All categories

3 years ago

Someone please help me with this chart, I don’t understand it.

Mathematics

1 answer:

lidiya [134]3 years ago

6 0

Answer:

The second one is (3 x 3 x 3 x 3) x (3 x 3)

Step-by-step explanation:

You might be interested in

In your own words, describe how to simply expressions.

Nata [24]

First you gotta remove all parentheses by multiplying the factors next remove all grouping if the terms are containing exponents then you combine all terms and lastly you combine the constants

7 0

2 years ago

Read 2 more answers

The weekly salaries of sociologists in the United States are normally distributed and have a known population standard deviation

Lostsunrise [7]

Answer:

$ME =z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

The confidence level is 0.98 and the significance is $\alpha=1-0.98 =0.02$ and $\alpha/2 =0.01$ and the critical value using the table is:

$z_{\alpha/2}= 2.326$

And replacing we got:

$ME=2.326 \frac{425}{\sqrt{22}}= 210.760\ approx 210.76$

Step-by-step explanation:

For this case we have the following info given:

$\sigma = 425$ represent the population deviation

$n =22$ the sample size

$\bar X =1520$ represent the sample mean

We want to find the margin of error for the confidence interval for the population mean and we know that is given by:

$ME =z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

The confidence level is 0.98 and the significance is $\alpha=1-0.98 =0.02$ and $\alpha/2 =0.01$ and the critical value using the table is:

$z_{\alpha/2}= 2.326$

And replacing we got:

$ME=2.326 \frac{425}{\sqrt{22}}= 210.760\ approx 210.76$

3 0

2 years ago

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n

Svetradugi [14.3K]

Complete Question

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150

Answer:

Normal sampling distribution can not be used

Step-by-step explanation:

From the question we are told that

The null hypothesis is $H_o : p = 0.015$

The alternative hypothesis is $H_a : p < 0.015$

The sample size is n= 150

Generally in order to use normal sampling distribution

The value $np \ge 5$

$np = 0.015 * 150$

$np = 2.25$

Given that $np < 5$ normal sampling distribution can not be used

8 0

3 years ago

PLEASE HELP!!!

AfilCa [17]

Answer:

$h=\dfrac{10\sqrt{3}}{2}=5\sqrt{3}\ cm.$

$JP=5\sqrt{3}\ cm$

Step-by-step explanation:

Connecting points O and E and points O and J, we get triangle EOJ. This triangle is equilateral triangle, because OJ=OE=JE=r=10 cm.

Since EP⊥IJ, then segment JP is the height of the triangle EOJ.

The height of the equilateral triangle can be found using formula

$h=\dfrac{a\sqrt{3}}{2},$

where a is the side length.

So,

$h=\dfrac{10\sqrt{3}}{2}=5\sqrt{3}\ cm.$

Therefore JP is 5√3 cm

3 0

2 years ago

Question 5 (1 point)

dexar [7]

Answer:

4.5

Step-by-step explanation:

8 + 7 + 6 + 5 + 4 + 3 + 2 + 1/8

36/8

=4.5

3 0

2 years ago