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Eddi Din [679]
3 years ago
8

Find the area of the semi circle please ​

Mathematics
2 answers:
Firlakuza [10]3 years ago
8 0

Answer:

58.4197cm^2

Step-by-step explanation

cirlce area formula: \pir^2

r= 6.1

\pi=3.14

Now plug numbers into the formula

area of circle:

= (3.14)(6.1^2)

=(3.14)(37.21)

= 116.8394cm^2  ------> area of full circle

now divide by 2, to find area of semi circle

116.8394/2= 58.4197cm^2

Novosadov [1.4K]3 years ago
6 0

\huge\fbox\red{A}\fbox\pink{n}\fbox\purple{S}\fbox\green{w}\fbox\blue{E}\fbox\orange{r}

area of semicircle is 58.572 m²

\huge \sf { \fcolorbox{green}{g}{Explanation :- }}

\huge \underline{\large \fbox \red{Concept : }}

We have to find the area of semicircle

So firstly we need to find the area of circle. because the semicircle of circle is the half of area of circle so first we find the area of circle then divide by 1 /2 then we get the area of semicircle.

We have given the radius of circle is 6.1 m

\huge\mathfrak\red{Solution : }

\mathfrak \green{Area \:  of \:  circle = π ×  ( radius) ²}

\mapsto  \:  \:  \small  \:  \:  \:  \:  \: \mathfrak \blue{ =  \frac{22}{7} \:  \times( 6.1 \: m)^{2} } \\

\mapsto  \small    \:  \:  \:  \:  \: \mathfrak \blue{ =  \frac{22}{7} \:  \times37.21 \: m {}^{2}  }  \\

\mapsto  \:  \small\:  \:  \:  \:  \:  \: \mathfrak \blue{ =  116.94 \: m {}^{2} }

Therefore,

\small \mathfrak \red{Area \:  of \:  semicircle = \frac{1}{2}   \times area  \: of  \: circle}

\small \mathfrak  \green{  =  \frac{1}{2} \times 116.94 \: cm \:  {}^{2} } \\

\mapsto  \:  \small\:  \:  \:  \:  \:  \: \mathfrak { =  58.472\: m {}^{2} }

{\underline  {\fbox\red{Hence, the area of semicircle is 58.572 m²}}}

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Step-by-step explanation:

It is given that,

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We need to find the distance x the diver must  swim up to the boat. It can be calculate by applying trigonometry as follows :

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3 years ago
The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standa
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Answer:

a) 0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

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c) 0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

Step-by-step explanation:

To solve this question, we need to understand 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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standard deviation 7.3.

This means that \mu = 56, \sigma = 7.3

(a) (5 points) Compute the probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

This is the pvalue of Z when X = 60. So

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

Z = \frac{60 - 56}{7.3}

Z = 0.55

Z = 0.55 has a pvalue of 0.7088

0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

(b) (5 points) Compute the 75th percentile for the age of United States Presidents on the day of inauguration.

This is X when Z has a pvalue of 0.75. So X when Z = 0.675.

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

0.675 = \frac{X - 56}{7.3}

X - 56 = 0.675*7.3

X = 61

The 75th percentile for the age of United States Presidents on the day of inauguration is 61.

(c) (5 points) Compute the probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

Now, by the Central Limit Theorem, we have that n = 4, s = \frac{7.3}{\sqrt{4}} = 3.65

This is the pvalue of Z when X = 60. So

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

By the Central Limit Theorem

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

Z = \frac{60 - 56}{3.65}

Z = 1.1

Z = 1.1 has a pvalue of 0.8643

0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

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