Answer: There are 17576000 ways to generate different license plates.
Step-by-step explanation:
Since we have given that
Numbers are given = 0 to 9 = 10 numbers
Number of letters = 26
We need to generate the license plate numbers.
Since there are repetition allowed.
We would use "Fundamental theorem of counting".
So, the number of different license numbers may be created as given as
Hence, there are 17576000 ways to generate different license plates.
There are a variety of methods of performing this subtraction, and corresponding different methods of regrouping.
If you're taught (as I was) to do the subtraction right-to-left, then the first regrouping you need to do is when you try to subtract 500 from 100. You must regroup the 8 thousands and 1 hundred to 7 thousands and 11 hundreds. Then, when you subtract 5 hundreds, you end with 7 thousands and 6 hundreds.
The next regrouping you need to do is when you try to subtract 6 ten-thousands from 5 ten-thousands. You must regroup the 2 hundred-thousands and 5 ten-thousands to 1 hundred-thousand and 15 ten-thousands. Then, when you subtract 6 ten-thousands, you end with 1 hundred-thousand and 9 ten-thousands.
The end result is
... 258,197 - 64,500 = 193,697
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If you use an abacus or soroban or similar tool to help you keep track of the numbers, you were likely taught to do the subtraction left-to-right. In this case, the first regrouping comes when you want to subtract 6 ten-thousands. Practitioners of this method know that -6 = -10 +4, so the number represented on the tool becomes (2-1) hundred-thousands and (5+4) ten-thousands plus the rest of the initial number, or 198,197 after subtracting the 6 ten-thousands.
The subtraction proceeds until you find you need to subtract 500 from 100. At this point, the tool is representing the partial result as 194,197. Again, if you practice this method, you know that -5 = -10 +5, so you reduce the thousands digit by 1 (to 3) and add 5 to the hundreds digit to get 193,697.
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An attempt is made to show the regroupings in the attachment. In each case there are two of them. However, working left-to-right, the result of the first subtraction of 6 ten-thousands is 19 ten-thousands, so you never actually write down anything else. Of course, if you're using an abacus or soroban, you don't write down anything—you simply change the position of the beads on the tool.
Answer:
For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.
Step-by-step explanation:
Assuming the following series:
We want to approximate the value for the series with less than 0.0001 of error.
First we need to ensure that the series converges. If we have a series where
[/tex] or
where
for all n if we satisfy the two conditions given:
1)
2) {} is a decreasing sequence
Then is convergent. For this case we have that:
And because
and
So then we satisfy both conditions and then the series converges. Now in order to find the approximation with the error required we can write the first terms for the series like this:
For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.