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3 years ago

PLS HELP ........................

Mathematics

1 answer:

lisov135 [29]3 years ago

3 0

Answer:

Step-by-step explanation:

First, check to see which graph has a line going through the point (2,3). B and D are the only ones that have lines going through (2,3) (A comes close but it is not quite).

Next, you need to see which line would be parallel to the equation 3x-y=2. When two lines are parallel, they have the same slope. You have to turn that equation into point-slope formula (y=mx+b with m being the slope and b being the y-intercept). First, subtract 3x from both sides. You will then get -y=-3x+2. Then y needs to be alone (right now it has a -1 attached). Divide both sides by -1 to get y=3x-2. The number in front of the x is the slope, in this case, 3 or 3/1 (we do not care about the y-intercept, it will not help us in this problem since we are looking for a line parallel to this equation and so our line will not have the same y-intercept as this other equation). Since the line in parallel to that equation, we know that this line also has a slope of 3/1. Find the line between B and D that has a slope of 3/1, you get B (the line goes up 3 over 1).

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A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

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a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$