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andrew-mc [135]
3 years ago
9

Evaluate the arithmetic series described:

Mathematics
1 answer:
Art [367]3 years ago
8 0

Answer:

Step-by-step explanation:

a 1296

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The population in a small city increased from 1250 people to 2780 people. What is the percent of increase in the number of peopl
Zarrin [17]
<h3>Answer: Choice B) 122%</h3>

==========================================

Work Shown:

A = 1250 is the initial amount

B = 2780 is the new amount

C = percent change = unknown for now

The formula to use is

C = [ (B-A)/A ] * 100

to calculate the percent change.

Basically we calculate the change (B-A) first and then divide that over the original amount A to compute the decimal form of the answer, which is then converted over to percentage form. The "100" tacked on at the end is what converts from decimal to percent form.

---------

C = [ (B-A)/A ] * 100

C = [ (2780 - 1250)/1250 ] * 100

C = (1530/1250)*100

C = 1.224 * 100

C = 122.4%

C = 122%

---------

The positive C value indicates we have a percent increase. If C were negative, then we'd have a percent decrease.

6 0
3 years ago
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The product of 4 and y​
lisabon 2012 [21]

Answer:

4y

Step-by-step explanation:

product means multiply

4y

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3 years ago
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3.
KIM [24]

I'm I'm I'm I'm I'd didishbddnbdx

3 0
3 years ago
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Lostsunrise [7]

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A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at
valina [46]

Answer:

P(Same)=\frac{61}{190}

Step-by-step explanation:

Given

Red = 5

White = 6

Black = 9

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

Total = 5 + 6 + 9

Total = 20

This is calculated as:

P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)

So, we have:

P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1}  + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1}  + \frac{9}{20} * \frac{9- 1}{20- 1}

P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19}  + \frac{9}{20} * \frac{8}{19}

P(Same)=\frac{20}{380} + \frac{30}{380}  + \frac{72}{380}

P(Same)=\frac{20+30+72}{380}

P(Same)=\frac{122}{380}

P(Same)=\frac{61}{190}

4 0
3 years ago
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