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2 years ago

9

After extensive measurements of the time necessary to complete the first homework assignment, a teacher determines that there is

a population mean of 100 and a stardard deviation of 20. If she samples a class of 60 students and calculates a mean of 96 minutes, what is the z statistic (round to the nearest 2 decimal places, don't forget a negative sign if necessary)?

1 answer:

MaRussiya [10]2 years ago

4 0

Answer: -1.55

Step-by-step explanation:

Given : Mean : $\mu=100$

Standard deviation : $\sigma=20$

Sample size : $n=60$

Sample mean : $\overline{x}=96$

The test statistic for the population mean is given by :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{96-100}{\dfrac{20}{\sqrt{60}}}=-1.54919333848\approx-1.55$

Hence, the value of z statistic = -1.55

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Step-by-step explanation:

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2 years ago

After simplifying, which expressions are equivalent? Select three options.

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Answer: Second option, third option and fifth option.

Step-by-step explanation:

In order to solve this exercise it is important to remember the multiplication of signs:

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Knowing that, you can distribute the sign and add the like terms. Repeat this procedure in each expression given in the exercise.

Therefore, you get:

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You can identify that the expressions $(2.5a + 1.6b) + (3.4a + 4b)$ , $(-3.9b + a) + (-1.7b + 4.9a)$ and $5.9a - 5.6b$ are equivalent after simplifying.

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1 year ago

Dean took out a 10-year loan for $40,000 at an APR of 4% compounded monthly. What will his balance be after he has made exactly

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Answer:

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To find : What will his balance be after he has made exactly half of his monthly payments?

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