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3 years ago

In the park, there are 25 pigeons, 15 squirrels, 5 rabbits, and 5 stray cats. Why

1 answer:

8 0

Answer: Line graph should be used to show how one variable changes over time not to show multiple categories or variables are at one specific point in time.

Explanation:

In maths, statistics, and related fields, graphs are used to visually display variables and their values. In the case of line graphs, these are mainly used to display evolution or change of a variable over time. For example, a line graph can show how the number of divorces changed from 1920 to 2010.

In this context, the number of different animals in the park cannot be represented through a line graph because this situation does not imply a variable changing over time. Moreover, this situation includes multiple variables or categories of animals and the data shows only one specific point in time, which can be better represented through a bar graph.

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The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a)

Lunna [17]

Answer:

The number of radians turned by the wheel in 2s is $\theta= 8\ radians$

The angular acceleration is $\alpha =14 rad/s^2$

Explanation:

The angular velocity is given as

$w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3$

Now generally the integral of angular velocity gives angular displacement

So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

This is mathematically evaluated as

$\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt$

$= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.$

$= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0$

$= 4 +4$

$\theta= 8\ radians$

Now generally the derivative of angular velocity gives angular acceleration

So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

This is mathematically evaluated as

$\frac{dw}{dt} = \alpha (t) = 2 + 3t^2$

so at t=2

$\alpha (2) = 2 +3(2)^2$

$\alpha =14 rad/s^2$

7 0

3 years ago

An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v

Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

$\ (x_{+ve}) = \frac{1}{2} \times 15 \times 20 \\\\$

$= \frac{1}{2} \times 300 \\\\= 150 \\\\$

Calculating the value of negative displacement upon the time t:

$(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\$

$= \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\$

$\to X= X_{+ve} + X_{-ve} \\\\$

$\to 150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25$

That's why its value lie in "between 20 s and 30 s".

6 0

4 years ago

Question points)

Marysya12 [62]

At number 5 (choice E).

6 0

3 years ago

Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

KIM [24]

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

The first distance is $d_1 = 4.0 \ km = 4000 \ m$

The first speed is $v_1 = 5.0 \ m/s$

The second distance is $d_2 = 1.0 \ km = 1000 \ m$

The second speed is $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is

$t_1 = \frac{d_1 }{v_1 }$

$t_1 = \frac{4000}{5}$

$t_1 = 800 \ s$

The time taken for second distance is

$t_1 = \frac{d_2 }{v_2 }$

$t_1 = \frac{1000}{4}$

$t_1 = 250 \ s$

The total time is mathematically represented as

$t = t_1 + t_2$

=> $t = 800 + 250$

=> $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

$v = \frac{d_1 + d_2}{t }$

=> $v = \frac{4000 + 1000}{1050 }$

=> $v = 4.76 \ m/s$

7 0

4 years ago

Tres personas, A, B, C, jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección ho

Yuri [45]

Answer:

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

Explanation:

Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:

$F_{A} + F_{B} + F_{C} = 0$ (1)

Si sabemos que $F_{A} = -3$ y $F_{B} = 5$ , entonces el valor de la fuerza que debe ejercer la persona C debe ser:

$F_{C} = -F_{A}-F_{B}$

$F_{C} = -(-3)-5$

$F_{C} = -2$

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

3 0

3 years ago