Which of the following could be the value of a in the relation below so that the relation is also a function

Mathematics

1 answer:

bija089 [108]2 years ago

3 0

Answer:

The class sociologist Max Weber formulated a three-component theory of stratification that saw political power as an interplay between “class”, “status” and “group power. ” Weber theorized that class position was determined by a person’s skills and education, rather than by their relationship

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What is the distance between the lines y = 2x and y = 2x+5? Round your answer to the nearest tenth.

Ilia_Sergeevich [38]

The distance between both lines is 2.24 units

Step-by-step explanation:

There is no straight method to find the distance between two lines. The distance can be found out by finding a point on one line and then finding the distance of that point from the other line. The y-coordinate of point is obtained by putting any value of x in equation . The x and y combined give us the point.

Given

$y=2x\\Putting\ x=1\\y=2(1)\\y=2\\So, the\ point\ on\ y=2x\ is\ (1,2)$

We have to find the distance of this point from y=2x+5

$y=2x+5\\Subtracting\ y\ from\ both\ sides\\y-y=2x+5-y\\2x-y+5=0$

The formula for finding distance of a point (x,y) from a line is:

$d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$

A=2

B=-1

C=5

Putting the values in the formula

$d=\frac{|(2)(1)+(-1)(2)+5|}{\sqrt{(2)^2+(-1)^2}}\\d=\frac{|2-2+5|}{\sqrt{4+1}}\\d=\frac{|5|}{\sqrt{5}}\\d=\frac{5}{\sqrt{5}}\\d=2.236\ units$

Rounding off will give: 2.24 units

The distance between both lines is 2.24 units

Keywords: Equations of lines

Learn more about distance in lines at:

#LearnwithBrainly

7 0

2 years ago

assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the

drek231 [11]

Answer:

65.3658 inches

Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value $x_{0}$ such that $P(X < x_{0}) = 0.76$ , but, $0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5)$ , i.e., $(x_{0}-63.6)/2.5$ is the 76th percentile of the standard normal distribution. So, $(x_{0}-63.6)/2.5 = 0.7063$ , $x_{0} =63.6+(2.5)(0.7063) = 65.3658$ . Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.