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Nesterboy [21]
3 years ago
14

A superstar major league baseball player just signed a new deal that pays him a record amount of money. The star has driven in a

n average of 110 runs over the course of his career, with a standard deviation of 31 runs. An average player at his position drives in 80 runs. What is the probability the superstar bats in fewer runs than an average player next year? Assume the number of runs batted in is normally distributed.

Mathematics
1 answer:
Likurg_2 [28]3 years ago
8 0

Answer:

  about 16.7%

Step-by-step explanation:

An appropriate probability calculator can tell you the probability of being below 80 for the given normal distribution. The calculator shown below says it is about 0.167.

__

The corresponding z-value is ...

  (x -μ)/σ = (80 -110)/31 = -30/31 ≈ -0.9677

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A model for the population in a small community after t years is given by P(t)=P0e^kt.
LUCKY_DIMON [66]
\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\\
\end{cases}

a)

so, if the population doubled in 5 years, that means t = 5.  So say, if we use an amount for "i" or P in your case, to be 1, then after 5 years it'd be 2, and thus i = 1 and A = 2, let's find "r" or "k" in your equation.

\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
I=\textit{initial amount}\to &1\\
r=rate\\
t=\textit{elapsed time}\to &5\\
\end{cases}
\\\\\\
2=1\cdot e^{5r}\implies 2=e^{5r}\implies ln(2)=ln(e^{5r})\implies ln(2)=5r
\\\\\\
\boxed{\cfrac{ln(2)}{5}=r}\qquad therefore\qquad \boxed{A=e^{\frac{ln(2)}{5}\cdot t}} \\\\\\
\textit{how long to tripling?}\quad 
\begin{cases}
A=3\\
I=1
\end{cases}\implies 3=1\cdot e^{\frac{ln(2)}{5}\cdot t}

\bf 3=e^{\frac{ln(2)}{5}\cdot t}\implies ln(3)=ln\left( e^{\frac{ln(2)}{5}\cdot t} \right)\implies ln(3)=\cfrac{ln(2)}{5} t
\\\\\\
\cfrac{5ln(3)}{ln(2)}=t\implies 7.9\approx t

b)

A = 10,000, t = 3

\bf \begin{cases}
A=10000\\
t=3
\end{cases}\implies 10000=Ie^{\frac{ln(2)}{5}\cdot 3}\implies \cfrac{10000}{e^{\frac{3ln(2)}{5}}}=I
\\\\\\
6597.53955 \approx I
3 0
3 years ago
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anygoal [31]

Answer:

5/6 = 10/12 and 3/4 = 9/12

Step-by-step explanation:

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klemol [59]

Answer:

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Step-by-step explanation:

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8 0
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I need help with these questions asap
AfilCa [17]

Answer:

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Step-by-step explanation:

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Answer:

6/15

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