Percent error = 1.2556%
Steps:
Percent Error =
Vobserved - Vtrue
Vtrue
=
250 - 246.9
246.9
=
3.1
246.9
= 1.2555690562981%
I can't see the graph but let's use logic
hmm, more than 10 cubic feet of topsoil so the first and 2nd options are wrong
let's ee the costs
3rd option
10*1=10
2*12=24
10+24=34 and 34<50, that is fine
4th option
3*10=30
2*12=24
30+24=54
54>50, nope, that is over cost
answer is 3rd one
the one with 1 cubic yard compost and 12 cubic yard topsoil
Answer:
You have to upload a pic of you're work?
Step-by-step explanation:
You are right. In circle equation we have =r² so the correct answer will be C od D. But we have to calculate the center of a circle.
![x^2+y^2-8x-6y+24=0\quad|-24\\\\(x^2-8x)+(y^2-6y)=-24](https://tex.z-dn.net/?f=x%5E2%2By%5E2-8x-6y%2B24%3D0%5Cquad%7C-24%5C%5C%5C%5C%28x%5E2-8x%29%2B%28y%5E2-6y%29%3D-24)
Complete the square:
![x^2-8x+a^2\implies a=\dfrac{8x}{2x}=4\implies x^2-8x+4^2\\\\\\y^2-6y+b^2\implies b=\dfrac{6y}{2y}=3\implies y^2-6y+3^2](https://tex.z-dn.net/?f=x%5E2-8x%2Ba%5E2%5Cimplies%20a%3D%5Cdfrac%7B8x%7D%7B2x%7D%3D4%5Cimplies%20x%5E2-8x%2B4%5E2%5C%5C%5C%5C%5C%5Cy%5E2-6y%2Bb%5E2%5Cimplies%20b%3D%5Cdfrac%7B6y%7D%7B2y%7D%3D3%5Cimplies%20y%5E2-6y%2B3%5E2)
And we have:
![(x^2-8x)+(y^2-6y)=-24\\\\(x^2-8x+4^2)-4^2+(y^2-6y+3^2)-3^2=-24\\\\ (x-4)^2+(y-3)^2-16-9=-24\\\\(x-4)^2+(y-3)^2-25=-24\quad|+25\\\\\boxed{(x-4)^2+(y-3)^2=1}](https://tex.z-dn.net/?f=%28x%5E2-8x%29%2B%28y%5E2-6y%29%3D-24%5C%5C%5C%5C%28x%5E2-8x%2B4%5E2%29-4%5E2%2B%28y%5E2-6y%2B3%5E2%29-3%5E2%3D-24%5C%5C%5C%5C%0A%28x-4%29%5E2%2B%28y-3%29%5E2-16-9%3D-24%5C%5C%5C%5C%28x-4%29%5E2%2B%28y-3%29%5E2-25%3D-24%5Cquad%7C%2B25%5C%5C%5C%5C%5Cboxed%7B%28x-4%29%5E2%2B%28y-3%29%5E2%3D1%7D)
So the answer is C (the same left side of equation).
Answer:
![\mu = 114, \sigma = 40](https://tex.z-dn.net/?f=%5Cmu%20%3D%20114%2C%20%5Csigma%20%3D%2040)
We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
And the standard deviation for the sampling distribution would be:
![\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B40%7D%7B%5Csqrt%7B100%7D%7D%3D%204)
So then the answer is TRUE
Step-by-step explanation:
Let X the random variable of interest and we know that the true mean and deviation for this case are given by:
![\mu = 114, \sigma = 40](https://tex.z-dn.net/?f=%5Cmu%20%3D%20114%2C%20%5Csigma%20%3D%2040)
We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
And the standard deviation for the sampling distribution would be:
![\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B40%7D%7B%5Csqrt%7B100%7D%7D%3D%204)
So then the answer is TRUE