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bija089 [108]
3 years ago
13

Can someone help me? My teacher said to put in simplisit form. and not to estimate.

Mathematics
1 answer:
maw [93]3 years ago
4 0

Sorry, but I don't really feel like doing all of them right now. Instead, heres a tool to help you out with that. Use the mixed numbers calculator and it'll simplify everything for you.

https://www.calculator.net/fraction-calculator.html

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There are a few ways you can go about solving this question. One way is to use the given information to find how many tons of flour can be processed in one hour. If 27 tons can be processed in 3 hours, we can do 27 divided by 3 to find that 9 tons of flour can be processed per hour. Then, if we want to see how many tons of flour can be processed in 8 hours, we can multiply 9 tons by 8 hours to get a total of 72 tons of flour.

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A researcher is going to estimate the average typing speed of students of a college. He selects a random sample of 20 students a
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Answer:

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

Step-by-step explanation:

We have the standard deviation 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 = 20 - 1 = 19

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 1.7291

The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have s = 5, n = 20. So

M = T\frac{s}{\sqrt{n}}

M = 1.7291\frac{5}{\sqrt{20}}

M = 1.933

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

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Answer:

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