4.15, 6.3, 8.45, 10.6, 12.75
You just subtract 4.15 from 6.3 and you get a difference of 2.15. Then you add it to 8.45 giving you the answer of 10.6 and then you add 2.15 to that getting 12.75.
Answer:
Let's define the cost of the cheaper game as X, and the cost of the pricer game as Y.
The total cost of both games is:
X + Y
We know that both games cost just above AED 80
Then:
X + Y > AED 80
From this, we want to prove that at least one of the games costed more than AED 40.
Now let's play with the possible prices of X, there are two possible cases:
X is larger than AED 40
X is equal to or smaller than AED 40.
If X is more than AED 40, then we have a game that costed more than AED 40.
If X is less than or equal to AED 40, then:
X ≥ AED 40
Now let's take the maximum value of X in this scenario, this is:
X = AED 40
Replacing this in the first inequality, we get:
X + Y > AED 80
Replacing the value of X we get:
AED 40 + Y > AED 80
Y > AED 80 - AED 40
Y > AED 40
So when X is equal or smaller than AED 40, the value of Y is larger than AED 40.
So we proven that in all the possible cases, at least one of the two games costs more than AED 40.
Answer:
C
Step-by-step explanation:
If you add all of the numbers above you will get 47 so
32+10+5=47
and of course there are 32 chocolates with orange filling
so the probability is 32/47
