The first thing you do is distribute. So multiply -7 to n and 6 to get -7n-42=-91 add over -42 to -91 to get -7n=-133 then divide by 7 on both sides and you get n=19
Answer:
-1/10
Step-by-step explanation:
(5/(-10))/5 = -(5/10)/5 = 1/10
(Hopefully this helped you out!)
Answer:
2.
Step-by-step explanation:
Slope = rise / run
Slope = y2-y1 / x2-x1
= (4-2) / (2 -1) = 2.
also 10-8 / 5-4 = 2.
3 3/8 × 3 = x
distribute
3×3 + 3×3/8=x
9+9/8=x
9+1+1/8=x
10 1/8=x
Answer:
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that the selection of the random pages will contain at least two errors is 0.2644
Step-by-step explanation:
From the information given:
Let q represent the no of typographical errors.
Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let
be the random variable that follows a Poisson distribution, then mean ![\mu = \dfrac{10}{500}= 0.02](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cdfrac%7B10%7D%7B500%7D%3D%200.02)
and the mean that the random selection of 50 pages will contain no error is ![\lambda = 50 \times 0.02 =1](https://tex.z-dn.net/?f=%5Clambda%20%3D%2050%20%5Ctimes%200.02%20%3D1)
∴
![Pr(q= 0) = \dfrac{e^{-1} (1)^0}{0!}](https://tex.z-dn.net/?f=Pr%28q%3D%200%29%20%3D%20%5Cdfrac%7Be%5E%7B-1%7D%20%281%29%5E0%7D%7B0%21%7D)
Pr(q =0) = 0.368
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that 50 randomly page contains at least 2 errors is computed as follows:
P(X ≥ 2) = 1 - P( X < 2)
P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2
![P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20%5B%20%5Cdfrac%7Be%5E%7B-1%7D%201%5E0%7D%7B0%21%7D%20%2B%5Cdfrac%7Be%5E%7B-1%7D%201%5E1%7D%7B1%21%7D%20%5D)
![P(X \geq 2) = 1 - [0.3678 +0.3678]](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20%5B0.3678%20%2B0.3678%5D)
![P(X \geq 2) = 1 -0.7356](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-0.7356)
P(X ≥ 2) = 0.2644
The probability that the selection of the random pages will contain at least two errors is 0.2644