assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviation of 30 pounds. suppose that 25 people squeeze into an elevator that is designed to hold 4300 pounds. 2a) what is the probability that the load (total weight) exceeds the design limit?

Based on the information in the graph, we can establish that a person who studied for six hours could have a grade of 204.

To calculate the qualification that a person who studied 6 hours would have, we must carry out the following mathematical procedure: Rule of three.

In this case, if a person who studied for two hours had a result of 68 points, then we can make the projection to find out how much someone who studied for six hours obtained.

According to the above, this person got 204 points after studying for 6 hours.

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Answer:

Step-by-step explanation:

Step 1: We know that the cosine of an angle in Quadrant III is negative.

Step 2: We convert the given angle A into radians by multiplying it by pi/180.

A = (7/√120) * (π/180) = (7π)/(12√5)

Step 3: We use the cosine formula to find the exact value of cos A.

cos A = cos((7π)/(12√5))

Step 4: We simplify the result using trigonometric identities.

cos A = (-1/3)√5

The probability that the rack ends up in alphabetical order is 1/9! or approximately 1.46 x 10⁻⁷.

The problem involves arranging 9 CDs out of 16 in alphabetical order in a CD rack. The total number of possible ways to arrange 9 CDs out of 16 is 16 choose 9, which equals 11440. Only one of these arrangements is in alphabetical order. Therefore, the probability of randomly selecting an alphabetical order is 1/11440, which is approximately 0.0000874 or 0.00874%. This is a very small probability, indicating that it is highly unlikely that the CDs will end up in alphabetical order by chance.

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The probability of getting a ticket when parking for a random amount of time that is uniformly distributed on .0; 4/ hours is 1/2.

Since the parking officials pass the same point every two hours, we can divide the possible parking times into two-hour intervals. If a car is parked for less than two hours, it will not be marked with chalk. If a car is parked for more than two hours, it will be marked with chalk and will receive a ticket if it remains parked for another two hours.

Since the possible parking times are uniformly distributed on .0; 4/ hours, the probability of parking for less than two hours is 2/4 = 1/2, and the probability of parking for more than two hours is also 1/2. Therefore, the probability of getting a ticket is the probability of parking for more than two hours, which is 1/2.

Alternatively, we can calculate the probability of getting a ticket by finding the area of the part of the uniform distribution that corresponds to parking for more than two hours. This area is a triangle with base 2 and height 1/2, so its area is (1/2)(2)(1/2) = 1/2. Therefore, the probability of getting a ticket is also 1/2.

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Answer:

First, we solve for the number of minutes in 7 and 3/5 hours by multiplying the number by 60 giving us,

                           (7 + 3/5) x (60) = 456 minutes

Spending 30 minutes for lunch will leave her with 426 minutes. Then, spending 42 minutes for switching of classes will finally give her 414 minutes.

We then divide this value by 6 (for her 6 classes) giving us 69 minutes. Thus, each class is 69 minutes long.

Step-by-step explanation:

Weight of wood block is about 15.5 ounces.

Given that:

Height of block [tex]= 4 \ \text{inches}[/tex]

Diameter of block [tex]= 3 \ \text{inches}[/tex]

1 cubic inch of wood [tex]= 0.55 \ \text{ounces}[/tex]

Find:

Weight of wood block

Computation:

Radius of block [tex]= 3 \div 2[/tex]

Radius of block [tex]= 1.5 \ \text{inches}[/tex]

Volume of toy [tex]= \pi r^2h[/tex]

Volume of toy [tex]= (3.14)(1.5)^2(4)[/tex]

Volume of toy [tex]= (3.14)(2.25)(4)[/tex]

Volume of toy [tex]= 28.26 \ \text{cubic inch}[/tex]

Weight of wood block [tex]= \text{Volume of toy} \times \ \text{0.55 ounces}[/tex]

Weight of wood block [tex]= 28.26 \times 0.55 \ \text{ounces}[/tex]

Weight of wood block [tex]= 15.5 \ \text{ounces}[/tex]

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An equation of the quadratic function that passes through the points (-2, 0), (2, - 16) and (6, 0) is: D. f(x) = (x - 2)² - 16.

In this exercise, you are required to determine the factored form of the given quadratic function that passes through the points (-2, 0), (2, - 16) and (6, 0).

In Mathematics, the vertex form of a quadratic function is represented by the following mathematical equation:

f(x) = a(x - h)² + k

Where:

Based on the information provided in the graph above, we can determine the value of a as follows:

f(x) = a(x - h)² + k

0 = a(-2 - 2)² - 16

0 = 16a - 16

16 = 16a

a = 1

Therefore, the required quadratic function is given by:

f(x) = a(x - h)² + k

f(x) = y = (x - 2)² - 16

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The length οf diameter οf the given sphere is 18m.

A sphere is a three-dimensiοnal οbject which is rοund in shape. It has nο edges οr vertices. The sphere can be defined in three axes, thοse are x-axis, y-axis and z-axis.

The given diagram is a sphere.

Given that,

The surface area οf the sphere is  1017.36 m²

Fοrmula fοr surface area οf the sphere = 4πr² where r is the radius and π=3.14

Equating both we get,

4πr² = 1017.36

[ dividing bοth sides by 4 ]

⇒πr² = (1017.36)/4

⇒πr² = 254.34

[putting the value of π]

⇒r² = 254.34/ 3.14

⇒r² = 81

⇒r =± 9

As radius cannot be a negative value so we take r=9

Sο the radius οf the sphere is 9 m

Diameter = 2× radius =2×9= 18m

Hence, diameter οf the given sphere is 18m.

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The significance level, denoted by alpha (α), is a predetermined threshold used in hypothesis testing to determine the probability of rejecting the null hypothesis when it is actually true.

The steps involved in determining the significance level are as follows:

Therefore, the significance level indicates the probability of rejecting the null hypothesis when it is actually true. A smaller significance level (e.g., 0.01) means that the researcher is less willing to accept the risk of rejecting the null hypothesis when it is actually true, while a larger significance level (e.g., 0.05) means that the researcher is more willing to accept that risk.

In conclusion, the significance level is a critical parameter in hypothesis testing that helps researchers make decisions about the null hypothesis. By setting a predetermined threshold for the probability of rejecting the null hypothesis, researchers can control the risk of Type I error (rejecting the null hypothesis when it is actually true) in their statistical analyses.

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a. n=100

The probability that a sample proportion will be within +/-6% of the population proportion for a sample of size 100 is 0.7469.

b. n=200  

The probability that a sample proportion will be within +/-6% of the population proportion for a sample of size 200 is 0.8214.

c. n=500  

The probability that a sample proportion will be within +/-6% of the population proportion for a sample of size 500 is 0.8977.

d. n=1000 Th

The probability that a sample proportion will be within +/-6% of the population proportion for a sample of size 1000 is 0.9227.

"With a larger sample size, there is a higher probability that the sample proportion will be within a certain range of the population proportion" is the advantage of having larger sample size.

To solve this problem, we need to use the formula for the standard error of the sample proportion:

SE = [tex]\sqrt{ p \times (1 - p) / n }[/tex]

where p is the population proportion, and n is the sample size.

a) For n = 100:

SE = [tex]\sqrt{0.45 \times (1 - 0.45) / 100 }[/tex] = 0.0499

The margin of error is +/-6%, which translates to +/-0.06 in proportion units.

We can find the z-scores corresponding to these margins of error using a standard normal distribution table:

[tex]z_1[/tex] = (0.45 - 0.06 - 0.45) / 0.0499 = -1.2024

[tex]z_2[/tex] = (0.45 + 0.06 - 0.45) / 0.0499 = 1.2024

The probability of a sample proportion within +/-6% of the population proportion is the area under the normal curve between these two z-scores:

P(-1.2024 < z < 1.2024) = 0.7469

b) For n = 200:

SE = [tex]\sqrt{ 0.45 \times (1 - 0.45) / 200 }[/tex] = 0.0353

[tex]z_1[/tex] = (0.45 - 0.06 - 0.45) / 0.0353 = -1.691

[tex]z_2[/tex] = (0.45 + 0.06 - 0.45) / 0.0353 = 1.691

P(-1.691 < z < 1.691) = 0.8214

c) For n = 500:

SE = [tex]\sqrt{ 0.45 \times (1 - 0.45) / 500 }[/tex] = 0.0249

[tex]z_1[/tex] = (0.45 - 0.06 - 0.45) / 0.0249 = -2.2088

[tex]z_2[/tex] = (0.45 + 0.06 - 0.45) / 0.0249 = 2.2088

P(-2.2088 < z < 2.2088) = 0.8977

d) For n = 1000:

SE = [tex]\sqrt{ 0.45 \times (1 - 0.45) / 1000 }[/tex] = 0.0177

[tex]z_1[/tex] = (0.45 - 0.06 - 0.45) / 0.0177 = -2.5424

[tex]z_2[/tex] = (0.45 + 0.06 - 0.45) / 0.0177 = 2.5424

P(-2.5424 < z < 2.5424) = 0.9227
Advantage of a larger sample size:
With a larger sample size, there is a higher probability that the sample proportion will be within a certain range of the population proportion.

This is because a larger sample size reduces the standard error, resulting in a more precise estimate of the population proportion.

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Question:-

The population proportion is 0.45 . What is the probability that a sample proportion will be within +/-6 of the population proportion for each of the following sample sizes? Round your answers to 4 decimal places. Use z-table.

a.  n=100

b.  n=200

c.  n=500

d.   n=1000

What is the advantage of a larger sample size?

With a larger sample, there is a - Select your answer (lower or higher) Item 5 probability will be within of the population proportion .

If at least one card is a heart then P is three-fοurths.

The study οf prοbability examines the likelihοοds οf results οccurring and is based οn the ratiο οf likely and imprοbable scenariοs.

Tο determine the likelihοοd οf at least οne card being a heart, we can determine the likelihοοd that neither card will be a heart and subtract that prοbability frοm 1.

The prοbability οf the first card nοt being a heart is 26/52 = 1/2, since there are 26 nοn-heart cards in the deck οut οf a tοtal οf 52 cards. Since the card is replaced, the prοbability οf the secοnd card nοt being a heart is alsο 1/2.

Therefοre, the prοbability οf neither card being a heart is (1/2) x (1/2) = 1/4.

The prοbability οf at least οne card being a heart is 1 - (1/4) = 3/4.

Hence, the answer is: three-fοurths.

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A scatterplot can help us visualize the relationship between two variables and determine if there is a correlation between them. If the scatterplot shows a clear pattern or trend, conclude that there is a relationship between the two variables.

Without access to the scatterplot or the newspaper report, it is difficult for me to provide a definitive answer. However, in general, to determine if the scatterplot supports the newspaper report about the number of semesters and starting salary, we need to examine the pattern in the scatterplot.

Suppose we see a clear trend or pattern, such as a linear relationship where an increasing number of semesters is associated with an increased starting salary or a non-linear relationship. In that case, we can say that the scatterplot supports the newspaper report.

However, if we see no discernible pattern or trend or significant variability in the data points makes it is difficult to draw any meaningful conclusions, then we cannot say that the scatterplot supports the newspaper report.

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Answer:

To compare the two functions, it is necessary to find the x values for which each function is positive.

For f(x), we can solve for when it is positive by setting the equation equal to zero and finding the x-intercepts.

-x^2 + 2x = 0

-x(x - 2) = 0

-x = 0 and x = 2

This tells us that f(x) is positive for (-inf, 0) and (2, inf).

Next, for g(x), we can solve for when it is positive by setting the equation equal to zero and finding the x-intercept.

log(2x+1) = 0

2x + 1 = 1

x = 0

This tells us that g(x) is positive for (0, inf).

Therefore, the interval on which both functions are positive is (0, 2).

Andres will practice with his band for a total of 6 hours and 45 minutes over these 6 days.

To add these times, we need to find a common denominator for 2, 4, and 8, which is 8.

1[tex]\frac{1}{2}[/tex] can be written as [tex]\frac{4}{8}[/tex],

1[tex]\frac{3}{4}[/tex] can be written as [tex]\frac{7}{8}[/tex],

and 2 can be written as [tex]\frac{16}{8}[/tex].

Now we can add the times:

[tex]\frac{4}{8} +\frac{7}{8} +\frac{16}{8} =\frac{27}{8}[/tex]

So Andres will practice for [tex]\frac{27}{8}[/tex] hours this week.

Next week, he will practice for the same number of hours on each day, so he will practice for a total of [tex]\frac{27}{8}[/tex] hours over the three days.

To find the total number of hours he will practice over the 6 days

[tex]\frac{27}{8} + \frac{27}{8} = \frac{54}{8}[/tex]

Simplifying,

[tex]\frac{54}{8} = 6\frac{3}{4}[/tex]

Andres will practice with his band for a total of 6 hours and 45 minutes over these 6 days.

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Answer:

Step-by-step explanation:

Therefore , the solution of the given problem of expressions comes out to be cost $357.

It is preferable to use moving integer variables, which can be rising, decreasing, or blocking, rather than estimates that are generated at random. They could only assist one another by exchanging resources, knowledge, or answers to problems. A declaration of truth equation may include the justifications, components, and mathematical comments for strategies like extra disapproval, production, and mixture.

Here,

Let's say Blair's previous PC cost "x" dollars.

Her new computer cost $5 less than twice as much as her previous one, according to the information provided, which can be written as:

=> Cost of Blair's new computer =  2x - 5

She also purchased a new PC, which cost her $709. In light of these two bits of knowledge, we can construct the following equation:

=> 2x - 5 = 709

In order to find "x," add 5 to both parts, divide by 2, and we get:

=> 2x = 714

=> x = 357

Blair's previous PC therefore cost $357.

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A sample size of 105 was used in the study with 99% of confidence interval.

A range of values that, with a certain degree of confidence, is likely to include the population mean is known as a confidence interval for a mean. The formula for calculating the confidence interval is:

CI = X ± Zα/2 × σ/√n

where X is the sample mean, σ is the population standard deviation, n is the sample size, and Zα/2 is the critical value of the normal distribution at α/2. Here, for a 99% confidence level, α = 0.01 and Zα/2 = 2.576).

We can use this formula to solve for n:

n = (Zα/2 × σ / E)²

where E is the margin of error, which is half of the confidence interval.

In this case, we have a 99% confidence interval with a range of 152 to 160 and a standard deviation of 15.

The margin of error (E) can be calculated as half of the range of the confidence interval:

E = (160 - 152) / 2 = 4

By adding these values to the previous formula, we obtain:

n = (2.576 × 15 / 4)² = 105

Therefore, we can conclude that a sample size of at least 105 was used in this study.

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Complete question is:

A 99% confidence interval for a population mean was reported to be 152 to 160. If standard deviation is 15, what sample size was used in this study? (Round your answer up to the next whole number.)

On solving the provided question we can say that So, the discriminant of the given quadratic equation is 109.

A quadratic equation is x ax2+bx+c=0, which is a single variable quadratic polynomial. a 0. Because this polynomial is of second order, the Basic Theorem of Algebra implies that it has at least one solution. Simple or complex solutions are possible. A quadratic equation is a quadratic equation. This means it has at least one word that must be squared. One of the most common solutions for quadratic equations is "ax2 + bx + c = 0." where a, b, and c are numerical coefficients or constants. where the variable "X" is unnamed.

The given quadratic equation is [tex]3x^2 - 5x - 7 = 0[/tex]

(a) Here, the equation is [tex]3x^2 - 5x - 7 = 0[/tex], so a = 3, b = -5, and c = -7. Therefore, the discriminant is:

[tex]b^2 - 4ac = (-5)^2 - 4(3)(-7) = 25 + 84 = 109[/tex]

So, the discriminant of the given quadratic equation is 109.

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Based on the information provided, we cannot conclude with certainty that the average pickup order has a higher amount than the average delivery order.

Based on the information given, we can conclude that there is a 95% chance that the true mean for pickup orders falls between $33 and $51, and the true mean for delivery orders falls between $21 and $41. However, we cannot conclude with certainty that the average pickup order has a higher amount than the average delivery order.

To determine if there is a significant difference between the means of the two groups, we need to perform a hypothesis test. We can set up our null hypothesis as "there is no significant difference between the means of the pickup and delivery orders" and our alternative hypothesis as "the average pickup order has a higher amount than the average delivery order."

We can use a two-sample t-test to test this hypothesis. Assuming equal variances, we would calculate the test statistic as

t = (xpickup - xdelivery) / (s / √n)

where xpickup and xdelivery are the sample means for pickup and delivery orders, s is the pooled standard deviation, and n is the sample size for each group.

If our calculated t-value is greater than the critical value at a chosen significance level (e.g. 0.05), we can reject the null hypothesis and conclude that there is a significant difference between the means of the two groups.

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Since 3 has the highest frequency, the mode is 3.Now, the curve is positively skewed if mean is bigger than mean.

Total number is a term used to refer to the sum of two or more individual elements. It is commonly used in the context of mathematics and can refer to the result of addition, subtraction, multiplication, or division. For example, the total number of apples in a basket could be five, the total number of days in a week could be seven, and the total number of people in a family could be four. In any case, the total number is the sum of all the individual elements that are being considered.

Presented to us is a distribution:

{4.5, 3, 1, 2, 4, 3, 6, 4.5, 4, 5, 2, 1, 3, 4, 3, 2}

We must determine the mean.

Mean is equal to the sum of all observations divided by the total number of observations.

Mean = 52/16

Mean = 3.25

We now determine the mode, which is the observation with the highest frequency, or how frequently it has occurred.

observations per unit of time

4.5 2

3 4

1 2

2 3

4 3

6 1

5 1

Since 3 has the highest frequency, the mode is 3.
Now, the curve is positively skewed if mean is bigger than mean.

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Complete questions as follows-
What is the mean of the given distribution, and which type of skew does it exhibit?
{4.5, 3, 1, 2, 4, 3, 6, 4.5, 4, 5, 2, 1, 3, 4, 3, 2}

WHAT IS THE MEAN? and WHAT TYPE OF SKEW EXHIBITS? negative, positive, symmetric, zero etc..

Answer:

[tex]\boxed{f^{-1}(x)=\frac{x-3}{-2}}[/tex]

Step-by-step explanation:

The inverse of the function satisfies the following:

[tex]f(a)=b \qquad \implies \qquad f^{-1}(b)=a[/tex]

Given the expression:

[tex]f(x)=-2x+ 3[/tex]

The inverse function can be calculated with the next steps:

[tex]\begin{aligned}f(x)=-2x+3 &\rightarrow y=-2x+3\\&\rightarrow y-3=-2x\\& \rightarrow x= \frac{y-3}{-2} \\&\therefore f^{-1}(x)=\frac{x-3}{-2}\end{aligned}[/tex]

Hope it helps.

[tex]\text{-B$\mathfrak{randon}$VN}[/tex]

0.625 is a rational number. A rational-number is the number that can be expressed as ratio of two integers with a denominator (not zero).

What is rational and irrational number?

A rational number is a number which can be expressed as the product of two integers with non-zero denominators.In other words, it is a number that can be written in the form of p/q,

where:

p and q -> integers

and q -> not equal to 0.

An irrational number, is just opposite, it cannot be expressed as the ratio of 2 integers.It is a number that cannot be written as a simple fraction, and its decimal expansion continues indefinitely. The most famous examples of irrational numbers are π and √2.

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3k+5k=8k
96/8=12
k=12
as we know blue-eyed are 3k and brown-eyed 5k so we need to put 12 instead of k
5x12=60

Answer:

A, C, E

Step-by-step explanation:

A, C, E

A monomial is a single number, or a single number multiplied by one or more variables.

A variable in a square root or in a denominator is not allowed in a monomial.

Answer:

  d.  9.4 feet

Step-by-step explanation:

You want the measure of long side XY of the right triangle that has its altitude dividing the hypotenuse into segments 3 and 8 feet in length.

In this geometry, all of the right triangles are similar. That means the ratio of long side to hypotenuse is the same for all of the triangles:

  XY/(3+8) = 8/XY

  XY² = (3+8)(8)

  XY = √88 ≈ 9.4

The length of segment XY is about 9.4 feet.

__

Additional comment

As we see here, the long side of the large triangle is the root of the product of the long segment of the hypotenuse with the whole hypotenuse:

  XY=√(8(8+3)) = √88

The same relation works on the other side of the triangle. The short side of the large triangle is the root of the product of the short segment of the hypotenuse and the whole:

  short side = √(3(11)) = √33

The relation for the altitude of the triangle is that it is the root of the product of the two hypotenuse segments:

  altitude = √(3·8) = √24

Answer: 1729 trout.

Step-by-step explanation:

We can use the logistic growth model to estimate the number of trout after 2 years:

N(t) = K / (1 + (K/N0 - 1) * e^(-rt))

where:

N(t) is the number of trout at time t

K is the carrying capacity of the lake (2700 trout)

N0 is the initial population (275 trout)

r is the growth rate (75% or 0.75, per year)

e is the base of the natural logarithm, approximately equal to 2.71828

To estimate the number of trout after 2 years, we can plug in t = 2 and solve for N(t):

N(2) = 2700 / (1 + (2700/275 - 1) * e^(-0.75*2))

N(2) = 2700 / (1 + 8.8 * e^(-1.5))

N(2) ≈ 1728.7

Therefore, we can estimate that the number of trout after 2 years will be approximately 1729 trout.

In hire-assistant, assuming that the candidates are presented in a random order, the probability of hiring exactly one time is 1/n, while the probability of hiring exactly n times is 1/n!.

In the Hire-Assistant problem, the objective is to hire the best candidate among a pool of candidates. The candidates are presented in random order, and you must decide whether to hire or reject a candidate after each interview. The hiring process ends when the best candidate is hired.

To answer the student question, let's first consider the probability of hiring exactly one time. This happens when the best candidate is the first one interviewed since no other candidate can be better than them. Since the candidates are in random order, the probability of this occurring is 1/n, where n is the total number of candidates.

Now let's consider the probability of hiring exactly n times. This means you would hire every candidate you interview. This only happens if the candidates are presented in ascending order of quality, so that each new candidate is better than the previous one. There is only 1 possible ordering like this out of the total n! possible orderings of candidates. So, the probability of hiring exactly n times is 1/n!.

In summary, the probability of hiring exactly one time is 1/n, while the probability of hiring exactly n times is 1/n!. These probabilities are derived from the random order of candidate presentations and the decision-making process of the Hire-Assistant problem.

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(a) The probability that both inspectors assign the same safety score is 0.80.

(b) The probability that the second inspector assigns a higher safety score than the first inspector is 0.11.

(c) The expectation of marginal probability mass function of the score assigned by the first inspector is 2.83 and variance is 0.83.

(d) The expectation of marginal probability mass function of the score assigned by the first inspector is 3.00 and variance is 0.77.

(e) The scores assigned by the two inspectors are not independent of each other.

(f) The covariance of the scores assigned by the two inspectors is -0.0148.

(a) The probability that both inspectors assign the same safety score is the sum of the joint probabilities where X and Y take the same value:

P(X = Y) = P(1,1) + P(2,2) + P(3,3) + P(4,4) = 0.09 + 0.15 + 0.24 + 0.32 = 0.80

(b) The probability that the second inspector assigns a higher safety score than the first inspector is given by:

P(Y > X) = P(2,1) + P(3,1) + P(4,1) + P(3,2) + P(4,2) + P(4,3) = 0.03 + 0.01 + 0.01 + 0.03 + 0.01 + 0.02 = 0.11

(c) The marginal probability mass function of the score assigned by the first inspector is obtained by summing the joint probabilities over all values of Y:

P(X = 1) = 0.09 + 0.02 + 0.01 + 0.00 = 0.12

P(X = 2) = 0.03 + 0.15 + 0.01 + 0.01 = 0.20

P(X = 3) = 0.01 + 0.03 + 0.24 + 0.02 = 0.30

P(X = 4) = 0.01 + 0.01 + 0.04 + 0.32 = 0.38

The expectation of X is given by:

E(X) = 10.12 + 20.20 + 30.30 + 40.38 = 2.83

The variance of X is given by:

Var(X) = E(X^2) - [E(X)]^2

= (1^20.12 + 2^20.20 + 3^20.30 + 4^20.38) - (2.83)^2

= 0.83

(d) The marginal probability mass function of the score assigned by the second inspector is obtained by summing the joint probabilities over all values of X:

P(Y = 1) = 0.09 + 0.03 + 0.01 + 0.00 = 0.13

P(Y = 2) = 0.02 + 0.15 + 0.01 + 0.01 = 0.19

P(Y = 3) = 0.01 + 0.03 + 0.24 + 0.02 = 0.30

P(Y = 4) = 0.01 + 0.01 + 0.04 + 0.32 = 0.38

The expectation of Y is given by:

E(Y) = 10.13 + 20.19 + 30.30 + 40.38 = 3.00

The variance of Y is given by:

Var(Y) = E(Y^2) - [E(Y)]^2

= (1^20.13 + 2^20.19 + 3^20.30 + 4^20.38) - (3.00)^2

= 0.77

(e) To determine if the scores assigned by the two inspectors are independent of each other, we need to check if the joint probability mass function can be expressed as the product of the marginal probability mass functions.

If the scores are independent, then P(X = i and Y = j) = P(X = i)P(Y = j) for all possible values of i and j.

Checking the probabilities in the given joint probability mass function, we can see that P(X = 1 and Y = 2) = 0.03, P(X = 1) = 0.12, and P(Y = 2) = 0.20.

Since P(X = 1 and Y = 2) is not equal to P(X = 1)P(Y = 2), the scores assigned by the two inspectors are not independent of each other.

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Answer:

Step-by-step explanation:

Using Pythagorean Theorem : A² = B²+C².

→x² = 5²+12² = 25+144 = 169.

→x = √169 = 13.