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ANTONII [103]
3 years ago
13

When you graduate college at the age of 20, you want to start saving up for retirement. If your investment pays a fixed APR of 8

.5% and you want to have $500,000 when you retire in 45 years, how much would you need to deposit, at the beginning of each month, to reach this goal?
Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
4 0

Answer:

A). $79.53

Step-by-step explanation:

GREYUIT [131]3 years ago
3 0

Answer: The answer is A

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alina1380 [7]

Answer:

D

Step-by-step explanation:

See in title it says thousands,

so multiply whatever number in graph by 1000

Rhode Islands shows 45 in graph, so:

45x1000=45,000

5 0
3 years ago
To rent a certain meeting room, a college charges a reservation fee of $41 and an additional fee of $7.90 per hour. The chemistr
Lemur [1.5K]

Answer:

41

Step-by-step explanation:

Rounding

Your cost just for the room is 41

You pay 7.90 for Your hours of being there

And your Budget is 96.30

the equation wants you for figure how many hours would you be there at the establishment that you would pay your full budget 96.30 or closer and You would only pay 41.00

4 0
3 years ago
7x + 9 = 10 - x please help meeeeeeee
nikklg [1K]

Answer:

7x + 9 = 10 - x \\ (7 + 1)x = 10 - 9 \\ 8x = 1 \\ x =  \frac{1}{8}

8 0
3 years ago
Read 2 more answers
Okay.. <br>so I'm confused<br>anyone willing to help me​
Firdavs [7]

1a2c3d4a5q that is fhev
6 0
3 years ago
Depreciation
hammer [34]

Answer:

a) y=-3950\cdot{x}+21500

b) f(x) =21500\cdot{(0.7953)^x}

c) Year 1: Linear $17550, exponential $17099

Year 4: Linear $5700, exponential $8601

d) Exponential model

e) The linear model depreciates the value quicker than exponential model long term around 4 years

Step-by-step explanation:

a) At year 0 the price is 21500 and at year 2 the price is 13600

WE can use points (0,21500) and (2,13600)

We can determine the gradient

m=(13600-21500)/(2-0)=-3950

We can use the point slow formula:

y-y_1=m\cdot{(x-x_1)}

y-21500=-3950\cdot{(x-0)}

y=-3950\cdot{x}+21500

b) We can use the following equation:

f(x) =a\cdot{(1+r)^x}

f(x) is the depreciation value after amount of time t, a is the new value, r is the rate of depreciation and x is the time.

13600=21500\cdot{(1+r)^2}

r=-0.2047

The depreciation rate is 20.47% and is negative because it decreases the new value of the car

c) Year 1:

y=-3950\cdot{1}+21500=17550

f(x) =21500\cdot{(0.7953)^1}=17099

Year 4:

y=-3950\cdot{4}+21500=5700

f(x) =21500\cdot{(0.7953)^4}=8601

d) Year 2

y=-3950\cdot{2}+21500=13600

f(x) =21500\cdot{(0.7953)^2}=13598

3 0
3 years ago
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