Answer:
Median = 116.
First quartile = 102.
Step-by-step explanation:
There is a total of 280 scores so the median will be the 140th score.
The first box gives 26 , second 34 = 60
The 3rd box contains 100 giving a total of 160 so the median lies in this third box.
140 - 26 - 34 = 80..
That gives a value of 80 / 5 = 16.
The median is 100 + 16 = 116.
The lower quartile is the 70th score
26 + 34 + 10 = 70.
So it is 100 + 10/5 = 102.
The region in question is the set

or equivalently,
![R = \left\{ (x, y) : 0 \le y \le 1 \text{ and } 0 \le x \le \sqrt[3]{y} \right\}](https://tex.z-dn.net/?f=R%20%3D%20%5Cleft%5C%7B%20%28x%2C%20y%29%20%3A%200%20%5Cle%20y%20%5Cle%201%20%5Ctext%7B%20and%20%7D%200%20%5Cle%20x%20%5Cle%20%5Csqrt%5B3%5D%7By%7D%20%5Cright%5C%7D)
Cross sections are taken perpendicular to the y-axis, which means each section has a base length equal to the horizontal distance between the curve y = x³ and the line x = 0 (the y-axis). This horizontal distance is given by
y = x³ ⇒ x = ∛y
so that each triangular cross section has side length ∛y.
The area of an equilateral triangle with side length s is √3/4 s², so each cross section contributes an infinitesimal area of √3/4 ∛(y²).
Then the volume of this solid is
![\displaystyle \frac{\sqrt3}4 \int_0^1 \sqrt[3]{y^2} \, dy = \frac{\sqrt3}4 \int_0^1 y^{2/3} \, dy = \frac{\sqrt3}4\cdot\frac35 y^{5/3} \bigg|_0^1 = \boxed{\frac{3\sqrt3}{20}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csqrt3%7D4%20%5Cint_0%5E1%20%5Csqrt%5B3%5D%7By%5E2%7D%20%5C%2C%20dy%20%3D%20%5Cfrac%7B%5Csqrt3%7D4%20%5Cint_0%5E1%20y%5E%7B2%2F3%7D%20%5C%2C%20dy%20%3D%20%5Cfrac%7B%5Csqrt3%7D4%5Ccdot%5Cfrac35%20y%5E%7B5%2F3%7D%20%5Cbigg%7C_0%5E1%20%3D%20%5Cboxed%7B%5Cfrac%7B3%5Csqrt3%7D%7B20%7D%7D)
I've attached some sketches of the solid with 16 and 64 such cross sections to give an idea of what this solid looks like.
Step-by-step explanation:
m(x + n) = n
mx + mn = n
mx = n - mn
x = (n - mn)/m
x(a + b) = b(c - x)
ax + bx = bc - bx
ax + bx + bx = bc
x(a + b + b) = bc
x(a + 2b) = bc
x = bc/(a + 2b)
mx = n(m + x)
mx = mn+ nx
mx - nx = mn
x(m - n) = mn
x = mn/(m-n)
The account in 2017 would hold y = $61.53