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xeze [42]
2 years ago
8

Rhianna is buying a car for $14,390. She has a $1000 trade-in allowance and will make a $1500 down payment. She will finance the

rest with a 4-year auto loan at 2.6% APR.
(a) How much money will she borrow in an auto loan?
(b) What will her monthly auto payment be?
(c) What is the total amount of interest she will pay?
(d) What is her total payment for the car?
(e) Rhianna is 22 years old. She buys 50/100/25 liability insurance, and collision and comprehensive insurance, each with $500 deductibles. What is her total annual premium?

Mathematics
1 answer:
antoniya [11.8K]2 years ago
8 0

Answer:

(a)

Original price for the Car Rhianna buy = $14,390.

Trade in allowance she has =$1000

So, the reduced price= $14,390-$1,000=$13,390.

Now, she make a $1500 down payment initially, so the remaining amount she pay using Auto-Loan is,    $13,390-$1500=$11,890

Therefore, $11,890 money will she borrow in an auto loan.

(b)

To find the monthly(M) auto payment be:

Since the monthly auto payment in 4 year with 2.6% APR(Annual percentage rate)

Now, Monthly Auto payment(M) be :

P=$11,890, Monthly Car payment per $1000 borrowed at 2.6% for 48 months = $21.958

∴ M=$11,890\cdot \frac{21.958}{1000}

On solving we get,  M=$261.082

Therefore, $261.082 will be her monthly auto payment.

(c)

Total amount of interest(A) she will pay ,

A=total payment in 4 years-total loan amount=(\$261.082\cdot 48)-11,890

=$12,531.936-$11,890=$641.936

So, the total amount of interest she will pay is, $641.936

(d)

Rhianna total payment for the car = total payment in 4 years+Down payment=(261.082\cdot 48)+$1500=$12531.936+$1500=$14,031.936

(e)

∵ Rhianna is 22 years female :

From the given Data in the picture:

Rating factor= 1.2  

Liability insurance 50/100/25 , Bodily injury 50/100

Premium= $310 and Property damage 25 : Premium $175  

Collision insurance=$500 deductible:  Premium $148  

Comprehensive insurance= $500 deductible : Premium $85  

Her total annual premium= 1.2\cdot (\$310+\$175+\$148+\$85) = \$861.60

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I think the points given here are plotted linearly: 

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Suppose X is the breaking strength (newtons) of a material, and X is normally distributed with
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Answer:

a) 0.997 is the  probability that the breaking strength is at least 772 newtons.

b) 0.974  is the probability that this material has a breaking strength of at least 772 but not more  than 820    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 800 newtons

Standard Deviation, σ = 10 newtons

We are given that the distribution of  breaking strength is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( breaking strength of at least 772 newtons)

P(x \geq 772)

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0.997 is the  probability that the breaking strength is at least 772 newtons.

b) P( breaking strength of at least 772 but not more  than 820)

P(772 \leq x \leq 820) = P(\displaystyle\frac{772 - 800}{10} \leq z \leq \displaystyle\frac{820-800}{10}) = P(-2.8 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2.8)\\= 0.977 - 0.003 = 0.974 = 97.4\%

P(772 \leq x \leq 820) = 97.4\\%

0.974  is the probability that this material has a breaking strength of at least 772 but not more  than 820.

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Answer:

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The total number of bugs in the collection is

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