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2 years ago

18. MP Reason Inductively The product of

Mathematics

1 answer:

timama [110]2 years ago

6 0

Answer:

- 2 and 12

Step-by-step explanation:

let the 2 integers be x and y , x > y , then

x - y = 14 → (1)

x + y = 10 → (2)

add the 2 equations term by term to eliminate y

2x = 24 ( divide both sides by 2 )

x = 12

Substitute x = 12 into (2)

12 + y = 10 ( subtract 12 from both sides )

y = - 2

As a check

x + y = 12 + (- 2) = 12 - 2 = 10

x - y = 12 - (- 2) = 12 + 2 = 14

xy = 12 × - 2 = - 24

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Christina paid $2 for 5/8 of a gallon of gas.How much does it cost per gallon

vladimir1956 [14]

Answer:

$3.20

Step-by-step explanation:

I divided the $2 by 8 to see how much 1/8 a gallon cost.

I got 0.40

Multiple this by 8 to get 8/8 or 1 gallon.

You get 3.2

so $3.20

Hope this helped. :)

5 0

3 years ago

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.

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