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____ [38]
3 years ago
5

What is the probability that the sum is a common multiple of 2 and 3? Please help me ASAP!

Mathematics
1 answer:
Rus_ich [418]3 years ago
3 0

Answer:

what is the question

Step-by-step explanation:

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a_n=72\left(\dfrac{1}{3}\right)^{n-1}\\\\\text{Put}\ n=1,\ n=2,\ n=3,\ n=4,\ n=5\ \text{to the equation}:\\\\n=1\to a_1=72\left(\dfrac{1}{3}\right)^{1-1}=72\left(\dfrac{1}{3}\right)^0=72(1)=72\\\\n=2\to a_2=72\left(\dfrac{1}{3}\right)^{2-1}=72\left(\dfrac{1}{3}\right)^1=72\left(\dfrac{1}{3}\right)=\dfrac{72}{3}=24\\\\n=3\to a_3=72\left(\dfrac{1}{3}\right)^{3-1}=72\left(\dfrac{1}{3}\right)^2=72\left(\dfrac{1}{9}\right)=\dfrac{72}{9}=8\\\\n=4\to a_4=72\left(\dfrac{1}{3}\right)^{4-1}=72\left(\dfrac{1}{3}\right)^3=72\left(\dfrac{1}{27}\right)=\dfrac{72}{27}=\dfrac{8}{3}\\\\n=5\to a_5=72\left(\dfrac{1}{3}\right)^{5-1}=72\left(\dfrac{1}{3}\right)^4=72\left(\dfrac{1}{81}\right)=\dfrac{72}{81}=\dfrac{8}{9}\\\\Answer:\ \boxed{72,\ 24,\ 8,\ \dfrac{8}{3},\ \dfrac{8}{9}}

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