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emmainna [20.7K]
2 years ago
5

PLEASE HELP ME PLEASE!!

Mathematics
1 answer:
Anna35 [415]2 years ago
8 0

Step-by-step explanation:

in such a case, where we have all the sides and need angles, the best tool is the extended Pythagoras (for general triangles and not just for right-angled ones) :

c² = a² + b² - 2ab×cos(C)

c is the side opposite of angle C.

the side opposite of B is CA.

so, we have

220² = 100² + 160² - 2×100×160×cos(B)

48,400 = 10,000 + 25,600 - 32,000×cos(B)

12,800 = -32,000×cos(B)

cos(B) = -12,800/32,000 = -0.4

B = 113.5781785...° ≈ 114°

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A test is used to assess readiness for college. In a recent​ year, the mean test score was 20.3 and the standard deviation was 4
Vlada [557]

Answer:

\mu = 20.3

\sigma = 4.9

And we can find the limits in order to consider values as significantly low and high like this:

Low\leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5

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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".

Solution to the problem

For this case we can consider a value to be significantly low if we have that the z  score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is  higher tor equal to 2.

For this case we have the mean and the deviation given:

\mu = 20.3

\sigma = 4.9

And we can find the limits in order to consider values as significantly low and high like this:

Low \leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5

High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1

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Answer:

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Step-by-step explanation:

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2 years ago
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