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tatuchka [14]
3 years ago
8

Gary deposits $1,500 into each of two savings accounts. Account I earns 5% annual simple interest. Account II earns 5% interest

compounded annually. Gary does not make any additional deposits or withdrawals. What is the sum of the balances of Account I and Account II at the end of 4 years?
Mathematics
1 answer:
hodyreva [135]3 years ago
7 0

Answer:

  • $3623.26

Step-by-step explanation:

<u>Given</u>

  • Each deposit = $1500
  • Interest rate= 5%
  • Time = 4 years
  • Compound number = annual

<u>Simple interest account</u>

  • B = 1500*(1 + 4*0.05) = 1500*1.2 = $1800

<u>Compound interest account</u>

  • B = 1500*(1 + 0.05)^4 = $1823.26

<u>Total balance</u>

  • $1800 + $1823.26 = $3623.26

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Answer:

Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

<em>D</em> = a person has the disease

<em>X</em> = the test is positive.

The information provided is:

P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99

Compute the probability that a person does not have the disease as follows:

P(D^{c})=1-P(D)=1-0.88=0.12

The probability of a person not having the disease is 0.12.

Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264

Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188

Compute the probability that a randomly selected person is tested negative  as follows:

P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})

           =0.0264+0.1188\\=0.1452

Thus, the probability of the test indicating that the person does not have the disease is 0.1452.

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faust18 [17]
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