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MrRa [10]
4 years ago
14

A person must enter a 4 digit code to gain access to his cell phone. He will enter codes until he is successful, however he cann

ot try more than 3 times or the phone will lock him out. Let S denote a successful attempt and F denote a failed attempt. What is the sample space for this random experiment?
A) {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
B) {S, FS, FFS}
C) {S, FS, FFS, FFF}
D) {S, SS, SSS}
E) {S, SF, SSF, SSS}
Mathematics
1 answer:
madreJ [45]4 years ago
7 0

Answer:

C) {S, FS, FFS, FFF}

Step-by-step explanation:

The person entering the code needs a success to access the cell phone or the person stops trying when he has failed at all 3 attempts. The possible outcomes are;

(1) The person is successful at first attempt {S}

(2) The person failed at first attempt, successful at second attempt {FS}

(3) The person failed at first and second attempts, successful at third attempt {FFS}

(4) The person failed at the three attempts {FFF}

The sample space is {S, FS, FFS, FFF}  

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At an interest rate of 8% compounded annually, how long will it take to double the following investments?
Paladinen [302]
Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,

recall your logarithm rules for an exponential,

\bf \textit{Logarithm of exponentials}\\\\
log_{{  a}}\left( x^{{  b}} \right)\implies {{  b}}\cdot  log_{{  a}}(x)\\\\
-------------------------------\\\\
\qquad \textit{Compound Interest Earned Amount}
\\\\


\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$100\\
P=\textit{original amount deposited}\to &\$50\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t
\\\\\\
\cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\


\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------\\\\


now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$1000\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t
\\\\\\


\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------

now, for the last, Principal is 1700, amount is then 3400,

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$3400\\
P=\textit{original amount deposited}\to &\$1700\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}

\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t
\\\\\\
\cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t
8 0
4 years ago
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