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Aliun [14]
3 years ago
12

A coach pays $100.44 for 36 baseballs. What is the cost of each baseball?

Mathematics
2 answers:
sergey [27]3 years ago
8 0

Answer:

2.79

Step-by-step explanation:

Take the total cost and divide by the number of baseballs

100.44 / 36

2.79 per baseball

Rainbow [258]3 years ago
8 0

Answer: 2.79

Step-by-step explanation: To find how much money each baseball costs, we need to find the unit price of 1 baseball.

Unit price means the cost per unit,

in this case, the cost per baseball.

Since we know that it costs $100.44 for 36 baseballs,

to find the cost for 1 baseball, we need to divide 36 into $100.44 or 100.44 divided by 36.

Dividing, 100.44 divided by 36 is 2.79.

So, the unit price for 1 baseball is $2.79.

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Complete Question

A 95% confidence interval of 19.3 months to 47.5 months has been found for the mean duration of? imprisonment, ??,of political prisoners of a certain country with chronic PTSD.

a. Determine the margin of error, E.

b. Explain the meaning of E in this context in terms of the accuracy of the estimate.

c. Find the sample size required to have a margin of error of 13 months and a 99% confidence level.? (Use 38 months. for standard deviation )

d. Find a 99% confidence interval for the mean duration of? imprisonment, ??, if a sample of the size determined in part? (c) has a mean of 36.5 months.

Answer:

a

 E =  14.1

b

In this context  E tell us that the true mean will lie within E = 14.1 of the sample mean

c

  n  =57

d

   23.514 < \mu < 49.486

Step-by-step explanation:

Considering question a

From the question we are told that

   The upper limit is  U =   47.5 months

    The lower limit is   L = 19.3 months

Generally the margin of error is mathematically represented as

       E =  \frac{U - L }{2}

 =>  E =  \frac{ 47.5 -  19.3 }{2}

 =>  E =  14.1

Considering question b

In this context  E tell us that the true mean will lie within E = 14.1 of the sample mean

Considering question c

Generally the sample size is mathematically represented  as

        n  = [ \frac{ Z_{\frac{\alpha }{2}  *  \sigma }}{ E} ]^2

Here  E is given as  E = 13

Given that the confidence level is  99% then the level of significance is  

      \alpha  =  (100 - 99 )\%

=>   \alpha  = 0.01

From the normal distribution table  the critical value  of  \frac{\alpha }{2}  is  

      Z_{\frac{\alpha }{2} } = Z_{\frac{0.01 }{2} } =  2.58

So

    n = [ \frac{2.58 *  38}{13}]^2

=>    n  =57

Considering question d

From the question

   The sample mean is   \= x  =  36.5

Generally the margin of error is mathematically represented as  

         E =  Z_{\frac{\alpha }{2} }  *  \frac{\sigma }{n}

=>      E =  2.58  *  \frac{38 }{57}

=>      E =  12.986    

Generally the 99% confidence interval  for  mean distribution is  mathematically represented as

    36.5  - 12.986  <  \mu  <   36.5  + 12.986

=>     23.514 < \mu < 49.486

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