What is the answer to the following question? 1.8 2.2 3.3 4.12 ?

A car rental agency has 150 cars. The owner finds that at a price of $48 per day, he can rent all the cars. For each $2 increase

hram777 [196]

Given that for each $2 increase in price, the demand is less and 4 fewer cars are rented.

Let x be the number of $2 increases in price, then the revenue from renting cars is given by
$(48 + 2x) \times (150 - 4x)=7,200+108x-8x^2$ .

Also, given that for each car that is rented, there are routine maintenance costs of $5 per day, then the total cost of renting cars is given by
$5(150-4x)=750-20x$

Profit is given by revenue - cost.
Thus, the profit from renting cars is given by
 $(7,200+108x-8x^2)-(750-20x)=6,450+128x-8x^2$

For maximum profit, the differentiation of the profit function equals zero.
i.e.
 $\frac{d}{dx} (6,450+128x-8x^2)=0 \\ \\ 128-16x=0 \\ \\ x= \frac{128}{16} =8$

The price of renting a car is given by 48 + 2x = 48 + 2(8) = 48 + 16 = 64.

Therefore, the rental charge will maximize profit is $64.

3 0

3 years ago

The Washington, DC, region has one of the fastest-growing foreclosure rates in the nation, as 15,613 homes went into foreclosure

Ilia_Sergeevich [38]

Answer:

a) 0.6226 = 62.26% probability that in a given year, fewer than 2 out of 100 houses in the Washington, DC area will go up for foreclosure.

b) 0.7837 = 78.37% probability that in a given year, fewer than 2 out of 100 houses in the nation will go up for foreclosure.

c) The proportion of foreclosures in the Nation is lower than in Washington, which means that with a sample size of 100, it is likely to have a small number(fewer than 2) of foreclosures than Washington DC.

Step-by-step explanation:

For each home, there are only two possible outcomes. Either it goes into foreclosure, or it does not. The probability of a home going into foreclosure is independent of other homes. This means that 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.

a. What is the probability that in a given year, fewer than 2 out of 100 houses in the Washington, DC area will go up for foreclosure?

The foreclosure rate is 1.31% for the Washington, DC area, which means that $p = 0.0131$

We wanto to find, with $n = 100$ :

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = C_{100,0}.(0.0131)^{0}.(0.9869)^{100} = 0.2675$

$P(X = 1) = C_{100,1}.(0.0131)^{1}.(0.9869)^{99} = 0.3551$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2675 + 0.3551 = 0.6226$

0.6226 = 62.26% probability that in a given year, fewer than 2 out of 100 houses in the Washington, DC area will go up for foreclosure.

b. What is the probability that in a given year, fewer than 2 out of 100 houses in the nation will go up for foreclosure?

Foreclosure rate of 0.87% for the nation, which means that $p = 0.0087$ . So

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

$P(X = 0) = C_{100,0}.(0.0087)^{0}.(0.9913)^{100} = 0.4174$

$P(X = 1) = C_{100,1}.(0.0087)^{1}.(0.9913)^{99} = 0.3663$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.4174 + 0.3663 = 0.7837$

0.7837 = 78.37% probability that in a given year, fewer than 2 out of 100 houses in the nation will go up for foreclosure.

c. Comment on the above findings.

The proportion of foreclosures in the Nation is lower than in Washington, which means that with a sample size of 100, it is likely to have a small number(fewer than 2) of foreclosures than Washington DC.

7 0

3 years ago

Teresa is simplifying an Step 1: (4x² Step 2: Step 3: 2x

Lesechka [4]

The answer is B

None of the rest equal up to the same thing

7 0

3 years ago

Someone please help me out

enyata [817]

Answer:

5/18

Step-by-step explanation: