<h3>
Answer: 680 different combinations</h3>
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Explanation:
If order mattered, then we'd have 17*16*15 = 4080 different permutations. Notice how I started with 17 and counted down 1 at a a time until I had 3 slots to fill. We count down by 1 because each time we pick someone, we can't pick them again.
So we have 4080 different ways to pick 3 people if order mattered. But again order doesn't matter. All that counts is the group itself rather than the individual or how they rank. There are 3*2*1 = 6 ways to order any group of three people, which means there are 4080/6 = 680 different combinations possible.
An alternative is to use the nCr formula with n = 17 and r = 3. That formula is
where the exclamation marks indicate factorials
Solve for y
minus x bot sides
4y=-x+12
divide 4
y=-1/4x+3
To solve this problem lets assume, that the number 60 is 100% - because it's the output value of the task. <span>We assume, that x is the value we are looking for. </span><span>If 100% equals 60, so we can write it down as 100%=60. </span><span>We know, that x% equals 120 of the output value, so we can write it down as x%=120. </span>Now we have two simple equations:
<u>1)100%=60</u>
<u>2) x%=120</u>
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that: 100%/x%=60/120.
Now we just have to solve the simple equation, and we will get the solution we are looking for.
<em>Solution for 120 is what percent of 60</em>
<em>100%/x%=60/120</em>
<span><em>(100/x)*x=(60/120)*x - </em></span><em>we multiply both sides of the equation by x</em>
<span><em>100=0.5*x - </em></span><em>we divide both sides of the equation by (0.5) to get x</em>
<span><em>100/0.5=x </em></span>
<span><em>200=x </em></span>
<em>x=200</em>
<span><em>now we have: </em></span>
<span><em>120 is 200% of 60</em></span>
Answer: Option D : All of the above.
Step-by-step explanation:
(1) The Cumulative Link Models assumptions can be summarized as, conditional on (x1, … xk)
(2) Whether normality of error u, and thus normality of y conditional on (x1, … xk) can be assumed, is an empirical matter
and (3) Normality of error u translates into normal sampling distributions of the OLS estimators are all correct statements describing linear regression.
You've certainly tried hard to describe it in enough detail so that
it can be answered without seeing the diagram. Sadly, it's not
enough.
We still don't know what's tangent to what, where point 'C' is, or
how any of this relates to a circle. You'll have to figure out some
way or us to see the diagram.