Brain enjoys hang gliding on his last flight, he jumped off a cliff 520 high he landed 450 from the base of the cliff how far di
1 answer:
Answer:
Brain travelled 687.68 from the cliff.
Step-by-step explanation:
Given that:
Height of the cliff = 520
Base of the cliff = 450
The distance will form the hypotenuse of right angled triangle.
Pythagorean theorem states square of hypotenuse of right angled triangle is equal to the sum of squares of base and height of triangle.
Using pythagorean theorem;
a = 520, b=450, c = hypotenuse
Taking square root on both sides
Hence,
Brain travelled 687.68 from the cliff.
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Answer:
(a) The Probability that at most three lines are in use is 0.70.
(b) The Probability that fewer than three lines are in use is 0.45.
(c) The Probability that at least three lines are in use is 0.55.
(d) The Probability that between two and five lines, inclusive, are in use is 0.70.
(e) The Probability that between two and four lines, inclusive, are not in use is 0.35.
(f) The Probability that at least four lines are not in use is 0.70.
Step-by-step explanation:
<u>The complete question is: </u>A mail-order company business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table.
X 0 1 2 3 4 5 6
P(X) 0.10 0.15 0.20 0.25 0.20 0.05 0.05
Calculate the probability of each of the following events.
(a) {at most three lines are in use}
(b) {fewer than three lines are in use}
(c) {at least three lines are in use}
(d) {between two and five lines, inclusive, are in use}
(e) {between two and four lines, inclusive, are not in use}
(f) {at least four lines are not in use}
Now considering the above probability distribution;
(a) The Probability that at most three lines are in use is given by = P(X 3)
P(X 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.10 + 0.15 + 0.20 + 0.25
= 0.70
(b) The Probability that fewer than three lines are in use is given by = P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= 0.10 + 0.15 + 0.20
= 0.45
(c) The Probability that at least three lines are in use is given by = P(X 3)
P(X 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
= 0.25 + 0.20 + 0.05 + 0.05
= 0.55
(d) The Probability that between two and five lines, inclusive, are in use is given by = P(2 X
5)
P(2 X
5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= 0.20 + 0.25 + 0.20 + 0.05
= 0.70
(e) The Probability that between two and four lines, inclusive, are not in use is given by = 1 - P(2 X
4)
1 - P(2 X
4) = 1 - [P(X = 2) + P(X = 3) + P(X = 4)]
= 1 - [0.20 + 0.25 + 0.20]
= 1 - 0.65 = 0.35
(f) The Probability that at least four lines are not in use is given by = 1 - Probability that at least four lines are in use = 1 - P(X 4)
1 - P(X 4) = 1 - [P(X = 4) + P(X = 5) + P(X = 6)]
= 1 - [0.20 + 0.05 + 0.05]
= 1 - 0.30 = 0.70
Answer:
Perimeter =