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.).
D I did this question a while back
You can solve this by putting the number of copies in the top part of the fraction. If you replace the one in the top part of 1/6 with 3, you get the fraction 3/6. Half of six is three and half of two is one, so both fractions are equal. You can test this be dividing the 3 in the top by 3 and the 6 in the bottom by three. Your result will be 1/2. Although there are three times ad many copies, each copy is three times as small, so it balances out.
Answer:
Step-by-step explanation:
it represents a linear function
Answer:
Step-by-step explanation:
2 1/2 + x = 5 1/3 Change the mixed numbers to improper fractions
5/2 + x = 16/3 The lowest common multiple is 6. Multiply by 6
5*3 + 6x = 16*2
15 + 6x = 32 Subtract 15 from both sides.
6x = 32 - 15
6x = 17 Divide by 6
6x/6 = 17/6
x = 2 5/6
Check
5/2 + 17/6 = 16/3
15/6 + 17/6 = 32/6
32/6 = 32/6 The question checks.
