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It is necessary to provide state of an art digital KIOSK for Rural India where citizens can register for Birth/Death Certificates, Insurance premium payments, Postal Schemes such as Investments, Money
... Transfer etc. The application will be easy and multilingual to be configured in Local Language. Provide suitable Analysis and Interface design for the same. **Mumbai University > Computer Engineering > Sem 8 > Human Machine Interation** **Marks:** 10 M **Difficulty:** Medium ...
mumbai university written 2 days ago by Ankit Pandey ♦♦ 10
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In the state Maharashtra, Water Distribution Company want to provide self-help portal for its customers.
... The portal consists of online meter logging facility, Bill Payments, VDS i.e Voluntary Deposit Scheme for Bill. Complaints and other facilities. Being a Subject Matter Experts (SME) provide the detailed analysis and for the same provide the Interface that will used by people in all Districts of Maha ...
mumbai university written 2 days ago by Ankit Pandey ♦♦ 10
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How reading is important in UI Design? Write your comments related to the quote "Poor Design may affect Reading".
... **Mumbai University > Computer Engineering > Sem 8 > Human Machine Interation** **Marks:** 10 M **Difficulty:** Medium ...
mumbai university written 2 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Evaluate $\int\int\int x^2 \ y \ z \ dx \ dy \ dz$ over the volume bounded by th
... Solution: Let $v=\iiint x^{2} \ d x \ d y \ d z$ Region of integration is volume bounded by the planes $x=0, y=0, z=0$ And $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ Put $x=a u, y=b v, z=c w$ $\therefore \quad \ d x \ d y \ d z=a \ b \ c \ d u . d v$ ![enter image description here][1] The inter ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Evaluate $\int_0^{\frac{a}{\sqrt2}} \int_0^{\sqrt{a^2-y^2}} log (x^2 + y^2) dx \
... Solution: let $\quad I=\int_{0}^{\frac{a}{\sqrt{2}}} \int_{y}^{\sqrt{a^{2}-y^{2}}} \log \left(x^{2}+y^{2}\right) d x \ d y$ $\begin{aligned} \text { Region of integration: } & y \leq x \leq \sqrt{a^{2}-y^{2}} \\ & 0 \leq y \leq \frac{a}{\sqrt{2}} \end{aligned}$ The line $x=y$ is inclined ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Using Beta function evaluate $\int^{\frac{\pi}{6}}_0 cos^6 3\theta \ sin^2 6 \th
... Solution: let $\quad I=\int_{0}^{\pi / 6} \cos ^{6} 3 \theta \cdot \sin ^{2} 6 \theta d \theta$ Put $\quad 3 \theta=t$ Diff. w.r.t $\theta$ $\quad d \theta=\frac{d t}{3} \quad$ limits: $\left[0, \frac{\pi}{2}\right]$ $\begin{aligned} \therefore \mathrm{I} &=\frac{1}{3} \int_{0}^{\pi / 2} \c ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Compute the value of
... Solution: let $I=\int_{0.2}^{1.4}\left(\sin x-\ln x+e^{x}\right) d x$ Dividing limits in six subintervals. $\quad \therefore \mathrm{n}=6 \quad \therefore \mathrm{h}=\frac{b-a}{n}=\frac{1.4-0.2}{6}=\frac{1}{5}$ $\begin{array}{|c|c|c|c|c|c|c|}\hline x_{0=0.2} & {x_{1}=0.4} & {x_{2}=0.6} & ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Use Taylors series method to find a solution of $\frac{dy}{dx} = 1 + x y, \ y (0
... **(1)** $\begin{aligned} \frac{d y}{d x} &=x y+1 \quad, x_{0}=0, y_{0}=0, \mathrm{h}=0.1 \\ f(x, y) &=1+x y \end{aligned}$ $\begin{array}{ll}{y^{\prime}=1+x y} & {y_{0}^{\prime}=1} \\ {y^{\prime \prime}=x y^{\prime}+y} & {y_{0}^{\prime \prime}=0} \\ {y^{\prime \prime \prime}=x y^{\ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Apply Runge Kutta method of fourth order to find an appropriate value of y when
... Solution: **(1)** $\frac{d y}{d x}=\frac{y-x}{y+x} \quad x_{0}=0, y_{0}=1, h=0.2$ $f(x, y)=\frac{y-x}{y+x}$ $k_{1}=h . f\left(x_{0}, y_{0}\right)=0.2 f(0,1)=0.2$ $k_{2}=h . f\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{1}}{2}\right)=0.2 . f(0.1,1.1)=0.1666$ $k_{3}=h . f\left(x_{0}+\frac{h}{2}, y_{ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Prove for an asteroid $x ^{2/3} + y ^{2/3} = a^{2/3}$, the line $\theta = \frac{
... Solution: Given curve: astroid $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$ The line $\theta=\pi / 6$ cuts the asroid in 1 st quadrant. ![enter image description here][1] C is the point on the curve which cuts the arc. Length of astroid in first quadrant: Put $\quad x=a \cos ^{3} t \quad$ and $\quad y=a \s ...
written 3 days ago by Ankit Pandey ♦♦ 10