Can someone help me with this
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.
This is t<span>he </span>Greatest Integer Function,<span> denoted by </span>![y = [x]](https://tex.z-dn.net/?f=y%20%3D%20%5Bx%5D)
<span>. For all real numbers, x, <em>the </em></span><em>greatest integer function</em><span><em> returns the largest </em></span><em>integer</em><span><em> less than or equal to x</em>. In essence, it rounds down a real number to the nearest </span>integer<span>.
a. FALSE. Because f(4.9) will round down to 4.
b. TRUE.
c. no it's not the absolute value function. that has V-shape
d. TRUE.
e. FALSE. because for every unique x-value you don't always get a unique y value. .... f(5)=5, f(5.1)=5, f(5.76)=5, etc.</span>
a. FALSE. Because f(4.9) will round down to 4.
b. TRUE.
c. no it's not the absolute value function. that has V-shape
d. TRUE.
e. FALSE. because for every unique x-value you don't always get a unique y value. .... f(5)=5, f(5.1)=5, f(5.76)=5, etc.</span>