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aev [14]
3 years ago
13

Since YouTube first became available to the public in mid-2005, the rate at which video has been uploaded to the site can be app

roximated by v(t) = 525,600(0.42t2+2.7t_1) hours of video per year (0.5 _ t _ 7),
where t is time in years since the start of 2005.Ê Use a definite integral to estimate the total number of hours of video uploaded from the start of 2006 to the start of 2012. (Round your answer to the nearest million hours of video.)
million hours of vie
Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
7 0
\int\limits^7_1 {525,600(0.42t^2+2.7t-1)} \, dx \\=525,600 \left \{\frac{0.42t^3}{3}+\frac{2.7t^2}{2}-t \right.  \left \} {{7} \atop {1}} \right. \\=525,600\{[0.14(343)+1.35(49)-7]-[0.14+1.35-1]\}\\=525,600(107.17-0.49)\\=525,600(106.68)\\=56,071,008
Total number of video uploaded is 56 million
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Answer: We do not reject the null hypothesis.

Step-by-step explanation:

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Given : Test statistic : z = -2.28

Significance level : \alpha=0.02

By using the standard normal distribution table ,

The p-value corresponds to the given test statistic ( two tailed ):-

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nata0808 [166]

Answer:

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=-4,\:y=3

Step-by-step explanation:

Given the system of the equations

x+3y=5;\:x=-6y+14

solving by elimination method

\begin{bmatrix}x+3y=5\\ x=-6y+14\end{bmatrix}

\mathrm{Arrange\:equation\:variables\:for\:elimination}

\begin{bmatrix}x+3y=5\\ x+6y=14\end{bmatrix}

x+6y=14

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\underline{x+3y=5}

3y=9

\begin{bmatrix}x+3y=5\\ 3y=9\end{bmatrix}

solve 3y=9 for y:

3y=9

\frac{3y}{3}=\frac{9}{3}

y=3

\mathrm{For\:}x+3y=5\mathrm{\:plug\:in\:}y=3

Solve x+3\cdot \:3=5 for x:

x+3\cdot \:3=5

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x=-4

Therefore,

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=-4,\:y=3

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