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m_a_m_a [10]
3 years ago
8

Please help me with this both question i will give you brainliest

Mathematics
2 answers:
NARA [144]3 years ago
8 0

Answer:

I got 40, but the picture cuts off so I don’t know if there’s another number.

Step-by-step explanation:

Just do 8 times 5 to get 40!

Hope this helped!

Inessa [10]3 years ago
3 0

\huge\red{\boxed{\tt{\colorbox{pink}{ANSWER:}}}}

40

just do 5 × 8

i remember this lesson when i was grade 4:)

<h3>#❀CARRY ON LEARNING❀</h3><h3>#❀KEEP ON LEARNING❀</h3><h3>#❀LEARNING IS FUN❀</h3><h3>#❀LETS LEARN❀</h3><h3>#❀NoToCopyPaste❀</h3><h3 /><h3>❀hope it helps(◕ᴗ◕✿)❀</h3><h3>❀Keep safe!❀</h3>

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2(x + 7) + 3x = 12<br> What is the first step in solving this equation for x?
yawa3891 [41]
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3 years ago
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Calls to a customer service center last on average 2.8 minutes with a standard deviation of 1.4 minutes. An operator in the call
Natali5045456 [20]

Answer:

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n-values of normal variable:

Suppose we have n values from a normally distributed variable. The mean of the sum of all the instances is M = n\mu and the standard deviation is s = \sigma\sqrt{n}

Calls to a customer service center last on average 2.8 minutes.

This means that \mu = 2.8

75 calls each day.

This means that n = 75

What is the expected total amount of time in minutes the operator will spend on the calls each day

This is M, so:

M = n\mu = 75*2.8 = 210

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

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