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Harlamova29_29 [7]

1 year ago

10

Glenna wants to rent a car for a trip to key west for one weekshe calls two car rental companies to get prices. mr kotters renta

ls rents a cadillac escalade for $99 for one week plus $0.11per mile over 100 miles. barbarino's rentals has the same vehicle for $75 per week and $0.15 per mile over 150 miles. if its 432 miles to key west, which company has the better deal? How many miles would Glenna drive before she would be spending the same amount at either company?

1 answer:

Korvikt [17]1 year ago

4 0

Answer: Barbarino's rentals has a better deal.

She has to drive 887.5 miles to spend the same amount at either company.

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

<em>Mr.kotters rentals (A) </em>

<em>$99 PER WEEK </em>
<em>$0.11per mile over 100 miles </em>

<em>Barbarino's rentals (B) </em>

<em>$75 per week </em>
<em>$0.15 per mile over 150 miles </em>

For "A"

Cost = 0.11 (432-100) + 99 = $135.52

For "B"

Cost= 0.15 (432-150) +75 = $117.3

Barbarino's rentals has a better deal, since $117.3(B) < $135.52 (A)

To find how many miles would Glenna drive before she would be spending the same amount at either company:

A =B

0.11 (M-100) + 99 =0.15 (M-150) +75 = $117.3

Solving for M (miles)

0.11 M -11+99 = 0.15 M -22.5+75

-11 +99 +22.5 -75 =0.15M -0.11 M

35.5 = 0.04M

35.5/0.04 = M

887.5 =M

She has to drive 887.5 miles to spend the same amount at either company.

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Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

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c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

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The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

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Part a

For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

Part b

For this case we can use the result from part a to conclude that:

$P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this assuming independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

Part c

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And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

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