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Ratling [72]
3 years ago
5

Drake wants to save $750 so that he can take a class on computer analysis for cars. The class is being held on various dates ove

r the next several months. Drake is planning to take the class 6 weeks from now, so he plans to save $125 each week.
Unfortunately, Drake had to take out a little money from his savings in the 3rd week. After 4 weeks, Drake has $470. He knows that he must adjust his plan in order to meet his goal. Drake came up with the following options:

Option A: Stay with saving the original amount each week but take the class a week later than originally planned.

Option B: Increase the amount of money he saves each week by $15 from his original plan.
Mathematics
2 answers:
LenaWriter [7]3 years ago
7 0

Answer:

the correct answer is C both options will work in this scenario.

Step-by-step explanation:

allochka39001 [22]3 years ago
6 0
<span>Drake can meet his $750 planned savings goal with either concept. After 4 weeks, he was $30 less than where he planned to be ($450 instead of $480). If he simply added $250 after two more weeks, he would be at $720 instead of $750, and would have to wait one more week to take the class. Conversely, if he added $15 to the next two weeks' savings, he would recoup the $30 he pulled out of the savings in those two weeks and would be back on the path to being able to take the course at the end of the 6-week period as originally planned.</span>
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Answer:

The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the population, we have that:

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The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.

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Answer:

https://youtu.be/iik25wqIuFo

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