Suppose two events a and b are mutually exclusive, with p(a) ≠ 0 and p(b) ≠ 0. by working through the following steps, you'll see why two mutually exclusive events are not independent. (a) for mutually exclusive events, can event a occur if event b has occurred? yes. by definition, mutually exclusive events can occur together. no. two events will never occur concurrently. yes. any two events can occur concurrently. no. by definition, mutually exclusive events cannot occur together. what is the value of p(a|b)? (b) using the information from part (a), can you conclude that events a and b are not independent if they are mutually exclusive? explain. yes. because p(a|b) ≠ p(a), the events a and b are not independent. no. two events being mutually exclusive always implies independence. no. because p(a|b) ≠ p(a), the events a and b may be independent. yes. two events being mutually exclusive always implies independence.

Answer with Step-by-step explanation:

We are given that two events A and B are mutually exclusive.

a.For mutually exclusive events,

Therefore, event A can not occurred if event B has occurred because two events can not occur together.

No, by definition mutually exclusive events cannot occur together.

b.When two events are independent

Then ,

If two events are mutually exclusive then

Then ,

Therefore,

Hence, we can concluded that events A and B are not independent if they are mutually exclusive.

Yes,

50- 250 is correct i did that one too

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.