The number of plastics of the size of 13 ft x 10ft will be 4.
<h3>What is the area?</h3>
The space occupied by any two-dimensional figure in a plane is called the area. The space occupied by the rectangle in a two-dimensional plane is called the area of the rectangle.
Given that:-
- A tennis match was delayed because of the rain. The officials were not prepared for the delay.
- They covered the 25ft by 20ft court with 13ft by 10ft plastic covers.
The number of the plastic required will be calculated as:-
When the total area of the tennis court will be divided by the area of the one plastic we will get the number of the plastic required.
Area of court = 25 x 20 = 500 square feet
Area of plastic = 13 x 10 = 130 square feet.
Number of the plastic = Area of court ÷ Area of plastic
= 500 ÷ 130
= 3.89 ≅ 4 plastics
Therefore the number of plastics of the size of 13 ft x 10ft will be 4.
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Their = 940
their and back = 1880
hope this helps
Answer:
by 28% or 280%
not completely sure
Step-by-step explanation:
Answer:
They can invite 136 while staying in budget
Step-by-step explanation:
Firstly, you know they will have to pay an $80 cleaning fee, therefore, you can subtract 80 from 5,000:
5,000-80 = 4,920
After doing this, you can take 4,920 and divide it by 36 (the price per person).
4,920/36 = 136.67
Because you can not have .67 of a person, your closest number you can use is 136.
leaving them with $24
(a) No. The probability of drawing a specific second card depends on the identity of the first card.
Two events are independent if knowledge on the first doesn't change anything about the probability distribution of the second event. This isn't the case: since you don't replace the first card, you have some extra knowledge about the second.
If, for example, you pick the 7 of hearts at the beginning, you are sure that the second card won't be the 7 of hearts. This extra piece of information depends on the first pick, and so the two pick are not independent: knowledge about the first pick changed what you expect from the second pick.
(b)
The probability of finding ace on 1st card is 4/52 = 1/13, because there are four aces (one for each suit) out of 52 cards. If this happens, you're left with 51 cards, 4 of which are kings (one for each suit). So, the probability of picking a king among the remaining cards is 4/51. The probability of the two happening one after the other is
(c)
You can use the same logic as point (b)
(d)
We have 8 "good" cards for the first pick: any of the 4 aces or any of the 4 kings. So, the first pick has a success rate of 8/52 = 2/13. For the second pick we have 4 "good cards" (the 4 aces if we picked a king, the 4 kings if we picked an aces) out of the 51 remaining cards, so we have a success rate of 4/51. So, the probability of picking an ace and a king, in any order, is