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2 years ago

Pierce currently has $10,000. What was the value of his money five years ago if he has earned 5 percent interest each year?

Mathematics

2 answers:

tia_tia [17]2 years ago

4 0

The first answer is correct, if you go back by 5% each year you will see that.

Yuri [45]2 years ago

3 0

Answer:

$ 7,835.26

Step-by-step explanation:

Let x be the value of his money five years ago,

Given,

Annual rate of interest = 5 %,

Time = 5 years,

So, the amount after 5 years,

$A=x(1+\frac{5}{100})^5$

$=x(1+0.05)^5$

$=x(1.05)^5$

According to the question,

$x(1.05)^5=10000$

$\implies x = 7835.262\approx 7835.26$

Hence, the value of his money five years ago was $ 7,835.26.

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1 year ago

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2 years ago

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The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev

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Answer:

(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716

(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;

Mean, $\mu$ = 30.05 inches and Standard deviation, $\sigma$ = 0.2 inches

Let X = A sheet selected at random from the population

Here, the standard normal formula is ;

Z = $\frac{X - \mu}{\sigma}$ ~ N(0,1)

(a) The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65)

P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)

P(X < 30.65) = P( $\frac{X - \mu}{\sigma}$ < $\frac{30.65 - 30.05}{0.2}$ ) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425

= 0.9985

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P(Y < 3.2) = P( $\frac{Y - \mu}{\sigma}$ < $\frac{3.2 - 0}{1}$ ) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688

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Dima020 [189]

Answer:

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Answer:

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