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Otrada [13]
3 years ago
11

Solve. 4y - 3 = 5y + 2 Enter your answer in the box. y = O

Mathematics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

y=-5

Step-by-step explanation:

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3 years ago
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Can someone plz help me. There is an equation and the cost too.
bonufazy [111]

Answer:

C = 40p

Cost for 15 people: 600 dollars

Step-by-step explanation:

The amusement park charges $40 per person.

That can be represented as 40 × <em>P</em> or 40<em>P</em>.

<em>C</em> is equal to the cost of admission for <em>P</em> people.

So, <em>C</em> = 40<em>P</em>

<em />

<em>C </em>= 40<em>P</em>

<em>C</em> = 40(15)

<em>C</em> = 600

The cost of admission for 15 people is 600 dollars.

Hope that helps.

5 0
3 years ago
Find the sum:
ruslelena [56]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Which describes the rotation? A. clockwise 1/2 turn B. counterclockwise 1/2 turn C. counterclockwise 1/4 turn D. clockwise 1/4 t
Irina-Kira [14]
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4 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

<em>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.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

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

<u>95% confidence interval for</u> \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].

7 0
3 years ago
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