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katrin2010 [14]
3 years ago
11

HELPPP MY MATH HOMEWORK!!!

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0
D and then A I would say
statuscvo [17]3 years ago
7 0

I dont know how to answer number 12 but number 13 is the first one 3x + 15 = 17

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M
den301095 [7]

Answer:

Step-by-step explanation:

m∠JXM + m∠NXP = 180° - m∠MXN = 90°

7 0
3 years ago
A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student
yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

The information provided is as follows:

n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43

The critical value of <em>z</em> for 98% confidence level is,

z_{\alpha/2}=z_{0.02/2}=2.326

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

     =(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0
3 years ago
$2.25
DiKsa [7]
2.50 that is the answer

8 0
3 years ago
Order 74.585, 74.586, 74.486, and 74.587 in decreasing order.
Veseljchak [2.6K]

Answer:

74.587, 74.586, 74.585, 75.486

Step-by-step explanation:

yes

6 0
2 years ago
Read 2 more answers
What is the range of the given function?
asambeis [7]

Answer:

{y | y = –7, –1, 0, 9}

Choice B is correct

Step-by-step explanation:

The range of a function y = f(x) refers to the set of y - values for all values of x for which the function is real and defined.

The set of y values from the given table is;

–7, –1, 0, 9

Therefore, the range of the given function is;

{y | y = –7, –1, 0, 9}

8 0
3 years ago
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