H(x)=4x-2; find h(-9)
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Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.
Step-by-step explanation:
Since we have given that
Integers between 10000 and 99999 = 99999-10000+1=90000
n( divisible by 3) =
n( divisible by 5) =
n( divisible by 7) =
n( divisible by 3 and 5) = n(3∩5)=
n( divisible by 5 and 7) = n(5∩7) =
n( divisible by 3 and 7) = n(3∩7) =
n( divisible by 3,5 and 7) = n(3∩5∩7) =
As we know the formula,
n(3∪5∪7)=n(3)+n(5)+n(7)-n(3∩5)-n(5∩7)-n(3∩7)+n(3∩5∩7)
Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.