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Nuetrik [128]
2 years ago
6

does somebody knows how to do this one?

Mathematics
1 answer:
vlada-n [284]2 years ago
6 0

Answer:

Step-by-step explanation:

To find the area subtract the area of the semicircle from the area of the rectangle.

Although the line isn’t there, if you imagine there is one, then you will see that you form a rectangle which is the same line as the semicircle’s diameter.

The area of rectangle is:

⇒ A = lw

⇒ A = (14)(8)

⇒ A = 112cm^{2}

The area of the semicircle;

⇒ A = \frac{1}{2}\pi r^{2}

⇒ A = \frac{1}{2}\pi (7)^{2}

*Note here that the radius is half the diameter, so it is 7cm, not 14cm

⇒ A = 76.97cm^{2}

Finally subtract the two areas;

⇒ A = 112 - 76.97

⇒ A = 35.03cm^{2}

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Part a)

The simple random sample of size n=36 is obtained from a population with

\mu = 74

and

\sigma = 6

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

\mu = 74

The standard error of the means becomes the standard deviation of the sampling distribution.

\sigma_ { \bar X }  =  \frac{ \sigma}{ \sqrt{n} }  \\ \sigma_ { \bar X }  =  \frac{ 6}{ \sqrt{36} }  = 1

Part b) We want to find

P(\bar X \:>\:75.9)

We need to convert to z-score.

P(\bar X \:>\:75.9)  = P(z \:>\: \frac{75.9 - 74}{1} )  \\  = P(z \:>\: \frac{75.9 - 74}{1} ) \\  = P(z \:>\: 1.9) \\  = 0.0287

Part c)

We want to find

P(\bar X \: < \:71.95)

We convert to z-score and use the normal distribution table to find the corresponding area.

P(\bar X \: < \:71.95)  = P(z \: < \: \frac{71.9 5- 74}{1} )  \\  = P(z \: < \: \frac{71.9 5- 74}{1} ) \\  = P(z \: < \:  - 2.05) \\  = 0.0202

Part d)

We want to find :

P(73\:

We convert to z-scores and again use the standard normal distribution table.

P( \frac{73 - 74}{1} \:< \: z

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