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grin007 [14]
3 years ago
9

If 100C2 has a value of 4,950, what is the value of 100C98? A. 4,950 B. 99,000 C. 49,500 D. 9,900

Mathematics
2 answers:
diamong [38]3 years ago
4 0
<span>If 100C2 has a value of 4,950, what is the value of 100C98?
10C2 is the number of groups of 2 that can be formed from 100 objects.
Note: Each time a group of 2 is selected you automatically form
a second group of 98. Therefore: 10C2 = 10C98 = 4950 is your Answer</span>
Diano4ka-milaya [45]3 years ago
3 0
100C2 is a count of the number of sets of two you can drawfrom a set of 100 items. But every time you draw a set of 2 from 100 you haveautomatically determined a set of 98 that you did not select. so the answer is A. 4950
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Answer:

The equation of the line is: y = 0.6x + 0.6

Step-by-step explanation:

Equation of a line:

The equation of a line has the following format:

y = mx + b

In which m is the slope and b is the y-intercept.

Two points:

We have these following two points in this exercise:

x = -6, y = -3, so (-6,-3)

x = 4, y = 3, so (4,3)

Finding the slope:

Given two points, the slope is given by the change in y divided by the change in x.

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So

m = \frac{6}{10} = 0.6

Then

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3 = 0.6*4 + b

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The equation of the line is: y = 0.6x + 0.6

4 0
3 years ago
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The 9th and the 12th term of an arithmetic progression are 50 and 65 respectively. find the common difference
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<h3>How to estimate the common difference of an arithmetic progression?</h3>

let the nth term be named x, and the value of the term y, then there exists a function y = ax + b this formula exists also utilized for straight lines.

We just require a and b. we already got two data points. we can just plug the known x/y pairs into the formula

The 9th and the 12th term of an arithmetic progression exist at 50 and 65 respectively.

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a + 11d = 65 ...............(2)

subtract them, (2) - (1), we get

3d = 15

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a + 8 * 5 = 50

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Therefore, the first term is 10 and the common difference is 2.

To learn more about common differences refer to:

brainly.com/question/1486233

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4 0
2 years ago
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