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den301095 [7]
2 years ago
14

I need serious help grades go in tonighttt

Mathematics
1 answer:
attashe74 [19]2 years ago
8 0
Rewrite it from least to greatest and set it up like shown

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Blanca runs 8 laps around the track each day to train for an endurance race. She times each lap to practice her pacing for the r
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Answer:

A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds.  4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds

Step-by-step explanation:

The given table is presented as follows;

\begin{array}{ccc}Day \ 1 \ lap \ times \ (seconds)&Day \ 2 \ lap \ times \ (seconds)&Day \ 3 \ lap \ times \ (seconds)\\83&87&85\\92&90&86\\91&92&91\\89&91&93\\94&92&91\\93&95&89\\88&90&88\\84&85&84\end{array}The number of laps in the range 82 to 84 seconds = 1

The number of laps in the range 84 to 86 seconds = 4

The number of laps in the range 86 to 88 seconds = 2

The number of laps in the range 88 to 90 seconds = 4

The number of laps in the range 90 to 92 seconds = 6

The number of laps in the range 92 to 94 seconds = 5

The number of laps in the range 94 to 96 seconds = 2

The number of laps in the range 96 to 98 seconds = 0

Therefore, the histogram that represents Blanca's lap times for the three days of practice is described as follows;

A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds.  4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds

5 0
2 years ago
Can I get help with finding the Fourier cosine series of F(x) = x - x^2
trapecia [35]
Assuming you want the cosine series expansion over an arbitrary symmetric interval [-L,L], L\neq0, the cosine series is given by

f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx

You have

a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)
a_0=-\dfrac{2L^2}3

a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}

and so

a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}

So the cosine series for f(x) periodic over an interval [-L,L] is

f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx
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Help plsssssss someone
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The answer would be C.
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