Answer:
Step-by-step explanation:
Tiles (6" by 6") can be installed at the rate of 50 per worker hour. Since the tile is a square, its area is
L^2 = 6^2 = 36 inches square
So in an hour, a worker installs 50 tiles whose area is 36 inches square. Total square meters installed per hour will be 50 × 36 = 1800 inches square per hour.
If three workers are assigned to this task, their work rate will be
1800 × 3 = 5400 inches square per hour.
The floor of a large lobby area (80’ by 40’) is to be covered with these tiles. We would convert from feet to inches.
1 foot = 12 inches
80 feet = 80 × 12 = 960 inches
40 feet = 40 × 12 = 480 inches
Area of the lobby = 90 × 480 = 460800 inches square.
If the 3 workers install 5400 inches square in 1 hour,
Then then they will install 460800 inches square in
460800/5400 = 85.33 hours
Converting to the nearest whole day, it becomes
85.33/24 = 3.55
Approximately 4 days
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
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Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
The independent variable is:
cold coffee
because: just an educated guess
sorry this is a little hard but i hope this helps
What are the opposite rays? I can’t see them
