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Whitepunk [10]
3 years ago
15

The scale on a map reads 2 inches = 5 miles. What is the actual distance between two cities on the map that are 6.5 inches apart

?
Mathematics
2 answers:
Eva8 [605]3 years ago
5 0

Answer

16.25 miles  is your answer

Step-by-step explanation:

so for every 2 inces is 5 millesyou divide 6.5 by two then multiplie you answer by 5.

tell me if  

Temka [501]3 years ago
5 0

Answer: 16.25 miles

Step-by-step explanation:

2 inches=5 miles

6.5 inches=

6.5*5=32.5

32.5/2=16.25

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Find the missing angle according to the Triangle Exterior Angle Theorem
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Answer:

x=128°

Step-by-step explanation:

90+38=128

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The length of a rectangle is 24 units. Can the perimeter x of the rectangle be 60 units when its width y is 11 units? (1 point)
zlopas [31]
The first choice is correct. No, the rectangle cannot have x=60 and y=11 because x=48+2y.

Perimeter is x=2w+2y. Therefore x=2(24)+2(11) which equals 70 not 60.
7 0
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The sum of two numbers is 24 and their difference is 2. What are the numbers?
Musya8 [376]
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8 0
3 years ago
Read 2 more answers
if Michael had four oranges and Timmy had 5 oranges and they combined them together how many oranges do they have in total
bija089 [108]
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4 0
3 years ago
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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}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}

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

\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \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) 
\\ = P(z \ \textgreater \  1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\  \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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