Calculate the probability of each of the following events. (a) {at most three lines are in use} .72 Correct: Your answer is correct. (b) {fewer than three lines are in use} .47 Correct: Your answer is correct. (c) {at least three lines are in use} .53 Correct: Your answer is correct. (d) {between two and five lines, inclusive, are in use} .72 Correct: Your answer is correct. (e) {between two and four lines, inclusive, are not in use} .63 Correct: Your answer is correct. (f) {at least four lines are not in use} . Incorrect: Your answer is incorrect.

(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:

The complete question is: 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

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.