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andreev551 [17]

3 years ago

6

A bike rental company charges its customers p dollars per day to rent a bike, where 5≤p≤25. The number of bikes rented per day c

an be modeled by the linear function n(p)=600−24p. How much should the company charge each customer per day to maximize revenue?

1 answer:

Alisiya [41]3 years ago

4 0

Answer:

FRAT HOUSE ΓΧΒ!!! GAMMA CHI BETA RULES!!!!!!!!!!!!

Step-by-step explanation:

JK um I got 12

ΓΧΒ Out!

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Which inequality describes all the solutions to 5(3 − x) < −2x + 6?

Marrrta [24]

Given:

The inequality is

$5(3-x)$

To find:

The inequality that solution describes all the solutions to the given inequality.

Solution:

We have,

$5(3-x)$

Using distributive property, we get

$5(3)+5(-x)$

$15-5x$

$15-6$

$9$

Divide both sides by 3.

$\dfrac{9}{3}$

$3$

It can be written as

$x>3$

Therefore, the value of x is greater than 3. So, the required inequality is either $x>3$ or $3$ .

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D) as age increases, weight increases by 5

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Clarence set his watch 16 seconds behind , and it falls behind 2 seconds every day . How many days behind has it been since Clar

harkovskaia [24]

Answer:

15 days.

Step-by-step explanation:

46 - 16 = 30.

It falls 2 seconds behind each day so the number of required days

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3 years ago

Read 2 more answers

Sophia wants to take out a loan of $6,000 with interest that compounds monthly. Use the formula A = P(1 + rn)n∗t to find which o

Naddika [18.5K]

Answer:

d

Step-by-step explanation:

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a. 6000x (1 + 0.04/12)^24 = 6498.86

b. . 6000x (1 + 0.03/12)^36 =6564.31

c. . 6000x (1 + 0.01/12)^48 = 6244.76

d. . 6000x (1 + 0.05/12)^60 =6630.47

4 0

3 years ago

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago