From any one vertex, you can draw a line that is perpendicular<span> to the opposite base — that's the </span>altitude<span> to this base. Any </span>triangle<span> has three </span>altitudes<span> and three bases. You can use any one </span>altitude<span>-base pair to </span>find<span> the area of the </span>triangle<span>, via the formula A = (1/2)bh </span>
Answer:
<h3>
a) Mean=1.2</h3><h3>b) The standard deviation is
![\sigma=3.1](https://tex.z-dn.net/?f=%5Csigma%3D3.1)
</h3>
Step-by-step explanation:
<h3>Given that the discrete probability distribution below :</h3><h3><u>Outcome Probability
</u></h3>
0 0.35
1 0.36
2 0.14
3 0.08
4 0.04
5 0.02
6 0.01
<h3>a) To find the mean of this distribution :</h3><h3>The formula is
![E(X)=\sum X.P(X)](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum%20X.P%28X%29)
</h3><h3><u>X P(X) XP(X) </u>
![X^2P(X)](https://tex.z-dn.net/?f=X%5E2P%28X%29)
<u /></h3>
0 0.35 0 0
1 0.36 0.36 0.36
2 0.14 0.28 1.12
3 0.08 0.24 2.16
4 0.04 0.16 2.56
5 0.02 0.1 2.5
6 0.01 0.06 2.16
_________________________________________________
![\sum X^2P(X)=10.86](https://tex.z-dn.net/?f=%5Csum%20X%5E2P%28X%29%3D10.86)
_________________________________________________
- Now substitute the value in the formula we get
![E(X)=\sum X.P(X)](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum%20X.P%28X%29)
![E(X)=1.2](https://tex.z-dn.net/?f=E%28X%29%3D1.2)
<h3>Therefore Mean=1.2</h3><h3>b) To find Standard Deviation :</h3><h3>The formula is
![\sigma=\sqrt{X^2P(X)-(XP(X))^2}](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7BX%5E2P%28X%29-%28XP%28X%29%29%5E2%7D)
</h3>
- Substitute the values i the formula we have
![\sigma=\sqrt{10.86-(1.2)^2}](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B10.86-%281.2%29%5E2%7D)
![=\sqrt{10.86-1.44}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B10.86-1.44%7D)
![=\sqrt{9.42}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B9.42%7D)
![=3.0692](https://tex.z-dn.net/?f=%3D3.0692)
<h3>Therefore the standard deviation is
![\sigma=3.1](https://tex.z-dn.net/?f=%5Csigma%3D3.1)
</h3>
Step-by-step explanation: The cosecant function is graphed in the given figure. we are to find the period of the function.
The period of a function is the distance travelled by the curve of the function in one complete revolution.
We can see that in the given figure, the distance between two consecutive points is given by
Therefore, the period of the cosecant function is
Thus, the correct option is (B) \pi.
Answer:
![t \approx 36.1\,min](https://tex.z-dn.net/?f=t%20%5Capprox%2036.1%5C%2Cmin)
Step-by-step explanation:
The time constant for the isotope decay is:
![\tau = \frac{8\min}{\ln 2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B8%5Cmin%7D%7B%5Cln%202%7D)
![\tau \approx 11.542\,min](https://tex.z-dn.net/?f=%5Ctau%20%5Capprox%2011.542%5C%2Cmin)
Now, the decay of the isotope is modelled after the following expression:
![m (t) = m_{o}\cdot e^{-\frac{t}{\tau} }](https://tex.z-dn.net/?f=m%20%28t%29%20%3D%20m_%7Bo%7D%5Ccdot%20e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D)
The time is now cleared with some algebraic handling:
![\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)
![t = -\tau \cdot \ln \frac{m(t)}{m_{o}}](https://tex.z-dn.net/?f=t%20%3D%20-%5Ctau%20%5Ccdot%20%5Cln%20%5Cfrac%7Bm%28t%29%7D%7Bm_%7Bo%7D%7D)
Finally, the time need for the element X to decay to 43 grams is:
![t = - (11.542\,min)\cdot \ln\left(\frac{43\,g}{980\,g} \right)](https://tex.z-dn.net/?f=t%20%3D%20-%20%2811.542%5C%2Cmin%29%5Ccdot%20%5Cln%5Cleft%28%5Cfrac%7B43%5C%2Cg%7D%7B980%5C%2Cg%7D%20%5Cright%29)
![t \approx 36.1\,min](https://tex.z-dn.net/?f=t%20%5Capprox%2036.1%5C%2Cmin)