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Gala2k [10]
3 years ago
12

The number of turtles in a pond is four more than twice the number of crocodiles. Which statement makes this comparison using th

e correct variable expression? If there are c crocodiles, then there are 4c + 2 turtles. If there are c crocodiles, then there are c + 2 + 4 ⁢ turtles. If there are c crocodiles, then there are 2c + 4 ⁢turtles. If there are c crocodiles, then there are c2+4 turtles.
Mathematics
2 answers:
vovangra [49]3 years ago
8 0
There are 2c+4 turtles
umka2103 [35]3 years ago
8 0
It would be 2c+4 turtles
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Answer:

a = 1

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Step-by-step explanation:

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Write a word problem for the equation. <br> 1/2 (x-4) = 1/3x + 2
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Answer:

Half the age of Carlos four years ago is 2 years more than one-third his current age.

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We want to write a word problem for the equation.

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3 years ago
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77julia77 [94]

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4 0
3 years ago
Read 2 more answers
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sergij07 [2.7K]
Answer: 25,600
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7 0
2 years ago
The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds
Bumek [7]
Answer: 11.5% 

Explanation:


Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds. 

Now, we get the z-score for 720 seconds by the following formula:

\text{z-score} =  \frac{x - \mu}{\sigma}

where 

t = \text{time for the auditor to finish his work } = 720 \text{ seconds}&#10;\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}&#10;\\ \sigma = \text{standard deviation } = 100 \text{ seconds}

So, the z-score of 720 seconds is given by:

\text{z-score} = \frac{x - \mu}{\sigma}&#10;\\&#10;\\ \text{z-score} = \frac{720 - 600}{100}&#10;\\&#10;\\ \boxed{\text{z-score} = 1.2}

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

P(t \ \textgreater \  720) &#10;\\ = P(z \ \textgreater \  1.2)&#10;\\ = 1 - P(z \leq 1.2)&#10;\\ = 1 - 0.885&#10;\\  \boxed{P(t \ \textgreater \  720)  = 0.115}

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice. 
8 0
3 years ago
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