in table data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:High Low Q1 Q3 IQR Median Mean σ
College 50 5 7.5 15 7.5 11 13.8 6.4
<span>High School16 0 9.5 14.5 5 13 10.7 5.3 </span>Both spreads are best described with the standard deviation.
Answer:
D) 1
Step-by-step explanation:
Every whole number (including the ones on a die) are either even or odd, nothing else. So it is guaranteed for the number to be even or odd.
Answer:
Yes , function is continuous in [0,2] and is differentiable (0,2) since polynomial function are continuous and differentiable
Step-by-step explanation:
We are given the Function
f(x) =
The two basic hypothesis of the mean valued theorem are
- The function should be continuous in [0,2]
- The function should be differentiable in (1,2)
upon checking the condition stated above on the given function
f(x) is continuous in the interval [0,2] as the functions is quadratic and we can conclude that from its graph
also the f(x) is differentiable in (0,2)
f'(x) = 6x - 2
Now the function satisfies both the conditions
so applying MVT
6x-2 = f(2) - f(0) / 2-0
6x-2 = 9 - 1 /2
6x-2 = 4
6x=6
x=1
so this is the tangent line for this given function.
Answer:
x= 2
Step-by-step explanation:
5 -2(x +5)= 3x -15
Expand:
5 -2x -2(5)= 3x -15
5 -2x -10= 3x -15
Simplify:
-2x -5= 3x -15
Bring x terms to 1 side, constant to the other:
3x +2x= 15 -5
5x= 10
Divide both sides by 5:
x= 10 ÷5
x= 2