How do you write 7231÷24
2 answers:
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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.
To solve this problem you must apply the proccedure shown below:
1. You have the following expression given in the problem above:
![\sqrt[3]{216 x^{27} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B216%20x%5E%7B27%7D%20%7D%20)
2. Rewriting the expression we have:
![\sqrt[3]{6^3 x^{27} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B6%5E3%20x%5E%7B27%7D%20%7D%20)
3. You have that
and the exponent 
are divisible by index 
. Therefore, you have:
![\sqrt[3]{216 x^{27} } =6 x^{9}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B216%20x%5E%7B27%7D%20%7D%20%3D6%20x%5E%7B9%7D%20)
Therefore, as you can see,
the answer is the option, which is:
1. You have the following expression given in the problem above:
2. Rewriting the expression we have:
3. You have that
Therefore, as you can see,
the answer is the option, which is: