A mosquito is walking at random on the nonnegative number line. She starts at $1$. When she is at $0$, she always takes a step $1$ unit to the right, but, from any positive position on the line, she randomly moves left or right $1$ unit with equal probability. What is the expected number of times the mosquito will visit $0$ before the first time she visits $4$?

The expected number of times is 4.

Step-by-step explanation:

Looking at the question, we see that this follows a geometric distribution because it is asking for the expected number of trials hat will bring about the FIRST SUCCESS. The probability of success is

Since it is a geometric distribution, we know that the expected value of a random variable X, E(X) that follows a geometric distribution is given as:

E(X) = 1/p where p is the probability of success.

Therefore, the expected number of times will be

E(X) = 1/(1/p) = 1/(1/4) = 4.

Hence, the expected number of times is 4.

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.

x=-7

step-by-step explanation:

x+3=-10

-10+3=-7

x=-7