The sum of all the even integers between 99 and 301 is 20200
To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.
In this case, the first even integer will be 100 and the last even integer will be 300.
nth term of the AP = first term + (n-1) x common difference
⇒ 300 = 100 + (n-1) x 2
Therefore, n = (200 + 2 )/2 = 101
That is, there are 101 even integers between 99 and 301.
Sum of the 'n' terms in an AP = n/2 ( first term + last term)
= 101/2 (300+100)
= 20200
Thus sum of all the even integers between 99 and 301 = 20200
Learn more about arithmetic progressions at brainly.com/question/24592110
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An operation that is done to a value and it stays the same, it is called the "identity property".
Some quick examples:
Adding any number with zero will result in the same number.
Multiplying any number with one will result in the same number.
Hope this helps!
This is one of those problems where you don't actually have to know anything. The only answer that is any combination of the two given numbers is the last one:
.. 8.4672*10¹³ miles
_____
That result is the product of 5.88*10¹² and 14.4.
9% is less than 0.4. 0.4 is really 4/10 which is 40% 40 is greater than 9 therefore 9% is less than 0.4! So your answer is 9%<0.4! I hope this helped!
Answer: 2%, second option is correct.
Step-by-step explanation:
To state 1/50 in percent, divide 1 by 50, then multiply by 100
=( 1 ÷ 50) x 100
= 0.02 x 100
= 2%
I hope this helps, please mark as brainliest.
