It needs to be a number multiplicated by 10 with a round power. With your number, s
you should write it like that : 3.482000000x10^9 (if your zeros are significant, otherwise 3.482x10^9) OR 3.482000000G (for Giga=1×10^9)
It is simple and easy to find out about how many number combinations there are if numbers cant repeat you simply look at which combinations that you are aiming for and find out the different type of ways that you can do to get that same equal numbers that you were hoping to find is simple like 1 + 9 = 10 as same as 5 + 5 =10 its not that hard. HOPE THIS HELPS!!!
The two consecutive odd numbers that have a product of 15 are 3 and 5
There are 100 random 2 digits numbers [00 , 99]
There are 34 divisible by 3 {00, 03, 06, 09, … 93, 96, 99}
There are 20 divisible by 5 {00, 05, 10, … 90, 95}
However we must avoid counting numbers twice so we need to subtract those divisible by 15.
There are 7 divisible by 15 {00, 15, 30, 45, 60, 75, 90}
This means there are 34 + 20 - 7 = 47 2-digit numbers that are divisible by 3 or 5.
If you pick a random 2-digit number then P(divisible by 3 or 5) = 47/100 = 0.47
Answer:
-20x+15 is your answer
Step-by-step explanation:
Solve by distributing and combing like terms
I think l and z are corresponding.<span />
