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Mathematics

1 answer:

andriy [413]3 years ago

8 0

In the first one divide -3 by both sides and when you get an answer plug it back in for x to check and got second one multiple bu four on both sides and you will get an answer. And put that answer on the graph.

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A building contains two elevators, one fast and one slow. The average waiting time for the slow elevator is 3 min. and the avera

cupoosta [38]

Answer: The expected waiting time is $1\dfrac{2}{3}\ min$

Step-by-step explanation:

Since we have given that

Average waiting time for slow elevator = 3 min

Average waiting time for fast elevator = 1 min

probability that a person choose the fast elevator = $\dfrac{2}{3}$

Probability that a person choose the slow elevator = $\dfrac{1}{3}$

So, the expected waiting time would be

$E[x]=\sum xp(x)=3\times \dfrac{1}{3}+1\times \dfrac{2}{3}\\\\=1+\dfrac{2}{3}\\\\=\dfrac{3+2}{3}\\\\=\dfrac{5}{3}\\\\=1\dfrac{2}{3}\ min$

Hence, the expected waiting time is $1\dfrac{2}{3}\ min$

8 0

2 years ago

The population of a town in 2007 is 113, 505

ELEN [110]

from 2007 to 2012 is only 5 years, so we can see this as a compound interest with a rate of 1.2% per annum for 5 years, so

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$113505\\ r=rate\to 1.2\%\to \frac{1.2}{100}\dotfill &0.012\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=113505\left(1+\frac{0.012}{1}\right)^{1\cdot 5}\implies A=113505(1.012)^5\implies A\approx 120481$