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olga nikolaevna [1]
3 years ago
11

When Rahul began 7th grade, he put his savings of $3,000 in an account that compounded interest annually. He hoped to have $6,00

0 by the time he graduates high school in six years. What interest rate is required for him to reach his goal?
 

An interest rate of
% is required.
Mathematics
1 answer:
DENIUS [597]3 years ago
7 0

Answer:

12%

Step-by-step explanation:

You are going to want to use a modified compound interest formula, as shown below. This version of the compound interest formula is what you use to solve for the interest rate.

r=n[(\frac{A}{P})^{1/nt}-1]

r = interest rate (decimal)

n = numbers of times compounded (annually)

A = total amount

P = principal amount

<em>t = time (years)</em>

Now, lets plug in the values:

r=1[(\frac{6,000}{3,000})^{1/(1)(6)}-1]

r=.12246

Next, multiply <em>r </em>by 100 to get our answer:

.12246(100)= 12.246

The last step is to round our answer to the nearest whole number:

12.246 -> 12

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SOVA2 [1]

Answer:

<u>Question 1:</u>

<u>(a) </u>P(x<60) = 0.9236

<u>(b) </u>P(x>16) = 0.9564

<u>(c) </u>P(16<x<60) = 0.88

<u>(d) </u>P (x>60) = 0.0764

<u />

<u>Question 2:</u>

<u>(a) </u>P(x<3) = 0.0668

<u>(b) </u>P(x>7) = 0.0062

<u>(c) </u>P(3<x<7) = 0.927

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Step-by-step explanation:

<u>Question 1:</u>

x = no. of mg of porphyrin per deciliter of blood.

μ = 40

σ = 14

(a) We need to compute P(x<60). We need to find the z-score using the normal distribution formula:

z = (x - μ)/σ

P(x<60) = P((x - μ)/σ < (60 - 40)/14)

             = P(z < 20/14)

             = P(z<1.43)

Using the normal distribution probability table we can find the value of p at z=1.43.

P(z<1.43) = 0.9236

so, P(x<60) = 0.9236

(b) P(x>16) = P(z>(16-40)/14)

                 = P(z>-1.71)

                 = 1 - P(z<-1.71)

                 = 1 - 0.0436

    P(x>16) = 0.9564

(c) P(16<x<60) = P((16-40)/14) < x < (60-40)/14)

                        = P(-1.71 < z < 1.43)

This probability can be calculated as: P(z<1.43) - P(z<-1.71)

                                         P(16<x<60) = 0.9236 - 0.0436

                                         P(16<x<60) = 0.88

(d) P(x>60) = 1 - P(x<60)

     we have calculated P(x<60) in part (a) so,

    P(x>60) = 1 - 0.9236

    P (x>60) = 0.0764

<u>Question 2:</u>                              

μ = 4.5 mm

σ = 1.0 mm

In this question, we will again compute the z-scores and then find the probability from the normal distribution table.

(a) P(x<3) = P(z<(3-4.5)/1)

                = P(z<-1.5)

     P(x<3) = 0.0668

(b) P(x>7) = 1 - P(x<7)

               = 1 - P(z<(7-4.5)/1)

               = 1 - P(z<2.5)

               = 1 - 0.9938

    P(x>7) = 0.0062

(c) P(3<x<7) = P(x<7) - P(x<3)

  we have computed both of these probabilities in parts (a) and (b) so,

    P(3<x<7) = 0.9938 - 0.0668

    P(3<x<7) = 0.927

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