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natka813 [3]
2 years ago
5

1 slope y-intercept equation o 6 5 4 3 2 2 3 5 j-intercept​

Mathematics
1 answer:
mars1129 [50]2 years ago
3 0

Answer:

Step-by-step explanation:

(-1,1) and (0,-2) are points on the line

slope of line = (-2-1)/(0-(-1)) = -3

y-intercept = -2

y = -3x - 2

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pashok25 [27]
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3 years ago
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muminat

Answer:

y = 3x + 3

Step-by-step explanation:

when you check, only that equation delivers the correct x and y pairs.

x = 0, => y = 3×0 + 3 = 3

x = 1, y = 3×1 + 3 = 3+3 = 6

x = 2, y = 3×2 + 3 = 6 + 3 = 9

x = 3, y = 3×3 + 3 = 9 + 3 = 12

it all fits.

6 0
2 years ago
Please help need answers now !
Degger [83]

Answer:

(2x -4) (Find your answer) then add your answers and divide it by 180

8 0
3 years ago
Which is the correct classification for the triangle?
kondaur [170]
Acute means that has 3 angles less than 90 degrees

obtuse means at lesat 1 angle is more than 90 degrees

right triangle means 1 angle=90 degrees

s we solve for thierd angle
all angles add to 180 so
35+45+x=180
80+x=180
subtract 80
x=100
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obtuse angle is answer
4 0
3 years ago
Read 2 more answers
Six different​ second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic re
ArbitrLikvidat [17]

Answer:

Range = 149-131=18

s^2 =\frac{(131-139.5)^2 +(137-139.5)^2 +(138-139.5)^2 +(141-139.5)^2 +(141-139.5)^2 +(149-139.5)^2}{6-1}=35.1s =\sqrt{35.1}=5.9

B. ​Ideally, the standard deviation would be zero because all the measurements should be the same.

Step-by-step explanation:

For this case we have the following data:

131 137 138 141 141 149

For this case the range is defined as Range = Max-Min

And for our case we have Range = 149-131=18

First we need to calculate the average given by this formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}=\frac{837}{6}=139.5

We can calculate the sample variance with the following formula:

s^2 = \frac{\sum_{i=1}^n (X_i -\bar x)^2}{n-1}

And if we replace we got:

s^2 =\frac{(131-139.5)^2 +(137-139.5)^2 +(138-139.5)^2 +(141-139.5)^2 +(141-139.5)^2 +(149-139.5)^2}{6-1}=35.1

And the standard deviation is just the square root of the variance so then we got:

s =\sqrt{35.1}=5.9

If the​ subject's blood pressure remains constant and the medical students correctly apply the same measurement​ technique, what should be the value of the standard​ deviation?

For this case the variance and deviation should be 0 since we not evidence change then we not have variation. And for this case the best answer is:

B. ​Ideally, the standard deviation would be zero because all the measurements should be the same.

8 0
3 years ago
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