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3 years ago

What is the approximate area of the shaded sector?

Mathematics

1 answer:

kodGreya [7K]3 years ago

3 0

Answer:

C. $302 \: {cm}^{2}$

Step-by-step explanation:

$Area \: of \: sector = \frac{ \theta}{360 \degree} \times \pi {r}^{2} \\ \\ = \frac{135 \degree}{360 \degree} \times \pi {r}^{2} \\ \\ = \frac{135 }{360 } \times 3.14 \times {(16)}^{2} \\ \\ = \frac{108,518.4 }{360 } \\ \\ = 301.44 \\ \\Area \: of \: sector \approx \: 302 \: {cm}^{2}$

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Can someone answer and explain ILL MARK BRAINLIEST

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It’s c because slope usually has to deal with numbers

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3 years ago

Solve 1/2x-4=7+1/6x

Westkost [7]

Answer:

x=11/2

Step-by-step explanation:

1/2x-4=7+1/6x

1/2x-1/6x=7+4

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3x-x=11 As 6 divided by 2 is 3 and 3 multiplied by x is 3x.

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4 years ago

Read 2 more answers

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Answer:

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Step-by-step explanation:

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3 years ago

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. (5 points)

lara31 [8.8K]

F(x)=(x-8)/(x+7). g(x)=(-7x-8)/(x-1). Plug in g(x) into f(x), f(g(x))=[(-7x-8)/(x-1)-8]/[-7x-8)/(x-1)+7], which can be simplified as (-7x-8-8x+8)/(-7x-8+7x-7)=-15x/-15=x. Plug in f(x) into g(x), g(f(x))=[-7*(x-8)/(x+7)-8]/[(x-8)/(x+7)-1]=(-7x+56-8x-56)/(x-8-x-7)=-15x/-15=x, as desired.

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3 years ago

In 1998, as an advertising campaign, the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of

Phoenix [80]

Answer:

A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Step-by-step explanation:

We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.

Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of chocolate chips = $\frac{\sum X}{n}$ = 1238.2

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 94.3

n = sample of car drivers = 16

$\mu$ = population mean number of chips in a bag

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $1238.2-2.131 \times {\frac{94.3}{\sqrt{16} } }$ , $1238.2+2.131 \times {\frac{94.3}{\sqrt{16} } }$ ]

= [1187.96, 1288.44]

Therefore, a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

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3 years ago