Need answers fast pleasee :(

Given f(x)= 3x-1 and g(x)= 2x-3, for what value of x does g(x) =f (2)?

Tasya [4]

Answer:

At x = 4 ⇒ g(x) = f(2)

Step-by-step explanation:

* In the functions f(x) = 3x - 1 and g(x) = 2x - 3, the domain is

the values of x and the range is the values of y

∵ g(x) = f(2)

* Then you will use the range of f(x) at x = 2 to find the domain of g(x)

∵ The domain of f(x) = 2

∵ f(x) = 3x - 1

∴ f(2) = 3(2) - 1 = 6 - 1 = 5

* The range of f(x) = 5

∵ f(2) = g(x)

∴ 5 = g(x)

∵ g(x) = 2x - 3

∴ 2x - 3 = 5 ⇒ add 3 to both sides

∴ 2x = 8 ⇒ divide both sides by 2

∴ x = 4

* At x = 4 ⇒ g(x) = f(2)

4 0

3 years ago

A student solves the compound inequality 16 < 2x + 4 = 18 and finds the

Greeley [361]

answer is in this link: brainly.com/question/17406759

4 0

3 years ago

1500 divided by 106.43

Illusion [34]

14.0937706
rounded to nearest tenth?
14.1

Hope this helps :-)

6 0

3 years ago

Read 2 more answers

The computer that controls a bank's automatic teller machine crashes a mean of 0.60.6 times per day. What is the probability tha

OLEGan [10]

Answer:

0.6045 = 60.45% probability that, in any seven-day week, the computer will crash more than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The computer that controls a bank's automatic teller machine crashes a mean of 0.6 per day.

7 day week, so $\mu = 7*0.6 = 4.2$

What is the probability that, in any seven-day week, the computer will crash more than 3 times?

Either it crashes 3 or less times, or it crashes more than 3 times. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 3) + P(X > 3) = 1$

We want P(X > 3). So

$P(X > 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

$P(X = 3) = \frac{e^{-4.2}*(4.2)^{3}}{(3)!} = 0.1852$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0150 + 0.0630 + 0.1323 + 0.1852 = 0.3955$

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.3955 = 0.6045$

0.6045 = 60.45% probability that, in any seven-day week, the computer will crash more than 3 times.

3 0

4 years ago

Help!!!!!!<br> I need To solve this in 2 mins!!!<br> 2 mins left to Test PLS HELP!!

Montano1993 [528]

Answer:

The equation is y=0.05x

Step-by-step explanation:

It only stays in the first quadrant because it does not move down or left just up and right. It also doesnt go below zero because of how price works.

8 0

3 years ago