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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First, let's create the ratios of sweetened to unsweetened. To do this, you would have to subtract the number of people that preferred unsweetened from the total.
Westside mall: 15:30 or

Eastside mall: 13:26 or

Now, you would divide to both ratios.
15 ÷ 30 = 0.5
13 ÷ 26 = 0.5
They have the same sum, meaning,
they are the same.
Westside Mall shoppers are just as likely to prefer unsweetened tea as Eastside Mall shoppers. I hope this helps!
Answer:
x=10.8
Step-by-step explanation:
9²+6²=c²
81+36=c²
117=c²
√117=10.81
x=10.81
Answer:
-9
Step-by-step explanation:
Answer:
The area of the triangle is 7.5 unit²
Step-by-step explanation:
Here we have the coordinates given as
(-3, 5) (-3, 8) and (2, 5)
Let us call the points A = (-3, 5)
B = (-3, 8) and
C = (2, 5)
Therefore the lengths of the sides of the triangle are
AB = a = 
AC = b = 
BC = c = 
Therefore the Area can be derived from Heron's formula which is
A =
where s = semi perimeter or (a + b + c)/2
Therefore, plugging the values, we get
s = (3 + 5 + √34)/2 = 6.9155≈6.92
A =
= 7.499≈7.5.