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pshichka [43]
3 years ago
10

A popular online music store allowed shoppers to listen to an entire album before buying it. Of the people who listened to the w

hole album, the store tracked what fraction of the album each person actually bought. The line plot displays the fraction of the album each shopper bought. What fraction of the shoppers bought less than of the album?
Mathematics
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

IDK this stuff is confusing

Step-by-step explanation:

Plz like and follow

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99.00000000000 x 9/11
julia-pushkina [17]

Step-by-step explanation:

=99*9/11

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_____

=891/11

=81

3 0
3 years ago
Factor the following expressions
goblinko [34]

Answer:

i hope <em>this </em><em>will</em><em> help</em><em> </em>you

7 0
2 years ago
Please help me with this please 2
kondaur [170]
Put this into a scientific calculator
1/3 x pi x 18 squared x 16
8 0
3 years ago
Mary earns $5 per hour as a waitress. Last week she took home her regular earnings of $5 per hour plus $182 in tips for a total
Greeley [361]
The answer would be 24
4 0
3 years ago
Read 2 more answers
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
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