step 1
compute the average: add the values and divide by 6
Average =(44+ 46+40+34+29+41)/6=39

step 2
Compute the deviations from the average
dev: (44-39)=5,
dev: (46-39)=7
dev: (40-39)=1
dev: (34-39)=-5
dev: (29-39)=-10
dev: (41-39)=2

step 3
Square the deviations and add
sum (dev^2): 5^2+7^2+1^2+-5^2+-10^2+2^2
sum (dev^2): 25+49+1+25+100+4-----> 204

step 4
Divide step #3 by the sample size=6
(typically you divide by sample size-1 to get the sample standard deviation,
but you are assuming the 6 values are the population,
so
no need to subtract 1, from the sample size.
This result is the variance
Variance =204/6=34

step 5
Standard deviation = sqrt(variance)
standard deviation= √(34)------> 5.83

the answer is
5.83

4 0

3 years ago

PLEASE HELP!!!!

Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

if "a" is positive, then opens UP
if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

the Axis Of Symmetry (x-value) is: x = $\frac{-b}{2a}$
y-value is found by plugging in the AOS for "x" in the equation

****************************************************************************************

1) y = x² + 11x + 24

a = +1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-11}{2(1)}$ = $-\frac{11}{2}$

y = $(-\frac{11}{2})^{2}$ + 11 $(-\frac{11}{2} )$ + 24 = $-\frac{25}{4}$

vertex $(-\frac{11}{2}$ , $-\frac{25}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

******************************************************************************************

2) y = -x² - 6x - 8

a = -1 so the parabola opens DOWN

x = $\frac{-b}{2a}$ = $\frac{-(-6)}{2(-1)}$ = $\frac{6}{-2}$ = -3

y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1

vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

******************************************************************************************

3) y = x² - 2x + 3

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-(-2)}{2(1)}$ = $\frac{2}{2}$ = 1

y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2

vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). There are no x-intercepts

******************************************************************************************

4) y = x² + 4x + 4

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-4}{2(1)}$ = -2

y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0

vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). The x-intercept is the vertex.

Compared to the other four graphs, this is most likely the equation for the rain gauge!

******************************************************************************************

5) y = 3x² + 21x + 30

a = +3 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-21}{2(3)}$ = $-\frac{7}{2}$

y = 3 $(-\frac{7}{2})^{2}$ + 21 $(-\frac{7}{2} )$ + 30 = $-\frac{27}{4}$

vertex $(-\frac{7}{2}$ , $-\frac{27}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

*******************************************************************************************

4 0

3 years ago

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