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jonny [76]
3 years ago
6

David put $1,000 into an account that yielded 1.3% interest, compounded quarterly. The money was left in the account for 20 year

s. How much is in the account when he withdraws it?
Mathematics
1 answer:
taurus [48]3 years ago
4 0

Based on the amount left in the account and the interest rate, the amount in the account after 20 years is <u>$1,296.38.</u>

<h3>Periodic Interest </h3>

You need to convert the interest from an annual interest rate to a quarterly one.

= 1.3% / 4

= 0.325%

<h3>Number of periods</h3>

= 20 years x 4

= 80 quarters

<h3>Amount after 20 years</h3>

= Amount x ( 1 + rate) ^ number of periods

= 1,000 x ( 1 + 0.325%)⁸⁰

= $1,296.38

In conclusion, the account would have $1,296.38.

Find out more about compound interest at brainly.com/question/25857644.

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Oduvanchick [21]

Complete question is;

Scores of an IQ test have a​ bell-shaped distribution with a mean of 100 and a standard deviation of 13. Use the empirical rule to determine the following.

​(a) What percentage of people has an IQ score between 87 and 113​?

​(b) What percentage of people has an IQ score less than 74 or greater than 126​? ​

(c) What percentage of people has an IQ score greater than 139​?

Answer:

A) 68% of the people had an IQ score between 87 and 113

B) 5% of the people had IQ scores less than 74 and greater than 126.

C) 0.15% had an IQ score greater than 139

Step-by-step explanation:

We are given;

Mean; x¯ = 100

Standard deviation; σ = 13

According to the empirical rule in statistics;

>> 68% of the data lies within one standard deviation of the mean

>> 95% of the data lies within two standard deviations of the mean

>> 99.7% of the data lies within three standard deviations

A) We want to find the percentage of people with an IQ score between 87 and 113.

Within one standard deviation, we have;

x¯ ± σ

>> (100 - 13), (100 + 13)

>> (87, 113) which is the range we are looking for.

Thus, 68% of the people had an IQ score between 87 and 113

B) we want to find percentage of people has an IQ score less than 74 or greater than 126.

Let's first find those who had between 74 and 126.

Let's use two standard deviations within the mean.

x¯ ± 2σ

>> (100 - (2×13)), (100 + (2×13))

>> (74, 126) which is the range we are looking for.

So percentage that have IQ scores between 74 and 126 is 95%

Thus;

Percentage of those who had less than 74 and greater than 126 is;

P = 1 - 95%

P = 5%

C) we want to find the percentage of people that had an IQ score greater than 139.

Let's use the z-score formula;

z = (x¯ - μ)/σ

z = (139 - 100)/13

z = 39/13

z = 3

This means it is 3 standard deviations above the mean.

Thus;

Since it's one side outside the mean, then;

Percentage of people with an IQ score greater than 139 = (1 - 99.7%)/2 = 0.3%/2 = 0.15%

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