An individual has one of three genotypes called dd, dD, and DD, respectively, for a gene associated with disease X. The probability that an individual has genotype dd is 0.64; the probability that an individual has genotype dD is 0.32; and the probability that an individual has genotype DD is 0.04. The probability that an individual with the dd genotype is affected with disease X is 0.05. The probability that an individual with the dD genotype is affected with disease X is 0.80. The probability that an individual with the DD genotype is affected with disease X is 0.99. What is the probability that an individual is affected with disease X? Given that an individual has disease X, what is the probability that the individual is genotype dD?

the answer is y = 11.54

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idk wbu

step-by-step explanation:

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)