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3 years ago

Find log_{3} [27*\sqrt[4]{3} ] wihout using a calculator. Please show all work!

1 answer:

4 0

$\bf \textit{logarithm of factors}
\\\\
log_a(xy)\implies log_a(x)+log_a(y)\\\\
-------------------------------\\\\
log_3(27\sqrt[3]{3})\qquad 
\begin{cases}
27=3\cdot 3\cdot 3\\
\qquad 3^3\\
\sqrt[3]{3}=\sqrt[3]{3^1}\\
\qquad 3^{\frac{1}{3}}
\end{cases}\implies log_3\left( 3^3\cdot 3^{\frac{1}{3}} \right)
\\\\\\
log_3\left( 3^{3+\frac{1}{3}} \right)\implies log_3\left( 3^{\frac{4}{3}} \right)\implies \cfrac{4}{3}\cdot log_3(3)\implies \cfrac{4}{3}\cdot 1\implies \cfrac{4}{3}$

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2 years ago

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Misha Larkins [42]

Answer:

a. $P(E_1/A)=0.0789$

b. $P(E_2/A)=0.395$ \

c. $P(E_3/A)=0.526$

Step-by-step explanation:

Let $E_1,E_2,E_3$ are the events that denotes the good drive, medium drive and poor risk driver.

$P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20$

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a.We have to find the probability Mr.Brophy is a good driver

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We have to find $P(E_1/A)$

Using the Bayes theorem

$P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

Substitute the values then we get

$P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}$

$P(E_1/A)=0.0789$

b.We have to find the probability Mr.Brophy is a medium driver

$P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395$

c.We have to find the probability Mr.Brophy is a poor driver

$P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526$

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