Answer:
A) 12.1 , 12.01
B) infinitely large
Step-by-step explanation:
Give that the function f(x)=x2 describes the area of the square, f(x), in square inches, whose sides each measure x inches. If x is changing.
Using the formula
(f(a + h) - f(a))/h
as x changes from 6 inches to 6.1 inches
a = 6, h = 0.1
= [f(6.1) - f(6)]/0.1
= (6.1^2 - 6^2)/0.1
= (37.21 - 36)/0.1
= 1.21/0.1
= 12.1
and when x changes from 6 inches to 6.01 inches.
a = 6, And h = 0.01
Using the same formula
(f(a + h) - f(a))/h
= [f(6.01) - f(6)]/0.01
= (6.01^2 - 6^2)/0.01
= (36.120 - 36)/0.01
=0.1201/0.01
= 12.01
B) the instantaneous rate of change of the area with respect to x at the moment when x = 6 inches is infinitely large since h = 0
Given:
Uniform distribution of length of classes between 45.0 to 55.0 minutes.
To determine the probability of selecting a class that runs between 51.5 to 51.75 minutes, find the median of the given upper and lower limit first:
45+55/2 = 50
So the highest number of instances is 50-minute class. If the probability of 50 is 0.5, then the probability of length of class between 51.5 to 51.75 minutes is near 0.5, approximately 0.45. <span />
You times the x and y by 3 good luck
68
I hope you understand this
Answer:
5995 hugs
Step-by-step explanation:
This is a combination problem: we want to make groups of two monkeys, where a pair of monkey 1 and 2 is the same pair of monkey 2 and 1 (it is the same hug). So, if the total number of monkeys at the zoo is 110, we have a combination of 110 choose 2:
C(110,2) = 110! / (2! * (110-2)!) = 110 * 109 / 2 = 5995
So the total number of hugs is 5995
