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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1×10+7×1+6×0.1+0×0.01+1×0.001 is the answer
Answer: Undefined
The x coordinates are the same, so a vertical line forms. All vertical lines have undefined slopes.
We can see it through the slope formula
m = (y2-y1)/(x2-x1)
m = (3-(-6))/(2-2)
m = (3+6)/(2-2)
m = 9/0
We cannot divide by zero, so the result is undefined.
Answer:
2/45
Step-by-step explanation:
Note this selection is done without replacement hence after each selection sample size will reduce
Given data
Samples 3 chocolate chip bars,
2 peanut butter bars,
1 lemon bar, and
4 raisin bars
Sample size S= [3+2+1+4]= 10
Probability that first selection is lemon = 1/10
Probability that second selection is raisin = 4/9
Hence the probability that the first bar Iesha selects will be lemon and the second will be raisin= 1/10*4/9
= 4/90= 2/45
<span>1. Y = 94x-39/4</span>
<span>2. Y = 10x-25</span>
<span>3. Y = 1/4x</span>
<span>4. Graph moves 8 units to the left</span>
<span>5. Y=3x+2</span><span>
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