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Ahat [919]
3 years ago
5

Car A has an initial value of 30,000 and depreciates at a rate of 20% per year. Car B has an initial value of 20,000 and depreci

ates at a rate of 15% per year.Car B will become greater than the value of car A during the ninth,seventh,third, or fifth year
Mathematics
2 answers:
tiny-mole [99]3 years ago
7 0

It’s the 7th year

Plato

Doss [256]3 years ago
4 0

Answer:

The value of Car B will become greater than the value of car A during the fifth year.

Step-by-step explanation:

Note: See the attached excel file for calculation of beginning and ending values of Cars A and B.

In the attached excel file, the following are used:

Annual Depreciation expense of Car A = Initial value of Car A * Depreciates rate of Car A = 30,000 * 20% = 6,000

Annual Depreciation expense of Car B from Year 1 to Year 6 = Initial value of Car B * Depreciates rate of Car B = 20,000 * 15% = 3,000

Annual Depreciation expense of Car B in Year 7 =  Beginning value of Car B in Year 7 = 2,000

Conclusion

Since the 8,000 Beginning value of Car B in Year 5 is greater than the 6,000 Beginning value of Car A in Year 5, it therefore implies that the value Car B becomes greater than the value of car A during the fifth year.

Download xlsx
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Is this right I need help with this
Margarita [4]
Y axis is counting by 2's
it started at 2 and went up to 6
Y = 4

X axis is counting by 2'2
starts at one goes to 3
X = 2

Slope is 4/2 because rise over run
4/2 = 2
Slope is simplified to 2. 
5 0
3 years ago
Find volume with full working out pls asap
ZanzabumX [31]

Answer:

868 m³

Step-by-step explanation:

Volume = Difference of cones

V = ⅓(pi×r²h)

Volume of cone with base Circle O:

h = 8.2+8.2 = 16.4 m

V1 = ⅓(pi×7.6²×16.4)

V1 = 991.97

Volume of cone with base Circle A:

V2 = ⅓(pi×3.8²× 8.2)

V2 = 124 m³

Volume = V1 - V2

991.97 - 124 = 868 m³

6 0
3 years ago
In 2002, there were 972 students enrolled at Oakview High School. Since then, the number of students has increased by 1.5% each
Rudiy27

Answer:

N(t) = 972(1.015)^{t}

Growth function.

The number of students enrolled in 2014 is 1162.

Step-by-step explanation:

The number of students in the school in t years after 2002 can be modeled by the following function:

N(t) = N(0)(1+r)^{t}

In which N(0) is the number of students in 2002 and r is the rate of change.

If 1+r>1, the function is a growth function.

If 1-r<1, the function is a decay function.

In 2002, there were 972 students enrolled at Oakview High School.

This means that N(0) = 972

Since then, the number of students has increased by 1.5% each year.

Increase, so r is positive. This means that r = 0.015

Then

N(t) = N(0)(1+r)^{t}

N(t) = 972(1+0.015)^{t}

N(t) = 972(1.015)^{t}

Growth function.

Find the number of students enrolled in 2014.

2014 is 2014-2002 = 12 years after 2002, so this is N(12).

N(t) = 972(1.015)^{t}

N(12) = 972(1.015)^{12}

N(12) = 1162

The number of students enrolled in 2014 is 1162.

7 0
3 years ago
What is the answer to <br>11-9m=-70
Fynjy0 [20]

Answer:

m=9

Step-by-step explanation:

11-9m=-70

-11        -11

-9m  =  -81

divide both sides by -9 and you get m=9

7 0
3 years ago
Read 2 more answers
A spinner has 10 equally sized sections, 5 of which are gray and 5 of which are blue. The spinner is spun twice. What is the pro
zhenek [66]

Answer:

P(Gray\ and\ Blue) = \frac{1}{4}

Step-by-step explanation:

Given

Sections = 10

n(Gray) = 5

n(Blue) = 5

Required

Determine P(Gray and Blue)

Using probability formula;

P(Gray\ and\ Blue) = P(Gray) * P(Blue)

Calculating P(Gray)

P(Gray) = \frac{n(Gray)}{Sections}

P(Gray) = \frac{5}{10}

P(Gray) = \frac{1}{2}

Calculating P(Gray)

P(Blue) = \frac{n(Blue)}{Sections}

P(Blue) = \frac{5}{10}

P(Blue) = \frac{1}{2}

Substitute these values on the given formula

P(Gray\ and\ Blue) = P(Gray) * P(Blue)

P(Gray\ and\ Blue) = \frac{1}{2} * \frac{1}{2}

P(Gray\ and\ Blue) = \frac{1}{4}

3 0
3 years ago
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