Answer:
H0: p(1980) = p(2010) ; H1 : P(1980) > P(2010)
0.02
Step-by-step explanation:
Given that:
Sample size in both 1980 and 2010 = 1000 samples :
Proportion in favor :
P(1980) = 0.66
P(2010) = 0.64
To test the hypothesis :
Null hypothesis :
Proportion in favor are the same in both years
Null hypothesis = H0 = p(1980) = p(2010)
Alternative hypothesis :
Proportion in favor in 1980 is greater than that in 2010
Alternative hypothesis = H1 : P(1980) > P(2010)
The sample statistic :
P(1980) - p(2010)
0.66 - 0.64
= 0.02
Step-by-step explanation:
i dont no but thx for the points
Sum would be equal to 720.
so, x-60 + x-40 + 130 + 120 + 110 + x-20 = 720
3x + 240 = 720
3x = 720 - 240
x = 480/3
x = 160
Angle F = x-20 = 160 - 20 = 140
In short, Your Answer would be 140
Hope this helps!
The correct answer to this problem is the third statement which is "Alexia completed 20 math problems and 10 science problem". The following conditions are stated in the problem:
Alexia can answer 3 minutes on each math problem
Alexia can answer 4 minutes on each science problem
She worked more than 60 minutes
We have the inequality 3x + 4y > 60 where x represents the total number of math problem while y represents the total number of the science problem.
in our answer, we have x=20and y=10
3*20 + 4*10 > 60
60 + 40 > 60
100 > 60
Therefore, our answer is correct.
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>
---------------------------------
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).
