Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.25 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 30. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 30.

mean = 7.5, standard deviation = 2.3717

Step-by-step explanation: the experiment here ( randomly selecting green peas pods) was done 30 times, meaning number of time experiment was performed is 30.

For an experiment performed n number of times with a probability of success p, the probability of any random variable (p(x=r)) is represented by a binomial distribution {X~ (n, p)}.

Where X is the random variable, p is probability of success, 1-p is the probability of failure and n is the number of times experiment was performed.

For this question of ours, n =30 and p = 0.25, q = 1-p = 1 - 0.25 = 0.75 = probability of failure.

The mean (x - bar) for a binomial distribution = np

Hence x - bar = 30 × 0.25 = 7.5

Standard deviation (σ) =√npq

Standard deviation = √30×0.25×0.75

Standard deviation = √5.625

Standard deviation = 2.3717

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

37 = ending with 7

unit's digit in the cube..3

56 = ending with 6

unit's digit in the cube..2

133 = ending with

unit's digit in the cube..7

hope it : ( : )