Which inequality is true when the value of x is -3?

Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr

Minchanka [31]

Answer:

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]p_1$ represent the real population proportion for 1

$\hat p_1 =0.768$ represent the estimated proportion for 1

$n_1=92$ is the sample size required for 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.646$ represent the estimated proportion for 2

$n_2=95$ is the sample size required for 2

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

$ME= \frac{0.2753-(-0.0313)}{2}=0.1533$

And we know that the margin of erro is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

We have all the values except the value for $z_{\alpha/2}$

So we can find it like this:

$0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}$

And solving for $z_{\alpha/2}$ we got:

$z_{\alpha/2}=2.326$

And we can find the value for $\alpha/2$ with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98$ or 98%

7 0

3 years ago

Chen completed the division problem below.

alekssr [168]

<u>Answer:</u>

B. He multiplied the dividend by 100 instead of 10.

<u>Step-by-step explanation:</u>

Chen had to divide $1.926$ by $0.9$ . To make it easier to divide, he thought of multiplying both the dividend and the divisor by 10.

Chen multiplied the divisor by 10 but mistakenly multiplied the dividend by 100 instead of 10.

So his whole division problem was changed.

Therefore, the correct answer option is B. He multiplied the dividend by 100 instead of 10.

6 0

3 years ago

Read 2 more answers

There are 550 students at Martin Middle School. One bundred and eighty- seven students walk to school. what percent of the stude

Anvisha [2.4K]

For a question like this, all you have to do is remember that a percentage is simply a fraction expressed as a decimal. So, all we have to do is:
$\frac{187}{550}$ , and then multiply our decimal answer by 100 to get the percentage:
$\frac{187}{550}$ = 0.34
0.34 x 100 = 34
Therefore 34% of the students walk to school.
I hope this helped and remember to please try and understand the maths that got the answer, not just writing the answer down :))

4 0

3 years ago

Help on $3.70÷2=no clue

Daniel [21]

3.7/2 = 1.85
2 goes into 3 1 time.
you have a remainder of 1
That one gets a 7 after it
now you have 17
2 goes into 17 8 times
remainder of 1
bring down the 0 that's there, but is not shown because it's not important in standard form or decimals.
You get 10 and that's 5
so you have 1.85
1.85$

3 0

3 years ago

Read 2 more answers

What is the reduced fraction of 25 and 60?

tino4ka555 [31]

25/60?
That would be 5/12

3 0

3 years ago