Which type of investment return can change the most from day to day?

If p is inversely proportional to the square of q, and p is 18 when q is

aniked [119]

Answer:

P=648

Step-by-step explanation:

6 0

3 years ago

More is coming but help help help #10 points

vovangra [49]

Answer:

Its the last one

Step-by-step explanation:

The angle R is supposed to be the letter in the middle which is known as the vertex in an angle,It leads to the angle in degrees forming a proper angle.

It not much information because its hard to explain but,hope this helps!!!.

6 0

3 years ago

Please help and explain how you solved it. Thanks.

galina1969 [7]

Answer:

This is 0.14 to the nearest hundredth

Step-by-step explanation:

Firstly we list the parameters;

Drive to school = 40

Take the bus = 50

Walk = 10

Sophomore = 30

Junior = 35

Senior = 35

Total number of students in sample is 100

Let W be the event that a student walked to school

So P(w) = 10/100 = 0.1

Let S be the event that a student is a senior

P(S) = 35/100 = 0.35

The probability we want to calculate can be said to be;

Probability that a student walked to school given that he is a senior

This can be represented and calculated as follows;

P( w| s) = P( w n s) / P(s)

w n s is the probability that a student walked to school and he is a senior

We need to know the number of seniors who walked to school

From the table, this is 5/100 = 0.05

So the Conditional probability is as follows;

P(W | S ) = 0.05/0.35 = 0.1429

To the nearest hundredth, that is 0.14

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B9%7Dx%20-%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20-1" id="TexFormula1" title="\frac{2

egoroff_w [7]

$\large \mathfrak{Solution : }$

let's solve for x :

$\dfrac{2}{9} x - \dfrac{1}{3} = - 1$

$\dfrac{2}{9} x = - 1 + \dfrac{1}{3}$

$\dfrac{2}{9} x = \dfrac{ - 3 + 1}{3}$

$x = \dfrac{ - 2}{3} \times \dfrac{9}{2}$

$x = - 3$

I hope it helped...

7 0

3 years ago

Read 2 more answers

Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (

Softa [21]

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

The correct equation is

$y+1=\sqrt{-2y-3}$

Solve for y

squared both sides

$(y+1)^2=(-2y-3)$

$(y^2+2y+1)=(-2y-3)$

$y^2+2y+1+2y+3=0$

$y^2+4y+4=0$

$(y+2)(y+2)=0$

$y=-2$

Verify

substitute the value of y in the original expression

$-2+1=\sqrt{-2(-2)-3}$

$-1=1$ ----> is not true

therefore

There are no true solutions to the equation.

4 0

3 years ago