AaBbCc x AaBbCc (use for all 3 questions) 1. What is the probability that this individual will: AABBcc 2. What is the probability that this individual will: AaBBcc 3. What is the probability that this individual will: AaBbCc

1. P = 1/64

             2. P = 1/32

             3. P = 1/8

Step-by-step explanation:

In genetics, an ofspring inherits one copy of an allele of each parent, which can be described in the tables below called Punnett Square:

For crossing Aa x Aa:

       A      a

A    AA   Aa

a    Aa    aa

For crossing Bb x Bb:

        B      b

B     BB    Bb

b     Bb    bb

For crossing Cc x Cc:

       C      c

C    CC    Cc

c     Cc     cc

We can separate them because they are assorted independently.

For offspring with genotype AABBcc, probability will be:

P(AA) = 1/4

P(BB) = 1/4

P(cc) = 1/4

As all three probabilities has to happen at the same time, it is a "E" rule:

P(AABBcc) =

P(AABBcc) = 1/64

Probability for the individual to be AABBcc is 1/64 or 1.56%.

Genotype AaBBcc:

P(Aa) = 2/4 = 1/2

P(BB) = 1/4

P(cc) = 1/4

P(AaBBcc) =

P(AaBBcc) = 1/32

Probability for the individual to be AaBBcc is 1/32 or 3.12%

Genotype AaBbCc:

P(Aa) = 1/2

P(Bb) = 1/2

P(Cc) = 1/2

P(AaBbCc) =

P(AaBbCc) = 1/8

Probability for the individual to be AaBbCc is 1/8 or 12.5%.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15