Application layer sends & receives data for particular applications using Hyper Text Transfer Protocol (HTTP), and Simple Mail Transfer Protocol (SMTP).

Application layer sends & receives data for particular applications using Hyper Text Transfer Protocol (HTTP), and Simple Mail Transfer Protocol (SMTP). Correct Answer True

Application layer sends & receives data for particular applications using HyperText Transfer Protocol (HTTP), and Simple Mail Transfer Protocol (SMTP). Hence, data encryption for HTTP and SMTP is important.
Bissoy MCQ

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
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 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
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 $${{\text{x}}_1}\left( 0 \right)
The general solution of the differential equation, $$\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 -