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3 years ago

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Specific data such as age, income, family size, and occupation, used for marketing purposes, is called

2 answers:

True [87]3 years ago

5 0

<span>B. demographics.

</span>Specific data such as age, income, family size, and occupation, used for marketing purposes, is called demographics.
NOT:
A. a secondary market.
C. survey statistics.
<span>D. the target market.</span>

djyliett [7]3 years ago

3 0

<h3><u>Answer;</u></h3>

B. demographics.

Specific data such as age, income, family size, and occupation, used for marketing purposes, is called <em><u>demographics.</u></em>

<h3><u>Explanation;</u></h3>

<em><u>Demographics are features of a particular population. It involves the study of a population based on factors such as age, sex, race, family size and occupation, etc.</u></em>
<em><u>For marketing purposes firms or companies use specific data such as age, income, family size, religion, race, nationality and occupation. </u></em><em><u>This may be referred to as demographic segmentation, which is used by the companies to get the right population using their products. </u></em>

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A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\

zloy xaker [14]

Answer:

t = 1.77 s

Explanation:

The equation of a traveling wave is

y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

v = λ f

Where the frequency and period are related

f = 1 / T

we substitute

v = λ / T

let's develop the initial equation

y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

2π /λ = 1.65

λ = 2π / 1.65

λ = 3.81 m

2π / T = 4.64

T = 2π / 4.64

T = 1.35 s

we seek the speed of the wave

v = 3.81 / 1.35

v = 2.82 m / s

Since this speed is constant, we use the uniformly moving ratios

v = d / t

t = d / v

t = 5 / 2.82

t = 1.77 s

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3 years ago

ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

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2 years ago

Prove(show) ''T=2π√(l/g)''

Nonamiya [84]

Answer:

Time period for Simple pendulum, $T=2\pi\sqrt{\frac{l}{g}$

Explanation:

The Simple Pendulum

Consider a small bob of mass $m$ is tied to extensible string of length $l$ that is fixed to rigid support. The bob is oscillating in the plane about verticle.

Let $\theta$ is the angle made by string with vertical during oscillation.

Vertical component of the force on bob, $F=-mg\sin\theta$

Negative sign shows that its opposing the motion of bob.

Taking $\theta$ as very small angle then, $\sin\theta\sim\theta$

$F=-mg\theta$

Let $x$ is the displacement made by bob from its mean position ,

then, $\theta=\frac{x}{l}$

so, $F=-mg\frac{x}{l}$ ........(1)

Since, pendulum is in hormonic motion,

as we know, $F=-kx$

where $k$ is the constant and $k=m\omega^{2}$

$F=-m\omega^2x$ .........(2)

From equation (1) and (2)

$-m\omega^2x=-mg\frac{x}{l}$

$\omega=\sqrt{\frac{g}{l}}$

Since, $\omega=\frac{2\pi}{T}$

$\frac{2\pi}{T}=\sqrt{\frac{g}{l}$

$T=2\pi\sqrt{\frac{l}{g}}$

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3 years ago

What part of the plant takes in carbon dioxide?

murzikaleks [220]

The answer is number 2 stomata.

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3 years ago

Read 2 more answers

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

Lana71 [14]

Answer:

a

$m = 0.169 \ kg$

b

$|v_{max} |= 0.5653 \ m/s$

Explanation:

From the question we are told that

The spring constant is $k = 14 \ N/m$

The maximum extension of the spring is $A = 6.0 \ cm = 0.06 \ m$

The number of oscillation is $n = 30$

The time taken is $t = 20 \ s$

Generally the the angular speed of this oscillations is mathematically represented as

$w = \frac{2 \pi}{T}$

where T is the period which is mathematically represented as

$T = \frac{t}{n}$

substituting values

$T = \frac{20}{30 }$

$T = 0.667 \ s$

Thus

$w = \frac{2 * 3.142 }{ 0.667}$

$w = 9.421 \ rad/s$

this angular speed can also be represented mathematically as

$w = \sqrt{\frac{k}{m} }$

=> $m =\frac{k }{w^2}$

substituting values

$m =\frac{ 15 }{(9.421)^2}$

$m = 0.169 \ kg$

In SHM (simple harmonic motion )the equation for velocity is mathematically represented as

$v = - Awsin (wt)$

The velocity is maximum when $wt = \(90^o) \ or \ 1.5708\ rad$

$v_{max} = - A* w$

=> $|v_{max} |= A* w$

=> $|v_{max} |= 0.06 * 9.421$

=> $|v_{max} |= 0.5653 \ m/s$

5 0

3 years ago