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Nastasia [14]
3 years ago
6

A company policy requires that, for every 30 employees, there must be 3 supervisors. If there are 261 supervisors at the company

, how many employees does the company have? A. 2,349 B. 1,305 C. 2,610 D. 3,132
Mathematics
1 answer:
Luda [366]3 years ago
5 0
Every 30 employees..........................................3 supervisors
          ?..............................................................261 supervisors
number of employees=(261x30)/3=2610
choice C
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Vanyuwa [196]
36:39 in simplest form is 12:13
3 0
3 years ago
Read 2 more answers
Pls help I will fail 10th grade
dsp73

Answer:

B) -6

Step-by-step explanation:

Placing a minus sign outside absolute value bars gives you a negative result For example, –|6| = –6, and –|–6| = –6.

3 0
2 years ago
D) A shopkeeper bought two radios for Rs 6,000. He sold one of them at a profit of 20℅
Orlov [11]

Answer:

2400 and 3600

Step-by-step explanation:

Radio 1 cost = x

Radio 2 cost= 6000- x

  • x*1.2+(6000-x)*0.9= 6000*1.02
  • 1.2x-0.9x+5400=6120
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8 0
3 years ago
The quotient of 365.085 and 79.8
Paul [167]

Answer:

4.575

Step-by-step explanation:

365.085/79.8=4.575

5 0
3 years ago
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distribute
balu736 [363]

Answer:

The area of the shaded region is 0.9082.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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, which is the area of the shaded region, is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 100, \sigma = 15, X = 120

Area of the shaded region:

pvalue of Z

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

Z = \frac{120 - 100}{15}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

So

The area of the shaded region is 0.9082.

4 0
3 years ago
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