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Alex17521 [72]
3 years ago
14

Suppose that extra fingers and toes are caused by a recessive trait, but it appears in only 60% of homozygous recessive individu

als. Two heterozygotes conceive a child. What is the probability that this child will have extra fingers and toes? 0.10 0.25 0.33 0.15 0.05
Mathematics
1 answer:
Ket [755]3 years ago
8 0

Answer:

0.15 = 15%

Step-by-step explanation:

Let's call the recessive allele x, and the dominant allele X.

The homozygotes have the pair xx, and the heterozygotes have the pair Xx.

If two heterozygotes conceive a child, there are four configurations that the child can have:

xx, xX, Xx, XX

the only configuration that allows the child the possibility to have extra fingers and toes is the xx, as this aspect is a recessive trait, so the probability of the child being xx is 25% (1 option between 4 options, all of them with equal probabilities).

If the child is xx, there is a chance of 60% of the child having extra fingers and toes. So, the final probability is:

25% * 60* = 0.25 * 0.6 = 0.15 = 15%

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Determine which of the lines are parallel and which of the lines are perpendicular. Select all of the statements that are true.
slega [8]

Answers:

Line A is parallel to line D.

Line A is perpendicular to line C.

Line C is perpendicular to line D.

=====================================================

Explanation:

Let's use the slope formula to calculate the slope of the line through (-1,-17) and (3,11)

(x_1,y_1) = (-1,-17) \text{ and } (x_2,y_2)  = (3,11)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{11 - (-17)}{3 - (-1)}\\\\m = \frac{11 + 17}{3 + 1}\\\\m = \frac{28}{4}\\\\m = 7\\\\

The slope of line A is 7

-------------

Now let's find the slope of line B.

(x_1,y_1) = (0,4) \text{ and } (x_2,y_2)  = (7,-5)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{-5 - 4}{7 - 0}\\\\m = -\frac{9}{7}\\\\

-------------

Now onto line C.

(x_1,y_1) = (7,1) \text{ and } (x_2,y_2)  = (0,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 1}{0 - 7}\\\\m = \frac{1}{-7}\\\\m = -\frac{1}{7}\\\\

-------------

Lastly we have line D.

(x_1,y_1) = (-1,-6) \text{ and } (x_2,y_2)  = (1,8)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{8 - (-6)}{1 - (-1)}\\\\m = \frac{8 + 6}{1 + 1}\\\\m = \frac{14}{2}\\\\m = 7\\\\

------------------------------

Here's a summary of the slopes we found

\begin{array}{|c|c|} \cline{1-2}\text{Line} & \text{Slope}\\\cline{1-2}\text{A} & 7\\\cline{1-2}\text{B} & -9/7\\\cline{1-2}\text{C} & -1/7\\\cline{1-2}\text{D} & 7\\\cline{1-2}\end{array}

Recall that parallel lines have equal slopes, but different y intercepts. This fact makes Line A parallel to line D.

Lines A and C are perpendicular to one another, because the slopes 7 and -1/7 multiply to -1. In other words, -1/7 is the negative reciprocal of 7, and vice versa. These two lines form a 90 degree angle.

Lines C and D are perpendicular for the same reasoning as the previous paragraph.

Line B unfortunately is neither parallel nor perpendicular to any of the other lines mentioned.

You can use a graphing tool like Desmos or GeoGebra to verify these answers.

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i think it's c but I'm rly not sure on this one...

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