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4 years ago

PLEASE HELP ILL REWARD BRAINLIST

Mathematics

2 answers:

kodGreya [7K]4 years ago

7 0

Answer: (2,1)

Hope it helps <3

Murrr4er [49]4 years ago

7 0

I hope this works for you :)

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Find the area of the triangle.<br> If necessary, round your answer to the nearest hundredth.

g100num [7]

Answer:

A = 1824 $in^{2}$

Step-by-step explanation:

$A=\frac{bh}{2} +bh=\frac{(48)(12)}{2} +(48)(32)=288+1536=1824in^{2}$

Hope this helps

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2 years ago

What is the answer to this question?

kirza4 [7]

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4 years ago

How can they been proven congruent ?

telo118 [61]

The two angles marked are congruent.
That is one pair of congruent angles.   ------->    A
Then the vertical angles are congruent.
That is a second pair of congruent angles.    -------> A
Finally, we have the two sides marked congruent.
That is one pair of congruent sides.    ------> S

Put together what we have above, and it gives us AAS.

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3 years ago

What is 24/25 divided by 3/4 and why?

NeTakaya

Answer:

1 7 /25

Step-by-step explanation:

bc i did the math

5 0

3 years ago

Read 2 more answers

Examine the required sample size needed to be able to

Xelga [282]

Answer:

$n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573$

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

$ME=2.33\frac{500}{\sqrt{2000}}=26.05$

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)

$\sigma=500$ represent the population standard deviation assumed

n represent the sample size (variable of interest)

Confidence =98% or 0.98

ME = 10 represent the margin of error desired

The margin of error is given by this formula:

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

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.33$ , replacing into formula (b) we got:

$n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573$

So the answer for this case would be n=13573 rounded up to the nearest integer

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

$ME=2.33\frac{500}{\sqrt{2000}}=26.05$

3 0

4 years ago