Answer:
252.m^5.n^5
Step-by-step explanation:
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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Answer: the correct options are b and d.
Step-by-step explanation:
Let us first define what mutually exclusive events are. If two events are mutually exclusive, it means that they cannot happen at the same time. For example, getting a head and a tail at the same time are mutually exclusive.
Considering a single spin of the spinner, the events that are mutually exclusive are
b)landing on a shaded sector and landing on a 3. This is because 3 is unshaded. You can either land on 3 or an unshaded sector at a time
d)landing on an unshaded sector and landing on a number less than 2. This is because the numbers in the unshaded sectors are greater than 2.
Need more information. Is this a triangle? Can you attach a picture?