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notsponge [240]
4 years ago
6

Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).

Mathematics
1 answer:
Fofino [41]4 years ago
7 0
<h3>Answer: (40/3)pi</h3>

This is the same as (40pi)/3

====================================

Explanation:

A = area of the smaller circle OB

A = pi*r^2

A = pi*3^2

A = 9pi

Circle OB has an exact area of 9pi square cm

We don't want the whole circle area, since we only are focusing on a 120 degree slice. Note how 120 degrees is 120/360 = 1/3 of a full circle; therefore we want 1/3 of the area to get the area of the slice

area of slice DOB = (1/3)*(area of circle OB)

area of slice DOB = (1/3)(9pi)

area of slice DOB = 3pi

We'll use this later so let's call this M

M = 3pi

----------------

B = area of larger circle OA

B = pi*r^2

B = pi*7^2

B = 49pi

We take 1/3 of this for similar reasons as mentioned earlier. So the area of pizza slice COA is (1/3)*(49pi) = 49pi/3

We'll use this later so let N = 49pi/3

----------------

Subtract the values of N and M.

Shaded area = N - M

Shaded area = (49pi/3) - (3pi)

Shaded area = (49pi/3) - (9pi/3)

Shaded area = (49/3)pi - (9/3)pi

Shaded area = (49/3 - 9/3)pi

Shaded area = (40/3)pi

side note: I think you computed 49/3 = 16.333 repeating which was the result for the area of circle OA; however it's not the final answer

another note: leave 40/3 as a fraction and don't convert it to 13.333 repeating. This is because the fraction 40/3 is exact while 13.333 is approximate

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Answer:

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Step-by-step explanation:

-3x^2+27x=42

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x^2-2x-7x+14=0

x*(xi2)-7(x-2)=0

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x-2=0

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3 years ago
Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean
earnstyle [38]

Answer:

t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167

p_v =2*P(t_{23}

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

Step-by-step explanation:

Data given

n_1 =10 represent the sample size for group 1

n_2 =15 represent the sample size for group 2

\bar X_1 =1085 represent the sample mean for the group 1

\bar X_2 =1034 represent the sample mean for the group 2

s_1=52 represent the sample standard deviation for group 1

s_2=61 represent the sample standard deviation for group 2

We are assuming that the two samples are normally distributed with equal variances and that means:

\sigma^2_1 =\sigma^2_2 =\sigma^2

System of hypothesis

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

The statistic is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The degrees of freedom are given by:

n_1+n_2 -2

And the pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

Replacing we got:

S^2_p =\frac{(10-1)(52)^2 +(15 -1)(61)^2}{10 +15 -2}=3323.043

And the deviation would be:

S_p=57.646

The degrees of freedom are:

df=10+15-2=23

The statistic would be:

t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167

The p value would be

p_v =2*P(t_{23}

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

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