Roberto is having his house painted. The job takes three days, and he pays the painter the same hourly rate

Dominique decided to build his sons a fort in the backyard. He put a roof on the fort in the shape of a triangular prism.

VARVARA [1.3K]

Answer:

B. 216 ft

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Simplify (6x^-2) (0.5x)^4

loris [4]

Answer:

(3/8)x^2

Step-by-step explanation:

Rewriting the given expression as follows simplifies the solution. Note that 6x^(-2) could be written as:

6

--------

x^2

Also, (0.5x)^4 = 0.0625x^4.

Then we have:

6*0.0625*x^4 0.375

---------------------- = ------------ * x^(4-2), or (3/8)x^2

x^2 1

Note that x^4 / x^2 = x^2.

6 0

3 years ago

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for random sampleof 1044 pe

Evgesh-ka [11]

Answer:

A) D

B) B

C) C

D) B

Step-by-step explanation:

The complete question is:

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for random sampleof 1044 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.44 hours with a standarddeviation of 0.57 hour. (A.) A histogram of time spent eating and drinking each day is skewed right. Use the result to explain why a large samplesize is needed to construct a confidence interval for the mean time spent eating and drinking each day.A. The distribution of the sample mean will never be approx. normalB. The distribution of the sample mean will always be approx. normalC. Since the distribution of time eating and drinking each is normally distributedthe sample must be large so that the distribution of the sample mean will be apprx. NormalD. Since the distribution of time spent eating and drinking each day is not normally distributed (skewed right), the sample must be large so that the distribution of the sample mean will be approx. normal.

(B) In 2010, there were over 200 million people nationally age 15 or older. Explain why this, along with the fact that that the data were obtained using a random sample, satisifies the requirements for constructing a confidence interval.A. The sampe size is greater than 5% of the population.B. The sample size is less than 10% of the populationC. The sample size is greater than 10% of the populationD. The sample size is less than 5% of the population.

(C.) Determine and interpret a 95% confidence interval for themean amount of time Americans age 15 older spend eating and drink each day.A. the nutritionist is 95% confident that the mean amount of time spent eatingand drinking per day is between blank and blank hours.B. the nutritionist is 95% confident that the amount of time spent eating or drinking per day forany individuals is between blank and blank hours.C. There is a 95% probability that the meant amount of time spent eating or drinking per day is between blank and blank hours.D. The requirements for constructing a confidence interval are not satisfied.

(D) Could the interval be used to estimate the mean amount of time a 9 year old spends eating and drinking each day?A. No, the interval is about individual time spent eating or drinking per day and cannot be used to find the mean time spent eating or drinking per day.B. No the intreval is about people age 15 or older. The mean amount of time spent eatinf or drinking per day for 9 year olds may differ.C. Yes, the interval is about individual time spent eating or drinking per dat and can be used to find the mean amount of time a 9 year old spends eating and drinking each day.D. Yes, the interval is about the mean amount of time spent eating or drinking per day for the 9 year olds.E. A confidence interval could not be constructed in part (C)D

A) a large sample size ensures normal distribution

B) a good sample size is around 10% of the population

C) confidence interval is a range of value and defines probability that mean of the population will lie with in the range

D) The data given is for 15 years and older so it can't be used for 9 years and younger

3 0

3 years ago

Belief in UFOs A survey found that of people believe that they have seen a UFO. Choose a sample of people at random. Find the pr

Mkey [24]

Answer:

b is the answer am not sure you know

8 0

3 years ago

Read 2 more answers

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper c

aalyn [17]

Answer:

a) This means that there is a 33% probability that one car chosen at random will have less than 49.5 tons of coal.

b) There is a 0.04% .probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal

Step-by-step explanation:

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

$Z = \frac{X - \mu}{\sigma}$

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.

In this problem, we have that:

The actual weights of coal loaded into each car are normally distributed, with mean = 50 tons and standard deviation = 0.9 ton. This means that $\mu = 50, \sigma = 0.9$

(a) What is the probability that one car chosen at random will have less than 49.5 tons of coal? (Round your answer to four decimal places.)

This is the pvalue of Z when $X = 49.5$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{49.5 - 50}{0.9}$

$Z = -0.44$

$Z = -0.44$ has a pvalue of 0.33.

This means that there is a 33% probability that one car chosen at random will have less than 49.5 tons of coal.

(b) What is the probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal? (Round your answer to four decimal places.)

Now we have to use the standard deviation of the sample, since we are working with the sample mean. That is

$s = \frac{\sigma}{\sqrt{n}} = \frac{0.9}{\sqrt{35}} = 0.15$

Now, we find pvalue of Z when $X = 49.5$

$Z = \frac{X - \mu}{s}$

$Z = \frac{49.5 - 50}{0.15}$

$Z = -3.33$

$Z = -3.33$ has a pvalue of 0.0004.

This means that there is a 0.04% .probability that 35 cars chosen at random will have a mean load weight of less than 49.5 tons of coal

7 0

3 years ago