I think it is -5.9. Sorry if I'm wrong.
Answer:
Step-by-step explanation:
Given
The attached histogram
Required
The average work hours by 20 salespeople
First, we get the total hours worked as follows:
Divide by number of salespeople
21. In an election involving only two candidates, every vote will be counted against one candidate or the other. The total of votes for the two candidates must add to the total of votes cast. The numbers 60% and 1/3 = 33 1/3% add to 93 1/3%, not 100%, so cannot be right.
22. 500/2000 = 1/4 (fraction) = 0.25 (decimal) = 25%
... 1200/2000 = 3/5 (fraction) = 0.60 (decimal) = 60%
... 300/2000 = 3/20 (fraction) = 0.15 (decimal) = 15%
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The ratio of each number to 2000 gives you the fraction. You need to reduce it by removing common factors from numerator and denominator. Some calculators will do this for you. You can start by removing an equal number of zeros from numerator and denominator, for example 500/2000 = 5/20. Then you make use of your knowledge of multiplication tables to identify and remove any additional common factors. In this example, 5/20 = 5/(5·4) = 1/4.
The decimal form can be found a couple of ways. One is to simply recognize the decimal form of the fraction you have. (You have memorized that 1/4 = 0.25, for example.) Another is to multiply numerator and denominator by a number that makes the denominator be a power of 10, such as 100. For example, 1/4 = (1·25)/(4·25) = 25/100. Now, you use place value to put this fraction into decimal form. Twenty-five hundredths is 0.25.
Yet another way to find the decimal value of a fraction is to divide the numerator by the denominator using long division (or your calculator).
If you have done the decimal part by making a denominator of 100, you have already found the percentage. The "%" symbol means "/100", so 25/100 is 25%. They are two ways of writing the same thing.
Moving the decimal point two places to the right and adding a "%" sign is another way to get to a percentage from a decimal. (0.463 = 46.3%, for an example not related to this problem.)
