Female III-4 is pregnant via male III-5. The owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (IV-3). If you need help with how to approach calculating these probabilities, use the Hint.

the following are required for IV-3 to have condition

Explanation:

-II-4 passes an X b chromosome to III-4 (probability = 1/2).

-If III-4 has the genotype X B X b (accounted for by the above probability), then she passes an X b chromosome to IV-3 (probability = 1/2).

-III-5 passes a Y chromosome to IV-3 (probability = 1/2).

All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8).

Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.).

x = 10.1

step-by-step explanation:

all of the angles of a tirangle add up to 180º

180 - 65 = 115 (since we already know the value of one angle)

(2x + 7) + (8x + 7) = 115

(combins like terms) 10x + 14 = 115

(isolate the variable subtract 14 from both sides): 10x = 101

(find the variable divide both sides by 10): x = 101/10

x = 10.1

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.