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8090 [49]
1 year ago
9

Consider the equation showing the distributive property 84+93=3(28+□)

Mathematics
1 answer:
arlik [135]1 year ago
4 0
Distributive property

We have that the distributive property states that the multiplication is done for all the terms inside the parenthesis. Then we have:

Since

3 · 28 = 84

and

3 · □ = 93

Since 3 · 31 = 93, then

□ = 31

<h2>Answer: 31</h2>

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Given: <br><br> ∆SPM, PK⊥ SM<br><br> SP = 25, SM = 28, PK = 9<br><br> Find: m∠S, m∠M, PM
nevsk [136]

Answer:

∠S = 21.10°

∠M = 79.45°

PM = 9.16

Step-by-step explanation:

Here Pythagorean theorem and trigonometry suffice to solve our problem.

From the Pythagorean theorem we get:

KM^2+9^2=PM^2 and

(25-KM)^2+9^2=25^2.

We solve for KM in the second equation and get:

(25-KM)=\sqrt{25^2-9^2}

\therefore KM=25-\sqrt{25^2-9^2} =\boxed{1.68}

Now since

SK+KM=25\\\\ SK=\boxed{23.32}

Therefore

{\angle}S=Tan^{-1}(\frac{PK}{SK}) = Tan^{-1}(\frac{9}{23.32})=21.10^o

and

{\angle}M=Tan^{-1}(\frac{PK}{KM}) = Tan^{-1}(\frac{9}{1.68})=79.45^o.

And finally again from the Pythagorean theorem:

PM^2=PK^2+KM^2=9^2+1.68^2

\therefore PM=\sqrt{9^2+1.68^2} =9.16.

Thus,

∠S = 21.10°

∠M = 79.45°

PM = 9.16.

7 0
3 years ago
The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books check
Verdich [7]

Answer:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =150 represent the population standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

\bar X = \frac{740+920}{2} = 830

and the margin of error is given by :

ME = \frac{920-740}{2}= 90

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

5 0
3 years ago
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