Answer:
<h2>A.

</h2>
Step-by-step explanation:
The given formula is

Where
and 
To solve for
, first we need to move 

Then, we move the term 1

Finally, we move the factor 

Replacing given values, we have

Thereofre, the right answer is A.
Step-by-step explanation:
<em> </em><em>REFER</em><em> </em><em>TO</em><em> </em><em>THE</em><em> </em><em>ATTACHMENT</em><em> </em>
<em>HEY</em><em> </em><em>IN</em><em> </em><em>WHICH</em><em> </em><em>CLASS</em><em> </em><em>YOU</em><em> </em><em>ARE</em><em> </em><em>ROSA</em>
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.
Y=5/4x-4
5/4 is the slope but to pass through the point it needs to cross the y axis at -4
Answer:
75?
Step-by-step explanation:
i hope this is the answer