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horrorfan [7]
3 years ago
12

108 divided by 9 using distributive property

Mathematics
1 answer:
Lostsunrise [7]3 years ago
7 0

108 = 81 + 27

Step 1: 81 / 9 = 9

Step 2: 27 / 9 = 3

Step 3: 3 + 9 = 12

108 / 9 = 12

Hope this helps! :D

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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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This happens for k=0, so the probability is:

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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.

P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002

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MaRussiya [10]

Answer:

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