the answer is 1/8. Because you divide how many pieces there are by that same number.
Answer:
Step-by-step explanation:
Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).
Let's calculate the common points:
![x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1](https://tex.z-dn.net/?f=x%5E2%3D%5Csqrt%7Bx%7D%5Cqquad%5Ctext%7Bsquare%20of%20both%20sides%7D%5C%5C%5C%5C%28x%5E2%29%5E2%3D%5Cleft%28%5Csqrt%7Bx%7D%5Cright%29%5E2%5C%5C%5C%5Cx%5E4%3Dx%5Cqquad%5Ctext%7Bsubtract%7D%5C%20x%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5Cx%5E4-x%3D0%5Cqquad%5Ctext%7Bdistribute%7D%5C%5C%5C%5Cx%28x%5E3-1%29%3D0%5Ciff%20x%3D0%5C%20%5Cvee%5C%20x%5E3-1%3D0%5C%5C%5C%5Cx%5E3-1%3D0%5Cqquad%5Ctext%7Badd%201%20to%20both%20sides%7D%5C%5C%5C%5Cx%5E3%3D1%5Cto%20x%3D%5Csqrt%5B3%5D1%5Cto%20x%3D1)
The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.
We have integrals:
![\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cint%5Climits_%7B0%7D%5E1%28%5Csqrt%7Bx%7D%29dx-%5Cint%5Climits_%7B0%7D%5E1%28x%5E2%29dx%3D%28%2A%29%5C%5C%5C%5C%5Cint%28%5Csqrt%7Bx%7D%29dx%3D%5Cint%5Cleft%28x%5E%5Cfrac%7B1%7D%7B2%7D%5Cright%29dx%3D%5Cdfrac%7B2%7D%7B3%7Dx%5E%5Cfrac%7B3%7D%7B2%7D%3D%5Cdfrac%7B2x%5Csqrt%7Bx%7D%7D%7B3%7D%5C%5C%5C%5C%5Cint%28x%5E2%29dx%3D%5Cdfrac%7B1%7D%7B3%7Dx%5E3%5C%5C%5C%5C%28%2A%29%3D%5Cleft%28%5Cdfrac%7B2x%5Csqrt%7Bx%7D%7D%7B2%7D%5Cright%5D%5E1_0-%5Cleft%28%5Cdfrac%7B1%7D%7B3%7Dx%5E3%5Cright%5D%5E1_0%3D%5Cdfrac%7B2%281%29%5Csqrt%7B1%7D%7D%7B2%7D-%5Cdfrac%7B2%280%29%5Csqrt%7B0%7D%7D%7B2%7D-%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%281%29%5E3-%5Cdfrac%7B1%7D%7B3%7D%280%29%5E3%5Cright%29%5C%5C%5C%5C%3D%5Cdfrac%7B2%281%29%281%29%7D%7B2%7D-%5Cdfrac%7B2%280%29%280%29%7D%7B2%7D-%5Cdfrac%7B1%7D%7B3%7D%281%29%7D%2B%5Cdfrac%7B1%7D%7B3%7D%280%29%3D2-0-%5Cdfrac%7B1%7D%7B3%7D%2B0%3D1%5Cdfrac%7B1%7D%7B3%7D)
Answer:
Step-by-step explanation:
7. 
6. 
5. ![\displaystyle 1000[0,85]^8 = 272,490525 ≈ \$272,49](https://tex.z-dn.net/?f=%5Cdisplaystyle%201000%5B0%2C85%5D%5E8%20%3D%20272%2C490525%20%E2%89%88%20%5C%24272%2C49)
4. ![\displaystyle 1000 = a \\ -15\% + 100\% = 1 - r; 85\% = 1 - r \\ 8\:years = time\:[t]](https://tex.z-dn.net/?f=%5Cdisplaystyle%201000%20%3D%20a%20%5C%5C%20-15%5C%25%20%2B%20100%5C%25%20%3D%201%20-%20r%3B%2085%5C%25%20%3D%201%20-%20r%20%5C%5C%208%5C%3Ayears%20%3D%20time%5C%3A%5Bt%5D)
3. ![\displaystyle /text{We need to use the "Exponential Decay" formula} - f(t) = a[1 - r]^t, where a > 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%2Ftext%7BWe%20need%20to%20use%20the%20%22Exponential%20Decay%22%20formula%7D%20-%20f%28t%29%20%3D%20a%5B1%20-%20r%5D%5Et%2C%20where%20a%20%3E%200)
2. 
1. 
I am joyous to assist you anytime.
one or more functions ? i think lol
Answer:
Step-by-step explanation:
We want to determine a 90% confidence interval for the true population mean textbook weight.
Number of sample, n = 22
Mean, u = 64 ounces
Standard deviation, s = 5.1 ounces
For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
64 ± 1.645 × 5.1/√22
= 64 ± 1.645 × 1.087
= 64 ± 1.788
The lower end of the confidence interval is 64 - 1.788 = 62.21 ounces
The upper end of the confidence interval is 64 + 1.788 = 65.79 ounces
Therefore, with 90% confidence interval, the true population mean textbook weight is between 62.21 ounces and 65.79 ounces
