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Strike441 [17]
2 years ago
9

. sipho buys a car for r169 000. he pays a deposit of 15% and then makes monthly installments of 1% of the leftover amount for 9

years. i) what amount must sipho pay for the deposit? ii) what is the monthly payment that sipho makes? iii) how much does sipho pay in total for the car?
Mathematics
1 answer:
Aliun [14]2 years ago
5 0

The amount for the deposit is $25350, monthly payment is $1436.5 and the total amount that Sipho pays is $178492.

Given price of car $169000, deposit of 15%, monthly installments being 1% of left over price and number of years be 9 years.

We have to find the amount of deposit, monthly installments and the total amount paid by Sipho.

We have been told that deposit is 15% of total amount. So the deposit will be 15%*169000=$25350.

Monthly installment is 1% of the left over amount.

left amount=169000-25350=$143650.

Monthly installment being =143650*1/100

=$1436.5

Total amount paid by Sipho=deposit +installment amount  * numberof installments

Number of installments=12*9

=108

Total amount=25350+1436.5*108

=25350+155142

=$178492

Hence deposit is $25350, monthly payment being $1436.5, total amount be $178492.

Learn more about installments  at brainly.com/question/2151013

#SPJ4

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Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times
Elina [12.6K]

Answer:

The  probability is   P(A') =  0.485

Step-by-step explanation:

Let assume that the number of computer produced by factory C is  k = 1  

 So  From the  question we are told that

       The number produced by  factory A is  4k =  4

        The  number produced by factory B is  7k  = 7

        The  probability of defective computers from A is  P(A) =  0.04

        The  probability of defective computers from B is  P(B)  =  0.02

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Now the probability of factory A producing a defective computer out of the 4 computers produced is  

       P(a) =  4 *  P(A)

substituting values

        P(a) =  4 * 0.04

        P(a) = 0.16

The probability of factory B producing a defective computer out of the 7 computers produced is  

       P(b) = 7  *  P(B)

substituting values

        P(b) =  7 * 0.02

        P(b) = 0.14

The probability of factory C producing a defective computer out of the 1 computer produced is  

       P(c) = 1  *  P(C)

substituting values

        P(c) =  1 * 0.03

        P(b) = 0.03

So the probability that the a computer produced from the three factory will be defective is  

     P(t) =  P(a) +  P(b) +  P(c)

substituting values

     P(t) =   0.16  + 0.14 +  0.03

     P(t) =   0.33

Now the probability that the defective computer is produced from factory A is

      P(A') =  \frac{P(a)}{P(t)}

       P(A') =  \frac{ 0.16}{0.33}

        P(A') =  0.485

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Step-by-step explanation:

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