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Scrat [10]
3 years ago
7

a sum of 1000 invested at an jntrest rate 12% per year. Find the amounts in the account after 3 years if intrest is compounded a

nnualy, semiannually, quarerterly, monthly, and daily.
Mathematics
1 answer:
Lina20 [59]3 years ago
4 0

Given Information:

Annual interest rate = r = 12%

Principal amount = P = $1000

Number of years = t = 3  

Required Information

Accumulated amount = A = ?

Answer:

Annual compounding = A = $1404.93

Semi-annuall compounding = A = $1418.52

Quarterly compounding = A = $1425.76

Monthly compounding = A = $1432.30

Daily compounding = A = $1433.14

Step-by-step explanation:

The accumulated amounts in terms of compound interest is given by  

A = P(1 + i)^N

Where  P is the initial amount invested and A is the accumulated amount.

For annual compounding:

i = 0.12

N = 3

A = 1000(1 + 0.12)^3 \\\\A =  \$ 1404.93

For semiannually compounding:

i = 0.12/2 = 0.06

N = 2*3 = 6

A = 1000(1 + 0.06)^6 \\\\A =  \$ 1418.52

For quarerterly compounding:

i = 0.12/4 = 0.03

N = 4*3 = 12

A = 1000(1 + 0.03)^12 \\\\A =  \$ 1425.76

For monthly compounding:

i = 0.12/30 = 0.004

N = 30*3 = 90

A = 1000(1 + 0.004)^90 \\\\A =  \$ 1432.30

For daily compounding:

i = 0.12/365 = 0.0003287

N = 365*3 = 1095

A = 1000(1 + 0.0003287)^1095 \\\\A =  \$ 1433.14

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Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let x be a rand
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

<u />

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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