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dusya [7]
3 years ago
13

Wanda reads 180 pages in 2 3/4 h. How many pages can she read in 5 1/2 h?

Mathematics
2 answers:
lakkis [162]3 years ago
8 0

Answer:

are there any options? if not then i think it's 360 ( sorry if i'm wrong)

Lostsunrise [7]3 years ago
6 0

Answer:

360

Step-by-step explanation:

We find Wanda's reading speed in pager per hour, which is 180/2 3/4 = 180/2.75 = 65.4545...

To find how many pages she reads in 5 1/2 hours, multiply 65.4545... by 5 1/2, or 5.5 to get 360. She reads 360 pages in 5 1/2 hours

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A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen
tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the  average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% 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 average desired retirement age = 55 years

            \sigma = sample standard deviation = 3.4 years

            n = sample of seniors = 101

            \mu = true mean retirement age of all college students

<em>Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 96% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.114 < t_1_0_0 < 2.114) = 0.96  {As the critical value of t at 100 degree

                                               of freedom are -2.114 & 2.114 with P = 2%}  

P(-2.114 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.114) = 0.96

P( -2.114 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

P( \bar X-2.114 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ) = 0.96

<u>96% confidence interval for</u> \mu = [ \bar X-2.114 \times {\frac{s}{\sqrt{n} } } , \bar X+2.114 \times {\frac{s}{\sqrt{n} } } ]

                                           = [ 55-2.114 \times {\frac{3.4}{\sqrt{101} } } , 55+2.114 \times {\frac{3.4}{\sqrt{101} } } ]

                                           = [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0
3 years ago
You are curious about the average number of yards Matthew Stafford throws for each game for the Detroit Lions. You randomly sele
Damm [24]

Answer:

The margin of error for this estimate is of 14.79 yards per game.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

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The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample.

You randomly select 20 games and see that the average yards per game is 273.7 with a standard deviation of 31.64 yards.

This means that n = 20, s = 31.61

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M = 2.093\frac{31.61}{\sqrt{20}}

M = 14.79

The margin of error for this estimate is of 14.79 yards per game.

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