What is the least possible sum of the digits displaying the time on a 12-hour digital clock?

1 answer:

4 0

<span>So if you think about it, this problem isn't that hard.So say it's 12:59, the total sum of those digits would be 71. if it was 1:11, then the sum would be 3. if it would be 3:47, the sum would be 50. So now, you have to find the time where it would be the least possible sum. That time would be 1:00 because 1+0+0=1.
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Answer:

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Step-by-step explanation:

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Answer:

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The table gives estimates of the world population, in millions, from 1750 to 2000. year population year population 1750 790 1900

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The poopulation exponential model is given by

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Where, P(t) is the population after year t; Po is the initial population, t is the number of years from the starting year; k is the groth constant.

Given that the population in 1750 is 790 and the population in 1800 is 970, we obtain the population exponential equation as follows:

$970=790e^{50k} \\ \\ \Rightarrow e^{50k}=1.228 \\ \\ \Rightarrow 50k=\ln{1.228}=0.2053 \\ \\ \Rightarrow k=0.0041$

Thus, the exponential equation using the 1750 and the 1800 population values is $P(t)=790e^{0.0041t}$

The population of 1900 using the 1750 and the 1800 population values is given by

$P(t)=790e^{0.041\times150} \\ \\ =790e^{0.6158}=790(1.8511) \\ \\ =1,462$

The population of 1950 using the 1750 and the 1800 population values is given by

$P(t)=790e^{0.041\times200} \\ \\ =790e^{0.821}=790(2.2729) \\ \\ =1,796$

From the table, it can be seen that the actual figure is greater than the exponential model values.

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