Answer:
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Answer:
Kenny is 18 years old Andrew is 10 years old Tim is 30 years old and Cameron is 5 years old
Step-by-step explanation:
Let a be Kenny's age , b be Andrew's age , c be Tim's age , d be Cameron's age
( b < a , b<c , d<b)
Kenny is 8 years older than Andrewn = > a = b + 8 (1)
Tim is 3 times as old as Andrew => c = 3 . b (2)
Cameron is 5 years younger than Andrew => d = b-5 (3)
The combined age of the four is 63 => a+b+c+d = 63 (4)
(1),(2),(3),(4) => b+8 + b + 3b + b-5 = 63 <=> 6b + 3 = 63 <=> b = 10
=> a= 10+8 =18, c = 3 . 10 = 30, d= 5
In case of dogs the value 10 is the minimum value. So all the values lie above 10. In total there were 100 dogs.
So for dogs, we can say number of dogs above the value of 10 pound are 100.
In case of Cats, 10 lies at the position of median. Median is the central value and 50% values lie above the median value. So number of cats with weight above 10 pound is 50.
Thus, we can conclude that there were 50 more dogs than the cats with weight over 10 pounds. So option C gives the correct answer.
Hell. fire blazing. pure hell<3
Answer:
The answer to your question is 8/3 u²
Step-by-step explanation:
Data
length of the side = 2/3
Surface area = ?
Process
1.- Calculate the area of one face
Area = side x side
-Substitution
Area = 2/3 x 2/3
-Result
Area = 4/9
2.- Calculate the area of the cube (a cube has 6 faces)
Surface area = 4/9 x 6
= 24/9
-Simplification
Surface area = 8/3 u²
