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makvit [3.9K]
4 years ago
13

Elijah earns $200 for 8 hours what is the unit rate per hour

Mathematics
1 answer:
Fiesta28 [93]4 years ago
6 0

Answer:

$25 per hour

Step-by-step explanation:

You would divide 200 by 8 and you'll get 25

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En un estudio estadistico se van a usar unicamente variables cuantitativas discretas para ingresarlas a un programa de analisis
notka56 [123]

Answer:

la variable que se debe ingresar en el programa corresponde a: c. edad

Step-by-step explanation:

Las variables cuantitativas son aquellas que toman valores numéricos y se clasifican en variables cuantitativas discretas que son las que sólo pueden asumir un número limitado de valores en un determinado rango, como por ejemplo, el número de carros que posee una persona y variables cuantitativas continuas que pueden tomar cualquier valor en un rango específico, como por ejemplo, el peso de un objeto. De acuerdo a estas definiciones, la respuesta es que la variable que se debe ingresar en el programa corresponde a: edad porque es una variable discreta dado que se registra en números enteros y no acepta cualquier valor en un intervalo específico.

Las otras opciones no son correctas porque la nacionalidad y el nivel de escolaridad no son variables cuantitativas y la altura es una variable cuantitativa continua.

5 0
3 years ago
here are approximately 3.6 million births in a country each year. Find the birth rate in units of births per minute.
Hatshy [7]

Answer:

6.8 approximately 7 births/mins

Step-by-step explanation:

3600000/525600

If we have 3600000 births in a year, and in a year we have 6omin * 24hrs*365days a year we therefore, have the equation below

Amount of births in a year / (amount of mins in a year)

= 6.8births per mins

3 0
3 years ago
Can someone help find missing side?
Anna35 [415]

Step-by-step explanation:

19 + 73 + 90 = 182

182 - 180 = 2

4 0
3 years ago
Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help
ki77a [65]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2867785

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}


Make a trigonometric substitution:

\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}


so the integral (i) becomes

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}


Now, substitute back for t = arcsin(x²), and you finally get the result:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}          ✔

________


You could also make

x² = cos t

and you would get this expression for the integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}          ✔


which is fine, because those two functions have the same derivative, as the difference between them is a constant:

\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}

\mathsf{=\dfrac{\pi}{4}}         ✔


and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.


I hope this helps. =)

6 0
3 years ago
Which equation represents an inverse variation?
Genrish500 [490]

Answer:

its y=2x so yaa thats ittt

8 0
2 years ago
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