-3=z-8 Solve and answer the equation

Both sets of values have an average of 13. Is Set A's standard deviation smaller, larger, or about the same as Set B's? (Note: T

Softa [21]

Answer:

Set A's standard deviation is larger than Set B's

Step-by-step explanation:

Standard deviation is a measure of variation. One way to judge the value of standard deviation is by looking at the range of the data. In general, a dataset with a smaller range will have a smaller standard deviation.

The range of data Set A is 25-1 = 24.

The range of data Set B is 18-8 = 10.

Set A's range is larger, so we expect its standard deviation to be larger.

__

The standard deviation is the root of the mean of the squares of the differences from the mean. In Set A, the differences are ±12, ±11, ±10. In Set B, the differences are ±5, ±3, ±1. We don't actually need to compute the RMS difference to see that it is larger for Set A.

Set A's standard deviation is larger than Set B's.

5 0

3 years ago

In triangle ABC, angle A is 25 degree, and Angle B is 105 degrees. what is the measure of angle c

Scilla [17]

Angle c= 60°

This is because 105+25+x=180°
Hope this helps and good luck!! :)

8 0

3 years ago

Sunday by 11:59pm

beks73 [17]

Answer:

The amount of markup is 88 dollars.

Step-by-step explanation:

Given that the whole sale price of bicycle is: $400

$P = \$400$

Mark up is the profit that the seller of the product makes after selling the product more in price more than its cost or buying price.

Given that mark up percentage is 22%

It can be written mathematically as:

$Mark\ up = 22\%\ of\ Price\\Mark\ up = 0.22 * P\\Mark\ up = 0.22 * 400\\= \$88$

Hence,

The amount of markup is 88 dollars.

7 0

3 years ago

A water wheel has a radius measuring between 13 feet and 24 feet. The wheel is able to turn 7π9 radians from its starting positi

madam [21]

Answer:

31.76 ft and 58.64 ft

Step-by-step explanation:

The radius measures between 13 feet and 24 feet.

The wheel is able to turn 7π/9 radians before getting stuck.

We need to find the range of distances that the wheel could spin before getting stuck. That is, the length of arc.

Length of an arc is given as:

$L = \frac{\theta}{2\pi} * 2\pi r$

where θ = central angle = 7π/9 radians

r = radius of the circle

Therefore, for 13 feet:

$L = \frac{7\pi}{18 \pi} * 2 * \pi * 13\\\\L = 31.76 ft$

For 24 feet:

$L = \frac{7\pi}{18 \pi} * 2 * \pi * 24\\\\L = 58.64 ft$

The wheel could spin between 31.76 ft and 58.64 ft before getting stuck.

4 0

3 years ago

Suppose that scores on the mathematics part of a test for eighth-grade students follow a Normal distribution with standard devia

riadik2000 [5.3K]

Answer:

We need an SRS of scores of at least 153.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large an SRS of scores must you choose?

This is at least n, in which n is found when $M = 20, \sigma = 150$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$20 = 1.645*\frac{150}{\sqrt{n}}$

$20\sqrt{n} = 1.645*150$

$\sqrt{n} = \frac{1.645*150}{20}$

$\sqrt{n} = 12.3375$

$(\sqrt{n})^{2} = (12.3375)^{2}$

$n = 152.2$

Rounding to the next whole number, 153

We need an SRS of scores of at least 153.

8 0

3 years ago