Please help me <33333

Mathematics

1 answer:

mariarad [96]3 years ago

8 0

Comment
You have to begin by declaring what g(f(x)) means. It means that wherever you see an x in g(x) you put in f(x).

It will look like this to start with
g(f(x)) = (f(x) + 5) / (f(x)

Now substitute into this for g(-3)

g(x^2 + 5) = (x^2 + 5 + 5)/(x^2 + 5) It's time to use some numbers.
g(- 3) = ((-3)^2 + 10)/( (-3)^2 +5)
g(-3) = ( 9 + 10 ) / ( 9 + 5)
g(-3) = (19)/14 <<<<<< answer.
C <<<< answer.

You might be interested in

The data set represents the number of miles Mary jogged each day for the past nine days. 6, 7, 5, 0, 6, 12, 8, 6, 9

qwelly [4]

The outlier is 0 because it is the only non-positive number

5 0

3 years ago

Read 2 more answers

What is lim x→3 x^2 - x - 6/ x - 3

Kisachek [45]

<h3>Answer: D) 5</h3>

=======================================================

Explanation:

If we plugged x = 3 into the expression, then we'd get x-3 = 3-3 = 0 in the denominator. That's not allowed. But we can simplify first

x^2-x-6 factors to (x-3)(x+2). The key here is that (x-3) is a factor. It cancels with the x-3 in the denominator

So, $\frac{x^2-x-6}{x-3} = \frac{(x-3)(x+2)}{x-3} = x+2$

Allowing us to say,

$\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x-3}\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}(x+2)\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 3+2\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 5\\\\\\$

5 0

3 years ago

Empirical rule (statistics)Help please!!!

emmasim [6.3K]

The empirical rule you're referring to is the 68-95-99.7 rule, which asserts that for a normal (bell-shaped) distribution, approximately 68% of the distribution lies within 1 standard deviation of the mean; 95% lies within 2 standard deviations of the mean; and 99.7% lies within 3 standard deviations of the mean.

Let $X$ be the random variable denoting vehicle speeds along this highway. We want to find $P(61$ . To use the rule, we need to rephrase this probability in terms of the mean and standard deviation.

Notice that $61=70-9=70-3\cdot3$ , and $79=70+9=70+3\cdot3$ . In other words, 61 and 79 both lie exactly 3 standard deviations away from the mean, so $P(61$ .