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Mademuasel [1]
4 years ago
10

Today a car is valued at $42000. The value is expected to decrease at a rate of 8% each year. Choose the equation that can be us

ed to solve the problem.
A.) y=42000(1+.08)^6
B.) y=42000(1+6)^8
C.) y=42000(1−.08)^6
D.) y=42000(1−.8)^6
And then: What is the value of the car expected to be 6 years from now?
This is the only question I don't understand, someone help please!
Mathematics
2 answers:
Scrat [10]4 years ago
3 0
I think the correct answer is

C.) y=42000(1−.08)^6

as for what it will cost in six years:

25466.91

not sure if you have to include the decimal or round it.
ivolga24 [154]4 years ago
3 0

Answer:

Option (c) is correct.

y=42000(1-0.08)^6 equation that can be used to solve the given problem and the value of the car expected to be 6 years from now is $25466.91

Step-by-step explanation:

Given :   Today a car is valued at $42000. The value is expected to decrease at a rate of 8% each year.

We have to choose the equation that can be used to solve the given problem.

We know

Depreciation formula given as,

A=P(1-\frac{r}{100} )^n

Where A is the amount value after depreciation.

P is present value

r is depreciate rate.

n is time period.

Given : P = 42000  and r= 8% also given time period is 6.

Substitute , we have,

y=42000(1-\frac{8}{100} )^6

Simplify , we have,

y=42000(1-0.08)^6

Simplify we get,

y=25466.91

Thus, y=42000(1-0.08)^6 equation that can be used to solve the given problem and the value of the car expected to be 6 years from now is $25466.91

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Answer:

A) check attached picture

B) 272

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Step-by-step explanation:

From the information given ;

A) The Venn diagram of the information can be found in the attached picture.

B) Number of students involved in both team sports and music.

n(student in sports team) = 385

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Therefore,

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Adding the following:

n(music alone) + n(team sport alone) + n( both music and team sport) + n(neither music nor team sport) = Total number of students

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P(involved in neither) = 87 / 650 = 0.1338

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0.1738 + 0.2738 = 0.4478

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