Bissoy.com
Doctors Medicines MCQs Q&A

The job of...... Is to translate the the array of dots into text.

The job of...... Is to translate the the array of dots into text. Correct Answer OCR

OCR (Optical Character Recognition) শুধু দাগই বোঝে না, বিভিন্ন বর্ণের পার্থক্যও বুঝতে পারে। ওসিআর সফটওয়্যার দিয়ে হাতে লেখা টাইপ করা ও ছাপার হরফের লেখাকে যন্ত্র পাঠযোগ্য লেখায় রূপান্তর করা যায়।

Related Questions

Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$
Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$      with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$   and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$   is
The error in $${\left. {\frac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)} \right|_{{\text{x}} = {{\text{x}}_0}}}$$   for a continuous function estimated with h = 0.03 using the central difference formula $${\left. {\frac{{\text{d}}}{{{\text{dx}}}}{\text{f}}\left( {\text{x}} \right)} \right|_{{\text{x}} = {{\text{x}}_0}}} = \frac{{{\text{f}}\left( {{{\text{x}}_0} + {\text{h}}} \right) - {\text{f}}\left( {{{\text{x}}_0} - {\text{h}}} \right)}}{{2{\text{h}}}},$$        is 2 × 10-3. The values of x0 and f(x0) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately
Consider a system governed by the following equations:
$$\frac{{{\text{d}}{{\text{x}}_1}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_2}\left( {\text{t}} \right) - {{\text{x}}_1}\left( {\text{t}} \right);\,\frac{{{\text{d}}{{\text{x}}_2}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_1}\left( {\text{t}} \right) - {{\text{x}}_2}\left( {\text{t}} \right)$$
The initial conditions are such that $${{\text{x}}_1}\left( 0 \right)
The rate of the heterogenous catalytic reaction A(g) + B(g) → C(g) is given by $$ - {{\text{r}}_{\text{A}}} = \frac{{{\text{k}}.{{\text{K}}_{\text{A}}}.{{\text{p}}_{\text{A}}}.{{\text{P}}_{\text{B}}}}}{{1 + {{\text{K}}_{\text{A}}}.{{\text{P}}_{\text{A}}} + {{\text{K}}_{\text{C}}}.{{\text{p}}_{\text{C}}}}},$$      where KA and KC are the adsorption equilibrium constants. The rate controlling step for this reaction is
Let $$\nabla \cdot \left( {{\text{f}}\overrightarrow {\text{v}} } \right) = {{\text{x}}^2}{\text{y}} + {{\text{y}}^2}{\text{z}} + {{\text{z}}^2}{\text{x}},$$      where f and v are scalar and vector fields respectively. If $$\overrightarrow {\text{v}} = {\text{y}}\overrightarrow {\text{i}} + {\text{z}}\overrightarrow {\text{j}} + {\text{x}}\overrightarrow {\text{k}} ,$$     then $$\overrightarrow {\text{v}} \cdot \nabla {\text{f}}$$   is
The general solution of the differential equation, $$\frac{{{{\text{d}}^4}{\text{y}}}}{{{\text{d}}{{\text{x}}^4}}} - 2\frac{{{{\text{d}}^3}{\text{y}}}}{{{\text{d}}{{\text{x}}^3}}} + 2\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = 0$$       is
The solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = 0$$   with boundary conditions
$${\text{i}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 0;\,{\text{ii}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 1{\text{ is}}$$
If {x} is a continuous, real valued random variable defined over the interval (-$$\infty $$, +$$\infty $$) and its occurrence is defined by the density function given as:
$${\text{f}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}$$     where 'a' and 'b' are the statistical attributes of the random variable {x}. The value of the integral $$\int_{ - \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$
Let $$\frac{{\text{a}}}{{\text{b}}}: - \frac{{\text{b}}}{{\text{a}}} = {\text{x}}:{\text{y}}{\text{.}}$$    If $$\left( {{\text{x - y}}} \right) = $$  $$\left\{ {\frac{{\text{a}}}{{\text{b}}}{\text{ + }}\frac{{\text{b}}}{{\text{a}}}} \right\}{\text{,}}$$   then x is equal to -