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3 years ago

Use a calculator to find the standard deviation of this data set: 8, 14, 12, 9, 16

Mathematics

2 answers:

muminat3 years ago

6 0

Answer : 23.6

diamong [38]3 years ago

6 0

Answer:3.3

Step-by-step explanation:use Math-way on these

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1.Prove each statement. <br><br><br> step by step PLEASE!!!

Leya [2.2K]

Answer:

Step-by-step explanation:

a. sin^-1(sin theta) = theta

(1/sin)(sin theta) = theta

<em>the sines cross out</em> theta = theta

b. cos(cos^-1x) = x

cos(x/cos) = x

<em>the cosines cross out </em>x = x

5 0

3 years ago

Consider this system of linear equations:

Hatshy [7]

I believe the answer is D. because -12 + -1 equal -13

5 0

3 years ago

One number is 6 greater than another. the product of the number is 55. find the numbers

Basile [38]

Here's how to do this problem:

6+ x= 55
-6 -6
x=49

5 0

3 years ago

EMERGENCY PLEASE HELP STRUGGLING! The triangles in each pair are similar.

Ludmilka [50]

Answer: The missing length is 120 {option C}

Step-by-step explanation: What we have in the question is a triangle placed on top of another triangle and as shown in the attached diagram we can separate them into triangles BFC and GFH.

A close observation shows that line BC is parallel to line GH. Hence we have two similar triangles, and we can determine their similarity ratios as follows;

BF/FC = GF/FH

Similarly BF/BC = GF/GH

We can also express the following ratio

FC/FH = BC/GH

Therefore to calculate the missing side, which is GF, we can use the ratio

FC/FH = FB/FG

30/100 = 36/FG

By cross multiplication we now have 30 (FG) = 100 (36)

30FG = 3600

Divide both sides of the equation by 30

FG = 120

5 0

2 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

3 0

3 years ago