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valentinak56 [21]

2 years ago

14

There are 20 cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all t

he cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all 20 cities?

1 answer:

blagie [28]2 years ago

6 0

Answer:

The answer is C/85,000

Step-by-step explanation:

You might be interested in

Solve each equation and leave the answer in terms of “i”

almond37 [142]

17)
x² + 8 = -8
x² = -8 - 8
x² = 16
x = ±√-16
x = ±√16i
x = <span>±4i
</span>
18)
x² + 5 = -3
x² = -3 - 5
x² = -8
x = ±√-8
x = ±√8i
x = ±2√2i

19)
x² + 3 = 0
x² = -3
x = ±√-3
x = <span>±</span>√3i

hope this helps, God bless!

3 0

3 years ago

The length of a rectangle is four times the width minus three. If the perimeter of the rectangle is 64 meters, what is the lengt

Nastasia [14]

Answer:

L = 25 m

Step-by-step explanation:

L = 4W - 3

perimeter = 2(L + W)

64 = 2(4W - 3 + W)

divide both sides by 2:

32 = 5W - 3

add 3 to each side:

5W = 35

divide both sides by 5:

W =7

L = 4(7) - 3 = 25

6 0

3 years ago

A summer camp has divided its campers into 8 groups of 9 campers.how many campers are at the summer camp?

Virty [35]

there are 72 campers at the summer camp because if we just multiply 8 times 9 equals 72

5 0

3 years ago

Read 2 more answers

using the giving information to write an equation. let X represent the number described in the exercise.then solve the equation

ANEK [815]

A number is multiplied by 8 and the result is 28.

A number= x.

8(x)= 28

Divide both sides by 8.

28/8= 3.5

x= 3.5

I hope this helps!
~kaikers

5 0

3 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

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

X = 205

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

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

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

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

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

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

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

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0

3 years ago