Find the sample variance and standard deviation. 6, 52, 16, 51, 35, 23, 28, 28, 32, 26

Sample variance is 199.79 and standard deviation 14.13

Step-by-step explanation:

Number of values: n=10

Mean: M=(6+52+16+51+35+23+28+28+32+26)/10

M=(297)/10→M=29.7

Sample Variance: v

v=[(6-29.7)^2+(52-29.7)^2+(16-29.7)^2+(51-29.7)^2+(35-29.7)^2+(23-29.7)^2+(28-29.7)^2+(28-29.7)^2+(32-29.7)^2+(26-29.7)^2]/(10-1)

v=[(-23.7)^2+(22.3)^2+(-13.7)^2+(21.3)^2+(5.3)^2+(-6.7)^2+(-1.7)^2+(-1.7)^2+(2.3)^2+(-3.7)^2]/(9)

v=[561.69+497.29+187.69+453.69+28.09+44.89+2.89+2.89+5.29+13.69]/(9)

v=[1,798.1]/(9)

v=199.7888888

v=199.79

Standard deviation: s

s=sqrt(v)

s=sqrt(199.7888888)

s=14.13466974

s=14.13

black lol

step-by-step explanation:

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i can tell you jamal now has 3 cookies xd

tommy has 2.

this made me lol.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3