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notka56 [123]
3 years ago
8

I’m a group of 55 people 33 are coffee drinkers, 12 are tea drinkers. Assuming that no one in this group drinks both tea and cof

fee, what is the probability that a person piked random from the group does not drink tea or coffee
Mathematics
2 answers:
Scilla [17]3 years ago
5 0
10/55 of the people do not drink coffee or tea.
tester [92]3 years ago
4 0
33+12 = 45 drink either tea or coffee (but not both)

55 - 45 = 10 people drink neither

Divide 10 over 55 and we get 10/55 = 2/11

Use a calculator to find that 2/11 = 0.1818 approximately which converts to 18.18% when you move the decimal point over 2 spots to the right

-------------------------------------------------

Answer as a fraction = 2/11
Answer in decimal form = 0.1818 (approximate)
Answer as a percentage = 18.18% (approximate)<span />
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Anon25 [30]

\huge\text{Hey there!}

\mathsf{\dfrac{10^{15}}{10^4}}

\mathsf{10^{15}}\\\mathsf{=10\times10\times10\times10\times10\times10\times10\times10\times10\times10\times 10\times10\times10\times10\times10}\\\mathsf{= \boxed{\bf 1,000,000,000,000,000}}

\mathsf{\dfrac{\bold{1,000,000,000,000,000}}{10^4}}

\mathsf{10^4}\\\mathsf{= 10\times10\times10\times10}\\\mathsf{10\times10=\bf 100}\\\mathsf{100\times100}\\\mathsf{= \boxed{\bf 10,000}}

\mathsf{\dfrac{1,000,000,000,000,000}{\bf 10,000}}

\mathsf{\dfrac{1,000,000,000,000,000}{10,000}}\\\\\mathsf{= 1,000,000,000,000,000\div 10,000}\\\\\mathsf{= \boxed{\bf 100,000,000,000}}

\mathsf{10^{11}}\\\mathsf{10\times10\times10\times10\times10\times10\times10\times10\times10\times10\times10}\\\mathsf{= \boxed{\bf 100,000,000,000}}

\mathsf{100,000,000,000= 100,000,000,000= 10^{11}}

\boxed{\boxed{\large\text{Answer: BASICALLY }\mathsf{\dfrac{10^1^5}{10^4}\large\text{ is EQUAL to or EQUIVALENT to}}}}\\\boxed{\boxed{\mathsf{10^{11}}\large\text{ because they both give you the result of \bf 100,000,000,000}}}\huge\checkmark

\large\textsf{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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