Number 3 i am so stuck help plz
2 answers:
Okay, think of it as a pie. They ate 3/4 of the pie, leaving 1/4. Then that 1/4 was cut into 6/6, and 1/6 was eaten. So there were 5/6 left of that 1/4. Your pie has 4 quarters, and each is made up of 6 mini pieces, so 6 x 4 = 24. There were 24 little 1/6 pieces at the beginning when your lasagna pie was whole. Now, at the end, you have 5 of them left. So you have five of the original 24—
.
You might be interested in
Check the picture below.
thus, if one side on the square is x - x², then the area for the square is (x - x²)(x - x²), or (x - x²)²
now... what is the "x" value for the region, let's check their intersection.
![\bf \begin{cases} y=x\\ y=x^2 \end{cases}\implies x=x^2\implies 0=x^2-x\implies 0=x(x-1) \\\\\\ x= \begin{cases} 0\\ 1 \end{cases}\\\\ -------------------------------\\\\ \displaystyle \int\limits_{0}^{1}\ (x-x^2)^2\cdot dx\implies \int\limits_{0}^{1}\ (x^2-2x^3+x^4)\cdot dx \\\\\\ \cfrac{x^2}{2}-\cfrac{2x^4}{4}+\cfrac{x^5}{5}\implies \left. \cfrac{x^2}{2}-\cfrac{x^4}{2}+\cfrac{x^5}{5} \right]_{0}^{1}\implies \left[ \cfrac{1}{2}-\cfrac{1}{2}+\cfrac{1}{5} \right]-[0]\implies \cfrac{1}{5}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ay%3Dx%5C%5C%0Ay%3Dx%5E2%0A%5Cend%7Bcases%7D%5Cimplies%20x%3Dx%5E2%5Cimplies%200%3Dx%5E2-x%5Cimplies%200%3Dx%28x-1%29%0A%5C%5C%5C%5C%5C%5C%0Ax%3D%0A%5Cbegin%7Bcases%7D%0A0%5C%5C%0A1%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B0%7D%5E%7B1%7D%5C%20%28x-x%5E2%29%5E2%5Ccdot%20dx%5Cimplies%20%5Cint%5Climits_%7B0%7D%5E%7B1%7D%5C%20%28x%5E2-2x%5E3%2Bx%5E4%29%5Ccdot%20dx%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bx%5E2%7D%7B2%7D-%5Ccfrac%7B2x%5E4%7D%7B4%7D%2B%5Ccfrac%7Bx%5E5%7D%7B5%7D%5Cimplies%20%5Cleft.%20%5Ccfrac%7Bx%5E2%7D%7B2%7D-%5Ccfrac%7Bx%5E4%7D%7B2%7D%2B%5Ccfrac%7Bx%5E5%7D%7B5%7D%20%5Cright%5D_%7B0%7D%5E%7B1%7D%5Cimplies%20%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D-%5Ccfrac%7B1%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B5%7D%20%5Cright%5D-%5B0%5D%5Cimplies%20%5Ccfrac%7B1%7D%7B5%7D)
thus, if one side on the square is x - x², then the area for the square is (x - x²)(x - x²), or (x - x²)²
now... what is the "x" value for the region, let's check their intersection.
Answer:
a. ; b. The 95% interval is USD [45.82, 56.86].
Step-by-step explanation:
The mean amount a customer spends at the grocery store
The mean is the sum of all of the data divided by the number of observations for this case, that is, n = 50.
The sum of all of the data is: 5.24+11.97+13.59+15.52+15.69+19.96+20.54+22.59+27.25+27.29+28.25+30.00+31.63+33.01+37.21+40.36+40.90+42.83+45.66+46.67+47.49+49.32+50.12+51.71+52.2 1+53.57+56.16+56.36+56.88+58.65+59.42+60.20+60.48+62.23+62.50+67.95+70.45+73.08+74.15+74.48+74.78+75.75+76.27+76.47+78.92+84.09+84.91+85.06+88.45+88.91 = 2567.17.
The mean is
Then, <em>the mean amount a customer spends at the grocery store is</em> USD 51.34.
The 95% confidence interval for the mean amount spent
For the 95% confidence interval for the mean, we have the value for the <em>mean = </em>51.34, as we have just found, the value for the<em> standard deviation</em> is already given as USD 20.00. There are 50 observations, so n = 50.
Moreover, for a 95% confidence, the value for the corresponding z-score is z = 1.96 (this is the confidence coefficient). Then, the formula for finding the "limits" values for the confidence interval is as follows:
For the upper limit
For the lower limit
Thus
The 95% confidence is USD [45.82, 56.86].
Mathematically
There is a probability of 0.95 that the values are between [45.82, 56.86], both inclusive.
We have to remember here that we are dealing with a sample. This sample has a standard deviation, called standard error of the sample, and is .
So, the population mean is, with a 95% confidence, between the values [45.82, 56.86], both inclusive.
We can see the graph below that represents the former.
