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3 years ago

Graph the function g(x) = 1/3 - 4/3x

1 answer:

6 0

Solutions

g(x) = $\frac{1}{3}$ - $\frac{4}{3}$ x

 g(x) = - $\frac{4}{3}$ x + $\frac{1}{3}$

x = -5

= g(x) = - $\frac{4}{3}$ × (-5) + $\frac{1}{3}$

= $\frac{20}{3}$ + $\frac{1}{3}$

= $\frac{21}{3}$

= 7

(-5,7)

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x = 7

 = g(x) = - $\frac{4}{3}$ × (7) + $\frac{1}{3}$

= $\frac{-28}{3}$ + $\frac{1}{3}$

= $\frac{-27}{3}$ = -9

(-7,9)

x = (-7,9) y = (-5,7)

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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:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

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

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

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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