The punnet square is used to predict phenotypes.
50% of the offsprings would have long stems, while the other 50% would have short stems.
See attachment for the SS vs ss cross
From the attachment, we have the following observations
--- number of squares
Because S is dominant over s;we can write Ss as S.
So, we have:
The percentage of each offspring calculated as follows is:
Hence, 50% of the offsprings would have long stems, while the other 50% would have short stems.
Read more about the punnet square at:
Answer:
Hope this helps
Answer:
neither A nor B will occur simultaneously, as they are mutually exclusive.
Step-by-step explanation:
To compute the probability that either event A occurs, or B occurs, or both occur, you can use the principle of the union of events. The probability of the union of two events A and B (denoted as A ∪ B) can be calculated as:
�
(
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∪
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)
=
�
(
�
)
+
�
(
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)
−
�
(
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∩
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)
P(A∪B)=P(A)+P(B)−P(A∩B)
In this case, you're given:
�
(
�
∁
)
=
0.40
P(A
∁
)=0.40, which means the probability of the complement of A (i.e., the probability that A does not occur).
�
(
�
∣
�
)
=
0.80
P(B∣A)=0.80, which is the conditional probability of B occurring given that A has occurred.
Let's break it down:
�
(
�
∁
)
P(A
∁
) is the probability that event A does not occur, which is
1
−
�
(
�
)
1−P(A).
�
(
�
∣
�
)
P(B∣A) is the conditional probability that event B occurs given that A has occurred.
So, you can calculate
�
(
�
)
P(A) and
�
(
�
)
P(B) as follows:
�
(
�
)
=
1
−
�
(
�
∁
)
=
1
−
0.40
=
0.60
P(A)=1−P(A
∁
)=1−0.40=0.60
Now, you can use the formula for the union of events to calculate
�
(
�
∪
�
)
P(A∪B):
�
(
�
∪
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)
=
�
(
�
)
+
�
(
�
)
−
�
(
�
∩
�
)
P(A∪B)=P(A)+P(B)−P(A∩B)
But before we calculate
�
(
�
∩
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)
P(A∩B), note that events A and B are independent, so
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(
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∩
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)
=
�
(
�
)
⋅
�
(
�
∣
�
)
P(A∩B)=P(A)⋅P(B∣A).
�
(
�
∩
�
)
=
�
(
�
)
⋅
�
(
�
∣
�
)
=
0.60
⋅
0.80
=
0.48
P(A∩B)=P(A)⋅P(B∣A)=0.60⋅0.80=0.48
Now, plug this value into the formula:
�
(
�
∪
�
)
=
0.60
+
�
(
�
)
−
0.48
P(A∪B)=0.60+P(B)−0.48
Solve for
�
(
�
)
P(B):
�
(
�
)
=
�
(
�
∪
�
)
+
0.48
−
0.60
P(B)=P(A∪B)+0.48−0.60
�
(
�
)
=
�
(
�
∪
�
)
−
0.12
P(B)=P(A∪B)−0.12
Now, you have the equation:
�
(
�
∪
�
)
=
0.60
+
�
(
�
∪
�
)
−
0.12
−
0.48
P(A∪B)=0.60+P(A∪B)−0.12−0.48
Simplify:
�
(
�
∪
�
)
=
0.60
−
0.12
−
0.48
P(A∪B)=0.60−0.12−0.48
�
(
�
∪
�
)
=
0.00
P(A∪B)=0.00
So, the probability that either event A occurs, or B occurs, or both occur is 0.00. This means that neither A nor B will occur simultaneously, as they are mutually exclusive.
Answer:
32,000
Step-by-step explanation:
Answer:
If Ozzie is 8 and 6 years older than McKenzie, then McKenzie is 2 years old. So if Theo is 3 times older than McKenzie, then Theo is 6 years old.
Theo's age is calculated by multiplying McKenzie's age by 3. Given that McKenzie is 2 years old (Ozzie's age minus 6), so Theo is 6 years old.
The problem requires the application of basic arithmetic operations, specifically addition and multiplication, to solve an age problem. We know that Ozzie is 8 years old. Since Ozzie is 6 years older than McKenzie, we subtract 6 from Ozzie's age to get McKenzie's age, which is 2. Theo is 3 times older than McKenzie. Thus, to get Theo's age, we multiply McKenzie's age (2) by 3, which results in 6 years old.
#SPJ12
Answer:
9 cm
Step-by-step explanation:
c^2=81
Take the square root of both sides.
The square root of c^2 is c.
The square root of 81 is 9.
c=9
Answer:
C = 9 centimeters
Step-by-step explanation:
First, look at the area of a square, which formula is c^2 or in standard format - s^2. Thus, we can say, c^2 = 81. Then, we can simplify, and put c = √81. Since 81 is a perfect square, 9 * 9 = 81. Thus the answer is 9 centimeters.
Answer:
answer Z
Step-by-step explanation:
Look for a graph that contains the following zeros: x = 1, x = 2 , x= 3, following the info derived by the binomial factors that the function contains. Also look ate the fact that the function in question has for leading term positive , then this function must go towards plus infinity when x becomes large. This is the case for the graph option Z (the last graph of the group)