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3 years ago

Find the Retail Price. Given: Original Price: $92; Markup: 12%

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

3 0

Answer: The retail price is $103.04

Work: I just did $92, and multiplied it by the percentage (.12) and I received the answer $11.04 (Markup) Then, I add $11.04 to the $92, and I get the answer $103.04.

I hope this helps, and Happy Holidays! :)

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For f(x) = x^2 and g(x) = x^2 + 2, find the following composite functions and state the domain of each

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3 years ago

A point on the top of the Leaning Tower of Pisa is shifted about 13.5 ft horizontally compared to the tower's base. To the neare

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Answer:

$4.15^0.$

Step-by-step explanation:

Given,

Leaning Tower of Pisa is shifted, P = 13.5 ft

Height of Leaning Tower of Pisa, B = 186 ft

Now, degree of tilt of the tower from vertical

$\tan \theta = \dfrac{P}{B}$

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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

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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