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3 years ago

Find area of blue shaded region

Mathematics

1 answer:

RUDIKE [14]3 years ago

8 0

Answer:

$\approx \: 7.33 \: {cm}^{2}$

Step-by-step explanation:

Area of the blue shaded region

$= \frac{ \theta}{360 \degree} \times \pi r_1^2 - \frac{ \theta}{360 \degree} \times \pi r_2^2 \\ \\ = \frac{ \pi\theta}{360 \degree} ( r_1^2 - r_2^2) \\ \\ ( r_1 = 4 \: cm, \: \: r_2 = 3\: cm, \: \: \theta = 120 \degree) \\ \\ = \frac{ 3.14 \times 120 \degree}{360 \degree} ( 4^2 - 3^2) \\ \\ = \frac{ 3.14 }{3} \times 7 \\ \\ = \frac{21.98}{3} \\ \\ = 7.32666667 \\ \\ \approx \: 7.33 \: {cm}^{2}$

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Answer:

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Step-by-step explanation:

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Answer: y= 4x -27

Step-by-step explanation:

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4 years ago

Create an exponential to describe $100 at 2% interest, compounded annually, for x years. y=100(.98)^x y=100(.8)^x y=100(1.2)^x y

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The exponential to describe $100 at 2% interest, compounded annually, for x years is $y=100(1.02)^{x}$

<h3>Solution:</h3>

Given that $ 100 at 2 % interest , compounded annually for "x" years

The formula for compound interest, including principal sum, is:

$A=P\left(1+\frac{r}{n}\right)^{n t}$

A = the future value of the investment/loan, including interest

P = the principal investment amount (the initial deposit or loan amount)

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per unit t

t = the time the money is invested or borrowed for

Here in this sum,

P = $ 100

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Let "y" be the amount after "x" years

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4 years ago

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58

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Answer:

Step-by-step explanation:

Hello!

You want to test two samples of batteries to see is the mean voltage of these battery types are different.

Sample 1 (type K)

n₁= 37

sample mean x₁[bar]= 8.54

standard deviation S₁= 0.225

Sample 2 (Type Q)

n₂= 58

sample mean x₂[bar]= 8.69

standard deviation S₂= 0.725

1. The test hypothesis are:

H₀: μ₁ = μ₂

H₁: μ₁ ≠ μ₂

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I hope it helps!

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4 years ago

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AlekseyPX

Answer:

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