- How much zeros has the number $1000!$ at the end?
yes it depends on $2$ and $5$ Note that there are plenty of even numbers Also note that $25\times 4 = 100$ which gives two zeros Also note that there $125\times 8 = 1000$ gives three zeroes and $5^4 \times 2^4 = 10^4$ Each power of $5$ add one extra zero So, count the multiple of $5$ and it's power less than $1000$
- What does it mean when something says (in thousands)
I'm doing a research report, and I need to determine a companies assets So I found their annual report online, and for the assets, it says (in thousands) One of the rows is: Net sales $ 26,234
- Solution Verification: How many positive integers less than $1000$ have . . .
A positive integer less than $1000$ has a unique representation as a $3$-digit number padded with leading zeros, if needed To avoid a digit of $9$, you have $9$ choices for each of the $3$ digits, but you don't want all zeros, so the excluded set has count $9^3 - 1 = 728$ Hence the count you want is $999 - 728 = 271$
- How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7?
The totient of $210$ - the number of values between $1$ and $210$ that are relatively prime to $210$ - is $(2-1)(3-1)(5-1)(7-1)=48$
- Numbers in a list which are perfect squares and perfect cubes of . . .
Cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, and 1728 44 squares and 12 cubes Numbers with both perfect squares and cubes in common : 1, (1^2 and 1^3) 64, (8^2 and 4^3) and 729 (27^2 and 9^3)
- Why is 1 cubic meter 1000 liters? - Mathematics Stack Exchange
Can anyone explain why $1\ \mathrm{m}^3$ is $1000$ liters? I just don't get it 1 cubic meter is $1\times 1\times1$ meter A cube It has units $\mathrm{m}^3$ A liter is liquid amount measurement 1 liter of milk, 1 liter of water, etc Does that mean if I pump $1000$ liters of water they would take exactly $1$ cubic meter of space?
- Find the sum of all the multiples of 3 or 5 below 1000
First of all, stop thinking on the number $1000$ and turn your attention to the number $990$ instead If you solve the problem for $990$ you just have to add $993, 995, 996$ $999$ to it for the final answer This sum is $(a)=3983$ Count all the #s divisible by $3$: From $3$ to $990$ there are $330$ terms
- probability - 1 1000 chance of a reaction. If you do the action 1000 . . .
So for your example, it would be 1-((1–1 1000)^1000) which equals 1-(0 999^1000), which turns out to be about 0 63230457, or 63 230457% There is a lot of confusion about this topic, as intuitively, you would think that if the odds are 1 1000 playing 1000 times would guarantee a win
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