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OlgaM077 [116]
3 years ago
5

3n - 5p + 2n = 10p, solve for n

Mathematics
1 answer:
Alex17521 [72]3 years ago
4 0
Combine (subtract) 3n and 2n to get 1n or n. Then you have n -5n =10n add 5n to 10n to get 15n. divide the 15 from n and your answer is n=15
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Answer:

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Step-by-step explanation:

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Pls help to get the answer
soldier1979 [14.2K]
When they say quotient, they want a fraction, so the answer for that would be
\frac{6}{10}
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The school can food drive has collected an average of 171 cans of food per month from january to june. the balance at the end of
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Answer:

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Step-by-step explanation:

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3 years ago
Using traditional methods it takes 92 hours to receive an advanced flying license. A new training technique using Computer Aided
DanielleElmas [232]

Answer:

The value of the test statistic is z = 1.39.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

Step-by-step explanation:

Using traditional methods it takes 92 hours to receive an advanced flying license.

This means that at the null hypothesis, it is tested if the mean is of 92, that is:

H_0: \mu = 92

Test if there is evidence that the technique lengthens the training time

At the alternative hypothesis, it is tested if the mean is more than 92, that is:

H_1: \mu > 92

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

92 is tested at the null hypothesis:

This means that \mu = 92

A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.

This means that n = 70, X = 93, \sigma = \sqrt{36} = 6

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{93 - 92}{\frac{6}{\sqrt{70}}}

z = 1.39

The value of the test statistic is z = 1.39.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.

Looking at the z-table, z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

8 0
3 years ago
Pls help with this math problem
valentinak56 [21]
-1.5. -1 1/2. -3/2






yeah so its this i’m 99.98% sure
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