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just olya [345]
4 years ago
13

what factor should be considered when choosing between the median or mean as a description of the center of a data set?

Mathematics
2 answers:
fiasKO [112]4 years ago
5 0

The <em><u>correct answer</u></em> is:

You should consider whether there is an outlier.

Explanation:

In order to find the mean, we add together all of the data points and divide by the number of data points. If there is an outlier, a number much higher or much lower than the rest of the data, this will cause the sum of the data to be affected, which will cause the mean to either be higher or lower than it would without the outlier.

The median, however, is not affected much by outliers, as the median is just the middle value in a set of data.

Nataliya [291]4 years ago
3 0
<span>when choosing between the median or mean as a description of the center of a data set, you consider whether there is an outlier in the data set.</span>
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The yearly Penn rock-paper-scissors tournament has come down to the final series between Zach and Shaya. They play a best-of-thi
Zepler [3.9K]

Answer:

p=41.79%

Step-by-step explanation:

First let us remember the calculation of a probability:

p=\frac{number of favorable outcomes}{total number of possible outcomes}

In this case let's calculate the number of possibilities of winning the tournament when either one of them, Zach(Z) or Shaya(S), wins the first game (number of favorable outcomes). That is:

Given that Z or S won the first game, the player has another 6 possibilities of winning among 12 remaining games, which is a combination:

number of favorable outcomes = 12C6 = 924

Therefore, the winner of the first game has 924 opportunities of winning 6 other games among 12 games and be the tournament winner.

Now, we must find the number of total possible outcomes, which we may analize as it follows:

Assuming that the first winner Z or S loses the tournament we have several outcomes:

1. After winning the first game, the player doesn't win again, then we have 0 possbilities among 7 other games, this is because the next player would win the tournament winning 7 games in a row. We then have the combination 7C0.

2. The player wins just one more game, but doesn't win again. Now we have 1 possibility among 8 games, given that the other player would win the other 7 games. We then have the combination 8C1.

3. The player wins just two more games, but doesn't win again,  now we have 2 possibilities among 9 games, given that the other player would win the other 7 games. We then have the combination 9C2.

And so on until we reach the outcome where the player wins 5 more games, but doesn't win again,  then we have 5 possibilities among 12 games, given that the other player would win the other 7 games. We then have the combination 12C5.

Now, we must also consider all the outcomes where the first winner wins the tournament, that is the first combination we calculated: 12C6

Therefore we obtain the following:

total number of possible outcomes = 7C0+8C1+9C2+10C3+11C4+12C5+12C6

total number of possible outcomes = 2,211

We have now the elements to calculate the probability:

p=\frac{924}{2,211}=0.4179

Hence, the probability that the winner of the first game wins the tournament is 41.79%

7 0
4 years ago
Solve for x: 3(x + 1) = −2(x − 1) − 4.<br> explain if u can
shutvik [7]
First, you distribute so 
3x+3=-2x+2-4 then Combine like terms
x=-5
6 0
3 years ago
Factored form with it’s solution x^2 + 2x -3 =0
Maru [420]

Answer:

(x + 3) (x - 1

Step-by-step explanation:

<em>here's</em><em> your</em><em> solution</em>

<em> </em><em> </em><em>=</em><em>></em><em> </em>x^2 + 2x - 3

=>. By using middle term split

=> x^2 + 3x - x - 3

=> taking common

=> x( x + 3) - 1(x + 3)

=> (x + 3) (x - 1)

Hope it helps

3 0
3 years ago
1
adelina 88 [10]

Step-by-step explanation:

16 matches=100%

what about,12 matches=?

you then cross multiply

12/16×100%=75%

=75%

4 0
3 years ago
Find the missing side of the triangle. Round to the nearest tenth.
Nimfa-mama [501]
5.5 because In front of a 30 degree angle is a cathete of half of Hypotenuse.
8 0
3 years ago
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