Answer:
add 15 by 19 by 12 to get 46 than cut it in half to get 23!
Hope i helped please mark brainiest I need it!
Answer:
Plot -2 2/3 two dots to the left of -2. Plot 5/6 one dot to the left of 1.
Step-by-step explanation:
Answer:
(a) 
(b) 
<em>(b) is the same as (a)</em>
(c) 
(d) 
(e) 
Step-by-step explanation:
Given

Solving (a): Probability of 3 or fewer CDs
Here, we consider:

This probability is calculated as:

This gives:


Solving (b): Probability of at most 3 CDs
Here, we consider:

This probability is calculated as:

This gives:


<em>(b) is the same as (a)</em>
<em />
Solving (c): Probability of 5 or more CDs
Here, we consider:

This probability is calculated as:

This gives:


Solving (d): Probability of 1 or 2 CDs
Here, we consider:

This probability is calculated as:

This gives:


Solving (e): Probability of more than 2 CDs
Here, we consider:

This probability is calculated as:

This gives:


A because i did this 1 month ago
k12
You should ask your parent/guardian
or teacher if needed! Merry Christmas!