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 
and the mean that the random selection of 50 pages will contain no error is 
∴

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 ≥ 2) = 0.2644
The probability that the selection of the random pages will contain at least two errors is 0.2644
Possible genetic combinations of offspring are:
<span>
- A1A3
- A2A3
- A1A4
- A2A4
</span>
Therefore, each genotype has a probability of 1/4
The probability of having thesame genotype the second time is 1/4 * 1/4
= 1/16
The probability that both offsprings will have the same genotype is 1/16
Omg ur keeping me awake with these notifications!! LOL :-D. JK
Arc RQ is a [minor arc] only because it is less than half the circle, which is 180 [PQ]
** remember that 180° always = half a circle
2(5X-2) (4X-3) is the correct answer
Answer:
perimeter of the figure: 
Step-by-step explanation:
perimeter = sum of all the sides.
→ 
→ 
→ 
→ 