Find BC if B(8,-7) and C1-4,-2).
1 answer:
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Answer: (a). at x = 0, its a removable discontinuity
and at x = 1, it is a jump discontinuity
(b). at x = -3, it is removable discontinuity
also at x = -2, it is an infinite discontinuity
(c). at x = 2, it is a jump discontinuity
Step-by-step explanation:
in this question, we would analyze the 3 options to determine which points gave us discontinuous in the category of discontinuity as jump, removable, infinite, etc.
(a). given that f(x) = x/x² -x
this shows a discontinuous function, because we can see that the denominator equals zero i.e.
x² - x = 0
x(x-1) = 0
where x = 0 or x = 1.
since x = 0 and x = 1, f(x) is a discontinuous function.
let us analyze the function once more we have that
f(x) = x/x²-x = x/x(x-1) = 1/x-1
from 1/x-1 we have that x = 1 which shows a Jump discontinuity
also x = 0, this also shows a removable discontinuity.
(b). we have that f(x) = x+3 / x² +5x + 6
we simplify as
f(x) = x + 3 / (x + 3)(x + 2)
where x = -3, and x = -2 shows it is discontinuous.
from f(x) = x + 3 / (x + 3)(x + 2) = 1/x+2
x = -3 is a removable discontinuity
also x = -2 is an infinite discontinuity
(c). given that f(x) = │x -2│/ x - 2
from basic knowledge in modulus of a function,
│x│= │x x ˃ 0 and at │-x x ∠ 0
therefore, │x - 2│= at │x - 2, x ˃ 0 and at │-(x - 2) x ∠ 2
so the function f(x) = at│ 1, x ˃ 2 and at │-1, x ∠ 2
∴ at x = 2 , the we have a Jump discontinuity.
NB. the figure uploaded below is a diagrammatic sketch of each of the function in the question.
cheers i hope this helps.
Answer:
or 0.2 or 20%
Step-by-step explanation:
For only two tests to be needed this means that the first test would need to come back as negative and the second test would be to come back as positive. Therefore, to find the probability of this happening we first need to find the probability of each individual test and multiply them together.
The first test needs to come back negative, there are four negative individuals out of the total 5 that are in the group. Therefore, the probability of the first test is 4/5.
Now we remove the individual that has just been tested and we are left with 4 total subjects in the group, of which only 1 is positive. Therefore, the probability of the second test is 1/4. Now we need to multiply these two probabilities together to get the probability of only needing two tests.
or 0.2 or 20%