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Alex Ar [27]
2 years ago
9

Davis has 141 figures in his action figure collection. One-third of the figures are sports related. How many of the figures in D

avis’ collection are NOT sports related
Mathematics
1 answer:
Lelu [443]2 years ago
7 0

We have been given that Davis has 141 figures in his action figure collection. One-third of the figures are sports related.

To find the number of figures that are not sports related we will find 2/3 of 141 as 2/3 of the figures will be not related to sports.

\text{ Figures not related to sports}=141\times\frac{2}{3}

\text{ Figures not related to sports}=47\times2

\text{ Figures not related to sports}=94

Therefore, 94 figures in Davis’ collection are not sports related.

 

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An arena receives $1,700 per event for 12 concerts. If the costs for this sponsorship total $14,000, what is the profit margin f
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Answer:

total profit margin for all 12 concert is $6,400.

profit margin for all one concert is  $533.33

Step-by-step explanation:

Amount of money received for one event = $1700

Total no of concert = 12

Total money received for 12 event = Amount of money received for one event * Total no of concert

Total money received for 12 event  = $1700 *12 = $20,400

Total cost of sponsorship = $14,000

Profit margin is the difference money invested and money earned.

here total investment is $14,000

Total money earned is  $20,400

Therefore profit margin = money earned - total investment

= $20,400 -  $14,000 = $6,400

Therefore total profit margin for all 12 concert is $6,400.

However to calculate profit margin for one concert we can simply divide the total profit margin for all 12 concert by no of concert (i.e 12)

profit margin for one concert = total profit margin for all 12 concert/total no of concert  = $6,400/12 = $533.33

profit margin for all one concert is  $533.33

6 0
2 years ago
A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s
denpristay [2]

Answer:

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

Step-by-step explanation:

Notation

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=40-1=39

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

3 0
2 years ago
Suppose a scheduled flight must average at least 60% occupancy to be profitable. A sample of 120 flights on a particular route w
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Answer: see the graphic

Step-by-step explanation:

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C. Type II error does not show that the flight is profitable

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Answer:

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Step-by-step explanation:

Total borrowed=$2,100

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X=$2,478 - $2,100

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Simple interest=P×R×T

Where,

P= principal=$2,100

R=Rate=?

T=Time=3 years

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Simple interest=P×R×T/100

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$378=$63R

R=$378/$63

R=6

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Rate=6%

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