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GREYUIT [131]
3 years ago
6

What is the equation in math N+8=-10

Mathematics
2 answers:
sp2606 [1]3 years ago
4 0

Your answer is...

N = -18

-18 + 8 = -10

dybincka [34]3 years ago
4 0

The objective in this question is to get rid of the 8 on the left. When you do that you will have all the numbers on the right, and all the variables on the left. The equation will be solved. If the solution is not what you want, then be a little bit more careful about how you post a question.

Add 8 to both sides.

n +8 - 8 = - 10 - 8

n = - 18 Think money when you do the question. If you owe 10 dollars and you spend 8 more, you go more and more in debt. That's the same way with this question. You go more and more negative.


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Step-by-step explanation:

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A study was conducted to estimate the difference in the mean salaries of elementary school teachers from two neighboring states.
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Answer:

(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70

(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698

The confidence interval would be -3301.70 \leq \mu \leq 501.698 and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

Step-by-step explanation:

We have the following info given by the problem

\bar X_1 = 28900 the sample mean for the salaries of teachers in Indiana

s_1 = 2300 the sample deviation for the salary of  teachers in Indiana

n_1 =10 the sample size from Indiana

\bar X_2 = 30300 the sample mean for the salaries of teachers in Michigan

s_2 = 2100 the sample deviation for the salary of  teachers in Michigan

n_2 =14 the sample size from Michigan

We want to find a confidence interval for the difference in the two means and the formula for this case is given by;

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom for this case are:

df = n_1 +n_2 -2= 10+14-2 =22

The confidence is 95%so then the significance is \alpha=0.05 and the \alpha/2 =0.025, we need to find a critical value in the t distribution with 22 degrees of freedom who accumulates 0.025 of the area on each tail and we got:

t_{\alpha/2}= 2.07

And now replacing in the formula for the confidence interval we got:

(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70

(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698

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