Ask question

Ask question

All categories

2 years ago

9

Using two six-sided number cubes, each labeled with the numbers 1 through 6, event A is defined by rolling a sum greater than 8W

hich of the following shows the sample space of event A?

1 answer:

lara [203]2 years ago

3 0

Answer:{(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}

Step-by-step explanation:

You might be interested in

Plz help i have been asking this question too much

inessss [21]

Answer:

6

Step-by-step explanation:

1/6 of 36 is 6

2/3 of 36 is 24

6+24=30

36-30=6

<3

8 0

3 years ago

Read 2 more answers

What is the slope of the line represented by the equation y = 4/5x-3?

dsp73

Answer:

C

Step-by-step explanation:

option C.

When in the form $y= mx +q$ the slope of a line is given by the number multiplying x

3 0

3 years ago

Read 2 more answers

Please answer! Thank you! :)

Aleonysh [2.5K]

Answer:

i think is n = 4. Sorry if wrong.

Step-by-step explanation:

5 0

3 years ago

Keith bought 8 new baseball trading cards to add to his collection. The next day his dog ate half of his collection. There are n

Juliette [100K]

Answer:

56

Step-by-step explanation:

5 0

3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago