The given data is the following:
Student Trial 1 Trial 2 Trial 3 Average
----------- -------- -------- -------- ------------
1 66.0 66.5 68.5 67.0
2 67.5 64.0 70.5 67.3
3 60.3 60.5 60.5 61.0
4 55.0 58.0 59.0 57.3
Let us check the reported averages.
Student 1:
Average = (66.0 + 66.5 + 68.5)/3 = 67.0 Correct
Student 2:
Average = (67.5 + 64.0 + 70.5)/3 = 67.3 Correct
Student 3:
Average = (60.3 + 60.5 + 60.5)/3 = 604 Incorrect
Student 4:
Average = (55.0 + 58.0 + 59.0)/3 = 57.3 Correct
Answer: Student 3
The area of the ellipse
is given by

To use Green's theorem, which says

(
denotes the boundary of
), we want to find
and
such that

and then we would simply compute the line integral. As the hint suggests, we can pick

The line integral is then

We parameterize the boundary by

with
. Then the integral is


###
Notice that
kind of resembles the equation for a circle with radius 4,
. We can change coordinates to what you might call "pseudo-polar":

which gives

as needed. Then with
, we compute the area via Green's theorem using the same setup as before:






Answer: The difference of 6 times a number m and 2
Step-by-step explanation:
The way to write this expression is:
The difference of 6 times a number m and 2.
Answer:
Ordering a soft drink is independent of ordering a square pizza.
Step-by-step explanation:
20% more customers order a soft drink than pizza, therefore they cannot be intertwined.
Given: P(A)=0.5 & P(B)=.7
P(A∩B) = P(A) × P(B)
= 0.5 × .7
= 0.35
P(A∪B) = P(A) + P(B) - P(A∩B)
= 0.5 + .7 - 0.35
= 0.85
P(AΔB) = P(A) + P(B) - 2P(A∩B)
= 0.5 + .7 - 2×0.35
= 0.5
P(A') = 1 - P(A)
= 1 - 0.5
= 0.5
P(B') = 1 - P(B)
= 1 - .7
= 0.3
P((A∪B)') = 1 - P(A∪B)
= 1 - 0.85
= 0.15