Answer:
a. p(orange) = 5/14
b. p(green) = 3/14
c. p(red) = 1/7
d. p(brown) = 2/7
e. p(brown or red) = 3/7
Step-by-step explanation:
1. You have a 14 pencils. Two pencils are red, 5 pencils are orange, 3 pencils are green and 4 pencils are brown.
p(color) = (number of pencils of that color)/(total number of pencils)
p(color) = (number of pencils of that color)/14
a. If a pencil is picked at random, what is the probability that the pencil
will be orange?
p(orange) = 5/14
b. If a pencil is picked at random, what is the probability that the pencil
will be green?
p(green) = 3/14
c. If a pencil is picked at random, what is the probability that the pencil will be red?
p(red) = 2/14 = 1/7
d. If a pencil is picked at random, what is the probability that the pencil
will be brown?
p(brown) = 4/14 = 2/7
e. If a pencil is picked at random, what is the probability that the pencil
will be brown or red?
brown: 4
red: 2
brown or red: 4 + 2
p(brown or red) = 6/14 = 3/7
Step-by-step explanation:
The number of tickets sold by Dan the first day is x.
Tickets sold on the 2nd day: x+3
on the 3rd day: x+6
on the 4th day: x+9
on the 5th: x+12
Since the total of tickets sold by Dan is 40:
x+(x+3)+(x+6)+(x+9)+(x+12)=40
5x+30=40
5x=10
x=2
number of tickets sold on the third day: 8
Answer:
2.5
Step-by-step explanation:
The answer is already in its simplest form. You can not add the two because the have different variables. Hope this helps ;)))
The two expressions are equivalent because they have the same value regardless of the number of substituted in for y