Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 100 million dollars

Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

Step-by-step explanation:

We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.

Let X = incomes for the industry

So, X ~ N()

Now, the z score probability distribution is given by;

         Z = ~ N(0,1)

where, = mean income of firms in the industry = 95 million dollars

            = standard deviation = 5 million dollars

So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)

    P(X < 100) = P( < ) = P(Z < 1) = 0.8413   {using z table]

Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).

b

step-by-step explanation: