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3 years ago

Find the mean for the amounts: $17,485; $14,978; $13,592; $14,500; $18,540; $14,978. Round to the nearest dollar.

Mathematics

1 answer:

Phoenix [80]3 years ago

8 0

Answer:$15,679

Step-by-step explanation:

Mean is calculated as total sum of the amount divided by number of the amount.

From the given question, we add the amount together and divide by six:

Mean = $17,485 + $14,978+ $13,592 +$14,500 + $18,540 +$14,978/6

=$94,073/6

=$15,678.8

To the nearest dollar will be $15,679 since 0.8 is greater than 0.5, we round up.

I hope this helps.

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Ok so I'll demonstrate in programming language python.

import math
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3 years ago

Last year Salma opened an investment account with $8200. At the end of the year the amount in the account had increased by 6.5%.

nikklg [1K]

Answer:

61,500

Step-by-step explanation:

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3 years ago

What is 3 numbers between 2.15 and 2.17

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2.153, 2.164, 2.167
Those are just a 3 of thousands of possibilities.

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4 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

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3 years ago

A ball was kicked into the air from a balcony 20

Ainat [17]

<span>Max ht is at the vertex of the parabola, at t = -b/2a
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hope this helps

</span>

5 0

4 years ago

Read 2 more answers