If two of these shapes are randomly chosen, one after the other without replacement, what is the probability that both chosen will be triangles?

Step-by-step explanation:

From the comments in your question; we have

Stars = 4

Triangles = 5

Circles = 3

Squares = 3

Required  

Determine the probability of both shapes being triangles

First, calculate the total

Next, calculate the probability of the first selected shape being a triangle;

P(T₁) = Number of triangles divided by total number of shapes

Next, calculate the probability of the second selected shape being a triangle;

P(T₂) = Number of triangles divided by total number of shapes

Because it's probability without replacement. the number of triangle left is 4 and the number of shapes left is 14;

So:

Hence:

Hence, the required probability is

2/21

Step-by-step explanation:

There are 4 stars, 5 triangles, 3 circles, and 3 squares

For the first drawing there are 5 triangles and a total of 15 shapes.

p(first triangle) = 5/15 = 1/3

Since there is no replacement, for the second drawing, and one triangle has been taken, now there are 4 triangles left and 14 total shapes left.

p(second triangle) = 4/14 = 2/7

p(two triangles) = p(first triangle) * p(second triangle)

p(two triangles) = 1/3 * 2/7

p(two triangles) = 2/21

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}