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alekssr [168]
3 years ago
5

The radius of a circle is 6 inches. What is the area of a sector bounded by a 90° arc?

Mathematics
1 answer:
hjlf3 years ago
7 0

Answer:

[1] Let's make the problem easier to start. Imagine that they had said to find the area of a sector bounded by a 360* arc (the whole circle). Could we have found the area then? 

Sure we could! We know that A = pi*r^2, so that would just be:

A = pi*(6)^2 = 36*pi sq. miles.

Easy. But of course, we weren't so lucky... they want 135*... drat!

[2] Well, let's get a little closer, could we find the area if the angle was 180*? Sure! That's just half the circle. After all:

180*/360* = 1/2.

We know the whole circle is 36*pi sq. miles. So, the area bounded by a 180* arc would just be half of this: 18*pi sq. miles. 

"But that's not the question," you scream!! Alright, alright, calm down... let's bring it all together.

[3] We know the area of the whole circle [1]. We also know that if we can figure out the fraction of the circle the problem is easy [2]. So, what fraction of a circle would match an arc of 135*?

Well, we can see that

135*/360* = 0.375    or    3/8

So the area is just 0.375 of the whole circle:

0.375* 36*pi sq. miles = 13.5*pi sq. miles

Read more on Brainly.com - brainly.com/question/2978439#readmore

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What is 0.75 in a fraction lowest terms?
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What is the value of x?<br><br> 27(x+4)=−6<br><br><br> −35<br><br> −25<br><br> 25<br><br> 35
Oksanka [162]

x = -38/9 !!!


27(x+4)=−6

Step 1: Simplify both sides of the equation.

27(x+4)=−6

(27)(x)+(27)(4)=−6 (Distribute)

27x+108=−6

Step 2: Subtract 108 from both sides.

27x+108−108=−6−108

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6 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

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3 years ago
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Likurg_2 [28]

Answer:

Step-by-step explanation:

0.912 (C) is the probability that a student who saw an increase in their grade had studied

8 0
3 years ago
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