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TEA [102]
3 years ago
7

PLEASE HURRY WILL GIVE BRAINLIEST A truck can be rented from Company A for $90 a day plus $0.30 per mile. Company B charges $70

a day plus $0.80 per mile to rent the same truck. Find the number of miles in a day at which the rental cost for company A and Company B are the same
Mathematics
1 answer:
rjkz [21]3 years ago
4 0

Answer:

  40 miles a day

Step-by-step explanation:

The difference in fixed charge is $20; the difference in mileage charge is $0.50 per mile. So it will take $20/$0.50 = 40 miles for the extra mileage cost to eat up the savings in fixed cost. At that mileage, both companies will charge the same amount:

  A: 90 + 0.30·40 = 102

  B: 70 + 0.80·40 = 102

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11%

Step-by-step explanation:

if he made 8/9 attempts, that leaves 1/9 and 1/9 is .11 so its 11% that he missed

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Select the correct answer.<br> Which graph shows a function and its inverse?
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Answer:

See below

Step-by-step explanation:

If a function is bijective and 1-to-1, then it will have an inverse function. Consequentially, they will be symmetrical about the line y=x, which is a diagonal line passing through the origin at a 45 degree angle.

None of the graphs look correct though, but it also seems that some options are cut out, so make sure to choose the correct graph given the characteristics I've previously described.

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Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

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P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

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GarryVolchara [31]

Answer:19

Step-by-step explanation:

Divide 190 by 10

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