Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
Look at the solution ( again me)
The value of 5.7 x 10⁸ is 570,000,000 CHOICE C.
The value of 8⁴ is 8 * 8 * 8 * 8 = 4,096 CHOICE C.
Prime factorization of 1,260
1260 ÷ 2 = 630
630 ÷ 2 = 315
315 ÷ 3 = 105
105 ÷ 3 = 35
35 ÷ 5 = 7
2 x 2 x 3 x 3 x 5 x 7 CHOICE D.
Composite numbers are numbers that have more factors other than 1 and itself.
91 / 1 = 91
91 / 7 = 13 CHOICE C. 91 is a composite number.
13*18
x*y =234
x+y=31
we know the 2 # have to be double digits and both between 10-20
find 2 values that * to a number that ends in 4. Those are the ones value
1*4 11 and 14
2*2 12 and 12
3*8 13 and 18
only 13 and 18 add up to 31 and multiply to 234
3/5 of 42000 is 25200 so the amount of people who did not vote is 16800 because the beginning amount of 42000 subtracted from 25200 gives you the final answer
If that's confusing the answer is 16800 votes didn't pick the winner
