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ankoles [38]
3 years ago
7

A new car is purchased for $ 41 , 000 $41,000 and over time its value depreciates by one half every 3 years. What is the value o

f the car 17 years after it was purchased, to the nearest hundred dollars?
Mathematics
1 answer:
Nat2105 [25]3 years ago
5 0

Answer:

$800

Step-by-step explanation:

A = P (0.5)^{t}

Because it's every 3 years, you need to find how many 3 years in the 17-year-period of time.

17/3 = 5.666...

P is initial amount:

A = $41, 000 (0.5)^{5.666...}

A = 41, 000 (0.01968...)

A = 807.13692...

A ≈ $800

Rounded to the nearest hundred dollars.

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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal
Sever21 [200]

Answer:

35 m/s

Step-by-step explanation:

Let gravitational acceleration g = 9.8m/s2. We can calculate the amount of time for the ball to fall vertically a distance of 1.6 m

1.6 = 9.8t^2/2

t^2 = 2*1.6/9.8 = 0.327

t = \sqrt{0.327} = 0.571s

This is also the time it takes for the ball to fly horizontally, assume constant horizontal speed, we can calculate the speed it takes to travel across 20m in 0.571s

v = 20 / 0.571 = 35 m/s

4 0
2 years ago
What do you know to be true about the values of a and b?
gizmo_the_mogwai [7]

Answer:

Can't be determined

Step-by-step explanation:

You just don't have enough info!

6 0
2 years ago
JULIE HAS 10 YARDS OF RIBBON. SHE DIVDES THE RIBBON INTO 3 EQUAL PIECES AND USES 2 OF THE PIECES ON GIFTS. HOW MUCH RIBBON DOES
Aliun [14]

Answer:

6.66 yards of ribbon.

Step-by-step explanation:

If she has 10 yards of ribbon and divides it equally into three parts, those 3 parts would each be 3.33 yards long or 3 (1/3) yards long.

If she uses two of those parts on gifts, then she would use 3.33 x 2 or 6.66 yards of ribbon.

8 0
3 years ago
If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (
AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

P=\frac{3}{36}

P=\frac{1}{12}

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

5 0
3 years ago
Which of the following functions is graphed below
Delicious77 [7]
Darn it i was looking for the same thing 
5 0
3 years ago
Read 2 more answers
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