Which pairs of variables would most likely have a positive association? Choose all that apply.
1 answer:
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Answer:
12 students
Step-by-step explanation:
Given
Required
Determine the number of students that eat lunch once a week
In Sets;
If out of 9, at least 7 eats one a week then
9 - 7 eats lunch once a week
<em>In other words;</em>
2 out of 9 students eat lunch once a week
Number of students in this category is then calculated as thus;
1/6d + 2/3 = 1/4(d - 2)
First, simplify
to 
/ Your problem should look like: + 
= 
(d - 2)
Second, simplify
to 
/ Your problem should look like: + =
Third, multiply both sides by 12 (the LCM of 6,4) / Your problem should look like: 2d + 8 = 3(d - 2)
Fourth, expand. / Your problem should look like: 2d + 8 = 3d - 6
Fifth, subtract 2d from both sides. / Your problem should look like: 8 = 3d - 6 - 2d
Sixth, simplify 3d - 6 - 2d to d - 6 / Your problem should look like: 8 = d - 6
Seventh,add 6 to both sides. / Your problem should look like: 8 + 6 = d
Eighth, simplify 8 + 6 to 14 / Your problem should look like:14 = d
Ninth, switch sides. / Your problem should look like: d = 14
Answer: d = 14
First, simplify
Second, simplify
Third, multiply both sides by 12 (the LCM of 6,4) / Your problem should look like: 2d + 8 = 3(d - 2)
Fourth, expand. / Your problem should look like: 2d + 8 = 3d - 6
Fifth, subtract 2d from both sides. / Your problem should look like: 8 = 3d - 6 - 2d
Sixth, simplify 3d - 6 - 2d to d - 6 / Your problem should look like: 8 = d - 6
Seventh,add 6 to both sides. / Your problem should look like: 8 + 6 = d
Eighth, simplify 8 + 6 to 14 / Your problem should look like:14 = d
Ninth, switch sides. / Your problem should look like: d = 14
Answer: d = 14
Answer:
These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form, I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it" anyway,
Step-by-step explanation:
a)
next the question asks 1-cos 2A and this is total abuse of notation. the way this should be written is 1- cos( 2A) so we know that the A is part of the cosine functions input... btw.. in any computer program, it would never ever let you get away with that top form of the expression. :/ anyway... I keep ranting.. huh... sorry :P
1-cos(2A) is an odd form of the identity 1/2(1-cos(2A) = (A) the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely, totally wrong, but I just miss one small thing :/ anyway....
our 1-cos(2A) needs the 1/2 added to it. or if we move that 1/2 to the other side it looks like 2*(A) = 1-cos(2A) and this is that "odd" from of the identity that I was talking about.
next let's deal with sin(2A) it has an identity of 2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !
now our fraction looks like [2* (A)] / 2 sin(A)cos(A)
so cancel out one of the sines
2*sin(A) / 2 cos(A)
cancel the 2s
Sin(A) / Cos(A) = Tan(A)
nice it worked out :P
b)
by the above that we just worked out, then
Tan(15) = Sin(15) / Cos(15)
I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of
Sin(15) = (√3 - 1) / 2√2
Cos(A) = (√3 + 1) / 2√2
you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)
back to our problem, the 2√2 will cancel out
then we have
Tan(15) = (√3 - 1) /(√3 + 1)
in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact. Most math professors like that form.