User: tanya.tanyabarnwal

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Posts by tanya.tanyabarnwal

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Answer: A: Chapter 3 .
... **Solution:** Let $m$ and $\sigma$ be the mean and standard deviation of the distribution. $z=\cfrac{X-m}{\sigma}$ --------------(1) **Case 1:** For X=45 at $z_{1}$ $z=z_{1}=\cfrac{45-m}{\sigma}$ ----------------(2) $\therefore P(X < 45)= 30 \%$ $P(z < z_{1})=0.30$ ![enter image descrip ...
written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Chapter 3 .
... For a normal distribution 30% items are below 45 and 8% are above 64. Find the mean and variance of the normal distribution. ...
mumbai university written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Answer: A: Chapter 6 .
... **Solution:** $z=2x_{1}^{2}-7x_{2}^{2}+12x_{1}x_{2}$ Subject to, $2x_{1}+5x_{2} \le 98$ $x_{1}, \ x_{2} \ge 0$ We write the given problem as $f(x_{1}, \ x_{2})=2x_{1}^{2}-7x_{2}^{2}+12x_{1}x_{2}$ -------------(1) $h(x_{1}, \ x_{2})=2x_{1}+5x_{2}-98$ ----------------(2) Now, Kuhn-Tucker condit ...
written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Chapter 6 .
... Using Kuhn-Tucker condition, solve the following NLPP. Maximise, $z=2x_{1}^{2}-7x_{2}^{2}+12x_{1}x_{2}$ Subject to, $2x_{1}+5x_{2} \le 98$ $x_{1}, \ x_{2} \ge 0$ ...
mumbai university written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Answer: A: Chapter 3 .
... We must have $\sum p_{i}=1$ $0.1+K+0.2+2K+0.3+K=1$ $0.6+4K=1$ $4K=1-0.6=0.4$ $K=0.1$ $\therefore$ Put K=0.1 in the table, | X | -2 | -1 | 0 | 1 | 2 | 3 | |:------:|:---:|:--:|:---:|:----:|:---:|:-:| | P(X=x) | 0.1 | 0.1 | 0.2 | 0.2 | 0.3 | 0.1 | Now, Mean$=E(x)=\sum p_{i}x_{i} ...
written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Chapter 3 .
... The probability distribution of random variable x is given by: | X | -2 | -1 | 0 | 1 | 2 | 3 | |:------:|:---:|:--:|:---:|:----:|:---:|:-:| | P(X=x) | 0.1 | K | 0.2 | 2K | 0.3 | K | Find K, mean and variance. ...
mumbai university written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Answer: A: Chapter 2. Show that the matrix $A$ satisfies Cayley-Hamilton Theorem and hence
... **Solution:** $A=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ $|A|=1$ Characteristic equation is $|A-\lambda I|=0$ $\begin{vmatrix} 1-\lambda & 2 & -2 \\ -1 & 3-\lambda & 0 \\ 0 & -2 & 1-\lambda \end{vmatrix}=0$ $\lambda^{ ...
written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Answer: A: Chapter 6 .
... We have, $z=2x_{1}+x_{2}$ Maximise, $z'=-z=-2x_{1}-x_{2}-0s_{2}-0s_{3}-MA_{1}-MA_{2}$ -----------------(1) Subject to, $3x_{1}+x_{2}+0s_{2}+0s_{3}+A_{1}+0A_{2}=3$ ---------------(2) $4x_{1}+3x_{2}-s_{2}+0s_{3}+0A_{1}+A_{2}=6$ ---------------(3) $x_{1}+2x_{2}+0s_{2}+0s_{3}+0A_{1}+0A_{2}=3$ ------ ...
written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Answer: A: Chapter 5 .
... **Solution:** | Color | Male | Female | Total | |:-----:|:----:|:------:|:-----:| | Red | 10 | 40 | 50 | | White | 70 | 30 | 100 | | Green | 30 | 20 | 50 | | Total | 110 | 90 | 200 | **Step 1:** Null Hypothesis $(H_{0}) \Longrightarrow$ There is no relationship b ...
written 10 days ago by tanya.tanyabarnwal ♦♦ 0
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Answer: A: Chapter 2 .
... **Solution:** $A=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$ $|A|=32$ Characteristic Equation $|A- \lambda I|=0$ $\begin{vmatrix} 6-\lambda & -2 & 2 \\ -2 & 3-\lambda & -1 \\ 2 & -1 & 3-\lambda \end{vmatrix}=0$ $\lambda^ ...
written 10 days ago by tanya.tanyabarnwal ♦♦ 0

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