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frosja888 [35]
3 years ago
8

at the district competition for the 5th grade jump-rope of the year, The following number of jumps without stopping recorded per

student: 203, 245, 237, 233, 90, 100, 100, 277, 265, 264, 265, 285, 288, 291, 291, 290, 300, 224, 200, 257, 290, 279, 266, 288. what is the mean number of jumps, rounded to the nearest hundredth? what is the median number of jumps?
Mathematics
1 answer:
julia-pushkina [17]3 years ago
8 0

Based on the given information, the answer for the following questions are:
1) What is the mean? 242.833
2) What is the median of the data? 265


Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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Que es un variable? (De matemáticas)
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Answer:

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✐ En matemáticas, una variable es un símbolo que funciona como marcador de posición para expresiones o cantidades que pueden variar o cambiar; se utiliza a menudo para representar el argumento de una función o un elemento arbitrario de un conjunto. Además de los números, las variables se utilizan comúnmente para representar vectores, matrices y funciones.

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3 years ago
The graph of a line passes through the points (0, 5) and (-10, 0). What is the equation of the line?
natta225 [31]
First find the slope:
y-y/x-x
5-0/0-(-10
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cross multiply.
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5 0
3 years ago
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Find the supplement of the angle.<br> A=133 degrees.
andrey2020 [161]
The supplement is 47
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4 years ago
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The rate at which the temperature is dropping is 6t 5t2 degrees Celsius per hour t hours after sundown. How much had the tempera
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Using a differential equation, it is found that the temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

<h3>What is the differential equation that describes the temperature in t hours after sundown?</h3>

The rate at which the temperature is dropping is 6t + 5t^2 degrees Celsius per hour t hours after sundown, hence the <em>differential equation</em> is:

\frac{dT}{dt} = 6t + 5t^2

Applying <em>separation of variables</em>, we find the solution as follows:

\frac{dT}{dt} = 6t + 5t^2

dT = (6t + 5t^2) dt

\int dT = \int (6t + 5t^2) dt

T(t) = \frac{5t^3}{3} + 3t^2 + T(0)

In which T(0) is the temperature at sundown.

In 3 hours, the change will be of:

C = T(3) - T(0) = \frac{5t^3}{3} + 3t^2|_{t = 3}

Hence:

\frac{5(3)^3}{3} + 3(3)^2 = 45 + 27 = 72

The temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

You can learn more about differential equations at brainly.com/question/24348029

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