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maksim [4K]
3 years ago
8

Conner opens a new savings account and decides to deposits $100 on the first day of each month into the account which earns 1.0%

interest each month. How much is in the account to the nearest dollar, right after Conner makes his last deposit on the first day of the sixth year (the 61st month)?
Mathematics
2 answers:
Lilit [14]3 years ago
8 0

like 7,523.38



i think lol




source : https://www.bankrate.com/calculators/savings/simple-savings-calculator.aspx

Sveta_85 [38]3 years ago
3 0

Given:

monthly deposit, A=100

rate of interest, i = 1% per month = 0.01 per period

compounding period = 1 month (assumed).

Total payments :

n=60 payments of $100 at the end of the month (beginning of the next month) +

1 payment at the beginning of the first month.


Future value after 60 months

= amount due to 60 monthly payments at end of month + value of first payment

= A((1+i)^n-1)/i  + A*(1+i)^60

=100(1.01^60-1) + 100(1.01^60)

= 8166.967 + 181.670

= 8348.64 (to the nearest cent)

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svetoff [14.1K]

Answer:

5x − 4 = 3x + 8

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5x = 3x + 12

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2x = 12

2x <em>/ 2</em> = 12 <em>/ 2</em>

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3 years ago
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A random sample of 85 sixth-graders in a large city take a course designed to improve scores on a reading comprehension test. Ba
BartSMP [9]

Answer:

12.6 < \mu < 14.8

For this case we can find the margin of error like this:

ME= \frac{14.8-12.6}{2}= 1.1

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

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So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

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We know that the confidence interval is given by:

12.6 < \mu < 14.8

For this case we can find the margin of error like this:

ME= \frac{14.8-12.6}{2}= 1.1

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

12.6 +1.1=13.7

14.8-1.1=13.7

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

3 0
3 years ago
Is 226.08cm to the third power the correct response?
drek231 [11]
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So

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Add & Subtract (left to right)
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Diano4ka-milaya [45]

Answer:

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