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Lesechka [4]
3 years ago
15

A student's grades in the three tests in College Algebra are 85, 60, and 69. a) How many points does the student need on the fin

al to average 74? b) Find part (a), assuming that the final carries double the weight. a) The student must score on the final. (Type an integer or a decimal.) b) The student must score on the final, assuming that the final carries double the weight. (Type an integer or a decimal.) A company manufactures shaving sets for $ each
Mathematics
1 answer:
Anastaziya [24]3 years ago
7 0

Answer:

(a) The student need 82 points on the final to average 74.

(b) The student must score 41 points on the final, assuming that the final carries double the weight.

Step-by-step explanation:

We are given that a student's grades in the three tests in College Algebra are 85, 60, and 69.

As we know that the formula for calculating the average of numbers is given by;

          Average (Mean)  =  \frac{\text{Sum of all values}}{\text{Number of observations}}

(a) Let the points student need on the final test to average 74 be 'x'.

So, Average of all four test  =  \frac{85+60+69+x}{4}

                     74 = \frac{85+60+69+x}{4}

                     74 = \frac{214+x}{4}

                     214+x =74 \times 4

                      x = 296-214

                        x=82

Hence, the student needs 82 points on the final to average 74.

(b) It is given that the final carries double the weight, this means that let the points student need on the final test to average 74 be '2x'.

So, Average of all four test  =  \frac{85+60+69+2x}{4}

                     74 = \frac{85+60+69+2x}{4}

                     74 = \frac{214+2x}{4}

                     214+2x =74 \times 4

                      2x = 296-214

                        x=\frac{82}{2} = 41

Hence, the student must score on the final 41 points, assuming that the final carries double the weight.

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Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

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

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 = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.989.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26

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

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

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Answer:

13 ≥ x

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  • Pat = x

<u>Larry has at least 3 more than Pat</u>

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<u>Simplified</u>

  • 16-3  ≥ x + 3 - 3
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Step-by-step explanation:

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