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3 years ago

(MC)

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

6 0

Answer:

Square root 112 Units.

Step-by-step explanation:

See the image attached for the work.

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Two glasses of milk and 4 snack bars have a total of 76 carbohydrates (carbs), and 4 glasses of milk and 3 snack bars have a tot

galina1969 [7]

Answer:

10 carbs in snack bars

14 carbs in milk

Step-by-step explanation:

Glasses of Milk = m

Snack Bars = s

I am going to make this a system of equations:

2m + 3s = 58

4m + 2s = 76

I am going to simplify one of them for a s:

4m + 2s = 76

2s = 76 - 4m

s = 38 - 2m

I am going to insert this into the other equation:

2m + 3s = 58

2m + 3(38 - 2m) = 58

2m + 114 - 6m = 58

(-4)m + 114 = 58

(-4)m = -56

m = 14 calories

I am going to plug this m into one of the equations (doesn't matter which):

2(14) + 3s = 58

28 + 3s = 58

3s = 30

s = 10

You can plug these values back into each of the equations to make sure they work.

Hope it helps! UvU

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3 years ago

A candle is 6 in tall after burning for 1 hour. After 3 hours it is 5 1/2 in tall. Write a linear equation to model the height y

wel

6-x=y it will take 22 hours

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Kasi is building a patio. He has 136 bricks. He wants the patio to have 8 rows, each with the same number of bricks. How many br

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4 years ago

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Suppose that 16-ounce bags of chocolate chip cookies are produced with weights that follow a Normal distribution with mean weigh

tatyana61 [14]

I think the answer is C but I’m not sure

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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