Answer:
y = -1/2x - 1/2
Step-by-step explanation:
Slope-Intercept Form: y = mx + b
m - slope
b - y-intercept
Step 1: Define
y = -1/2x + 3/7
Point (-1, 0)
Step 2: Find parallel line
<em>Parallel lines have the same slope as the original but different y-intercepts.</em>
<em>m</em> = -1/2
y = -1/2x + b
0 = -1/2(-1) + b
0 = 1/2 + b
b = -1/2
Step 3: Write parallel line
y = -1/2x - 1/2
Since you did not attach any picture we cannot say for sure what is the correct answer, but we can discuss the options in order to find the most probable correct answer.
First of all, according to the Cavalieri's principle, an oblique cylinder has the same volume as a right cylinder with the same base surface area and same height.
A cross-section of an oblique cylinder will be a small right cylinder with the same base surface area and a height as small as possible.
I guess the oblique cylinder has height h and it is divided into many (probably 10) cross-sections.
Option A: <span>πr2h
This is exactly the volume of the right cylinder, therefore, unless you are given a cross-section of height h (which would be too easy), this won't be the correct answer.
Option B: </span><span>4πr2h
This is 4 times the right cylinder. Again, here the height of the cross-section should</span> be 4h, but it doesn't sound like a possible data (too easy again).
Option C: <span>1 10 πr2h
Here comes a n issue with the notation: I think the right number you meant to write is (1/10)</span>·πr2h and not 110·<span>πr2h.
If I am right, this means that your oblique cylinder of height h is divided into 10 cross-sections, and therefore the volume of each of these cross-sections will be a tenth of the volume of the oblique cylinder, which means </span>1/10·<span>πr2h.
Option D: </span><span>1 2 πr2h
Here, we have the same notation issue as before. I think you meant (1/2)</span>·<span>πr2h.
Here, your oblique cylinder height h should be divided into only 2 cross-sections. Now, we said the cross-section's height should be the smallest as possible, so an oblique cylinder divided only into two pieces doesn't sound good.
Therefore, the most probable correct answer will be C) </span>(1/10)·<span>πr2h</span>
Answer:
B
Step-by-step explanation:
The small triangle is half the size of the entire triangle. 27/2 = 13.5
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
