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Reika [66]
3 years ago
13

Find a polynomial for the sum of the shaded areas of the figure. A = 6, B = 4

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

The polynomial for the sum of the shaded  \pi r² - 20 \pi

Step-by-step explanation:

Given as :

The figure is shown which is of concentric circle with radius B , A , r

The radius B = 4 unit

The radius A = 6 unit

Let The sum of shaded portion = x unit

Now, The circumference of circle = 2 \pi R , where R is the radius

So, for circle with radius B.

The circumference = 2 \pi R = 2 \pi B

Or, The circumference = 2 \pi × 4 = 8 \pi

<u>Similarly</u>

For circle with radius A.

The circumference = 2 \piR = 2 \pi A

Or, The circumference = 2 \pi × 6 = 12 \pi

Now, <u>The area of circle with radius r is</u>

Area = \pi ×radius × radius

Or, Area = \pi r²

Now,

The sum of shaded region area = The area of circle with radius r - ( The circumference with radius B + The circumference with radius A )

Or, The sum of shaded region area =  \pi r² - ( 8 \pi + 12 \pi )

Or,  The sum of shaded region area =  \pi r² - 20 \pi

Hence The polynomial for the sum of the shaded area is \pi r² - 20 \pi  Answer

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lesya692 [45]

<span>The problem is to calculate the angles of the triangle. However, it is not clear which angle you have to calculate, so we are going to calculate all of them
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we know that
Applying the law of cosines
c²=a²+b²-2*a*b*cos C------> cos C=[a²+b²-c²]/[2*a*b]
a=12.5
b=15
c=11
so
 cos C=[a²+b²-c²]/[2*a*b]--->  cos C=[12.5²+15²-11²]/[2*12.5*15]
cos C=0.694------------> C=arc cos (0.694)-----> C=46.05°-----> C=46.1°

applying the law of sines calculate angle B
15 sin B=11/sin 46.1-----> 15*sin 46.1=11*sin B----> sin B=15*sin 46.1/11
 sin B=15*sin 46.1/11-----> sin B=0.9826----> B=arc sin (0.9826)
B=79.3°

calculate angle A
A+B+C=180------> A=180-B-C-----> A=180-79.3-46.1----> A=54.6°

the angles of the triangle are
A=54.6°
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3 years ago
Molly works at an aple orchard.she picked 1,078 apples last weekend and put them into bags that hold 12 apples each.how many bag
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She filled 89 bags with 10 apple remaining 

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3 years ago
The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of the girls
Elanso [62]

Answer:

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=22 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =12.9^2=166.41 represent the sample variance obtained

\sigma^2_0 =16.1^2 =259.21 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis: \sigma^2 \geq 259.21

Alternative hypothesis: \sigma^2

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

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suter [353]

Answer:

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Answer:

x = 43.2°

Step-by-step explanation:

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