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wariber [46]
2 years ago
9

13-28 Sketch the region enclosed by the given curves and find its area.

Mathematics
1 answer:
Semenov [28]2 years ago
7 0

Answer:

rea (A)  0.0682

Step-by-step explanation:

The sketch for the region enclosed by the given curves can be found in the image attached below.

From the image below;

The two curves intersect in the area of tan 9x = 2 sin 9x

Recall that:

making x the subject; then:

The subdivision of the domain is in two intervals

where;

Area (A) =

Area (A) = 0.06818

Area (A)  0.0682

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3 years ago
A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select
Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; \mu = 57,800  and  \sigma = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( \frac{X-\mu}{\sigma} <= \frac{58000-57800}{750} ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( \frac{X-\mu}{\sigma} < \frac{57000-57800}{750} ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = \frac{\sigma}{\sqrt{n} } = \frac{750}{\sqrt{100} } = 75

The z probability is given by;

           Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{58000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{57000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

7 0
3 years ago
6.
Degger [83]
Answer: True

---------------------------------

Work Shown:

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Leroy gave half of his crayons to a friend he then lost 2 he had 10 left how many crayons did he have to begin with
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24 crayons he started with

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Evaluate ( 4 upon 3 ) -2<br><br><br>please amswer fast​
Taya2010 [7]
<h2><u>Solution</u></h2>

\frac{4}{3}  - 2 \\  \\   = \frac{4 - 6}{ 3}  \\  \\  =  \frac{ - 2}{3}

<h3>Hope This Helps You ❤️</h3>
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