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3 years ago

Convert the mixed number, 1 3/4 , to a fraction. Reduce fraction if necessary

1 answer:

8 0

I suppose we make it an improper fraction? Multiply the whole number by the denominator and then add the numerator

1 * 4 + 3 = 7

Put it over the denominator

7/4

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Math help. you could just answer the inequality and I can do the rest but it would be helpful you you could answer both.

REY [17]

Answer:

Answer is first line

Step-by-step explanation:

-7(x+9) >= 37+3x

Going to solve for x:

-7x - 63 >= 37 + 3x

-7x - 3x >= 37 + 63

-10x >= 100

x <= 100/-10 (notice the change in equality because we divided by negative)

x <= -10

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3 years ago

7. Write the following proportion: 8 is to 64 as 2 is to X

Paul [167]

2 is to 4 because 64/8 is 8. 4/2 is 2. 2 x 2 is in proportion to 8 x 8

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4 years ago

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balu736 [363]

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3 years ago

A rectangular field is twelve times as long as it is wide. If the perimeter of the field is 1690 feet, what are the dimensions o

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Answer:

The width of the field is 65 feet.

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4 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.

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

5 0

4 years ago