To solve this, we need to understand Slope Intercept Form (SIF), as well as how to graph a line.
SIF is the standard equation of lines on graphs. It is "y=mx+b" where m is the slope and b is the y-intercept. The y-intercept is the value of y when x is 0.
To find the y-intercept (which we will need to form the equation), we should simply graph the line. This will let us visualize the y-intercept, and overall make it easier to understand.
To graph a line, we should start with the point we have (that being (3, 3)) and follow the slope with rise/run. This means in this case, we will go right 2 for every 1 up, or 2 left for every 1 down.
Below I have attached a graph to help you see how to graph this, which we will get our equation from. The highlighted area is our y-intercept. The red circle shows our original point (3,3), and the blue dots show our slope.
Using the graph, we can see the equation for this line is
y=1/2x+1.5.
Answer:
(b) (c) (a)
Step-by-step explanation:
Standard Normal distribution has a higher peak in the center, with more area in this región, hence it has less area in its tails.
Student's t-Distribution has a shape similar to the Standard Normal Distribution, with the difference that the shape depends on the degree of freedom. When the degree of freedom is smaller the distribution becomes flatter, so it has more area in its tails.
Student's t-Distributionwith 1515 degrees of freedom has mores area in the tails than the Student's t-Distribution with 2020 degrees of freedom and the latter has more area than Standard Normal Distribution
Answer: 9 over 12 is = 15 over x
Step-by-step explanation:
Answer:
The total cost of the ski hat and the ski boots is $112.35.
Step-by-step explanation:
First, we need to add the prices of the ski hat and the ski boots.
85.50 + 19.50 = $105
Now, we need to find 7% of $105, and then add it to the total.
105 x 0.07 = $7.35
105 + 7.35 = $112.35
Therefore, the total cost of the ski hat and the ski boots is $112.35.
Let me know if this helps!
Check the picture below.
since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.
1)
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%5C%5C%20h%3D4%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%283%29%5E2%284%29%7D%7B3%7D%5Cimplies%20V%3D12%5Cpi%20%5Cimplies%20V%5Capprox%2037.7)
2)
now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%283%29%5E3%7D%7B3%7D%5Cimplies%20V%3D36%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%20for%20a%20semisphere%7D%7D%7BV%3D18%5Cpi%20%7D%5Cimplies%20V%5Capprox%2056.55)
3)
well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.
4)
pretty much the same thing, we get the volume of the cone and its top, add them up.
