A.x=48 B.x=24 C.x=12 D.x= -18

Mathematics

1 answer:

Roman55 [17]2 years ago

6 0

Answer: “B” is the answer.

Explanation: I did it in my head so I am correct.

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The difference of two numbers is 44 1 /2 . If the smaller of the two numbers increases 7 times then the difference will be 10 3/

pshichka [43]

Let’s call the numbers x and y. Let’s also say that x is the smaller of the two. Since their difference (the answer you get when you subtract is 44 and a half) we know that y-x=44.5

Next we are told that if the smaller increases 7 times (that means 7x) their difference would be 10 3/14. Let’s write that as a fraction. It is 14x10+3 over 14. That is 143/14.

So this means y-7x=143/14.

We have two equations and two unknowns (a system of equations). We write one over the other and subtract them to get:

y-x=44.5
y-7x=143/14

Subtracting yields:
(y-y)+(-x - - 7x) = 44.5 - 143/14
6x = (89/2) - (143/14)
6x= 623/14 - 143/14
6x=480/14
X= (480/14)(1/6)
X=80/14
x= 40/7 = 5 5/7

Since y-x=89/2
We get y - 40/7 = 89/2
Y = 40/7 + 89/2
Y = 80/14 + 623/14
Y = 703/14 = 50 3/14

8 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.