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aalyn [17]
3 years ago
12

There are five identical blue books, two identical green books, and three identical black books. How many different patterns can

the books be arranged on a shelf?
Mathematics
1 answer:
dsp733 years ago
8 0

Answer:

2520 patterns

Step-by-step explanation:

In 'n' 10!  ways, books can be arranged. But, there are also 5! permutation of blue books 'n1', 2! permutation of identical green books 'n2', and 3! permutation identical black books 'n3'.

Therefore, for non identical arrangements:

\frac{n!}{n1!n2!n3!}

\frac{10!}{5!2!3!} = 2520

Therefore, the books can be arranged on a shelf in 2520 patterns

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Helppp it's timed
fomenos

Answer:

D.

Step-by-step explanation:

h(x)=f(x)g(x) means multiply the expression for f to the expression for g.

That is the problem is just asking you to do (11x-5)(-2x-4).

Let's use foil.

First:  11x(-2x)=-22x^2

Outer: 11x(-4)=-44x

Inner: -5(-2x)=10x

Last: -5(-4)=20

------------------------Add together!

-22x^2-34x+20

D.

6 0
3 years ago
Seventh grade
kupik [55]

-5+1+-2+9v+-5v

-3+1+4v

-2+4v

5 0
2 years ago
a triangle has a side length of 3/4 inch and a side length of 3in what could be the length in inches of the third side of the tr
eimsori [14]
I think the answer will be 6.75
6 0
3 years ago
At a raffle, 1500 tickets are sold at $2 each. There are four prizes given for $500, $250, $150, and $75. You buy one ticket. Us
wolverine [178]

Answer:

<u>The expected value of every ticket is a loss of $ 1.35</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of tickets sold at the raffle = 1,500

Price of each ticket = $ 2

Total prizes = 1 * $ 500 + 1 * $ 250 + 1 * $ 150 + 1 * $ 75

2. Use a probability distribution table to calculate the expected value of your gain. Your expected value is_____ ?

Let's answer the question using a probability distribution for the gains, this way:

  • Probability of 1st prize of $ 498 (500 - ticket) = 1/1,500
  • Probability of 2nd prize of $ 248 (250 - ticket) = 1/1,500
  • Probability of 3rd prize of $ 148 (150 - ticket) = 1/1,500
  • Probability of 4th prize of $ 73 (75 - ticket) = 1/1,500
  • Probability of losing $ 2 (ticket) = 1,496/1,500

Now, we calculate the mean for all the tickets (winners and non-winners), this way:

Expected value = [(1,496 * -2) + (1 * 498) + (1 * 248) + ( 1 * 148) + ( 1 * 73)]/1,500

Expected value = [- 2,992 +498 + 248 + 148 + 73)/1,500

Expected value = -2,025/1,500 = - 1,35

<u>The expected value of every ticket is a loss of $ 1.35</u>

8 0
3 years ago
Factor: (a+3)^2 -a(a+3)
lutik1710 [3]

Answer:

3(a +3)

Step-by-step explanation:

(a+3)^2 -a(a+3)

FOIL

a^2 +3a+3a+9 -a(a+3)

Distribute

a^2 +3a +3a +9 -a^2 -3a

Combine like terms

3a +9

FACTOR out a+3

(a+3)( a+3 -a)

(a+3) (3)

3(a+3)

6 0
2 years ago
Read 2 more answers
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