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deff fn [24]
3 years ago
12

Chris Evans drives 300 miles per week in his Honda Civic that gets 22 miles per gallon of gas. He

Mathematics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

  • old car $20,909
  • new car: $29,960

Step-by-step explanation:

At 300 miles per week, Chris drives 300×52 = 15,600 miles per year. His gas cost can be figured as ...

  (5 years)×(miles per year)÷(miles/gallon)×($ per gallon) = $273,000/(miles per gallon)

__

a) old car cost = repair cost + gas cost + insurance cost

  = 5($1200) + $273,000/22 + 5($500) ≈ $20,909 . . . over 5 years

__

b) new car cost = purchase cost + gas cost + insurance cost

  = $20,000 + $273,000/50 +5($900) = $29,960 . . . over 5 years

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A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get
Tems11 [23]
An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899  (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
P(x)=C(n,x)p^x(1-p)^{n-x}where
p=probability of success
n=number of trials
x=number of successesC(n,x)=\frac{n!}{x!(n-x)!}

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more) 
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%.  The work is similar to that of n=10.
8 0
3 years ago
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