Answer:
Using Matlab code for Fourier series to calculate for the function, see the attached
Step-by-step explanation:
Go through the picture step by step.
If the 1m turns into 3m then he cm increases also by 3.The 3 meters would equal 300 cm.
Check the picture below.
based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).
and that will give us two x-intercepts, at x = 3 and x = k.
since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.
now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.
let's find the y-intercept by setting x = 0 now,
![\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)](https://tex.z-dn.net/?f=%5Cbf%20y%3D0.5%28x-3%29%28x%2Bk%29%5Cimplies%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28x-3%29%28x%2Bk%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsetting%20x%20%3D%200%7D%7D%7By%3D%5Ccfrac%7B1%7D%7B2%7D%280-3%29%280%2Bk%29%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28-3%29%28k%29%5Cimplies%20%5Cboxed%7By%3D-%5Ccfrac%7B3k%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20triangle%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7Dbh%7D~~%20%5Cbegin%7Bcases%7D%20b%3D3%2Bk%5C%5C%20h%3Dy%5C%5C%20%5Cquad%20-%5Cfrac%7B3k%7D%7B2%7D%5C%5C%20A%3D1.5%5C%5C%20%5Cqquad%20%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B3%7D%7B2%7D%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29)

now, we can plug those values on A = (1/2)bh,
![\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-2%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-2%29%5Cleft%28-%5Ccfrac%7B3%28-2%29%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%281%29%283%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-1%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-1%29%5Cleft%28-%5Ccfrac%7B3%28-1%29%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7D%282%29%5Cleft%28%20%5Ccfrac%7B3%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5)
Answer:
a. 144 cubic foot
b. 12 ft
c. 18 ft
Step-by-step explanation:
Let the volume of prism A be V_a
and that of prism b be V_b
ATQ, V_a + V_b= 432
also V_a= 0.5 V_b
⇒1.5 V_b= 432
= V_b= 432/1.5= 288 cubic feet
therefore V_a= 144 cubic feet
volume of prism= area of base×height = V_a
24×h = 144
⇒h= 12 ft
A_b= 2/3×24
⇒base area of prism B= 16 sq.ft
now 16×h_b= 288⇒h_b= 288/16= 18 ft
Answer:
a.) f(x) =
where 90 < x < 120
b.) 
c.) 
d.) 
Step-by-step explanation:
Let
X be a uniform random variable that denotes the actual charging time of battery.
Given that, the actual recharging time required is uniformly distributed between 90 and 120 minutes.
⇒X ≈ ∪ ( 90, 120 )
a.)
Probability density function , f (x) =
where 90 < x < 120
b.)
P(x < 110) = 
= ![\frac{1}{30}[x]\limits^{110}_{90} = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B110%7D_%7B90%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20110%20-%2090%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2020%5D%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
c.)
P(x > 100 ) = 
= ![\frac{1}{30}[x]\limits^{120}_{100} = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B120%7D_%7B100%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20120%20-%20100%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2020%5D%20%3D%20%5Cfrac%7B2%7D%7B3%7D)
d.)
P(95 < x< 110) = 
= ![\frac{1}{30}[x]\limits^{110}_{95} = \frac{1}{30} [ 110 - 95 ] = \frac{1}{30} [ 15] = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%5Bx%5D%5Climits%5E%7B110%7D_%7B95%7D%20%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%20110%20-%2095%20%5D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%5B%2015%5D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)