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Anit [1.1K]
3 years ago
11

Two points from the line of best fit can be used to find the slope of the line, which can be used to find the equation of the li

ne.
Mathematics
2 answers:
Anton [14]3 years ago
8 0
Bvng is the answer I think
matrenka [14]3 years ago
6 0

Answer:

bvng

Step-by-step explanation:

You might be interested in
McKenzie makes and sells quilts at an art fair. The materials for each quilt cos t$30.50. She sells the quilts for $80. If "n" r
Nitella [24]

Answer:

Total profits = 49.50n

Step-by-step explanation:

Cost of each quilt = $30.50

Selling price of each quilt = $80

Profits = selling price - cost price

= $80 - $30.50

= $49.50

Profits = $49.50

If n = number of quilts she sells

which expression could be used to represent her total profits?

Total profits = profits of each quilt × number of quilt sold

= $49.50 × n

= 49.50n

Total profits = 49.50n

4 0
3 years ago
I’m not sure what to do, this was what I got. Someone please help and solve problem #14 in the picture. Thanks!
mars1129 [50]

Answer:

  • x = 8
  • AM = BM = 16

Step-by-step explanation:

Your problem setup is correct. The triangle is isosceles, so the marked segments are the same length.

  3x - 8 = 2x

You can solve this by subtracting 2x (from both sides) ...

  x - 8 = 0

Then adding 8 (to both sides):

  x = 8

MB = MA = 2x = 16.

4 0
3 years ago
A computer program in Shannon’s computer carries out a single mathematical operation in 1.5 × 10-6 seconds. How much time would
kykrilka [37]

1/1.5 * 10^-6 = 2500/x

x = 1.5 * 10^-6 * 2500

x = 0.00375 or 3.75 * 10^-3 second.

3 0
3 years ago
Consider a set of one-dimensional points: 6, 12, 18, 24, 30, 42, 48.
Vinvika [58]

Answer:

1) With initial centroids 10-40, final clusters are

first cluster (6,12,18,24,30)

second cluster (42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 378

2) With initial centroids 10-20, final clusters are

first cluster (6,12,18,24)  

second cluster (30,42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 348

Step-by-step explanation:

K-means works as follows

  • the points will be clustered according to their distance to the centroids.
  • Then centroids are updated as the cluster means.
  • This process continues until clusters doesn't change anymore

1)  <u><em>initial centroids</em></u> 10-40

first cluster (6,12,18,24)  mean:15

second cluster (30,42,48) mean:40

<u><em>new centroids</em></u> 15-40

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

<u><em>final centroids</em></u> 18-45

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

Sum of Squared errors = (18-6)^{2}+(18-12)^{2}+(18-18)^{2}+(18-24)^{2}+(18-30)^{2}+(45-42)^{2}+(45-48)^{2}=378

2)<u><em>initial centroids</em></u> 10-20

first cluster (6,12)  mean:9

second cluster (18,24,30,42,48)  mean:32.4

<u><em>new centroids</em></u> 9-32.4

first cluster (6,12,18) mean:12

second cluster (24,30,42,48) mean:36

<u><em>new centroids</em></u> 12-36

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

<u><em>final centroids</em></u> 15-40

first cluster (6,12,18,24) mean:15  

second cluster (30,42,48) mean:40

Sum of Squared errors = (15-6)^{2}+(15-12)^{2}+(15-18)^{2}+(15-24)^{2}+(40-30)^{2}+(40-42)^{2}+(40-48)^{2}=348

6 0
3 years ago
In each situation, write a recurrence relation, including base case(s), for the given function. briefly explain in words why thi
vova2212 [387]
A thumb-wrestling match requires two thumbs, so we can either suppose that we need at least two people (n\ge2), or allow one to thumb-wrestle one's self. In either case, we'd have t(1)=t(2)=1, so let's just say we need a minimum of two players.

If we add one more person to the set of players, then the first two people would need to play 1 additional match each. So t(3)=t(2)+2.

If we add one more person, then the first three people would again each have to play 1 more match with the new person. So t(4)=t(3)+3.

And so on, so that in general, the number of games needed for everyone to play exactly one match with everyone else is given recursively by

\begin{cases}t(2)=1\\t(n+1)=t(n)+n&\text{for }n\ge2\end{cases}
6 0
3 years ago
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