## User: tanya.tanyabarnwal

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#### Posts by tanya.tanyabarnwal

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... **Solution:** (i) p=10%=0.1 q=1-p=0.9 n=5 $P(X=Y)=^{n}C_{r}p^{r}q^{n-r}$ For at the most two are defective r=0, 1, 2 $P(X=0 \ or \ 1 \ or \ 2)=^{5}C_{0}(0.1)^{0}(0.9)^{5}+^{5}C_{1}(0.1)^{1}(0.9)^{4}+^{5}C_{2}(0.1)^{2}(0.9)^{3}$ $P(X=0 \ or \ 1 \ or \ 2)=0.59049+0.32805+0.0729=0.99144$ (ii) ...
written 5 weeks ago by tanya.tanyabarnwal0
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... $A=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ ...
written 5 weeks ago by tanya.tanyabarnwal0
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... Use the dual simplex method to save the following LPP: Minimize, $z=2x_{1}+x_{2}$ subject to, $3x_{1}+x_{2} \ge 3$ $4x_{1}+3x_{2} \ge 6$ $x_{1}+2x_{2} \le 3$ $x_{1}, \ x_{2} \ge 0$ ...
written 5 weeks ago by tanya.tanyabarnwal0
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... Justify if there is any relationship between sex and color for the following data: | Color | Male | Female | |:-----:|:----:|:------:| | Red | 10 | 40 | | White | 70 | 30 | | Green | 30 | 20 | ...
written 5 weeks ago by tanya.tanyabarnwal0
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... Check whether the following matrix is derogatory or non-derogatory. $A=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$ ...
written 5 weeks ago by tanya.tanyabarnwal0
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... (i) If 10% of the rivets produced by a machine are defective, find the probability that out of 5 randomly chosen rivets at the most two will be defective. (ii) If x denotes the outcome, when a fair dice is tossed, find the M.G.F. of x and hence find the mean and variance of x. ...
written 5 weeks ago by tanya.tanyabarnwal0
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... **Solution:** Consider, $\int_{0}^{2\pi}\cfrac{\cos{2 \theta}}{5+4 \cos {\theta}} d\theta$ Put $z=e^{j \theta}$ -----------(1) $dz=j e^{j \theta} d\theta$ $d\theta=\cfrac{dz}{jz}$ ---------------(2) $\cos {\theta} = \cfrac{e^{j \theta} + e^{-j \theta}}{2} =\cfrac{z+z^{-1}}{2}=\cfrac{z+\cfrac{1} ... written 5 weeks ago by tanya.tanyabarnwal0 1 answer 44 views 1 answer ... Using the residue theorem, Evaluate$\int_{0}^{2\pi}\cfrac{\cos{2 \theta}}{5+4 \cos {\theta}} d\theta$. ... written 5 weeks ago by tanya.tanyabarnwal0 1 answer 57 views 1 answers Answer: A: Chapter 4 . ... **Solution:**$\bar{X}=9.3, \ \mu=8.9, \ s=1.6, \ n=50$(i) Null Hypothesis$(H_{o}): \ \mu_{1}=\mu_{2}$Alternate Hypothesis$(H_{a}): \ \mu_{1} \ne \mu_{2}$(ii) Test statistic$Z=\cfrac{\bar{X}- \mu}{s/ \sqrt{n}}= \cfrac{9.3-8.9}{1.6 / \sqrt{50} }=1.7678\$ (iii) Level of significance at 5%, ...
written 5 weeks ago by tanya.tanyabarnwal0
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