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True [87]
3 years ago
10

Lynn plotted point G, 3 units to the left and 2 units above point F. Where did lynn plot point G?

Mathematics
1 answer:
kvasek [131]3 years ago
4 0

Answer:

G = (2,6)

Step-by-step explanation:

See Attachment for Complete Question

From the attachment, we can see that F is plotted at

F = (5,4)

On the x axis, G is 3 units to the left;

And on the y axis, G is 2 units above

So, the coordinates of G is calculated as:

G = (5-3,4+2)

G = (2,6)

Hence:

<em>G is plotted at (2,6)</em>

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Six different​ second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic re
Yanka [14]

Answer:

Range=14

\sigma^2 =32.4

\sigma = 5 .7

The standard deviation will remain unchanged.

Step-by-step explanation:

Given

Data: 136, 129, 141, 139, 138, 127

Solving (a): The range

This is calculated as:

Range = Highest - Least

Where:

Highest = 141; Least = 127

So:

Range=141-127

Range=14

Solving (b): The variance

First, we calculate the mean

\bar x = \frac{1}{n} \sum x

\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)

\bar x = \frac{1}{6} *810

\bar x = 135

The variance is calculated as:

\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2

So, we have:

\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]

\sigma^2 =\frac{1}{5}*[162]

\sigma^2 =32.4

Solving (c): The standard deviation

This is calculated as:

\sigma = \sqrt {\sigma^2 }

\sigma = \sqrt {32.4}

\sigma = 5 .7 --- approximately

Solving (d): With the stated condition, the standard deviation will remain unchanged.

5 0
3 years ago
1) Where is the midpoint of , when A (6, 7) and B (-9, -5)?
lord [1]

Answer:

(−1.5,1)

Step-by-step explanation:

Finding the distance, midpoint, slope, equation and the x y-intercepts of a line passing between the two points p1 (6,7) and p2 (-9,-5)

The distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √  ((X2-X1)2+(Y2-Y1)2)

d = √ (-9-6)2+(-5-7)2

d = √ ((-15)2+(-12)2)

d = √ (225+144)

d = √ 369

The distance between the points is 19.2093727122985

The midpoint of two points is given by the formula

Midpoint= ((X1+X2)/2,(Y1+Y2)/2)

Find the x value of the midpoint

Xm=(X1+X2)/2

Xm=(6+-9)/2=-1.5

Find the Y value of the midpoint

Ym=(Y1+Y2)/2

Ym=(7+-5)/2=1

The midpoint is: (-1.5,1)

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