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shepuryov [24]
3 years ago
5

What is the equation for the line in slope intercept form

Mathematics
2 answers:
san4es73 [151]3 years ago
7 0
Y=mx+b
where m is the slope and b is the y intercept
qwelly [4]3 years ago
6 0

Answer:


Step-by-step explanation:

so the the slope is going to be 2/1 because the line goes up 2 over 1 or you can say 2 and the y intercept is 10 because that is where it starts on the y axis so the answer to your equation is y=2x+10 hope this helps


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ANSWER THIS AND ILL LOVE YOU <br><br> find angles FAB and BAC
jeyben [28]

Answer:

m<FAB = 75°

m<BAC = 105°

Step-by-step explanation:

First, find the value of x.

(13x - 3)° = (3x + 2)° + 55° (exterior angle theorem of a ∆)

Solve for x

13x - 3 = 3x + 2 + 55

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13x - 3x = 57 + 3

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Plug in the value of x

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4 0
2 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

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For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

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P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

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