Answer:
Here we have the domain:
D = 0 < x < 1
And we want to find the range in that domain for:
1) y = f(x) = x
First, if the function is only increasing in the domain (like in this case) the minimum value in the range will match with the minimum in the domain (and the same for the maximums)
f(0) = 0 is the minimum in the range.
f(1) = 1 is the maximum in the range.
The range is:
0 < y < 1.
2) y = f(x) = 1/x.
In this case the function is strictly decreasing in the domain, then the minimum in the domain coincides with the maximum in the range, and the maximum in the domain coincides with the minimum in the range.
f(0) = 1/0 ---> ∞
f(1) = 1/1
Then the range is:
1 < x.
Notice that we do not have an upper bound.
3) y = f(x) = x^2
This function is strictly increasing, then:
f(0) = 0^2 = 0
f(1) = 1^2 = 1
the range is:
0 < y < 1
4) y = f(x) = x^3
This function is strictly increasing in the interval, then:
f(0) = 0^3 = 0
f(1) = 1^3 = 1
the range is:
0 < y < 1.
5) y = f(x) = √x
This function is well defined in the positive reals, and is strictly increasing in our domain, then:
f(0) = √0 = 0
f(1) = √1 =1
The range is:
0 < y < 1
Answer:
x= -2 and y= 8
Step-by-step explanation:
8x+5y=24-------A
y=-4x---------B
Putting the value of y from B in A
8x+5(-4x)=24
8x +(-20x)= 24
8x-20x=24
-12x=24
x= -24/12
x=-2--------C y =-4(-2)=8
Answer:
i dont know what you mean but
3x(4x-3) simplified is 12x - 4 or 12x+(-4)
Answer:
The claim that the scores of UT students are less than the US average is wrong
Step-by-step explanation:
Given : Sample size = 64
Standard deviation = 112
Mean = 505
Average score = 477
To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.
Solution:
Sample size = 64
n > 30
So we will use z test
Formula : 
Refer the z table for p value
p value = 0.9772
α=0.05
p value > α
So, we accept the null hypothesis
Hence The claim that the scores of UT students are less than the US average is wrong
Answer:
16 pies were neither apple nor peach.
Step-by-step explanation:
8/10, or 4/5 of the 80 apple pies were either apple or peach, which 4/5 of 80 is 64. The remaining pies will be neither apple nor peach. 80 - 64 = 16.
