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3 years ago

Point C is graphed on a number line at –7. Point D is 13 units away from point C on the number line.

2 answers:

7 0

well if it is 13 units away it could be both 13 to the right and 13 to the left so....you add 13 to -7:6 so you get 6 meaning c is an option and now subtract 13 from -7 giving you -20 so d is also an option answer: C and D

Illusion [34]3 years ago

5 0

Your answer is C: 6 because you always want to go awaay from negatives and go to the positives.

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3 years ago

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A sample of 49 observations is taken from a normal population with a standard deviation of 10. the sample mean is 55. determine

Anna35 [415]

The answer I think is D.

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3 years ago

Simplify the following ratios:

jenyasd209 [6]

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3 years ago

Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$

8 0

3 years ago

Nine out of ten students prefer math class over lunch. How many students do not prefer math if 200 students were asked?

fredd [130]

So, 9 out of 10 students prefer class, so, obviously, it would not be 100. It would be 150. So in conclusion, 50 students prefer lunch over math class, whilst 150 other students prefer math class over lunch.

3 0

3 years ago