Answer:
Step-by-step explanation:
p(2p-1)-10
Answer:
18.6 months
Step-by-step explanation:
Given that :
Best fit line from scatterplot :
y=-12.05x +224.26
x = Number of month
y = charge on battery
Number of months a typical battery uses before being dead completely :
When battery is dead completely ; charge =0, y = 0
y = -12.05x + 224.26
0 = - 12.05x + 224.26
12.05x = 224.26
x = 224.26 / 12.05
x = 18.610788
Hence, 18.6 months before battery is completely dead.
See the graph attached.
The midpoint rule states that you can calculate the area under a curve by using the formula:
![M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]](https://tex.z-dn.net/?f=M_%7Bn%7D%20%3D%20%5Cfrac%7Bb%20-%20a%7D%7B2%7D%20%5B%20f%28%5Cfrac%7Bx_%7B0%7D%20%2B%20x_%7B1%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7Bx_%7B1%7D%20%2B%20x_%7B2%7D%20%7D%7B2%7D%29%20%2B%20...%20%2B%20%20f%28%5Cfrac%7Bx_%7Bn-1%7D%20%2B%20x_%7Bn%7D%20%7D%7B2%7D%29%5D)
In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1
Therefore, you'll have:
![M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%20-%200%7D%7B4%7D%20%5B%20f%28%5Cfrac%7B0%20%2B%20%20%5Cfrac%7B1%7D%7B4%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7B%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%7D%7B2%7D%29%20%2B%20f%28%5Cfrac%7B%5Cfrac%7B3%7D%7B4%7D%20%2B%201%7D%20%7B2%7D%29%5D)
![M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%20f%28%5Cfrac%7B1%7D%7B8%7D%29%20%2B%20%20f%28%5Cfrac%7B3%7D%7B8%7D%29%20%2B%20%20f%28%5Cfrac%7B5%7D%7B8%7D%29%20%2B%20f%28%5Cfrac%7B7%7D%7B8%7D%29%5D)
Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³
Therefore:
![M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%28%5Cfrac%7B1%7D%7B8%7D%20-%20%28%5Cfrac%7B1%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B3%7D%7B8%7D%20-%20%28%5Cfrac%7B3%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B5%7D%7B8%7D%20-%20%28%5Cfrac%7B5%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B7%7D%7B8%7D%20-%20%28%5Cfrac%7B7%7D%7B8%7D%29%5E%7B3%7D%29%5D)
![M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%28%5Cfrac%7B1%7D%7B8%7D%20-%20%5Cfrac%7B1%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B3%7D%7B8%7D%20-%20%5Cfrac%7B27%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B5%7D%7B8%7D%20-%20%5Cfrac%7B125%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B7%7D%7B8%7D%20-%20%5Cfrac%7B343%7D%7B512%7D%29%5D)
M₄ = 1/4 · (2 - 478/512)
= 0.2666
Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately
0.267 square units.
Answer:
Because she is concluding that her class is an accurate representation of the overall school. The sample size in this case is too small and should increase her sample size(the amount of people surveyed) in order to get a better result.