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Assoli18 [71]
3 years ago
15

a dog trainer has 104ft of fencing that will be used to create a retangular work area for dogs. If the trainer wants to enclosed

an area of 532ft what will be the dimensions of the work area?
Mathematics
1 answer:
just olya [345]3 years ago
6 0
Answer:
36 ft by 16 ft
Step-by-step explanation:
To solve this problem, you need to find dimensions of a rectangle such that the perimeter is 104 ft and the area is 576 ft. The perimeter is twice the sum of length and width, so the sum of length and width is 52 ft.
The area is the product of length and width, so if w represents the width, we have ...
w(52 -w) = 576
w² -52w = -576 . . . . . eliminate parentheses, multiply by -1
w² -52w +26² = 26² -576 . . . . . . complete the square
(w -26)² = 676 -576 = 100
w = 26 ±√100 = {16, 36}
If the width is the short dimension, it is 16 feet. Then the length is 36 feet. Let me know if I’m right
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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
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Answer:

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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.

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