Answer:
Domain ![=(-\infty,\ \infty)](https://tex.z-dn.net/?f=%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
Range ![=(-\infty,\ \infty)](https://tex.z-dn.net/?f=%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
Step-by-step explanation:
Domain : Domain of a function
is the set of all possible values of
for which
exists.
Range : range of a function
is the set of all possible values of
.
Here ![f(x)=36-3x](https://tex.z-dn.net/?f=f%28x%29%3D36-3x)
can be any value from
to
.
![\forall\ x=a\ there\ exists\ f(x)\ such\ that\ f(a)=36-3a](https://tex.z-dn.net/?f=%5Cforall%5C%20x%3Da%5C%20there%5C%20exists%5C%20f%28x%29%5C%20such%5C%20that%5C%20f%28a%29%3D36-3a)
hence possible value of
can be any value between
and ![\infty](https://tex.z-dn.net/?f=%5Cinfty)
![domain =(-\infty,\ \infty)](https://tex.z-dn.net/?f=domain%20%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
let ![y=-f(x)](https://tex.z-dn.net/?f=y%3D-f%28x%29)
![y=36-3x\\3x=36-y\\\\\\x=\frac{36-y}{3}\\](https://tex.z-dn.net/?f=y%3D36-3x%5C%5C3x%3D36-y%5C%5C%5C%5C%5C%5Cx%3D%5Cfrac%7B36-y%7D%7B3%7D%5C%5C)
so
.
hence
can have any value between
.
Range ![=(-\infty,\ \infty)](https://tex.z-dn.net/?f=%3D%28-%5Cinfty%2C%5C%20%5Cinfty%29)
![(x^3+x^2+x+2) divide by (x^2-1)](https://tex.z-dn.net/?f=%28x%5E3%2Bx%5E2%2Bx%2B2%29%20divide%20by%20%28x%5E2-1%29)
We use long division to divide
There is no x term in x^2 -1 so we put 0x
x + 1
----------------------------
x^2+0x-1 x^3+ x^2 + x+ 2
x^3+0x^2-x
-----------------------------(subtract the bottom from top)
x^2 +2x + 2
x^2 +0x - 1
--------------------------------(subtract the bottom from top)
2x + 3
-----------------------------------------
Quotient : x+1
Remainder : 2x+3
Answer:
![(f+g)(x) = x^2 + 4x - 4](https://tex.z-dn.net/?f=%28f%2Bg%29%28x%29%20%3D%20x%5E2%20%2B%204x%20-%204)
Step-by-step explanation:
We are given these following functions:
![f(x) = 4x + 2](https://tex.z-dn.net/?f=f%28x%29%20%3D%204x%20%2B%202)
![g(x) = x^2 - 6](https://tex.z-dn.net/?f=g%28x%29%20%3D%20x%5E2%20-%206)
(f+g)(x)
We add the common terms. Thus:
![(f+g)(x) = f(x) + g(x) = 4x + 2 + x^2 - 6 = x^2 + 4x + 2 - 6 = x^2 + 4x - 4](https://tex.z-dn.net/?f=%28f%2Bg%29%28x%29%20%3D%20f%28x%29%20%2B%20g%28x%29%20%3D%204x%20%2B%202%20%2B%20x%5E2%20-%206%20%3D%20x%5E2%20%2B%204x%20%2B%202%20-%206%20%3D%20x%5E2%20%2B%204x%20-%204)
Thus:
![(f+g)(x) = x^2 + 4x - 4](https://tex.z-dn.net/?f=%28f%2Bg%29%28x%29%20%3D%20x%5E2%20%2B%204x%20-%204)
Answer:
c. 6.2 ± 2.626(0.21)
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 101 - 1 = 100
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.626
The confidence interval is:
![\overline{x} \pm M](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%5Cpm%20M)
In which
is the sample mean while M is the margin of error.
The distribution of the number of puppies born per litter was skewed left with a mean of 6.2 puppies born per litter.
This means that ![\overline{x} = 6.2](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%3D%206.2)
The margin of error is:
![M = T\frac{s}{\sqrt{n}} = 2.626\frac{2.1}{\sqrt{101}} = 2.626(0.21)](https://tex.z-dn.net/?f=M%20%3D%20T%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%202.626%5Cfrac%7B2.1%7D%7B%5Csqrt%7B101%7D%7D%20%3D%202.626%280.21%29)
In which s is the standard deviation of the sample and n is the size of the sample.
Thus, the confidence interval is:
![\overline{x} \pm M = 6.2 \pm 2.626(0.21)](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%5Cpm%20M%20%3D%206.2%20%5Cpm%202.626%280.21%29)
And the correct answer is given by option c.
Answer:
1.88 pounds
Step-by-step explanation:
First, find the total amount of sugar, since there were 2 shipments
22.56(2)
= 45.12
Then, divide this by 24:
45.12/24
= 1.88
So, each canister had 1.88 pounds of sugar