Answer: 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.).
This is a combination question
Hope this helps
The first choice is the contains only irrational number
(x, y) --> (x + 5, y - 1)
A(3, -1) --> A'(3 + 5, -1 - 1) --> A'(8, -2)
B(6, 1) --> B'(6 + 5, 1 -1) --> B'(11, 0)
C(2, 4) --> C'(2 + 5, 4 - 1) --> C'(7, 3)
D(-1, 3) --> D'(-1 + 5, 3 - 1) --> D'(4, 2)
<span>p=210e^(0.0069*20)
'e' is a mathematical constant equal to 2.718281828459
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<span>0.0069*20 = .138
e^.138 =
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<span>2.718281828459^.138 =
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1.1479755503
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210 * <span>1.1479755503 =
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241.074865563
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