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lukranit [14]
3 years ago
9

Customers arrat a service window according to a poisson process with an average of 0.2 per minute.

Mathematics
1 answer:
suter [353]3 years ago
3 0

Answer:

Step-by-step explanation:

Given that customers arrat a service window according to a poisson process with an average of 0.2 per minute

=  12 per hour.

a)  the probability that the time between two successive arrivals is less than 6 minutes

= For one hour more than 60/6 =10 customers arrive.

P(X>10) = 0.6528

b) Prob that there will be exactly three arrivals during a given 10-minute period

P(x=3) for average 2

= 0.1804

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The graph of g(x) = -x^2 is shifted 3 units left and 1 unit up. If this new graph is f(x), then what is the value of f(-1.5)
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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see.

to the left by 3 units, C = +3
up by 1  unit, D = 1

\bf g(x)=-x^2\\\\\\ g(x)=-1(1x+\stackrel{C}{0})^2+\stackrel{D}{0}\implies g(x)=-1(1x+3)^2+1
\\\\\\
g(x)=-(x+3)^2+1\impliedby f(x)\qquad \qquad f(-1.5)=-[(-1.5)+3]^2+1

and surely you know how much that is.
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4 years ago
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