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MissTica
3 years ago
6

Find the area of the shaded portion the one who answer first will get brainliest

Mathematics
2 answers:
slega [8]3 years ago
5 0

Answer:

≈ 77 cm²

Step-by-step explanation:

The area of the shaded portion is calculated as

area of outer sector - area of inner sector

= (πR² - π r²) × \frac{60}{360} ( R is the outer radius, r is the inner radius )

= (π × 14² - π × 7² ) × \frac{1}{6}

= (196π - 49π ) ÷ 6

= 147π ÷ 6

≈ 77 cm² ( nearest whole number )

kondaur [170]3 years ago
3 0

Answer:

A_s=76.97\ cm^2

Step-by-step explanation:

<u>Circular Sector</u>

It's the portion of a circle enclosed by two radii and an arc. The area of a sector is calculated as follows:

\displaystyle A=\frac {r^{2}\theta }{2}

Where r is the radius and θ is the central angle expressed in radians. The central angle will be converted to radians:

\theta=60*\frac{\pi}{180}=1.0472\ rad

The shaded region in the figure can be obtained by subtracting the smaller sector A2 area from the larger sector area A1:

A_s=A_1-A_2

The larger area is calculated with a radius of r1=14 cm:

\displaystyle A_1=\frac {14^{2}*1.0472 }{2}

\displaystyle A_1=102.626\ cm^2

The smaller area is calculated with r2=7 cm:

\displaystyle A_2=\frac {7^{2}*1.0472 }{2}

\displaystyle A_2=25.656\ cm^2

The shaded area is:

A_s=102.626\ cm^2-25.656\ cm^2

\boxed{A_s=76.97\ cm^2}

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a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958      

c) p_v =P(t_{34}>2.958)=0.0028    

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And the sample deviation is:

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s=4 represent the sample standard deviation

n=35 sample size  

\mu_o =15 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less or equal than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

Since we don't  know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958  

Part c P-value  

The degrees of freedom are given by:

df = n-1= 35-1=34

Since is a right tailed test the p value would be:  

p_v =P(t_{34}>2.958)=0.0028  

Part d Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.  

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