Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a service facility. Suppose they are normal random variables with means of 50 minutes, 60 minutes, and 40 minutes, respectively. The standard deviations are 15 minutes, 20 minutes, and 10 minutes, respectively. a Suppose X1, X2, and X3 are independent. All three repairs must be completed on a given object. What is the mean and variance of the total repair time for this object.

Answers

Answer 1

Answer:

Answer:

The mean of the total repair time is 150 minutes.

The variance of the total repair time is 725 minutes^2.

Step-by-step explanation:

To solve this problem, we have to use the properties of the mean and the variance. Our random variable is the sum of 3 normal variables.

In the case, for the mean, we have that the mean of the sum of 3 normal variables is equal to the sum of the mean of the 3 variables:

$y=x_1+x_2+x_3 \n\nE(y)=E(x_1+x_2+x_3)=E(x_1)+E(x_2)+E(x_3)\n\nE(y)=50+60+40=150$

For the variance, we apply the property for the sum of independent variables (the correlation between the variables is 0):

$V(y)=V(x_1)+V(x_2)+V(x_3)\n\nV(y)=s_1^2+s_2^2+s_3^2\n\nV(y)=15^2+20^2+10^2\n\nV(y)=225+400+100\n\nV(y)=725$

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The amount of cream sauce on the fettuccine at Al Fred-O's follows a Normal distribution, with a mean of 3.78 ounces and a standard deviation of 0.14 ounce. A random sample of 12 plates of fettuccine is selected every day and the sauce is measured. What is the probability that the mean weight will exceed 3.81 ounces

Answers

Answer:

22.66% probability that the mean weight will exceed 3.81 ounces

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = (\sigma)/(√(n))$

In this problem, we have that:

$\mu = 3.78, \sigma = 0.14, n = 12, s = (0.14)/(√(12)) = 0.04$

What is the probability that the mean weight will exceed 3.81 ounces

This probability is 1 subtracted by the pvalue of Z when X = 3.81. So

$Z = (X - \mu)/(\sigma)$

By the Central Limit Theorem

$Z = (X - \mu)/(s)$

$Z = (3.81 - 3.78)/(0.04)$

$Z = 0.75$

$Z = 0.75$ has a pvalue of 0.7734

1 - 0.7734 = 0.2266

22.66% probability that the mean weight will exceed 3.81 ounces

Answer:

Correct answer is 0.2290

Step-by-step explanation:

An animal shelter spends $2.00 per day to care for each bird and $6.00 per day to care for each cat. Maya noticed that the shelter spent $70.00 caring for birds and cats on Thursday. Maya found a record showing that there were a total of 27 birds and cats on Thursday. How many birds were at the shelter on Thursday ?

Answers

Answer:there were 23 birds and 4 cats

Step-by-step explanation:

Let x represent the number of birds at the shelter on Thursday.

Let y represent the number of cats at the shelter on Thursday.

Maya found a record showing that there were a total of 27 birds and cats on Thursday. This means that

x + y = 27

An animal shelter spends $2.00 per day to care for each bird and $6.00 per day to care for each cat. Maya noticed that the shelter spent $70.00 caring for birds and cats on Thursday. This means that

2x + 6y = 70 - - - - - - - - - - - - - - 1

Substituting x = 27 - y into equation 1, it becomes

2(27 - y) + 6y = 70

54 - 2y + 6y = 70

- 2y + 6y = 70 - 54

4y = 16

y = 16/4

y = 4

x = 27 - y = 27 - 4

x = 23

Q # 19 The data below shows the. number of hours a week on average group of students spend volunteering for community service project.What is a cumulative frequency table that represents the data.4 5 10 21 6 2 9 8 12 15 8 14 6 4 6 11 3 2 9 16 22 23

Answers

If those two are your only choices then the answer is none of the above.

Attached is a cumulative frequency table for your data.

If you take a look at the two tables given, the frequencies were not tallied properly. If the frequency column is wrong, then the cumulative frequency will be wrong.

The answer is then none of the above or find one that matches the table attached.

Evaluate the expression of 1/2 x3 when x = 4

Answers

Answer: 6

Step-by-step explanation:

1/2 x3 = x4/2

x=4

3 x 4 = 12

12/2 = 6

How to you simplify:

Answers

Answer:

Step-by-step explanation:

First you must use distributive property on the numberator. You multiply 4 with 8, then 4 with 2. It will look like this: 32-8. Then you solve, it equals 24. Then add the denominator, which equals 12. Lastly, divide 24 by 12, which gets you 2.

Answer:

Step-by-step explanation:

4(8-2)/3+9

First subtract 8 by 2

8-2=6

4(6)/3+9

Now multiply 4 and 6

4(6)=24

24/3+9

now add 3 and 9

3+9=12

24/12

now divide the top by the bottom

24/12=2

Hope This Helps :)

What is the quotient of 1 over (1 plus the square root of 3)?

Answers

You can rationalize the denominator by multiplying by 1 in the form of the conjugate of the denominator divided by itself.

$(1)/(1+√(3))=(1)/(1+√(3))\cdot (1-√(3))/(1-√(3))\n\n=(1-√(3))/(1^(2)-(√(3))^(2))=(1-√(3))/(1-3)\n\n=(√(3)-1)/(2)$

This eqvaluates to approximately 0.3660254037844386.

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