area of square = 9 x 9 = 81 square feet
area of 1 corner = 1/2 * 3 * 3 = 4.5 square feet
4.5 * 4 = 18 square feet total of all 4 corners
81 - 18 = 63 square feet total not covered by benches.
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Answer:
Step-by-step explanation:
To solve this, we would follow these simple steps. We have
unvrs :
The arithmetic mean, x-bar for the yellow paper group (Y) = 20.6
The arithmetic mean, x-bar for the green paper group (G) = 21.75
Recall that, H0: µY = µG
And from the data we have, we can see that
H0: µY< µG
We proceed to say that the
T-Test-statistic = -0.404
Also, the p-value: 0.349
From our calculations, we can see that the p-value > 0.05, and as such, we conclude that we will not reject H0. This is because there is not enough evidence to show that test that is printed on the yellow paper decreases anxiety at a 0.05 significance level.
A reasonable number between 18 and 23 is 20. 69 minutes is 3 times 23 minutes, so we multiply the 20 by 3, getting an estimate of 60 cars.
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.
Answer:
sin A = 12/13
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
sin theta = opposite side/ hypotenuse
sin A = 12/13