O Mr. Math S. Fun is looking for his missing pair of

A math teacher’s midterm is worth 200 points. The mean of the test is 143 and the standard deviation is 15.7. If the scores have

Klio2033 [76]

The answer is D: about 95%.

7 0

3 years ago

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In a study on the time that

krok68 [10]

The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

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To solve this question, we need to find the confidence interval for the amount of time it takes the students to get the degree.

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

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The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 80 - 1 = 79

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95% confidence interval

Standard level of confidence, we have to find a value of T, which is found looking at the t table, with 79 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.9905.

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The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9905\frac{2.2}{\sqrt{80}} = 0.5$

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

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The lower end of the interval is the sample mean subtracted by M. So it is 4.8 - 0.3 = 4.3 years.

The upper end of the interval is the sample mean added to M. So it is 4.8 + 0.3 = 5.3 years.

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The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

A similar question is given at brainly.com/question/24278748

6 0

3 years ago

2) AY RA O positive zero O negative O undefined

bezimeni [28]

Answer:

negative

hope it helps;)

8 0

3 years ago

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Based on a study of population projections for 2000 to 2050, the projected population of a group of people (in millions) can b

Olegator [25]

Answer:

(a) 0.107 million per year

(b) 0.114 million per year

Step-by-step explanation:

$A(t) = 11.19(1.009)^t$

(a) The average rate of change between 2000 and 2014 is determined by dividing the difference in the populations in the two years by the number of years. In the year 2000, $t=0$ and in 2014, $t=14$ . Mathematically,

$\text{Rate}=\dfrac{A(2014) - A(2000)}{2014-2000}=\dfrac{11.19(1.009)^14-11.19(1.009)^0} {14}$

$A(0)=\dfrac{12.69-11.19}{14}=\dfrac{1.5}{14}=0.107$

(b) The instantaneous rate of change is determined by finding the differential derivative at that year.

The result of differentiating functions of the firm $y=a^x$ (where $a$ is a constant) is $\dfrac{dy}{dx}=a^x\ln a$ . Let's use in this in finding the derivative of $A(t)$ .