Answer:
![n = 2.36](https://tex.z-dn.net/?f=n%20%3D%202.36)
Explanation:
The stress experimented by the circular bar is:
![\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cleft%5B%5Cfrac%7B2000%5C%2C%20lbf%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%280.5%5C%2Cin%29%5E%7B2%7D%7D%5Cright%5D%5Ccdot%20%5Cleft%28%5Cfrac%7B1%5C%2Ckpsi%7D%7B1000%5C%2Cpsi%7D%20%5Cright%29)
![\sigma = 10.186\,kpsi](https://tex.z-dn.net/?f=%5Csigma%20%3D%2010.186%5C%2Ckpsi)
The safety factor is:
![n = \frac{24\,kpsi}{10.186\,kpsi}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B24%5C%2Ckpsi%7D%7B10.186%5C%2Ckpsi%7D)
![n = 2.36](https://tex.z-dn.net/?f=n%20%3D%202.36)
Answer:
P ( 2.5 < X < 7.5 ) = 0.7251
Explanation:
Given:
- The pmf for normal distribution for random variable x is given:
f(x)=0.178 exp(-0.100(x-4.51)^2)
Find:
the fraction of individuals demonstrating a response in the range of 2.5 to 7.5.
Solution:
- The random variable X follows a normal distribution with mean u = 4.51, and standard deviation s.d as follows:
s.d = sqrt ( 1 / 0.1*2)
s.d = sqrt(5) =2.236067
- Hence, the normal distribution is as follows:
X ~ N(4.51 , 2.236)
- Compute the Z-score values of the end points 2.5 and 7.5:
P ( (2.5 - 4.51) / 2.236 < Z < (7.5 - 4.51 ) / 2.236 )
P ( -0.898899327 < Z < 1.337168651 )
- Use the Z-Table for the probability required:
P ( 2.5 < X < 7.5 ) = P ( -0.898899327 < Z < 1.337168651 ) = 0.7251
<h2>Answer:</h2>
<h3>Required Answer is as follows :-</h3>
- 3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²
- ¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²
- ¾ = m₁ × (V)²/m₂ × (V²/4)
<h3>Now,</h3>
- ∆m = m₂ - m₁
- ∆m = 16m₂/3 - m₁ = 13m₁/3
- Ratio = (13m₁/3)/ m₁
- Ratio = 13/3
Ratio = 13:3
<h3>Know More :-</h3>
Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg.