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andrew11 [14]
2 years ago
15

The probability that Brian buys a sandwich is 0.1. The probability that Brian gets the bus is 0.6. Assuming the events are indep

endent, what is the probability that Brian buys a sandwich and gets the bus?
Mathematics
2 answers:
Vsevolod [243]2 years ago
7 0

Answer:

Step-by-step explanation:

omeli [17]2 years ago
5 0

Answer:

6% chance

Step-by-step explanation:

I think it would be 0.1 * 0.6 which is 0.06

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2)

a)

\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
(4x^5\cdot x^{\frac{1}{3}})+(2x^4\cdot x^{\frac{1}{3}})-(7x^3\cdot x^{\frac{1}{3}})+(3x^2\cdot x^{\frac{1}{3}})\\\\+(9x^1\cdot x^{\frac{1}{3}})-(1\cdot x^{\frac{1}{3}})
\\\\\\
4x^{5+\frac{1}{3}}+2x^{4+\frac{1}{3}}-7x^{3+\frac{1}{3}}+9x^{1+\frac{1}{3}}-x^{\frac{1}{3}}

\bf 4x^{\frac{16}{3}}+2x^{\frac{13}{3}}-7x^{\frac{10}{3}}+9x^{\frac{4}{3}}-x^{\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{16}}+2\sqrt[3]{x^{13}}-7\sqrt[3]{x^{10}}+9\sqrt[3]{x^4}-\sqrt[3]{x}

b)

\bf \cfrac{4x^5+2x^4-7x^3+3x^2+9x-1}{x^{\frac{1}{3}}}\impliedby \textit{distributing the denominator}
\\\\\\
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\\\\\\
(4x^5\cdot x^{-\frac{1}{3}})+(2x^4\cdot x^{-\frac{1}{3}})-(7x^3\cdot x^{-\frac{1}{3}})+(3x^2\cdot x^{-\frac{1}{3}})\\\\+(9x^1\cdot x^{-\frac{1}{3}})-(1\cdot x^{-\frac{1}{3}})

\bf 4x^{5-\frac{1}{3}}+2x^{4-\frac{1}{3}}-7x^{3-\frac{1}{3}}+9x^{1-\frac{1}{3}}-x^{-\frac{1}{3}}
\\\\\\
4x^{\frac{14}{3}}+2x^{\frac{11}{3}}-7x^{\frac{8}{3}}+9x^{\frac{2}{3}}-x^{-\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{14}}+2\sqrt[3]{x^{11}}-7\sqrt[3]{x^{8}}+9\sqrt[3]{x^{2}}-\frac{1}{\sqrt[3]{x}}



3)

\bf \begin{cases}
f(x)=\sqrt{x}-5x\implies &f(x)x^{\frac{1}{2}}-5x\\\\
g(x)=5x^2-2x+\sqrt[5]{x}\implies &g(x)=5x^2-2x+x^{\frac{1}{5}}
\end{cases}
\\\\\\
\textit{let's multiply the terms from f(x) by each term in g(x)}
\\\\\\
x^{\frac{1}{2}}(5x^2-2x+x^{\frac{1}{5}})\implies x^{\frac{1}{2}}5x^2-x^{\frac{1}{2}}2x+x^{\frac{1}{2}}x^{\frac{1}{5}}

\bf 5x^{\frac{1}{2}+2}-2x^{\frac{1}{2}+1}+x^{\frac{1}{2}+\frac{1}{5}}\implies \boxed{5x^{\frac{5}{2}}-2x^{\frac{3}{2}}+x^{\frac{7}{10}}}
\\\\\\
-5x(5x^2-2x+x^{\frac{1}{5}})\implies -5x5x^2-5x2x+5xx^{\frac{1}{5}}
\\\\\\
-25x^3+10x^2-5x^{1+\frac{1}{5}}\implies \boxed{-25x^3+10x^2-5x^{\frac{6}{5}}}

\bf 5\sqrt{x^5}-2\sqrt{x^3}+\sqrt[10]{x^7}-25x^3+10x^2-5\sqrt[5]{x^6}
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