X = 4 ; x = 3 + i ; x = 3 - i

(If you get a zero that is adding or subtracting, you always need to write it twice but change the sign do they cancel out)

f(x) = (x-4)(x-3-i)(x-3+i)

Distributing the last two parenthesis first is always the best way to start off

(x-3-i)(x-3+i) has (x-3) in common so it can be separated to

(x-3)^2 + (-i)(+i)

(x^2 - 6x + 9) ; (-i)(+i) is always +1
(x^2 - 6x + 9) + 1
(x^2 - 6x + 10)

Now multiply this with (x-4)
x^3 - 6x^2 + 10x
- 4x^2 + 24x - 40

x^3 - 10x^2 + 34x - 40 = f(x)

3 0

3 years ago

Two cards are drawn from a standard deck without replacement. What is the probability that the first card is a spade and the sec

Verizon [17]

Answer:

The answer is 0.123.

Step-by-step explanation:

You want to find the intersection between P(Spade) and P(Red). Knowing that there's no replacement, the probability will be given by the formula:

⇒ P(Spade∩Red) = P(Spade)*P(Red/Spade)

⇒ P(Spade∩Red) = P(Red)*P(Spade/Red)

Either formula works, but I'm going to choose to work with the first formula.

Therefore: