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drek231 [11]
3 years ago
15

Calculate the slope of the line given the coordinates of the endpoints. (4, 10) , (6, 20) The slope is

Mathematics
1 answer:
labwork [276]3 years ago
7 0

Answer:

the slope should be 5

Step-by-step explanation:

y = 5x - 10

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What is the solution to the following equation?
lions [1.4K]

Answer:

D. x = 7

Step-by-step explanation:

NOTE : <u><em>there should be an equal sign somewhere in the given expression.</em></u>

suppose the equation is the following:

3(x-4)-5 = x-3

………………………………………………………

3(x - 4) - 5 = x - 3

⇔ 3x - 12 - 5 = x - 3

⇔ 3x - 17 = x - 3

⇔ 3x - 17 + 17= x - 3 + 17

⇔ 3x = x + 14

⇔ 2x = 14

⇔ x = 14/2

⇔ x = 7

8 0
2 years ago
Find k so that the distance from (–1, 1) to (2, k) is 5 units. k= k= *there are two solutions for 2*
dalvyx [7]

Answer:

k = -3

k =5

Step-by-step explanation:

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d = 5\\(-1,1) =(x_1,y_1)\\(2,k)=(x_2,y_2)\\

5=\sqrt{\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2}\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\

\mathrm{Solve\:}\:k-5=0:\quad k=5\\\\k =5 , k=-3

7 0
3 years ago
Read 2 more answers
Help ASAP !’please explain
patriot [66]

Answer:

First one is 15.Second one is 3

Step-by-step explanation:

1 inch:10 meters 150÷10:15 26÷2:13 6÷2:3

4 0
3 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.

(c)

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.

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

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.

8 0
3 years ago
Which choice best represents the balance of a savings account at the end of 6 years if the simple interest earned each year is 6
Amiraneli [1.4K]

Answer:

$507.00.

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 6.5%/100 = 0.065 per year,

then, solving our equation

I = 1300 × 0.065 × 6 = 507

I = $ 507.00

The simple interest accumulated

on a principal of $ 1,300.00

at a rate of 6.5% per year

for 6 years is $ 507.00.

4 0
3 years ago
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