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dsp73
3 years ago
12

FILL IN THE BLANK

Mathematics
2 answers:
IgorLugansk [536]3 years ago
3 0
Now, we know that he's charging $12 per session to each of his students, and he has 14 students currently, so his revenue is just 14 * 12 or 168 bucks.

now, let's take a peek as the session price goes up in jumps of 2, from 12, to 14, 16, 18 and so on, as each jumps happen, the students drop by 1, from 14, to 13, to 12 and so on.

\bf \begin{array}{ccccllll}
\stackrel{increase}{x}&\stackrel{new}{price}&students&\stackrel{revenue}{y}\\
------&------&------&------\\
0&12&14&168\\
1&14&13&182\\
2&16&12&192\\
3&18&11&198\\
\boxed{4}&20&10&\boxed{200}\\
5&22&9&198\\
6&24&8&192\\
7&26&7&182\\
8&28&6&168\\
..&..&..&..
\end{array}

notice, the revenue starts off at 168, goes up up, reaches 200 bucks and then starts to drop back down.

thus, that means the U-turn or vertex of that revenue function is at 4,200, namely h = 4, and k = 200

\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad  vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
\stackrel{c(x)}{y}=a(x-\stackrel{h}{4})\stackrel{k}{+200}
\\\\\\
\textit{now, we also know that }
\begin{cases}
x=2\\
y=192
\end{cases}\implies 192=a(2-4)^2+200
\\\\\\
-8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a
\\\\\\


\bf therefore\qquad \stackrel{ c(x)}{y}=-2(x-4)^2+200\impliedby \textit{let's expand that}
\\\\\\
y=-2(x^2-8x+16)+200\implies y=-2x^2+16x-32+200
\\\\\\
\stackrel{y}{c(x)}=-2x^2+16x+168
s344n2d4d5 [400]3 years ago
3 0

Answer:

c(x)= <u>-2 </u>x2 + <u>16</u>x + <u>168</u>

Step-by-step explanation:

#PLATOFAMILY

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

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Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

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Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

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