What goes into 9 12 6 3 and 4
1 answer:
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Answer:
Probability of N Divisible by 3 - 0.33
Probability of N Divisible by 5 - 0.2
Probability of N Divisible by 7 - 0.413
Probability of N Divisible by 15 - 0.066
Probability of N Divisible by 105 - 0.0095
Step-by-step explanation:
Given data:
Integer N {1,2,.....10^3}
Thus total number of ways by which 1000 is divisible by 3 i.e. 1000/3 = 333.3
Probability of N divisible by 3 {N%3 = 0 }
total number of ways by which 1000 is divisible by 5 i.e. 1000/5 = 200
Probability of N divisible by 5 {N%5 = 0 }
total number of ways by which 1000 is divisible by 7 i.e. 1000/7 = 142.857
Probability of N divisible by 7 {N%7 = 0 }
total number of ways by which 1000 is divisible by 15 i.e. 1000/15 = 66.667
Probability of N divisible by 15 {N%15 = 0 }
total number of ways by which 1000 is divisible by 105 i.e. 1000/105 = 9.52
Probability of N divisible by 105 {N%105 = 0 }
similarly for N is selected from 1,2.....(10)^k where K is large then the N value. Therefore effect of k will remain same as previous part.
You have two equations.
since the second is already isolated, sub in x-4 for every y in equation 1 so that
![x^{2} - 4 [(x-4)^{2}] =16 ](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20-%204%20%5B%28x-4%29%5E%7B2%7D%5D%20%3D16%0A%20)
expand, collect like terms, factor to find x, then plug x value back into original equation to find y
since the second is already isolated, sub in x-4 for every y in equation 1 so that
expand, collect like terms, factor to find x, then plug x value back into original equation to find y