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11Alexandr11 [23.1K]
3 years ago
9

a rectangle is placed around a semicircle as shown below. the width of the rectangle is 2cm. find the area of the shaded region.

use 3.14

Mathematics
1 answer:
Lelu [443]3 years ago
5 0
Since figure is not given, assume drawing a rectangle and label it from left top to right top as A and B. Similarly label it as C and D from bottom left to right. Draw a semicircle starting from C to D inside the rectangle.
You are asked to find area of shaded region which I assume is semicircular part.
Area of a circle is determined by Pi × r ^2
r is the radius of circle which is half of diameter. The diameter of circle is same as width of rectangle. So radius = 7/2 = 3.5 mm And area of shaded region is

1/2(Pi×r^2) = 1/2(3.14×3.5^2) = 19.2325 sq
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3 0
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A road has an incline of 10°. To the nearest foot, find the increase in altitude of a car after driving 4,000 feet along the roa
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3 years ago
Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the fiv
Dahasolnce [82]

Answer and Explanation:

Given : Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029

\sum P(X)=1

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

\mu=\sum xP(x)=2.5

The standard deviation of the random sample is

Variance=\sigma^2

\sigma^2=\sum x^2P(x)-\mu^2

\sigma^2=7.436-(2.5)^2

\sigma^2=7.436-6.25

\sigma^2=1.186

\sigma=\sqrt{1.186}

\sigma=1.08

Therefore, The mean is 2.5 and the standard deviation is 1.08.

5 0
3 years ago
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