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Setler [38]
3 years ago
8

Consider the inequality -5(x+7)<-10. Right in inequality representing the solution for x

Mathematics
1 answer:
scZoUnD [109]3 years ago
8 0
First, divide -5 on both sides:

(x + 7) < -2

Then subtract 7 on both sides:

x < -9

x < -9 is your final answer.
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PLZ I REALLY NEED HELP 15 POINTS
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To find the X intercept, replace y with 0 and solve for x:

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vTo estimate the parameter stated in question 2, the following was done. A simple random sample of 22 students who attended a re
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Answer:

The 90% confidence interval for the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session is (3.7, 12.3).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

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When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

A simple random sample of 22 students who attended a review session was selected, and the mean grade on test 7 for this sample of 22 students was 85 with a standard deviation of 9.8.

This means that:

\mu_A = 85, s_A = \frac{9.8}{\sqrt{22}} = 2.09

An independent simple random sample of 47 students who did not attend a review session was selected, and the mean grade on test 7 was 77 with a standard deviation of 10.6.

This means that:

\mu_N = 77, s_N = \frac{10.6}{\sqrt{47}} = 1.55

Distribution of the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session.

Mean:

\mu = \mu_A - \mu_N = 85 - 77 = 8

Standard deviation:

s = \sqrt{s_A^2 + s_B^2} = \sqrt{2.09^2 + 1.55^2} = 2.6

Confidence interval:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = zs

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M = 1.645*2.6 = 4.3

The lower end of the interval is the sample mean subtracted by M. So it is 8 - 4.3 = 3.7

The upper end of the interval is the sample mean added to M. So it is 8 + 4.3 = 12.3

The 90% confidence interval for the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session is (3.7, 12.3).

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