Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75probability that a pea has green pods. assume that the offspring peas are randomly selected in groups of 10.complete parts (a) through (c) below. a. find the mean and the standard deviation for the numbers of peas with green pods in the groups of 10.the value of the mean is muequalsnothingpeas.

The value of the mean and standard deviation values of the hybridization experiment is :

Given the Parameters :

To obtain the mean , we use the relation :

Mean, μ = np

Mean, μ = (10 × 0.75) = 7.5

To obtain the Standard deviation, we use the relation :

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The mean is and standard deviation is .

Step-by-step explanation:

The mean and standard deviation for the binomial distribution is given by :-

,where n is the number of trials and p is the probability of getting success in each trial.

Given : The probability that a pea has green pods :

Then , the standard deviation for the numbers of peas with green pods in the groups of 10  :-

Hence, the mean is and standard deviation is  .

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq