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2 years ago

9

The image represents what geometric construction? A) Copy a triangle construction B) Copy a segment construction C) Parallel lin

es construction D) Perpendicular bisector and midpoint

1 answer:

Umnica [9.8K]2 years ago

5 0

Answer:

d perpendicular bisected and midpoint

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What is the midpoint of the x-intercepts of f(x) = (x – 2)(x – 4)?

tatyana61 [14]

The x-intercepts are readily identified on the graph: (2,0) and (4,0). The midpoint of a line connecting those 2 points would be (3,0).

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2 years ago

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5. Lauren is saving for gymnastics camp, Camp costs $225 to attend. She has 40 percent of the money saved. How much money has sh

barxatty [35]

To solve this problem you have to convert the percentage to a decimal which would
be .40.

40% =40/100=.40

After doing so all you need to do is divide the amount needed by the percentage which gives you the amount she has saved.

225/.40= $90

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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.

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

5 0

3 years ago

What is the value of x? Also show work pls

drek231 [11]

Answer:

x = 34

Step-by-step explanation:

The angles are corresponding angles and corresponding angles are equal if the lines parallel

2x+3 = 71

Subtract 3 from each side

2x+3-3 = 71-3

2x = 68

Divide by 2

2x/2 = 68/2

x = 34

4 0

2 years ago

1/2 x + 3 = 2/3 x + 1

Vladimir [108]

Answer:

x = 12

Step-by-step explanation:

4 0

2 years ago