Final answer:
To determine the value of a whole number a such that .3/a, a/8, and 63% are ordered from least to greatest, we must convert all values to decimals and solve for a. The solution requires that a be greater than 5.04, and thus the smallest whole number fulfilling this condition is 6.
Step-by-step explanation:
To find a whole number a such that the numbers .3/a, a/8, and 63% are in order from least to greatest, we have to first convert all the numbers to the same format, which is usually decimals, so that they can be easily compared.
First, we convert 63% to a decimal by dividing by 100: 63% = 0.63.
Now, let's look at the two remaining expressions involving a:
For the numbers to be in order, we must have .3/a < 0.63 < a/8. Looking at the first inequality, .3/a < 0.63, multiplying both sides by 'a' yields .3 < 0.63a, which, upon dividing by 0.63, gives us a > approximately 0.476. So a needs to be larger than 0.476.
Next, examining the second inequality, 0.63 < a/8, multiplying both sides by 8 yields 5.04 < a. So a needs to be larger than 5.04. Since we are looking for a whole number, the smallest whole number larger than 5.04 is 6.
Therefore, the smallest whole number a that satisfies the conditions is 6 because .3/6 = 0.05, 6/8 = 0.75, and these are in the correct order with 63% = 0.63 which gives us 0.05 < 0.63 < 0.75.