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alexgriva [62]
3 years ago
9

How many different breakfasts can you have at the local diner if you can select 3 different egg dishes​ (scrambled, fried,​ poac

hed), 4 choices of meat​ (steak, ham,​ bacon, sausage), 5 breads​ (white, wheat,​ rye, muffin,​ bagel), 6 juices​ (tomato, grape,​ apple, orange, mixed​ berry, vegetable​ cocktail), and 4 beverages​ (water, milk,​ coffee, tea)?
Mathematics
1 answer:
faltersainse [42]3 years ago
6 0

To find the total number of combinations, multiply all the choices together:

3 eggs x 4 meats x 5 breads x 6 juices x 4 beverages:

3 x 4 x 5 x 6 x 4 = 1,440 different breakfasts

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A country's population in 1994 was 182 million. In 2002 it was 186 million. Estimate the population in 2004 using the exponentia
iogann1982 [59]
\bf =ae^{kt}\qquad 
\begin{cases}
1994\impliedby \textit{year 0, starting point}\\
t=0\qquad P=182
\end{cases}\implies 182=ae^{k0}
\\\\\\
182=a\cdot e^0\implies 182=a\cdot 1\implies 182=a
\\\\\\
thus\qquad P=182e^{kt}\\\\
-------------------------------\\\\

\bf P=182e^{kt}\qquad 
\begin{cases}
2002\impliedby \textit{8 years later}\\
t=8\qquad P=186
\end{cases}\implies 186=182e^{k8}
\\\\\\
\cfrac{186}{182}=e^{8k}\implies ln\left( \frac{93}{91} \right)=ln(e^{8k})\implies ln\left( \frac{93}{91} \right)=8k
\\\\\\
\cfrac{ln\left( \frac{93}{91} \right)}{8}=k\implies 0.0027\approx k\implies \boxed{P=182e^{0.0027t}}

what's the population in 2004?  well,  from 1994 to 2004 is 10 years later, so t = 10

plug that in, to get P for 2004
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