E^2x -2e^x -8=0 => e<span>^(2x) -2e^x -8=0

Temporarily replace e^x with y.

Then (y)^2 - 2y - 8 = 0. Factors are (y-4) and (y+2).

Roots are y = 4 and y= -2.

Now remembering that we temporarily replaced e^x with y, we let
y=4 = e^x. We need to solve for x. Taking the natural log of both sides, we get:

ln 4 = x (answer)

We have to discard the other root (y= -2), because we cannot take the ln of a negative number.

</span>

8 0

3 years ago

A bag contains 50 lottery balls, numbered 1-50. A ball is chosen at random, then a month of the year is selected. What is the pr

Morgarella [4.7K]

Answer:

4/75

Step-by-step explanation:

We have two sample spaces

1.Lottery balls

2. Month of the year

For lottery balls the sample space is S= 50

Multiples of 3 between 1 and 50 are

(3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48)= 16 in numbers

Hence the probability of choosing a multiple of 3

Pr(multiple of 3) = 16/50= 8/25

Also the sample space for months of the year is S= 12

Two months starts with letter m, March and may

Pr(of months with m) = 2/12= 1/6

the probability of choosing a multiple of 3 and a month starting with the letter M= 8/25*1/6= 8/150= 4/75

6 0

4 years ago