Fill in the table with the output (f(x)) values for each given input (x) value based on the given function. Х f(x)

Mathematics

2 answers:

Otrada [13]3 years ago

8 0

Answer:

f(0) = -6

f(1) = -4

f(2) = 2

f(3) = 12

Step-by-step explanation:

$f(x)=2x^{2} -6\\f(0)=2(0)^{2}-6=0-6=-6\\ f(1)=2(1)^{2}-6=2-6=-4\\ f(2)=2(2)^{2}-6=2(4)-6=8-6=2 \\f(3)=2(3)^{2}-6=2(9)-6 =18-6=12$

zzz [600]3 years ago

5 0

Answer:

Plug in the X values for the 2 X's in this equation and solve

Hope this helps out

You might be interested in

Determine whether the events are independent (I) or dependent (D). Place your answer in the blank provided. 1. Ramiro draws a ma

Stels [109]

Answer:

Independent events

Step-by-step explanation:

Given that:

Ramiro draws a marble from a jar without replacement and then flips a coin

Let $E_1$ be the event that Ramiro draws a marble without replacement and;

Let $E_2$ be the event of flipping a coin.

Let's have an analogy so that we can better understand the concept of independent and dependent events.

Consider a random experiment in which a marble is drawn from a jar without replacement and a fair coin is flipped together.

The event $E_1$ does not in any way affect the event $E_2$ of a head or a tail showing up in a flip of a coin.

Therefore, we say that $E_1$ and $E_2$ are independent events.

Suppose the event $E_1$ affects or influence the event $E_2$ , then we can say they are dependent events.

4 0

3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$