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mote1985 [20]
1 year ago
10

Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly. After c years, Calvin has $658. 80. Makay

la deposits $300 in a different savings account that accrues 6% interest compounded quarterly. After m years, Makayla has $613. 4. What is the approximate difference in the number of years that Calvin and Makayla have their money invested?.
Mathematics
1 answer:
Drupady [299]1 year ago
5 0

The approximate difference between Calvin and Makayla's number of years of money invested is 2 years.

<h3>What is compound interest?</h3>

Compound interest is the amount charged on the principal amount and the accumulated interest with a fixed rate of interest for a time period.

The formula for the final amount with the compound interest formula can be given as,

A=P\left(1+\dfrac{r}{t}\right)^{t}

Here, A is the final amount (principal plus interest amount) on the principal amount of P with the rate of r in the time period of t.

Calvin deposits $400 in a savings account and Interest rate is 5 present compounded monthly.

As the final amount Calvin has $658.80 in c years. Thus the monthly compound interest can be given as,

658.8=400\left(1+\dfrac{5}{12\times100}\right)^{12c}\\\dfrac{658.8}{400}=\left(\dfrac{1205}{1200}\right)^{12c}\\c\cong10\rm years

Now Makayla deposits $300 in a different savings account that accrues 6% interest compounded quarterly. The final amount she gets is $613.4 in m years. Thus,

613.4=300\left(1+\dfrac{6}{12\times100}\right)^{12m}\\\dfrac{613.4}{300}=\left(\dfrac{1206}{1200}\right)^{12m}\\c\cong12\rm years

Thus, the approximate difference in the number of years that Calvin and Makayla have their money invested is 2 years.

Learn more about the compound interest here;

brainly.com/question/24274034

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This measure is just 0.17 standard deviations from the mean, so we should not be surprised.

Step-by-step explanation:

Normal probability distribution

When the distribution is normal, we use the z-score formula.

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

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If the z-score is lower than -2, or higher than 2.5, the score of X is considered unusual.

Subtraction of normal variables:

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Let xS – xL represent the sampling distribution.

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Should we be surprised if the sample mean housebroken age for the small breed dogs is at least 2.5 months more than the sample mean housebroken age for the large breed dogs? Explain your answer.

We have to find the z-score for X = 2.5. So

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

Z = \frac{2.5 - 2}{2.915}

Z = 0.17

This measure is just 0.17 standard deviations from the mean, so we should not be surprised.

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