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MariettaO [177]
3 years ago
10

Find the equation of the line that has the same slope as y = 2x – 3 and goes through the point (–1, 3).

Mathematics
1 answer:
tankabanditka [31]3 years ago
6 0

Answer:

y=2x+5

Step-by-step explanation:

y=2x+b

3=2(-1)+b

3=-2+b

+2    +2

b=5

y=2x+5

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Which statement about the slope of the line is true?
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A. Create a set of 5 points that are very close together and record the standard deviation. Next, add a sixth point that is far
defon

Answer: The addition of the new point alters the previous standard deviation greatly

Step-by-step explanation:

Let the initial five points be : 2 3 4 5 and 6. In order to calculate the standard deviation for this data, we will need to calculate the mean first.

Mean = summation of scores/number of scores.

The mean is therefore: (2+3+4+5+6)/5 = 20/5 = 4.

We'll also need the sum of the squares of the deviations of the mean from all the scores.

Since mean = 4, deviation of the mean from the score "2" = score(2) - mean (4)

For score 3, it is -1

For 4, it's 0

For 5 it's 1

For 6 it's 2.

The squares for -2, -1, 0, 1, and 2 respectively will be 4, 1 , 0, 1, 4. Summing them up we have 10 i.e (4+1+0+1+4=10).

Calculating the standard deviation, we apply the formula:

√(summation of (x - deviation of mean)^2)/N

Where N means the number of scores.

The standard deviation = √(10/5) = 1.4142

If we add another score or point that is far away from the original points, say 40, what happens to the standard deviation. Let's calculate to find out.

i.e we now have scores: 2, 3, 4, 5, 6 and 40

We calculate by undergoing same steps.

Firstly mean. The new mean = (2+3+4+5+6+40)/6 = 60/6 = 10.

The mean deviations for the scores : 2, 3, 4, 5, 6 and 40 are -8, -7, -6, -5, -4 and 30 respectively. The squares of these deviations are also 64, 49, 36, 25, 16 and 900 respectively as well. Their sum will then be 1090. i.e. (64+49+36+25+16+900 = 1090).

The new standard deviation is then=

√(1090/6)

= √181.67

= 13.478.

It's clear that the addition of a point that's far away from the original points greatly alters the size of the standard deviation as seen /witnessed in this particular instance where the standard deviation rises from 1.412 to 13.478

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