Question 5 (1 point)
Find the area of the shaded region of this regular polygon. (Round to the nearest
hundredth)

Question 5 (1 Point)Find The Area Of The Shaded Region Of This Regular Polygon. (Round To The Nearesthundredth)

Answers

Answer 1
The polygon is a regular octagon. Each shaded triangle is isosceles, has vertex angle of 360°/8 = 45°, base angles 67.5°, and base = 8.
Height of triangle = 4tan(67.5°)
area of triangle = 0.5(base)(height) = 0.5(8)(4tan(67.5°))
Shaded region = 2*0.5(8)(4tan(67.5°)) = 32tan(67.5°) ≅180.71 cm^2

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A phone manufacturer offers its customers two different versions of the same phone. The low storage version costs $350, and the high storage version costs $450. let X represent the manufacturer's income on a randomly chosen purchase of this phone. Based on previous data, here's the probability distribution of X along with summary statistics:

Low capacity High capacity
х income ($) 350 450
P(X) 0.65 0.35
Mean: μx=$385
Standard deviation: σx= $48
The manufacturer is considering a $25 price increase on both versions of the phone. Assume that this increase will not change the probability that corresponds to each version. Let Y represent their income on a randomly chosen purchase of this higher priced phone. What are the mean and standard deviation of Y?

Answers

Answer:

Mean = $410

Standard deviation = $48

Step-by-step explanation:

From the given information:

                              Low capacity         High capacity

Y income ($)             350 + 25                450  + 25

                                       = 375                    = 475

P(Y)                                0.65                      0.35

[tex]U_Y = (375 \times 0.65) + (475 \times 0.35)[/tex]

[tex]U_Y = 243.75 + 166.25[/tex]

[tex]\mathbf{U_Y = 410}[/tex]

[tex]Var _{(Y)} = E(Y^2) - E(Y)^2[/tex]

[tex]Var _{(Y)} = \Big [(375)^2 \times 0.65 + (475)^2 \times 0.35 \Big ]- (410)^2[/tex]

[tex]Var _{(Y)} =170375 - 168100[/tex]

[tex]Var _{(Y)} = 2275[/tex]

[tex]\sigma _Y = \sqrt{Var _{(Y)}}[/tex]

[tex]\sigma _Y = \sqrt{2275}[/tex]

[tex]\mathbf{\sigma _Y \simeq 48}[/tex]

This question is based on the concept of statistics. Therefore, the mean of Y is [tex]U_y[/tex] = 410 and the standard deviation of Y is  [tex]\sigma_{(Y)}[/tex] =48.

Given:

Low capacity = $350

High capacity = $450

P(X)  of low capacity = 0.65

P(X)  of high capacity = 0.35

Mean: μx=$385

Standard deviation: σ(x)= $48

We need to calculate the the mean and standard deviation of Y (Y represent their income on a randomly chosen purchase of this higher priced phone).

According to the question,

From the given information, it is observe that,

                             Low capacity         High capacity

Y income ($)             350 + 25                450  + 25

                                      = 375                    = 475

P(Y)                               0.65                      0.35

Now, calculate the mean of Y,

[tex]U_y = (375 \times 0.65) + ( 475 \times 0.35)\\U_y = 243.75 + 166.25\\\bold{U_y = 410}[/tex]

Now, calculate the variance of Y,

[tex]Var_{(Y)} = E( Y^2) - E ( Y) ^2\\\\Var_{(Y)} = [(375)^2 \times 0.65 + (475)^2 \times 0.35] - (410)^2\\\\Var_{(Y)} = 170375 - 168100\\\\Var_{(Y)} = 2275\\[/tex]

Calculating the standard deviation of Y,

[tex]\sigma_{(Y)} = \sqrt{Var_{(Y)}} \\\\\sigma_{(Y)} = \sqrt{2275} \\\\\sigma_{(Y)} = 47.6 \approx 48[/tex]

Therefore, the mean of Y is [tex]U_y[/tex] = 410 and the standard deviation of Y is  [tex]\sigma_{(Y)}[/tex] =48.

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Say I have a question that said "Arjun had Rs. 10, and his friend Raghav borrowed Rs. 5, what percentage of Arjun's amount did Raghav borrow?". 10 is the whole. It is the total amount of money that they have. 5 is the part. You are looking for the percentage. The 100 will always stay the same. Put the values that you know into the proportions, and replace what you don't know with the variable x. Like so:

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Step-by-step explanation:

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