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ale4655 [162]
4 years ago
7

Five days a week, you carpool with 3 co-workers and take turns driving each week. It is 14 miles from your home to your office.

When you drive the carpool you must go an extra 4 miles to pick up your co-workers. If your car averages 18 miles per gallon, about how many gallons of gas should you save every 4 weeks by carpooling?
Mathematics
2 answers:
rusak2 [61]4 years ago
8 0

Answer:

I will save 21.11 gallons in 4 weeks of car pooling.

Step-by-step explanation:

If i have to go office alone then in 4 weeks i will be driving my car = 2×distance from office to home×number of days = 2×14×(4×5) = 28×20 = 560 km.

Now when i carpool with 3 co-workers then i have to drive my car only for 1 week out of 4 weeks.

Means i will be driving total distance = 2× ( distance form office to home + addition distance to pickup co-workers) × number of days

= 2 × (14+4) × 5= 180 km

Now the distance I drove less in 4 weeks = 560 - 180 = 380 km

My car averages 18 miles per gallon gas.

Therefore amount of gas I saved in 4 weeks = 380/18 = 21.11 gallons

elena55 [62]4 years ago
4 0
She saves 8.8888888 gallons of gas every week by carpooling.
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A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute
nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

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Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

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a. What is the probability of selecting a carton and finding no defective pens?

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b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

P(k\geq8)=1-P(k

P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\

P(k

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

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Answer:

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