Answer:
Graph A and C
Step-by-step explanation:
Because for every x point there is just one y point.
Answer:
y^3/(27 x^3)
Step-by-step explanation:
Simplify the following:
((3 x)/y)^(-3)
((3 x)/y)^(-3) = (y/(3 x))^3:
(y/(3 x))^3
Multiply each exponent in y/(3 x) by 3:
(y^3)/((3 x)^3)
Multiply each exponent in 3 x by 3:
y^3/(3^3 x^3)
3^3 = 3×3^2:
y^3/(3×3^2 x^3)
3^2 = 9:
y^3/(3×9 x^3)
3×9 = 27:
Answer: y^3/(27 x^3)
Problem 1
The error is 2 inches since her estimate is 2 inches off the true value.
We can think of it like this
4 feet = 4*12 = 48 inches
4 feet, 2 inches = 4 ft + 2 in = 48 in + 2 in = 50 inches
So she guesses he is 48 inches, but he's really 50 inches, so 50-48 = 2 inches is her error.
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Problem 2
Divide the error (2 inches) over the actual height (50 inches) to get
2/50 = 4/100 = 4%
The percentage error is 4%
This means she is 4% off the target.
Note how 4% of 50 = 0.04*50 = 2 which was the error we found back in problem 1.
Answer:
The probability is 
≅ 
Step-by-step explanation:
Let's analyze the question.
There are 15 students in the 8th grade.
The students are randomly placed into three different algebra classes of 5 students each.
We are looking for the probability that Trevor, Terry and Evan will be in the same algebra class.
One possible way to solve this question is to think about the product probability rule.
We can use it because we are in an equiprobable space. (And also the events are independent).
Let's set for example a class for Evan.
The probability that Evan will be in a class is 
Then for Terry there are 
places out of 
that puts Terry in the Evan's class.
We write 
Finally for Trevor there are 
places out of the remaining 
that puts Trevor in the same class with Evan and Terry.
Using the product rule we write :
The probability of the event is ≅
