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2 years ago

How much bigger is the Sum of first 50 even numbers than the sum of first 50 odd numbers?

Mathematics

1 answer:

Lunna [17]2 years ago

7 0

Answer:

Step-by-step explanation:

Sum Even numbers

n = 50

d = 2

a1 = 2

The last number is

an = a1 + (n-1)d

an = 2 + (50 - 1)*2

an = 2 + 49 * 2

an = 2 + 98

an = 100

Sum of the even numbers

Sum = (a1 + a50)*n/ 2

Sum = (2 + 100)*50/2

sum = 102 * 25

sum = 2550

Sum of the first 50 odd numbers

a1 = 1

n = 50

d = 2

l = ?

Find l

l = a1 + (n - 1)*2

l = 1 + 49*2

l = 99

Sum

Sum = (1 + 99)*50/2

Sum = 2500

The difference and answer is 2550 - 2500 = 50

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Answer:

$\bar X = 100.4$

And we can calculate the deviations from each value like this:

$|101.5-100.4 |=1.1$

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And the mean absolute deviation would be:

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Step-by-step explanation:

For this case we have the following dataset given:

101.5 98.7 95.4 92.3 109.8 104.7

We can calculate the mean with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X = 100.4$

And we can calculate the deviations from each value like this:

$|101.5-100.4 |=1.1$

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$|95.4-100.4 |=5.0$

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Answer:

130° is the correct answer

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∴∠QOR=180°−50°

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