
the denominator cannot be zero, because the division by zero is not defined, therefore:
![\begin{gathered} x^2-9=0 \\ \text{Solving for x:} \\ x^2=9 \\ \sqrt[]{x^2}=\sqrt[]{9} \\ x=\pm3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E2-9%3D0%20%5C%5C%20%5Ctext%7BSolving%20for%20x%3A%7D%20%5C%5C%20x%5E2%3D9%20%5C%5C%20%5Csqrt%5B%5D%7Bx%5E2%7D%3D%5Csqrt%5B%5D%7B9%7D%20%5C%5C%20x%3D%5Cpm3%20%5Cend%7Bgathered%7D)
Therefore the domain of (f o g)(x) is:
Answer:
x = 6
Step-by-step explanation:
Given
y =
x - 3 and y = 11, thus
x - 3 = 11 ( add 3 to both sides )
x = 14
Multiply both sides by 3 to clear the fraction
7x = 42 ( divide both sides by 7 )
x = 6
1. 2
that's all I can figure out
Answer:
$19.62
Step-by-step explanation:
To round 19.623 to the nearest cent consider the thousandths’ value of 19.623, which is 3 and less than 5. Therefore, the cents value of 19.623 remains 2.
$19.623 rounded to the nearest cent = $19.62
Answer:
A) 0.1612
B) 0.8031
C) 0.1969
Step-by-step explanation:
For each toss of the die, there are only two possible outcomes. Either it is a 3, or it is not. The probability of getting a 3 on each toss is independent from other tosses. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
In this problem we have that:
Five tosses, so 
The die has 6 values, from 1 to 6. The die is fair, so each outcome is equally as likely. The probability of a 3 appearing in a single throw is 
(a) twice
This is 


(b) at most once





(c) at least two times.
Either a 3 appears at most once, or it does at least two times. The sum of the probabilities of these events is decimal 1. So

We want 
So
