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3 years ago

9

A store requires you to pay 15% up front on special orders. If you plan to special order items worth $74.86, estimate how much y

ou will have to pay up front.

1 answer:

Lina20 [59]3 years ago

8 0

Answer:

you will have to pay about $11.23 more

you will pay total of $86.09

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The time a randomly selected individual waits for an elevator in an office building has a uniform distribution with a mean of 0.

Amiraneli [1.4K]

Answer:

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are 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.

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, of at least 30, can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 0.5, \sigma = 0.289$

What are the mean and standard deviation of the sampling distribution of means for SRS of size 50?

By the Central Limit Theorem

$\mu = 0.5, s = \frac{0.289}{\sqrt{50}} = 0.0409$

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

Does it matter that the underlying population distribution is not normal?

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

What is the probability a sample of 50 people will wait longer than 45 seconds for an elevator?

We have to use 45 seconds as minutes, since the mean and the standard deviation are in minutes.

Each minute has 60 seconds.

So 45 seconds is 45/60 = 0.75 min.

This probability is 1 subtracted by the pvalue of Z when X = 0.75. So

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

By the Central Limit Theorem

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

$Z = \frac{0.75 - 0.5}{0.0409}$

$Z = 6.11$

$Z = 6.11$ has a pvalue of 1

1-1 = 0

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

8 0

3 years ago

How do I do a question 5* “I, J, k, and L

7nadin3 [17]

Answer:

Step-by-step explanation:

5 (i) 243 ÷x = -81

243 ÷ -81 = x

x = -3

5 (j) 50 ÷ x = -50

50 ÷ -50 = x

x = -1

5(k) -92 * x = 184

x = 184 ÷ -92

x = -2

5(l) -800 ÷x = -20

-800 ÷ -20 = x

x = 40

3 0

3 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$

Then

$\dfrac{V_{\text{new flask}}}{V_{\text{flask}}} =\dfrac{8}{1}=8.$

Answer 2: correct choice is D.

8 0

3 years ago

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Evaluate the expression for the given values.<br> 12x+5y/3z, where x + 1/2, y = 6 and z = 3

Semmy [17]

Answer:

4

Step-by-step explanation:

Substitute: $\frac{\frac{12}{2} +5*6}{3*3}$

Cross out the common factor: $\frac{6+5*6}{3*3}$

Calculate the product or quotient: $\frac{6 +30}{3*3}$

Calculate the product or quotient: $\frac{6+30}{9}$

Calculate the sum or difference: $\frac{36}{9}$

Cross out the common factor: $4$

Answer: $4$

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2 years ago

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Please help me with this math problem:)

Anastaziya [24]

A y-intercept always has to have an x-value of 0 so the answer is D (the last option)

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2 years ago

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