how many terms of this sequence are less than 1950000, we can use the closed form equation and solve for n:

0.4483 * 2.9^(n-1) < 1950000

To find the exact value, we need to solve the inequality for n. However, without further calculations or approximations, we can conclude that there will be multiple terms before the sequence exceeds 1950000 since the common ratio is greater than 1. Thus, there are multiple terms less than 1950000 in this sequence.

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The points where the polynomial function has a local maximum can be found by using the ALEKS graphing calculator.

Explanation:

1st Part: The ALEKS graphing calculator can provide precise information about the points where a function has a local maximum.

2nd Part:

To find the points where the polynomial function has a local maximum, you can follow these steps using the ALEKS graphing calculator:

1. Enter the polynomial function f(x) = 4x^4 - 2x^3 - 8x^2 + 5x + 2 into the graphing calculator.

2. Set the viewing window to an appropriate range that covers the region where you expect to find local maximum points.

3. Use the calculator's features to identify the points where the function reaches local maximum values. These points will be the x-values (x-coordinate) along with their corresponding y-values (f(x)).

4. Round the x-values and their corresponding y-values to the nearest hundredth.

By following these steps, the ALEKS graphing calculator will help you determine all the points (x, f(x)) where the polynomial function has a local maximum.

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The graph of the function 1/67 f(x) can be obtained from the graph of y=f(x) by vertically compressing the graph of f(x) by a factor 67.

When we have a function of the form y = k * f(x), where k is a constant, it represents a vertical transformation of the graph of f(x). In this case, we have y = (1/67) * f(x), which means the graph of f(x) is vertically compressed by a factor of 67.

To understand why this is a vertical compression, let's consider an example. Suppose the graph of f(x) has a point (a, b), where a is the x-coordinate and b is the y-coordinate. When we multiply f(x) by (1/67), the y-coordinate of the point becomes (1/67) * b, which is much smaller than b since 1/67 is less than 1. This shrinking of the y-coordinate values causes a vertical compression of the graph.

By applying this vertical compression to the graph of f(x), we obtain the graph of 1/67 f(x). The overall shape and features of the graph remain the same, but the y-values are compressed vertically.

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The objective function in terms of x is Q = 4x² - 16x + 32.

To minimize the objective function Q = 2x² + 2y², where x + y = 4, we need to express the objective function in terms of x only. By substituting the value of y from the constraint equation into the objective function, we can rewrite it solely in terms of x.

Given that x + y = 4, we can rearrange the equation to express y in terms of x as y = 4 - x.

Substituting this value of y into the objective function Q = 2x² + 2y², we get:

Q = 2x² + 2(4 - x)²

Simplifying further:

Q = 2x² + 2(16 - 8x + x²)

Expanding:

Q = 2x² + 32 - 16x + 2x²

Combining like terms:

Q = 4x² - 16x + 32

Therefore, the objective function in terms of x is Q = 4x² - 16x + 32.

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There are no solutions to the given logarithmic equation that satisfy the conditions.

Let's solve the logarithmic equation step by step:

log₃(x) + 7 = 11 - log₃(x - 80)

Combine logarithms

Using the property logₐ(M) + logₐ(N) = logₐ(MN), we can combine the logarithms on the left side of the equation:

log₃(x(x - 80)) + 7 = 11

Simplify the equation

Using the property logₐ(a) = 1, we simplify the equation further:

log₃(x(x - 80)) = 11 - 7

log₃(x(x - 80)) = 4

Rewrite in exponential form

The equation logₐ(M) = N is equivalent to aᴺ = M. Applying this to our equation, we get:

3⁴ = x(x - 80)

Convert to a quadratic equation

Expanding the equation on the right side, we have:

81 = x² - 80x

Set the equation equal to 0

Rearranging the terms, we get:

x² - 80x - 81 = 0

Solve for x

To solve the quadratic equation, we can factor or use the quadratic formula. However, upon closer examination, it appears that the equation does not have any integer solutions.

Check for extraneous solutions

Since we don't have any solutions from the quadratic equation, we don't need to check for extraneous solutions in this case.

Therefore, there are no solutions to the given logarithmic equation that satisfy the conditions.

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The equation x² + y²- 4x = 0 can be expressed in polar coordinates as r - 4 * cos(θ) = 0, which corresponds to option B. r = 4 * cos(θ).

To write the equation x² + y² - 4x = 0 in polar coordinates (r, θ), we can use the following conversions:

x = r * cos(θ)

y = r * sin(θ)

Substituting these values into the equation x² + y² - 4x = 0:

(r * cos(θ))² + (r * sin(θ))² - 4(r * cos(θ)) = 0

Simplifying, we have:

r² * cos^2(θ) + r^² * sin^2(θ) - 4r * cos(θ) = 0

Using the trigonometric identity cos^2(θ) + sin^2(θ) = 1, we can simplify further:

r^2 - 4r * cos(θ) = 0

Factoring out an r, we get:

r(r - 4 * cos(θ)) = 0

Now we have the equation in polar coordinates (r, θ):

r - 4 * cos(θ) = 0

Therefore, the equation x² + y²- 4x = 0 can be written in polar coordinates as r - 4 * cos(θ) = 0, which corresponds to option B. r = 4 * cos(θ).

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The value of the determinant is -59. Given matrix is

[tex]\[ \left|\begin{array}{rrr} 3 & 5 & -5 \\ 1 & -2 & 3 \\ 1 & 3 & 2 \end{array}\right| \][/tex]

We use the method of minors to find the value of this determinant.

Applying the expansion along the first row, we get,

[tex]\[ \left|\begin{array}{rrr} 3 & 5 & -5 \\ 1 & -2 & 3 \\ 1 & 3 & 2 \end{array}\right| = 3\left|\begin{array}{rr} -2 & 3 \\ 3 & 2 \end{array}\right| - 5\left|\begin{array}{rr} 1 & 3 \\ 1 & 2 \end{array}\right| - 5\left|\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right| \][/tex]

Solving the determinants on the right-hand side, we get,

[tex]\[ \begin{aligned} \left|\begin{array}{rr} -2 & 3 \\ 3 & 2 \end{array}\right| &= (-2 \times 2) - (3 \times 3) = -13 \\ \left|\begin{array}{rr} 1 & 3 \\ 1 & 2 \end{array}\right| &= (1 \times 2) - (1 \times 3) = -1 \\ \left|\begin{array}{rr} 1 & -2 \\ 1 & 3 \end{array}\right| &= (1 \times 3) - (1 \times -2) = 5 \end{aligned} \][/tex]

Substituting these values in the original expression, we get,

[tex]\[ \left|\begin{array}{rrr} 3 & 5 & -5 \\ 1 & -2 & 3 \\ 1 & 3 & 2 \end{array}\right| = 3(-13) - 5(-1) - 5(5) = -39 + 5 - 25 = \boxed{-59} \][/tex]

Therefore, the value of the determinant is -59.

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Answer:

"If f:[a,b]→R is integrable then f is differentiable on [a,b]" is FALSE.

There is an example of a function that is integrable but not differentiable.

A popular example is the function $f(x) = |x|$.

This function is integrable on any bounded interval such as $[a,b]$ and yet not differentiable at the point $x=0$ .

Since the slope of the tangent line on the left is -1 and on the right is +1.

In other words, it is possible to have an integrable function that is not differentiable, so the statement is false.

Therefore, the circle F should be circled.

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An observation is considered an outlier if it is below Q1 – 1.5 (IQR) and above Q3 + 1.5 (IQR).

It is the concept of the box and whisker plot. It is used to identify the outlier data. Here, the outlier is calculated as below:

Q1 – 1.5 (IQR) and Q3 + 1.5 (IQR) are calculated as:

Q1= The first quartile

Q3= The third quartileI

QR= Interquartile RangeI

QR= Q3 – Q1

Let’s have an example to understand it better.Example:In the given data set:

{25, 37, 43, 47, 52, 56, 60, 62, 63, 65, 66, 68, 69, 70, 70, 72, 73, 74, 74, 75}

Here,Q1 = 56Q3 = 70I

QR = Q3 – Q1= 70 – 56= 14

To identify the outliers,Q1 – 1.5 (IQR) = 56 – 1.5(14)= 35

Q3 + 1.5 (IQR) = 70 + 1.5(14)= 91

The observation below 35 and above 91 is considered an outlier.

So, an observation is considered an outlier if it is below Q1 – 1.5 (IQR) and above Q3 + 1.5 (IQR). This formula is used in the identification of the outliers.

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In this game, two robbers, R1 and R2, have just robbed a bank and find themselves in a hotel room with a suitcase containing 100 million dollars. Each robber wants to have the entire amount for themselves and is armed with a pistol.

However, their shooting skills are not great, with R1 having a 20% chance of killing their target if they shoot, and R2 having a 40% chance. The game proceeds as follows: first, R1 decides whether to shoot. If R1 shoots, R2 (if still alive) then decides whether to shoot. If R1 chooses not to shoot, R2 decides whether to shoot. If both survive, they split the money equally.

In the extensive form of the game, the initial decision node represents R1's choice to shoot or not. If R1 chooses to shoot, it leads to a chance node where R2's decision to shoot or not is determined. If R1 decides not to shoot, it directly leads to R2's decision node.

The outcome of each decision node is the respective robber's survival or death. At the final terminal nodes, the money is divided equally if both survive, or the surviving robber takes the entire amount if the other robber is killed.

The extensive form allows for a comprehensive representation of the sequential decision-making process and the potential outcomes at each stage of the game.

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the corresponding eigenvalues are λ1 = 9 and λ2 = 7.

Let's start with the first eigenvector, v1 = [1, -2]:

Av1 = λ1v1

Substituting the values of A and v1:

[[-11, -6, 12], [7]] * [1, -2] = λ1 * [1, -2]

Simplifying the matrix multiplication:

[-11 + 12, -6 - 12] = [λ1, -2λ1]

[1, -18] = [λ1, -2λ1]

From this equation, we can equate the corresponding components:

1 = λ1  ---- (1)

-18 = -2λ1  ---- (2)

From equation (2), we can solve for λ1:

-18 = -2λ1

λ1 = -18 / (-2)

λ1 = 9

So, the first eigenvalue is λ1 = 9.

Now, let's move on to the second eigenvector, v2 = [-1, 1]:

Av2 = λ2v2

Substituting the values of A and v2:

[[-11, -6, 12], [7]] * [-1, 1] = λ2 * [-1, 1]

Simplifying the matrix multiplication:

[-11 - 6 + 12, 7] = [-λ2, λ2]

[-5, 7] = [-λ2, λ2]

From this equation, we can equate the corresponding components:

-5 = -λ2  ---- (3)

7 = λ2  ---- (4)

From equation (4), we can solve for λ2:

λ2 = 7

So, the second eigenvalue is λ2 = 7.

Therefore, the corresponding eigenvalues are λ1 = 9 and λ2 = 7.

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To find the real and imaginary parts of a complex number, write it in the form a + bi, where a is the real part and b is the imaginary part.

To find the real and imaginary parts of a complex number, you can use the following steps:1. Write the complex number in the form a + bi, where a is the real part and b is the imaginary part.

2. Identify the coefficient of the imaginary unit, "i." This coefficient is the value of "b" in the complex number.

3. The real part of the complex number is given by "a," and the imaginary part is given by "b."

For example, let's consider the complex number z = 3 + 2i.The real part, denoted as Re(z), is 3, and the imaginary part, denoted as Im(z), is 2.Therefore, Re(z) = 3 and Im(z) = 2.By following these steps, you can easily determine the real and imaginary parts of any complex number.

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the simplified expression of the given trigonometric equation sin(2[tex]\theta[/tex])/(2cos([tex]\theta[/tex])) is option (c) sin([tex]\theta[/tex]).

We have sin(2[tex]\theta[/tex]) in the numerator and 2cos([tex]\theta[/tex]) in the denominator. By using the trigonometric identity sin(2[tex]\theta[/tex]) = 2sin([tex]\theta[/tex])cos([tex]\theta[/tex]), we can simplify the expression. This identity allows us to rewrite sin(2[tex]\theta[/tex]) as 2sin([tex]\theta[/tex])cos([tex]\theta[/tex]). Canceling out the common factor of 2cos([tex]\theta[/tex]) in the numerator and denominator, we are left with sin([tex]\theta[/tex]) as the simplified expression. This means that the original expression sin(2[tex]\theta[/tex])/(2cos([tex]\theta[/tex])) is equivalent to sin([tex]\theta[/tex]).

To simplify the expression sin(2[tex]\theta[/tex])/(2cos([tex]\theta[/tex])), we can use the trigonometric identity:

sin(2[tex]\theta[/tex]) = 2sin([tex]\theta[/tex])cos([tex]\theta[/tex])

Replacing sin(2[tex]\theta[/tex]) in the expression, we get:

(2sin([tex]\theta[/tex])cos([tex]\theta[/tex]))/((2cos([tex]\theta[/tex]))

The common factor of (2cos([tex]\theta[/tex]) in the numerator and denominator cancel out, resulting in:

sin([tex]\theta[/tex]).

Therefore, the simplified expression is sin([tex]\theta[/tex]).

The correct answer is c. sin([tex]\theta[/tex]).

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The approximate probability that the mean tip for the random sample of 32 customers is greater than $15.50 is 0.0793.

We use the Central Limit Theorem, which states that for a sufficiently large sample size, the sampling distribution of the sample mean will approach a normal distribution, regardless of the shape of the original population distribution.

Given that the population distribution of tips is left-skewed with a mean of $15 and a standard deviation of $2, we can approximate the sampling distribution of the sample mean as a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

First, let's calculate the standard deviation of the sampling distribution (also known as the standard error):

Standard error = Population standard deviation / sqrt(sample size)

Standard error = $2 / sqrt(32) ≈ $0.3536

Next, we need to calculate the z-score, which measures the number of standard errors away from the mean:

z = (sample mean - population mean) / standard error

z = ($15.50 - $15) / $0.3536 ≈ 1.4142

Finally, we can use a standard normal distribution table or a calculator to find the probability that the z-score is greater than 1.4142. The approximate probability is 0.0793.

The approximate probability that the mean tip for the random sample of 32 customers is greater than $15.50 is approximately 0.0793.

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The value of tan^−1(tan(m)) where m = 17π/2 radians is undefined (∅) without further information about the value of k.

The inverse tangent function, often denoted as tan^−1(x) or atan(x), is a mathematical function that gives the angle whose tangent is equal to a given value. It is the inverse of the tangent function (tan(x)).

The value of tan^−1(tan(m)) can be calculated using the property of the inverse tangent function, which states that tan^−1(tan(x)) = x - kπ, where k is an integer that makes the result fall within the range of the inverse tangent function.

In this case, m = 17π/2 radians, and we need to find tan^−1(tan(m)). Let's calculate it:

m - kπ = 17π/2 - kπ

Since m = 17π/2 radians, we have:

tan^−1(tan(m)) = 17π/2 - kπ

The result is in terms of k, and we don't have any additional information about the value of k. Therefore, we cannot determine the exact numerical value of tan^−1(tan(m)) without knowing the specific value of k.

Hence, the value of tan^−1(tan(m)) is undefined (∅) without further information about the value of k.

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The graph of the parent absolute value function is decreasing over the interval (-∞, 0). The function exhibits a decreasing behavior as x moves from negative infinity towards zero, where the absolute value decreases.

The parent absolute value function is defined as f(x) = |x|. To determine where the graph of this function is decreasing, we need to identify the intervals where the function's slope is negative.

Let's analyze the behavior of the parent absolute value function:

For x < 0, the function can be rewritten as f(x) = -x. In this interval, the function is a linear function with a negative slope of -1. As x decreases, f(x) also decreases, indicating a decreasing behavior.

For x > 0, the function remains f(x) = x. In this interval, the function is a linear function with a positive slope of 1. As x increases, f(x) also increases, indicating an increasing behavior.

At x = 0, the function is not differentiable since the slope changes abruptly from negative to positive. However, it is worth noting that the function does not strictly decrease or increase at x = 0.

Therefore, we can conclude that the graph of the parent absolute value function is decreasing over the interval (-∞, 0).

In this interval, as x moves from negative infinity towards zero, the function values decrease. The farther away x is from zero (in the negative direction), the larger the absolute value, resulting in a decrease in the function values.

On the other hand, the graph of the parent absolute value function is increasing over the interval (0, ∞), as explained earlier.

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The domain of f⁻¹(x) = [2,∞) is in interval notation, where 2 is included as the inverse of the function at x = 2 will exist. The solution is:  

[tex]f^1(x) = x^2 - 4[/tex]  for the domain [2,∞)

Given function is f(x) = √(x-4+8)

= √(x+4) where x ≥ 4

We are to find the inverse of f(x).

The steps to find the inverse are as follows:

Replace f(x) by y, to get x in terms of y:

y = √(x+4)

Squaring both sides, we get:

y² = x + 4

which means, x = y² - 4

Replacing x by f⁻¹(x) and y by x in the above equation we get:

[tex]f^{-1}(x) = x^2 - 4[/tex]

where x ≥ √4 = 2.

So the domain of f⁻¹(x) = [2,∞) is in interval notation, where 2 is included as the inverse of the function at x = 2 will exist.

Hence, the solution is:  [tex]f^1(x) = x^2 - 4[/tex]  for the domain [2,∞)

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The function S maps subsets of X to bit strings of length 3. For each element in X, if it belongs to the subset Y, the corresponding bit in the string is set to 0; otherwise, it is set to 1. The value of S(X) will provide the bit string representation of all elements in X.

Given the set X={a, b, c}, the function S maps subsets of X to bit strings of length 3. Let's determine the value of S(X).

For element a, since a∈X, the corresponding bit s1 is set to 0.

For element b, since b∈X, the corresponding bit s2 is set to 0.

For element c, since c∈X, the corresponding bit s3 is set to 0.

Therefore, the value of S(X) is 0, 0, 0; representing that all elements a, b, and c are present in the set X.

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