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2 years ago

5

Find the standard deviation {25,4,15,19,20,18,7,21,24}

1 answer:

MakcuM [25]2 years ago

5 0

hello jaguar Yu I'm under the water

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Simplify the following expressions. Write your answers as a power. Complete problems 1 through 6.

sertanlavr [38]

The expression is given to be:

$\frac{5^4\cdot5^2}{5^3}$

Recall the multiplication rule of exponent given to be:

$x^a\cdot x^b=x^{a+b}$

Therefore, we can simplify the numerator to be:

$5^4\cdot5^2=5^{4+2}=5^6$

Hence, the expression becomes:

$\Rightarrow\frac{5^6}{5^3}$

Recall the division rule of exponent given to be:

$\frac{x^a}{x^b}=x^{a-b}$

Therefore, the expression becomes:

$\frac{5^6}{5^3}=5^{6-3}=5^3$

Evaluating the answer, we have:

$5^3=125$

Therefore, the answer is:

$\frac{5^4\cdot5^2}{5^3}=125$

3 0

1 year ago

A cellular service provider charged a customer $40 for 150 minutes of airtime. The same provider charged another customer $55 fo

defon

Answer:

Y=10x-250

Step-by-step explanation:

We can treat this problem like we are attempting to graph a line from 2 points (hopefully you know how to do that)

the two points are (40,150)(55,300)

7 0

3 years ago

A band must pay a $2500 entrance fee, It will split the cost between its 25 members. How much will each member pay?

dsp73

Answer $100 each

Step-by-step explanation:

2500/25

4 0

3 years ago

Read 2 more answers

Solve the following system by graphing. If there isn't a unique solution to the system, state the reason. y=-2x y=2x+4

natka813 [3]

Answer:

We have the system of equations:

y = -2*x

y = 2*x + 4.

To solve it by graphing, you need to graph both lines and see in which point the lines intersect each other.

You can see it in the image below. (blue is y = 2*x + 4, and green is y = -2*x)

We also can calculate the solution with math, let's do that:

y = -2*x

y = 2*x + 4.

We can replace the first equation in the second one, to get:

-2*x = 2*x + 4

Now let's solve this for x

-2*x -2*x = 4

-4*x = 4

x = 4/-4 = -1

Evaluating this in one of the equations we get:

y = -2*-1 = 2

Then the solution is the point (-1, 2), exactly what we can see in the graph.

3 0

3 years ago

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position

stira [4]

Answer:

$v(t) = (2t+1) \mathbb{i} +3t^2 \mathbb{j} +4t^3 \mathbb{k}$

$r(t) = (t^2+t) \mathbb{i} +(t^3+2) \mathbb{j} +(t^4 - 3) \mathbb{k}$

Step-by-step explanation:

The velocity vector is the integral of the acceleration vector i.e.

$v(t) = \int a(t) dt$

$v(t) = \int (2 \mathbb{i}+6t \mathbb{j} 12t^2 \mathbb{k}) dt$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + C_1$

When $t=0$ , $v(0) = \mathbb{i}$ . Inserting these values in $v(t)$ ,

$C_1= \mathbb{i}$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + \mathbb{i}$

$v(t) = (2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}$

The position vector is the integral of the velocity vector i.e.

$r(t) = \int v(t) dt$

$r(t) = \int ((2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}) dt$

$r(t) = (t^2+t) \mathbb{i}+t^3 \mathbb{j} t^4 \mathbb{k} + C_2$

When $t=0$ , $r(t) =2\mathbb{j}-3\mathbb{k}$ . Inserting these values in $r(t)$ ,

$C_2=2\mathbb{j}-3\mathbb{k}$

$r(t)= (t^2+t) \mathbb{i}+t^3 \mathbb{j}+ t^4 \mathbb{k} + 2\mathbb{j}-3\mathbb{k}$

$r(t) = (t^2+t) \mathbb{i}+(t^3+2) \mathbb{j}+ (t^4-3) \mathbb{k}$

5 0

4 years ago