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:


Answer: C
Step-by-step explanation:
On the big triangle, the length from the origin to point A is three blocks. On the new triangle it is only one block. And it is getting smaller, so it is a fraction.
1/3 over the y axis.
Its mean and range . Even if we remove 70 the median is the same. The mean changes also as 580/7 isnt the same as 510/6 and also the range changes as it decreases by 5. Hope it helped :)
Answer: Cost of each ball , $3.
Cost of each bat , $35 .
Total money spend , $131 .
She buys twice as many balls as bats.
Tax , $8 .
To Find :
The number of bats she buys .
Solution :
Let , x represent the number of bats she buys.
So , number of balls she buys is 2x .
Therefore , number of bats she buys is 3 .
Hence , this is the required solution .
Step-by-step explanation: