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ladessa [460]
2 years ago
7

What's the sum of the infinite geometric series where a1 = 240, and the common ratio is r = 1∕3 ?

Mathematics
1 answer:
ollegr [7]2 years ago
6 0

I think its B because it is always B

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Chris earns a fixed amount of money every month. At the end of 2 years, Chris had earned $86,688. He earned $ each year. He earn
Sophie [7]

Answer: He earned 43,344 a year and earned and earned 3612 a month

6 0
3 years ago
A minor-league hockey team had the following scores for the start of the season: 3, 6, 2, 1, 2, 3, 0, 4, 5, 1, 5, 4.
ale4655 [162]

If the hockey team scores 9 goals in their next game, it means that; the mean would be the most affected.

<h3>How to find mean, median and mode?</h3>

We are given the data set as;

3, 6, 2, 1, 2, 3, 0, 4, 5, 1, 5, 4.

Let us rearrange in ascending order to get;

0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6.

The mean = (0 + 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6)/12 = 3

Median = (3 + 3)/2 = 3

Mode = 1, 2, 3, 3, 5

Now, if they score 9 goals in their next game, it means that;

Mean = 45/13 = 3.46

Median = 3

Mode remains the same

Thus, mean will be the most affected.

Read more about Mean, Median and Mode at; brainly.com/question/14532771

#SPJ1

4 0
1 year ago
What is the opposite of -5.
frosja888 [35]
The answer should be positive 5 or 5
5 0
2 years ago
Roger served 5/8 pound of crackers, which was 2/3 of the entire box. what was the weight of the crackers originally in the box
ira [324]
5/8 = 2/3 the entire box

Half of 2/3 is 1/3, and 2/3 + 1/3 = 1

So find half of 5/8 and add it to 5/8.

5/8 * 1/2 = 5/16

5/8 + 5/16

Change 5/8 to a fraction with a denominator of 16:

10/16 + 5/16 = 15/16

So the original weight is 15/16 pounds.
4 0
2 years ago
Select True or False for each statement.
SSSSS [86.1K]

\left( \dfrac 1 {64} \right)^{- 5/6} =64^{5/6} = (\sqrt[6]{64})^5 = 2^5 =32

TRUE

\sqrt[5]{36^4}=36^{4/5}

which surely isn't 36.  FALSE

\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0

FALSE

For the fourth one we have a

\sqrt{98b} + \sqrt{2b}

which isnt

10\sqrt{b}

so this is FALSE.

\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}

= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}

=(\sqrt 5 + \sqrt 6)^2

No fractions in that one so FALSE.

3 0
2 years ago
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