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

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step-by-step explanation:

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)