First, calculate f(g(x))==> you plug (5x+4) in the value of x in f(x)
==>f(g(x))= 8-[10(5x+4)===>8-50x-40===>f(g(x))= -50x + 32
& f(g(-2))= -100+32 =68.
There is a mistake in your answers, it should be 68 & not 78
I think the answer to 1 is Friday
Answer:
The Probability = 0.20
Step-by-step explanation:
From the question stated, the first step to take is to find probability that an employee selected at random will need either eyeglasses or major dental work
Solution
Given
Now,
The exams showed tha the number of employees needed eyeglasses = 8%
Employees that needed major dental work = 15%
Employees that needed both eyeglasses and major dental work =3%
Thus,
The P(needed eyeglasses ) = 8% = 0.08
P(major dental work) = 15% = 0.15
P(eye glasses and major dental work) = 3% = 0.03
The probability that an employee selected at random will need either eyeglasses or major dental work is given as
= P(eye glasses ) + P(major dental work) - P(eyeglasses and major dental work)
= 0.08 + 0.15 - 0.03 = 0.20
Therefore the Probability = 0.20
Y=30x is the answer to this
Answer:
<h3>Part A</h3>
<u>A particular man on the seat, then 2 men are out of 4:</u>
- 4C2 = 4!/(4-2)!2! = 4*3*2/2*2 = 6
<u>4 women out of 7:</u>
- 7C4 = 7!/(7-4)!4! = 7*6*5*4!/3!4! = 35
<u>Total combinations:</u>
<h3>Part B</h3>
<em>Option 1</em>. 2 men excluded
3 men out of 3 and 4 women out of 7
<u>Total ways:</u>
<em>Option 2</em>. 2 women are excluded
<u>3 men out of 5 and 4 women out of 5</u>
- 5C3*5C4 =
- 5!/3!(5-3)! * 5!/4!(5-4)! =
- 5*4/2 * 5/1 = 10
<em>Option 3</em>. 1 man and 1 woman excluded
<u>3 out of 4 men and 5 out of 6 women:</u>
- 4C3*6C5 =
- 4!/3!(4-3)! * 6!/5!(6-5)! =
- 4*6 = 24
<h3>Part C</h3>
<u>No restrictions, so the number of ways:</u>
- 3C5 * 7C4 =
- 5!/3!(5-3)! * 35 =
- 4*5/2 * 35 =
- 10*35 = 350
