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e-lub [12.9K]
2 years ago
12

Liam had \$250$250dollar sign, 250. Then, he and his classmates bought a present for their teacher, evenly splitting the \$p$pdo

llar sign, p cost among the 242424 of them.
How much money does Liam have left?
Write your answer as an expression.
The answer is: 250-p/24
Mathematics
1 answer:
Nostrana [21]2 years ago
3 0

Answer: 250- 9/24

p

​

250, minus, start fraction, p, divided by, 24, end fraction dollars left.

Step-by-step explanation:

Liam had \$250$250dollar sign, 250. Then, he and his classmates bought a present for their teacher, evenly splitting the \$p$pdollar sign, p cost among the 242424 of them.                                                                                                                  

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The diagram shows several planes, lines, and points. Vertical plane P intersects horizontal plane R at line f. Vertical plane T
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Answer:

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a basketball player made 27 free throws in her last 45 tries. what is the experimental probability that she will make her next f
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2 years ago
Riley and Jace work at a dry cleaners ironing shirts. Riley can iron 25 shirts per hour, and Jace can iron 30 shirts per hour. R
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Answer:

  • 25r +30j = 460
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Step-by-step explanation:

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  25r +30j = 460

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1 year ago
A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
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Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

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for n = 3

 a_{3} = 3

for n = 4

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Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

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here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

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265+353=410<br>124+312=721<br>410+204=1676<br>132+230=1021<br>211+101=?????​
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