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Ok this is the deal, I work at a place where we have indoor units and outdoor units, so outside we are 88.2% occupied and 50% occupied on the inside units, what is the percentage occupied for the whole facility indoor and outdoor together?

2006-06-08 05:40:39 · 15 answers · asked by Anonymous in Education & Reference Homework Help

There are a total of 155 units 121 inside and 34 outside

2006-06-08 06:09:02 · update #1

15 answers

You can take the average of the two (88.2 + 50) / 2 and get 69.1. Given the information presented, this is your best answer.

Unfortunately, this answer is not likely correct. If the indoor units and outdoor units are the same size, this is true. But let's say the outdoor unit has a capacity of 100 and the indoor unit has a capacity of 500. You're certainly not 69% full, as there is lots of room for growth left.

What you really need here is a weighted average. If we take my numbers, 100 and 500, determine each's weight as a percentage of the total: 100/600 and 500/600, and multiply these by the percentages:

Indoors: (500/600) * .50 = .49 weighted
Outdoors: (100/600) * .882 = .017 weighted

Add these weighted averages together and you'll get a more accurate representation of how full you are. By my numbers, still 56% capacity instead of 69%.

2006-06-08 05:59:30 · answer #1 · answered by Mantis 6 · 0 0

You would need to know how many of both indoor and outdoor units there are.
Assuming there are the same amount of unit in each, then average the two numbers; (88.2+50)/2=69.1

2006-06-08 12:45:29 · answer #2 · answered by spyguy 3 · 0 0

Hello
You would have to know the total number of indoor and out door units and then you would add the two figures and that would be the final total , to find percent you would change the percent to a decimal , for example : to change 88.2 % to a decimal = 88.2 % x 100 = 8.82 or to save you from a lot of writing a lot of " pencil work " just " mental move " the decimal point " one place to the left , = 88.2 becomes 8.82 , 50 % would become .5 ( a n ote ) - in numbers involve ( 0 ) the zere doesn't count if it to the right of the decimal point it count only if it to the left of the point . ex 0.05 and .5 are two different number s

Note in figure percents just remember the little symbol means 100

2006-06-08 13:07:41 · answer #3 · answered by Anonymous · 0 0

no no no no

You can't answer this question based on percentages alone. If there are 10 indoor units, and 500 outdoor units, then the overall percentage of occupation will be much higher than the average of 88 and 50 percent.

2006-06-08 12:48:13 · answer #4 · answered by powhound 7 · 0 0

the answer depends on the ratio of indoor units and outdoor units
e.g. indoor units/ outdoor units = 1
69.1% occupied for the whole facility
if indoor units/ outdoor units = 0.4:0.6
65.28% occupied for the whole facility

2006-06-08 12:47:38 · answer #5 · answered by bgrt 3 · 0 0

From the information given there is no way to know. If you happen to have the same number of outside and inside units then the previous answers are correct. Something tells me that is not the case, though.

2006-06-08 12:45:33 · answer #6 · answered by Oh Boy! 5 · 0 0

Impossible to answer without knowing the area of each section, not just the percentage occupied.

2006-06-08 12:47:27 · answer #7 · answered by Loz T 4 · 0 0

if you have the same number of units inside as outside you just take an average between the two.. which would be 69.1%.. but if you have more inside than outside .. you will have to account for that .. or vice versa ..

2006-06-08 12:44:03 · answer #8 · answered by SigmundS of Yew 3 · 0 0

88.2+50=138.2

138.2/200=70%

My calculator rounds so check that.

2006-06-08 12:44:08 · answer #9 · answered by goober_head_13 3 · 0 0

88.2%=0.882
50%=0.50

0.882+0.50=1.382

1.382/2=0.69

0.69=69%

So you are 69% occupied

2006-06-08 12:44:11 · answer #10 · answered by fun_southern_sweetheart 2 · 0 0

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