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N76 [4]
3 years ago
13

Will give BRAINLIEST!!!

Mathematics
1 answer:
forsale [732]3 years ago
3 0

Answer:

y = 2x - 5

Step-by-step explanation:

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If f x equals x -3 + g x equals 1 - x 2 - 9 find ( f og) (x and give any restrictions on its domain
evablogger [386]
\begin{gathered} f(x)=x-3 \\ g(x)=\frac{1}{x^2-9} \\ (\text{fog)(x)}=\frac{1}{x^2-9}-3 \end{gathered}

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}

Therefore the domain of (f o g)(x) is:

D\colon(-\infty,-3)\cup(-3,3)\cup(3,\infty)

3 0
1 year ago
Solve: y= 7/3x-3 and y=11
Travka [436]

Answer:

x = 6

Step-by-step explanation:

Given

y = \frac{7}{3} x - 3 and y = 11, thus

\frac{7}{3} x - 3 = 11 ( add 3 to both sides )

\frac{7}{3} x = 14

Multiply both sides by 3 to clear the fraction

7x = 42 ( divide both sides by 7 )

x = 6

3 0
3 years ago
Read 2 more answers
PLEASE HELP MEEEEE<br><br><br> check the picture.
Alecsey [184]
1. 2
that's all I can figure out

7 0
3 years ago
Round $19.623 to the nearest cent
Alinara [238K]

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

5 0
3 years ago
Find the probability that in five tosses of a fair die, a 3 will appear (a) twice, (b) at most once, (c) at least two times.
Jobisdone [24]

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.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

Five tosses, so n = 5

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 p = \frac{1}{6} = 0.167

(a) twice

This is P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{5,2}.(0.167)^{2}.(0.833)^{3} = 0.1612

(b) at most once

P(X \leq 1) = P(X = 0) + P(X = 1)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.167)^{0}.(0.833)^{5} = 0.4011

P(X = 1) = C_{5,1}.(0.167)^{1}.(0.833)^{1} = 0.4020

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.4011 + 0.4020 = 0.8031

(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

P(X \leq 1) + P(X \geq 2) = 1

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.8031 = 0.1969

6 0
3 years ago
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