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Alekssandra [29.7K]
3 years ago
7

Eliza started her savings account with $100. Each month she deposits $25 into her account. Determine the average rate of change

in Eliza's account from the 2nd month to the 10th month. Show all work for full credit.
Mathematics
1 answer:
Afina-wow [57]3 years ago
6 0
First, lets create a equation for our situation. Let x be the months. We know four our problem that <span>Eliza started her savings account with $100, and each month she deposits $25 into her account. We can use that information to create a model as follows:
</span>f(x)=25x+100
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is: \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1} }. So lets replace the values in our formula to find the average of our function:
\frac{25(10)+100-[25(2)+100]}{10-2}
\frac{350-150}{8}
\frac{200}{8}
25

We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>
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When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

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The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

Z = \frac{X - \mu}{\sigma}

Z = \frac{1 - (-5)}{\sqrt{13}}

Z = \frac{6}{\sqrt{13}}

Z = 1.66

Z = 1.66 has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

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