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kolbaska11 [484]
3 years ago
10

Which of the following pairs of lines are perpendicular? Select all that apply

Mathematics
1 answer:
ad-work [718]3 years ago
4 0
Perpendicular lines will have negative reciprocal slopes

B. y = -4x+ 5  and y = 1/4x + 4
D. y = x + 4 and y = -x + 4
E. y = -3 and x = -3

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(1 3/4 + 2 3/8) + 5 7/8=
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Given a line segment that contains the points A,B,& C in order, if AB = x + 12, and Bo2.4, and AC = 26 = ?
natulia [17]

Answer:

x=11.6

Step-by-step explanation:

According to the angle addition postulate, if B lies on AC, than AB+BC=AC.

We can use this to create this equation.

x+12+2.4=26 Where x+12 is the value of AB and 2.4 is BC.

The next step is to combine like terms, so we add 12 to 2.4 to get

x+14.4=26. When trying to find the value of a variable, you must isolate it. To do this for this equation, we must subtract 14.4 from both sides which will give you.

x=11.6

Hope this helps. If you have any follow-up questions or I misunderstood the question, please tell me and I will fix it or explain more.

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7 0
3 years ago
Read 2 more answers
he time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the a
Eva8 [605]

Answer:

The answer is below

Step-by-step explanation:

The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the answers to 3 decimal places.

(a) What is the probability that more than three aircraft arrive within an hour?

(b) If 30 separate one-hour intervals are chosen, what Is the probability that no interval contains more than three arrivals?

(c) Determine the length of an interval of time (in hours) such that the probability that no arrivals occur during the interval is 0.1.

Solution:

a) A poisson distribution is given by the formula:

P(X=x)=\frac{e^{-\lambda }\lambda^x}{x!}

λ = 1 hour

Therefore:

P(X > 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(x = 3)]

P(X=0)=\frac{e^{-1}*1^0}{0!}=0.3679

P(X=1)=\frac{e^{-1}*1^1}{1!}=0.3679

P(X=2)=\frac{e^{-1}*1^2}{2!}=0.1839

P(X=3)=\frac{e^{-1}*1^3}{3!}=0.0613

P(X > 3) = 1 - [0.3679+0.3679 + 0.1839 + 0.0613] = 0.019

b) Assuming 30 1 hour intervals, hence:

P(X \leq 3)^{30}=[1-P(X\geq 30)]^{30}=(1-0.019)^{30}=0.5624

c) mean = 1 hour

mean = 1 / λ

1 = 1 / λ

λ = 1

The cumulative distribution function of a continuous variable is:

F(x)=1-e^{-\lambda x}

0.1=1-F(x)\\\\0.1=1-(1-e^{- x})\\\\0.1=e^{- x}\\\\-x=ln(0.1)\\\\-x = -2.3\\\\x=2.3\ hours

4 0
3 years ago
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