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Shalnov [3]
3 years ago
7

Mr. Perkins wants to rent a car for a day. It will cost the daily fee of $75 plus $0.55 per mile driven, which expression shows

the amount he will pay if m=the number of miles driven?
A. 75m+.55
B. 75+0.55m
Mathematics
1 answer:
Alla [95]3 years ago
8 0

Answer:

B. 75 + 0.55m

Step-by-step explanation:

We know that the constant is $75. You only pay this once.

Whereas $0.55 you pay per mile. Which means that you will multiply 0.55 by how many miles you drive.

This means that the equation will look like:

75 + 0.55m

Hope this helps! :)

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3 years ago
The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont
devlian [24]

Answer:

a) 0.39347

b) 0.22313

c) 0.95021

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)

and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition<em> P(A | B) = P(A∩B)P(B) </em>

but B⊂A, so  A∩B = B and  

\bf P(A | B) = P(B)P(B) = (P(B))^2

We have  

\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237

hence,

\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313

• If the signal has six red bulbs, what is the probability that atleast one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

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Now the probability that exactly k bulbs need replacement is

\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}

We also have:

<em>Probability that at least one of them needs replacement at the first inspection = 1 </em><em>-</em><em> probability that none of them needs replacement at the first inspection. </em>

This means that,

Probability that at least one of them needs replacement at the first inspection =  

\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021

4 0
3 years ago
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