Adding all the marks and dividing them by the total no. of students to get the average marks.
Avg. marks = (7+8+7+9+6)/5
Avg. marks = 37/5
Avg. marks = 7.4
7 9/12 = 7.75
2 11/12 = 2.917
So we can round both of these up.
7.75 rounds to 8
2.917 rounds to 3
8 + 3 = 11
So the estimated sum is 11.
7 9/12 + 2 11/12 = 10 2/3
So the actual sum is 10 2/3.
The estimated sum is a pretty good estimate to the original number.
Ratio of classes A, B, C = 4 : 6 : 5, respectively
4 + 6 + 5 = 15
So, class A represents 4/15 of the TOTAL population
Class B represents 6/15 of the TOTAL population
Class C represents 5/15 of the TOTAL population
So, weighted average of 3 groups combined = (4/15)(65) + (6/15)(80) + (5/15)(77)
= 260/15 + 480/15 + 385/15
= 1125/15
= 75
Answer: B
<span>t is the number of teachers and s is the number of students
</span>
<span>The school requires 3 teachers for every 93 students.
</span>
<span>So, 3t = 93 s ⇒⇒⇒ divide by 3
</span>
<span>∴ 3t/3 = 93s/3
</span>
<span>∴ t = 31 s
</span>
The <span>equation represents this relationship is ⇒⇒⇒ t = 31 s
</span>
The equation of the line through (0, 1) and (<em>c</em>, 0) is
<em>y</em> - 0 = (0 - 1)/(<em>c</em> - 0) (<em>x</em> - <em>c</em>) ==> <em>y</em> = 1 - <em>x</em>/<em>c</em>
Let <em>L</em> denote the given lamina,
<em>L</em> = {(<em>x</em>, <em>y</em>) : 0 ≤ <em>x</em> ≤ <em>c</em> and 0 ≤ <em>y</em> ≤ 1 - <em>x</em>/<em>c</em>}
Then the center of mass of <em>L</em> is the point 
with coordinates given by
where 
is the first moment of <em>L</em> about the <em>x</em>-axis, 
is the first moment about the <em>y</em>-axis, and <em>m</em> is the mass of <em>L</em>. We only care about the <em>y</em>-coordinate, of course.
Let <em>ρ</em> be the mass density of <em>L</em>. Then <em>L</em> has a mass of
Now we compute the first moment about the <em>y</em>-axis:
Then
but this clearly isn't independent of <em>c</em> ...
Maybe the <em>x</em>-coordinate was intended? Because we would have had
and we get
