Answer:
The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution 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 = 45 - 1 = 44
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 2
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 12.1 - 0.4 = 11.7 units
The upper end of the interval is the sample mean added to M. So it is 12.1 + 0.4 = 12.5 units
The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.
the -7 is found there twice, so it has a multiplicity of 2
the +7 is there thrice, so it has multiplicity of 3
If f(x) is given with points (0,5) and (4,3), it can be easily plotted in a coordinate system (see .jpeg image in attachment).If g(x) is a function defined by text:
Kyle started the summer having read 3 books but plans to read 6 books per month over the summer.Then, g(x) is given with:
g(x) = 6*x+3 (3 books read already, and 6 books will be read each month).h(x) is already given with:
h(x)=3*x+4
These all three functions are plotted in coordinate system and it can be seen that the lowest y-intercept has function g(x), and it is the value of 3 on y-axis. Others have 4 (h(x)) and 5 (f(x)).
Answer:
V = (About) 22.2, Graph = First graph/Graph in the attachment
Step-by-step explanation:
Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.
![\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cpi%20%5Cint%20_a%5Eb%5Cleft%28r%5Cright%29%5E2dy%5C%3A%7D%2C%5C%5C%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%7D)
The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.
![V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\](https://tex.z-dn.net/?f=V%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%2C%5C%5C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%5Cpi%20%5Ccdot%20%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1dy%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Athe%5C%3ASum%5C%3ARule%7D%3A%5Cquad%20%5Cint%20f%5Cleft%28x%5Cright%29%5Cpm%20g%5Cleft%28x%5Cright%29dx%3D%5Cint%20f%5Cleft%28x%5Cright%29dx%5Cpm%20%5Cint%20g%5Cleft%28x%5Cright%29dx%5C%5C%3D%20%5Cpi%20%5Cleft%28%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2dy-%5Cint%20_1%5E31dy%5Cright%29%5C%5C%5C%5C)
Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.
