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raketka [301]
2 years ago
15

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.

Mathematics
1 answer:
ale4655 [162]2 years ago
4 0

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

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Answer:

See below ~

Step-by-step explanation:

<u>Perimeter</u>

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<u>Area</u>

  • 2(4 x 2) + 2(8)
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Will eight be less or more than 2/4
daser333 [38]

Step-by-step explanation:

8 is more than 2/4 because the answer of 2/4 is 0.5 so 8 is more than 2/4

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Ramona wants to buy cheese pizza for her family and friends gathering. She stop by nine wonder pizza and figures out that they h
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6 0
3 years ago
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8x plus 8y equals 8. 8x plus 2y equals 28. Find x and y
alukav5142 [94]

Answer:

x,y = 13/3, -10/3

Step-by-step explanation:

8x+8y=8

8x+2y=28  Subtract eq1 by eq 2

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8x - 80/3=8

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Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.
Lilit [14]

For starters,

\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k

Consider the nth partial sum, denoted by S_n:

S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n

Multiply both sides by \frac34:

\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}

Subtract S_n from this:

\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)

Solve for S_n:

-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)

S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)

Now as n\to\infty, the exponential term will converge to 0, since r^n\to0 if 0. This leaves us with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}

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3 years ago
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