Answer: C. (1, 4)
Step-by-step explanation:
The point where the two lines meet or intersect is the solution to the system of equations graphed. And in this case, the lines intersect at (1, 4).
I suppose

The vectors that span
form a basis for
if they are (1) linearly independent and (2) any vector in
can be expressed as a linear combination of those vectors (i.e. they span
).
Compute the Wronskian determinant:

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of
is 3.
Write an arbitrary vector in
as
. Then the given vectors span
if there is always a choice of scalars
such that

which is equivalent to the system

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

so the vectors do span
.
The vectors comprising
form a basis for it because they are linearly independent.
Answer:

Step-by-step explanation:
- The cross section of a sphere is a circle with area
, where radius is given as 4 - The cross section of the cube is a square with area
, where a is side length of square
The circle area = the square area. Thus we have:

Event: Probability: A. Too much enamel 0.18 B. Too little enamel 0.24 C. Uneven application 0.33 D. No defects noted 0.47
let P(AC) = x, P(BC) = y, then P(A) + P(B) + P(C) - (x+y) = 1-0.47 = 0.53 x+y = 0.22
3. The probability of paint defects that results to <span>an improper amount of paint and uneven application? </span>
P(A U B U C) = 0.53
4. <span>the probability of a paint defect that results to</span>
<span>the proper amount of paint, but uneven application?</span>
P(C) - P(AC) - P(BC) = 0.47 - 0.22 = 0.25
A and B are disjoint so P(ABC) = 0, but you can have P(AC) and P(BC). you can't compute these separately here, but you can compute P(AC) + P(BC). By the way, P(AC) eg is just an abbreviated version of P(A∩C).
Answer:
D
Step-by-step explanation:
The answer is D the median is 1.5 and the mean is greater