Each student receives one of 4 calculator models and one of 3 types of ruler. How many possible outcomes are there if a student gets one ruler and one calculator?

24 possible outcomes

Step-by-step explanation:

Combination has to do with selection. For example, if r object is selected from a pool of n objects, the number if possible ways can be expressed according to the combination formula:

nCr = n!/(n-r)!r!

Applying this in question, if each student receives one of 4 calculator models and one of 3 types of ruler, the number of ways this can be done is:

4C1 × 3C1

4C1 = 4!/(4-1)!1! {If a student gets one calculator)

4C1 = 4×3×2/3×2

4C1 = 4ways

3C1 = 3!/(3-2)!1! {If a student gets a ruler}

3C1 = 3×2/1

3C1 = 6ways

Total number of possible outcomes if a student gets one ruler and one calculator will be 4×6 = 24ways

it has to be a.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}