This type of essay answer is pretty short.
You play out the equation and say what each part of the equation means. The vertex in this case, because it's a downward facing parabola, is the highest point on the graph. Now since they're kicking a ball, the vertex will be the highest point that the ball will go up to.
Hopefully this helps some. For full points you'll probably have to add a bit of padding and further explanation.
Answer:
and ![\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmax%7D%20%5Capprox%200.661)
Step-by-step explanation:
The equation of the isotope decay is:
![\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }](https://tex.z-dn.net/?f=%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%3D%20e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D)
14-Carbon has a half-life of 5568 years, the time constant of the isotope is:
![\tau = \frac{5568\,years}{\ln 2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B5568%5C%2Cyears%7D%7B%5Cln%202%7D)
![\tau \approx 8032.926\,years](https://tex.z-dn.net/?f=%5Ctau%20%5Capprox%208032.926%5C%2Cyears)
The decay time is:
(There is no a year 0 in chronology).
![t = 3335 \pm 13\,years](https://tex.z-dn.net/?f=t%20%3D%203335%20%5Cpm%2013%5C%2Cyears)
Lastly, the relative amount is estimated by direct substitution:
![\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }](https://tex.z-dn.net/?f=%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%3D%20e%5E%7B-%5Cfrac%7B3335%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D%5Ccdot%20e%5E%7B%5Cmp%5Cfrac%7B13%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D)
![\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmin%7D%20%3D%20e%5E%7B-%5Cfrac%7B3335%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D%5Ccdot%20e%5E%7B-%5Cfrac%7B13%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D)
![\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmin%7D%20%5Capprox%200.659)
![\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmax%7D%20%3D%20e%5E%7B-%5Cfrac%7B3335%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D%5Ccdot%20e%5E%7B%5Cfrac%7B13%5C%2Cyears%7D%7B8032.926%5C%2Cyears%7D%20%7D)
![\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D%20%5Cright%29_%7Bmax%7D%20%5Capprox%200.661)
Answer:
i believe that 25 ounces of jelly for 3.39$ is better to buy
:)
Answer: The confidence interval would be (0.2928,0.3672).
Step-by-step explanation:
Since we have given that
Total number of vehicles = 432
Number of SUVs sold = 142
So, it becomes,
![\hat{p}=\dfrac{x}{n}=\dfrac{142}{432}=0.33](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%3D%5Cdfrac%7Bx%7D%7Bn%7D%3D%5Cdfrac%7B142%7D%7B432%7D%3D0.33)
At 90% confidence interval,
z = 1.645
So, the confidence interval would be
![\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.33\pm 1.645\sqrt{\dfrac{0.33\times 0.67}{432}}\\\\=0.33\pm 0.0372\\\\=(0.33-0.0372,0.33+0.0372)\\\\=(0.2928,0.3672)](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%5Cpm%20z%5Csqrt%7B%5Cdfrac%7B%5Chat%7Bp%7D%281-%5Chat%7Bp%7D%29%7D%7Bn%7D%7D%5C%5C%5C%5C%3D0.33%5Cpm%201.645%5Csqrt%7B%5Cdfrac%7B0.33%5Ctimes%200.67%7D%7B432%7D%7D%5C%5C%5C%5C%3D0.33%5Cpm%200.0372%5C%5C%5C%5C%3D%280.33-0.0372%2C0.33%2B0.0372%29%5C%5C%5C%5C%3D%280.2928%2C0.3672%29)
Hence, the confidence interval would be (0.2928,0.3672).
Just act like you are subtracting normal numbers, and then an a decimal in front of the answer!