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vlabodo [156]
3 years ago
7

Which of the following could be used to calculate the area of a sector in the circle shown below

Mathematics
1 answer:
lara [203]3 years ago
8 0

Answer:

Therefore the Last option is correct

\textrm{Area of Sector ACD}=\pi (8\ in)^{2}\times (\frac{42}{360})

Step-by-step explanation:

Given:

Radius = r = 8 in

θ = 42°

To Find:

Area of Sector = ?

Solution:

We know that

\textrm{Area of Sector}=\pi (Radius)^{2}\times \frac{\theta}{360}

Substituting the given values in the formula we get

\textrm{Area of Sector ACD}=\pi (8\ in)^{2}\times (\frac{42}{360})

Which is the required Answer.

Therefore the Last option is correct

\textrm{Area of Sector ACD}=\pi (8\ in)^{2}\times (\frac{42}{360})

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Find the length of side x in simplest radical form with a rational denominator.<br> 45°<br> X<br> V6
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Answer:

x = 6

Step-by-step explanation:

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The population of a town after t years is represented by the function f(t)=7,248(0.983)t.
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The answer is a number that can be written as a ratio of two integers.  The answer is C.
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Solve for a.<br> 5 + 14a = 9a - 5
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Since you are solving for a, you want to have a on one side of the equation and the other terms on another side of the equation. It would be easiest to have all the terms with a on the left side of the equation, so that is what we will do.

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3 0
4 years ago
Read 2 more answers
What is the area of the composite figure below?
musickatia [10]

Answer:

33 sq cm.

Step-by-step explanation:

Let's calculate each shape serperately,

--------

For the square, the area is 25 since

5 × 5 = 25

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For the rectangle, the area is 8:

4 × 2 = 8

-----

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Have a good day :)

3 0
3 years ago
The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If 64 women are
shtirl [24]

Answer:

0.3569 is the probability that they have a mean pregnancy between 266 days and 268 days.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  268 days

Standard Deviation, σ =  15 days

We are given that the distribution of lengths of pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{64}} = \frac{15}{8}

P(pregnancy between 266 days and 268 days)

P(266 \leq x \leq 268) = P(\displaystyle\frac{266 - 268}{\frac{15}{8}} \leq z \leq \displaystyle\frac{268-268}{\frac{15}{8}}) = P(-1.0667 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1.067)\\= 0.5000 - 0.1431 = 0.3569 = 35.69\%

P(266 \leq x \leq 268) = 35.69\%

6 0
3 years ago
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