Answer:(p-4)/33
Step-by-step explanation:
Answer:
x = 2, y = -3
Step-by-step explanation:
Add them together and get
9x + 8y - 9x - 9y = -6 + 9
-y = 3
so y = -3.
Sub it into the first equation and get
9x + 8(-3) = -6
9x - 24 = -6
9x = -6 + 24 = 18
x = 18/9 = 2
Answer:
![\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}](https://tex.z-dn.net/?f=%5Ctextsf%7BMidpoint%20rule%7D%3A%20%5Cquad%20%5Cdfrac%7B2%5Cpi%7D%7B%5Csqrt%5B3%5D%7B2%7D%7D)
Step-by-step explanation:
<u>Midpoint rule</u>
![\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7D%20f%28x%29%20%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%5Capprox%20h%5Cleft%5Bf%28x_%7B%5Cfrac%7B1%7D%7B2%7D%7D%29%2Bf%28x_%7B%5Cfrac%7B3%7D%7B2%7D%7D%29%2B...%2Bf%28x_%7Bn-%5Cfrac%7B3%7D%7B2%7D%7D%29%2Bf%28x_%7Bn-%5Cfrac%7B1%7D%7B2%7D%7D%29%5Cright%5D%5C%5C%5C%5C%20%5Cquad%20%5Ctextsf%7Bwhere%20%7Dh%3D%5Cdfrac%7Bb-a%7D%7Bn%7D)
<u>Trapezium rule</u>
![\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7D%20y%5C%3A%20%5C%3A%5Ctext%7Bd%7Dx%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7Dh%5Cleft%5B%28y_0%2By_n%29%2B2%28y_1%2By_2%2B...%2By_%7Bn-1%7D%29%5Cright%5D%20%5Cquad%20%5Ctextsf%7Bwhere%20%7Dh%3D%5Cdfrac%7Bb-a%7D%7Bn%7D)
<u>Simpson's rule</u>
<u>Given definite integral</u>:
![\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
Therefore:
Calculate the subdivisions:
<u>Midpoint rule</u>
Sub-intervals are:
![\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cdfrac%7B1%7D%7B2%7D%5Cpi%2C%20%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cpi%20%2C%20%5Cdfrac%7B3%7D%7B2%7D%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cdfrac%7B3%7D%7B2%7D%5Cpi%2C%202%20%5Cpi%20%5Cright%5D)
The midpoints of these sub-intervals are:
Therefore:
![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%5Bf%20%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B3%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B5%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B7%7D%7B4%7D%20%5Cpi%20%5Cright%29%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%5B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%20%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%5Cright%5D%5C%5C%5C%5C%20%26%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B%5Csqrt%5B3%5D%7B2%7D%7D%5C%5C%5C%5C%26%20%3D%204.986967483...%5Cend%7Baligned%7D)
<u>Trapezium rule</u>
![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%5B%280%2B0%29%2B2%281%2B0%2B1%29%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cleft%5B4%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cpi%5Cend%7Baligned%7D)
<u>Simpson's rule</u>
<u />
<u />![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%5Capprox%20%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%280%2B4%281%29%2B2%280%29%2B4%281%29%2B0%5Cright%29%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%288%5Cright%29%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B4%7D%7B3%7D%20%5Cpi%5Cend%7Baligned%7D)
Answer:
The question is about the least amount to charge each policyholder as premium
The least premium is $484
Step-by-step explanation:
The least amount of premium to charge for this policy is the sum of the expected values of outcome of both instances of policyholder dying before the age of 70 and living after the age of 70 years
expected value of dying before 70 years=payout*probability=$24,200*2%=$484
Expected of living after 70=payout*probability=$0*98%=$0
sum of expected values=$484+$0=$484
Note that payout is nil if policyholder lives beyond 70 years
The premium of $573 means that a profit of $89 is recorded
