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aksik [14]
4 years ago
10

15 - (-3) - 4 -16 22 -8 14

Mathematics
1 answer:
uranmaximum [27]4 years ago
8 0
In this equation, you have to treat the number in the bracket first on the basis of BODMAS
15 - [-3]- 4
Note that when two minuses come together the product is a plus sign.
15 +3 - 4
You have to add before you subract
18 - 4 =14
Therefore, 15- [-3] - 4 = 14.
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If each small cube has a volume of 1 cm,
ella [17]

Answer:

The answer is 40 Centimeters.

Step-by-step explanation:

This answer is correct since there are 4 rows of 10, creating the Rectangular prism.  i hope this helped!

6 0
3 years ago
a railroad car container can hold 42,000 pounds. Mr.Evans wants to ship 90 ovens and some freezers in the same container. If eac
NeX [460]
The amount of freezers that can be shipped is 70. I'm not completely sure, because I do not know how many pounds the ovens are. But to get the answer, just simply divide 42000 by 600.
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Answer:

Option 3. step 1.

Step-by-step explanation:

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3 years ago
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A group of market women sell at least one of yam, plantain and maize.12 of them sell maize,10 sell yam and 14 sell plantain,5 se
mina [271]

Let the three items be M, Y and P.

n{M ∩ Y} only = 4-3 = 1

n{M ∩ P) only = 5-3 = 2

n{ Y ∩ P} only = 2

n{M} only = 12-(1+3+2) = 6

n{Y} only = 10-(1+2+3) = 4

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8 0
3 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}
\\ \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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