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Katen [24]
2 years ago
7

Mitchell purchased a car in 2008 for $25,980. The function y 25,980(0.88y represents the depreciated value of the car after t ye

ars since 2008. What is the approximate value of the car in 2012?
Mathematics
1 answer:
Tasya [4]2 years ago
7 0

Answer:

\$15,580.09

Step-by-step explanation:

we know that

The exponential decay function that represent the depreciated value of the car is given by

y=25,980(0.88^t)

where

y is the depreciated value of the car

t is the number of years since 2008

What is the approximate value of the car in 2012?

Find the value of t

t=2012-2008=4\ years

substitute in the equation

y=25,980(0.88^4)=\$15,580.09

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Calculate the (modeled) probability P(E) using the given information, assuming that all outcomes are equally likely.
Luba_88 [7]

The probability P(E) is 13/15.

According to the statement

we have given that the S = {1, 3, 5, 7, 9}, E = {1, 5, 7}

And we have to find the probability P(E).

So, For this purpose

Recall the formula for the probability of an event E in case when all outcomes are equally likely:

P(E)= n(E) /n(S)

in which S is the sample space.

But we have the S = {1, 3, 5, 7, 9},

so, n(S) = Sum of all outcomes / number of outcomes

n(S) = 1+3+5+7+9 /5

n(S) = 25 /5

n(S) = 5 and

For n(E) = Sum of all outcomes / number of outcomes

n(E) = 1+5+7 /3

n(E) = 13 /3

n(E) = 13 /3

Substitute these values in the above written formula then

P(E)= n(E) /n(S)

P(E)=  (13/3)/ 5

P(E)= 13 /15

So, The probability P(E) is 13/15.

Learn more about the PROBABILITY here brainly.com/question/25870256

#SPJ4

5 0
1 year ago
Which situation can be represented by this equation? <br> 2,500 + 40x = 7,300-60x
koban [17]

Answer:

x= 48

Step-by-step explanation:

2500+40x+60x=7300\\40x+60x=7300-2500\\100x=7300+2500\\100x=4800\\x=4800/100\\x=48

4 0
3 years ago
Write a verbal expression for:<br> 20+6
Helen [10]

what is the result of the addition of 20 and 6

8 0
2 years ago
Read 2 more answers
Which simplified fraction is equal to 0.1ModifyingAbove 7 with bar? StartFraction 9 Over 17 EndFraction StartFraction 8 Over 45
Kazeer [188]

Answer:

  8/45

Step-by-step explanation:

Define x to be the value of your fraction:

  x=0.1\overline{7}\\\\10x=1.7\overline{7}\qquad\text{multiply by 10}\\\\10x-x=1.7\overline{7}-0.1\overline{7}=1.6\\\\x=\dfrac{1.6}{9}=\dfrac{8}{45}\qquad\text{divide by 9; put in lowest terms}

____

When in doubt, you can use your calculator to see which fraction gives you 0.1777777778.

_____

We multiplied by 10^1 above, because there is 1 repeating digit. The power of 10 we use matches the number of repeating digits.

8 0
3 years ago
A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 8 reservations went to regular customer
wolverine [178]

Answer:

a) 32.04% probability that overbooking occurs.

b) 40.79% probability that the flight has empty seats.

Step-by-step explanation:

For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. The probability of a passenger arriving is independent of other passengers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Our variable of interest are the 8 reservations that went for the passengers with a 48% probability of arriving.

This means that n = 8, p = 0.48

A) Find the probability that overbooking occurs.

12 seats, 8 of which are already occupied. So overbooking occurs if more than 4 of the reservated arrive.

P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{8,5}.(0.48)^{5}.(0.52)^{3} = 0.2006

P(X = 6) = C_{8,6}.(0.48)^{6}.(0.52)^{2} = 0.0926

P(X = 7) = C_{8,7}.(0.48)^{7}.(0.52)^{7} = 0.0244

P(X = 8) = C_{8,5}.(0.48)^{8}.(0.52)^{0} = 0.0028

P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2006 + 0.0926 + 0.0244 + 0.0028 = 0.3204

32.04% probability that overbooking occurs.

B) Find the probability that the flight has empty seats.

Less than 4 of the booked passengers arrive.

To make it easier, i will use

P(X < 4) = 1 - (P(X = 4) + P(X > 4))

From a), P(X > 4) = 0.3204

P(X = 4) = C_{8,4}.(0.48)^{4}.(0.52)^{4} = 0.2717

P(X < 4) = 1 - (P(X = 4) + P(X > 4)) = 1 - (0.2717 + 0.3204) = 1 - 0.5921 = 0.4079

40.79% probability that the flight has empty seats.

4 0
2 years ago
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