Advanced Placement (AP), 03.02.2020 23:56 HopeBordelon4

What happens when amother and father has same group of blood . then whats the blood group of a child?

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If the Mother's blood type is B, and the father's blood type is O, then their child must be either B or O. For instance, if both mother and alleged father have type A blood, then the child can only have A or O blood. On the other hand, if the mother's blood type is A, and the father's is B, the child could be ANY of the four blood types A, B, AB or O so its ability to predict paternity is minimal in this case. When both father and mother contribute an A gene, their child has Type A blood. In this case, the child is "pure-bred" for the A gene. The child can only pass on an A gene to his offspring. If one parent contributes and A gene and the other contributes an O gene, the child will have A blood type, but, like his parents, he'll also have both A and O genes that he may pass along to his children. However, in 1 out of 4 instances, both parents will contribute their recessive O gene to their child. When that happens, instead of the child expressing the A blood type, he will instead express the O blood type. Scientists use the term genotype to refer to the combination of mother and father genes in the person's make up. AA, AO and OO are the 3 genotypes that may result from the mating of the parents in the example above. Scientists use the term phenotype to refer to what gene is actually expressed. In the example above, there are 2 phenotypes of Type A blood and Type O blood. The type O blood phenotype can only result from one genotype of OO. But Type A blood can result from two different genotypes of AA and AO.

Answer from: tylorroundy

Its both cuz they cant have a different gene if both parents are dominant or not

Answer from: Quest

The answer is (d) 7/2======we have $f(x) = \begin{cases} x & \text{ for } x \le 0 \\ x + 1 & \text{ for } x \ \textgreater \ 0 \end{cases}$ we also have an integral $\displaystyle\int_{-2}^1 x\, f(x)\; dx$ notice how function f splits up at x = 0. there's a left piece and a right pieceso we'll need to do the same thing with our integral: split it at x = 0we use a property of integrals that allows us to split up and sum: $\displaystyle\int_{-2}^1 x\, f(x)\; dx = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx$ notice that for the piecewise function f, the function has the equationf(x) = x when x ≤ 0. so for the integral $\int_{-2}^0 x\, f(x)\; dx$ , f(x) = x so it changes into $\begin{aligned} \int_{-2}^0 x\, f(x)\; dx & = \int_{-2}^0 x \cdot x\; dx \\ & = \int_{-2}^0 x^2\; dx \\ & = \left[\frac{x^3}{3}\right]_{-2}^0 & & \text{\footnotesize antidifferentiate} \\ & = 0 - \frac{(-2)^3}{3} \\ & = \frac{8}{3} \end{aligned}$ for the piecewise function f, the function has the equationf(x) = x+1 when x > 0. for the integral $\int_{0}^1 x\, f(x)\; dx$ , we have $\begin{aligned} \int_{0}^1 x\, f(x)\; dx & = \int_{0}^1 x(x+1)\; dx \\ & = \int_{0}^1 (x^2+x)\; dx \\ & = \left[ \frac{x^3}{3} + \frac{x^2}{2}\right]_0^1 \\ & = \frac{1}{3} + \frac{1}{2} - 0 = \frac{5}{6} \end{aligned}$ therefore, back to our splitted integral: $\begin{aligned} \displaystyle\int_{-2}^1 x\, f(x)\; dx & = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx \\ & = \frac{8}{3} + \frac{5}{6} \\ & = \frac{7}{2} \end{aligned}$
Can anyone me on this ap calc question? ?

Can anyone me on this ap calc question? ?

Answer from: Quest

Answer the discount rate

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