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3 years ago

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Mathematics

2 answers:

sweet-ann [11.9K]3 years ago

8 0

Answer:

h = 40°

Step-by-step explanation:

These 2 angles are congruent due to the Corresponding Angles Theorem.

Sedbober [7]3 years ago

4 0

Answer:

40°

Step-by-step explanation:

The lines are parallel. We know this because of the 2 arrows that are on the lines, indicating they are parallel.

Therefore, the 2 angles are congruent by the Corresponding Angle Theorem.

So, we can set the 2 angles equal to each other.

40=h

Therefore, angle h is 40 degrees.

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Ashley has a rectangle made out of paper that is 8 cm by12 cm. She folds it in half twice, first vertically and then horizontall

Svetach [21]

Answer:

$Area =24cm^2$

Step-by-step explanation:

Given

$L = 8cm$

$W = 12cm$

$r = 2$ -- folded twice

Required

The area of the new rectangle

When the length was folder, the new length is:

$l = L/2 = 8cm/2 = 4cm$

When the width was folder, the new width is:

$w = W/2 = 12cm/2 = 6cm$

So, the new area is:

$Area =l * w$

$Area =4cm * 6cm$

$Area =24cm^2$

8 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.

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