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____ [38]
3 years ago
12

The president of a certain university receives a salary that is three times the salary of one of the department heads. The total

of the two salaries is $240,000
Mathematics
1 answer:
marissa [1.9K]3 years ago
8 0

Answer:

Salary of President =   $180,000

Salary of Dept. Head =  $60,000

Step-by-step explanation:

The question is to find the salary of president and department head.

We let the salary of one of department heads be "x"

Since, the salary of president is THREE (3) TIMES a department head, we can say:

Salary of President = 3x

Now, we know the total of these 2 salaries is 240,000. Thus, we can write:

Salary of President + Salary of Dept. Head = 240,000

3x + x = 240,000

Now, we can solve for x:

3x + x = 240,000\\4x=240,000\\x=60,000

And now the president is "3x", so his salary is:

3(60,000) = 180,000

Thus,

President's salary = 180,000

Dept head's salary = 60,000

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Answer:

Step-by-step explanation:

Part A

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

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Answer:

Option (d) is correct.

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Step-by-step explanation:

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Thus, Option (d) is correct.

 

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