No matter the number of times you rolled the dice, the probability of getting a number is always 1/6. But here we can choose 4 numbers ( 1 to 4) hence the probability P( 1 or 2 or 3 or 4) = 4/6 2/3 = 0.6667 = 66.67% (A)
Answer:
I am about 95% sure that this is what they want. P.s (sorry I cant put the divide symbol)
<u><em>The variable is the number of shirts she bought or x</em></u>
and the equation is <u><em>(375+50) divided by (15+2) = x</em></u>
in the solve spot put
<em><u>(375+50) divided by (15+2)=x</u></em>
<em><u>425 divided by 17=x</u></em>
<em><u>x=25</u></em>
Then your answer would be <u><em>25</em></u>
Step-by-step explanation:
first add 50 to 375 so you get the full price which will give you 425. then to find out how many shirts she bought you must add 2 to 15 to get 17. then divide 425 by 17 which is 25. so she bought <em><u>25 shirts</u></em>.
Answer: I think it’s skewed right
Step-by-step explanation:
Problem 1
<h3>Answer: False</h3>
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Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
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Problem 2
<h3>Answer: True</h3>
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Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).
