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ArbitrLikvidat [17]
3 years ago
11

Find the area of the semicircle. Round your answer to the nearest hundredth. YEAH IVE BEEN STUCK ON THIS help please I need the

steps if possible:(

Mathematics
2 answers:
guapka [62]3 years ago
8 0

Answer:

=127.23

Step-by-step explanation:

we can start of by finding the area of a <em>CIRCLE</em>

18/2=9

9=radius

area of circle= pi*r squared

9 squared=81*pi=254.4690

<em>Now we can half the area of the circle to find out the area of the semi circle as a semi circle is just half a circle therefore half the area of a circle</em>

254.469/2=127.2345

127.2345=127.23

=127.23

Hope this helped

aivan3 [116]3 years ago
7 0

Answer:

your answer will be 127.17 sq in

Step-by-step explanation:

hope it helps you

have a wonderful day!!!

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HELP<br><br>jusko ang hirap<br><br><br><br>❗NONSENSE = REPORTED❗​
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1.A

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Step-by-step explanation:

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4 0
3 years ago
Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr
bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 109.0 inches, and a standard deviation of 12 inches.

This means that \mu = 109, \sigma = 12

Sample of 25.

This means that n = 25, s = \frac{12}{\sqrt{25}} = 2.4

What is the probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches?

This is the p-value of Z when X = 112. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{112 - 109}{2.4}

Z = 1.25

Z = 1.25 has a p-value of 0.8944.

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

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Each song cost $2

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Make me brainliest if this helps!

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