Find the number of possible ways one can choose marbles from a bag containing three marbles: a) if the order is important b) if the order is not important.

There are 3 marbles, that we assume are different.

Let's define them as marble 1, marble 2, and marble 3.

a) If the order is important, and we select the 3 marbles, then:

for the first marble, we have 3 options.

For the second marble, we have 2 options (because one is already taken)

for the third marble is only one option.

The total number of combinations is equal to the product between the numbers of options, so in this case, we have:

C = 3*2*1 = 6 combinations.

Now, if we only select 2 marbles from the bag, we have:

for the first marble, we have 3 options.

For the second marble, we have 2 options

Here the number of different combinations is:

C = 3*2 = 6 combinations.

And if we select only one marble from the bag, we have:

for the first marble, we have 3 options.

Then here are only 3 combinations.

the total number of combinations is:

C' = 6 + 6 + 3 = 15 different combinations.

b) If the order is not important, then there is only one combination when we draw the 3 marbles.

When we draw one marble there are 3 combinations (one for each marble)

When we draw two marbles, again there are 3 combinations (one for each marble that we do not draw)

then the total number of combinations in this case is:

C = 1 + 3 + 3 = 7

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.