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Anna35 [415]
2 years ago
12

Please help with this equationQuestions not that Hard its just Im really tired

Mathematics
2 answers:
ella [17]2 years ago
6 0

Answer:

X =5y

Step-by-step explanation:

1 Subtract yy from both sides.

6x=4x+11y-y

6x=4x+11y−y

2 Simplify  4x+11y-y4x+11y−y  to  4x+10y4x+10y.

6x=4x+10y

6x=4x+10y

3 Subtract 4x4x from both sides.

6x-4x=10y

6x−4x=10y

4 Simplify  6x-4x6x−4x  to  2x2x.

2x=10y

2x=10y

5 Divide both sides by 22.

x=\frac{10y}{2}

x=

2

10y

​

6 Simplify  \frac{10y}{2}

2

10y

​

  to  5y5y.

x=5y

x=5y

belka [17]2 years ago
6 0

\tt 6x+y=4x+11y\\\\6x-4x=11y-y\\\\2x=10y\\\\y=0.2x

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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

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

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!}

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

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

5 0
3 years ago
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