#1.) x=7
#3.)on the right its the top and bottom
#4.)keep adding 3 then for the 3rd you multiply <span><span><span /></span></span>
Step-by-step explanation:
the first one is 2 that appears most while the second is 20 and 3
The answer is 9/11.
<span>The addition rule is used to calculate the probability of one of the events from multiple pathways. If you want that only one of the events happens, you will use the addition rule. So, we need to calculate the probability of two different events - the one is that the object is green, and the other is that the object is a cube.
</span>1. T<span>he probability that the object is green:
There are 22 objects in total (7 blue cubes + 4 blue spheres + 5 green cubes + 6 green spheres = 22 objects).
There are 11 green objects (</span><span>5 green cubes + 6 green spheres = 11 green objects).
So, there are </span>11 green objects out of 22 objects. Therefore, t<span>he probability that the object is green is 11/22.
</span>2. T<span>he probability that the object is the cube:
There are 22 objects in total (7 blue cubes + 4 blue spheres + 5 green cubes + 6 green spheres = 22 objects).
There are 12 cube objects (7 blue cubes + </span><span>5 green cubes).
So, there are </span>12 cube objects out of 22 objects. Therefore, t<span>he probability that the object is green is 12/22.
Now, using the addition rule, we could calculate </span>the probability that the object <span>is green or a cube:
11/22 + 12/22 = 23/22
But, we counted 5 green cubes in both events and need to subtract them:
23/22 - 5/22 = 18/22 = 9/11
</span>
The easiest way is a successive division by 4:
1st division: 2032/4 = 508 and 0 (0 is the 1st Remaining)
2nd division 508/4 = 127 and 0 (0 is the 2nd Remaining)
3rd division 127/4 = 31 and 3 (3 is the 3rd Remaining)
4th division 31/4 = 7 and 3 (3, is the 4th Remaining)
5th division 7/4 = 1 and 3 (3 is the 5th Remaining)
Now you write the last quotient (the 5th) , that is 1 and add to it all remaining starting from the 5th , backward to the 1st remaining,
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