On her first three 5 point math quizzes amy scores were 4,3 and 5. she will take three more quizzes this semester what three scores would give her an average that is a whole number? a repeating decimal?

Score 1 = 4

Score 2 = 3

Score 3 = 5

Score 1 + Score 2 + Score 3 = 12

Amy plans to give three more 5 point quizzes

Part A

Total number of quizzes Amy would have taken by the end = 6

Average score = Sum of all Scores / Total Number of quizzes

Let sum of the last 3 quizzes be x

⇒ Average Score =

Hence, for the average to be a whole number, the sum of all her scores needs to be divisible by 6.

⇒ x+12 should be divisible by 6

Since 12 is divisible by 6, we need to make sure now that x is divisible by 6

The possible values for the scores are 0, 1, 2, 3, 4 and 5. Out of these values we will pick those 3 numbers for which the sum is a multiple of 6:

0+0+0=0

2+2+2=6

3+3+0=6

4+2+0=6

4+4+4=12

5+1+0=6

5+4+3=12

Hence, her average score would be a whole number, if she scores: 0,0,0 OR 2,2,2 OR 3,3,0 OR 4,2,0 OR 4,4,4 OR 5,1,0 OR 5,4,3

Part B

For the average to be a repeating decimal, we will definitely ignore the score combinations determined above as they lead to an average that is a whole number.

Out of 0,1,2,3,4,5, the sum of the three numbers (x) should be such that is a repeating decimal

So if a person scores 0,0,1 then would be 0.16666.... i.e. repeating decimal (and average in this case would be i.e. 2.166666.... , which is also a repeating decimal)

Another such combination can be 0,1,1, then would be 0.3333..... and average in this case would be i.e. 2.33333....

There can be many such combinations.

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.  

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let = true mean number of sales at location A.

= true mean number of sales at location B

So, Null Hypothesis, :  or       {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, :  or      {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

                        T.S. =    ~

where, = sample average of items sold at location A = 39

= sample average of items sold at location B = 49

= sample standard deviation of items sold at location A = 8

= sample standard deviation of items sold at location B = 4

= sample of days location A was observed = 18

= sample of days location B was observed = 13

Also,    =  = 6.64

So, test statistics  =    ~  

                               =  -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.