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hodyreva [135]
2 years ago
11

Use the minimum and maximum data entries and the number of classes to find the class width, the lower class limits, and the uppe

r class limits. min = 14, max = 121, 8 classes
Mathematics
1 answer:
deff fn [24]2 years ago
5 0

Answer:

The  class width is  C_w  \approx 13

Step-by-step explanation:

From the question we are told that

 The  upper class limits is max  =  121

  The  lower class limits is  min  =  14

   The number of classes is  n  =  8  \ classes

The class width is mathematically represented as  

       C_w  =  \frac{max  -  min}{n  }

substituting values

       C_w  =  \frac{121 -  14}{8  }

       C_w  = 13.38

      C_w  \approx 13

Since

     

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Answer:

2(x+2)

Step-by-step explanation:

1) 2 : the greatest common factor

2) 2x+4

2x+4=2 ⋅ 2 : Separate out 2

Hope that helps

3 0
2 years ago
What steps do I need to do to get 18×8.7<br><br>23×56.1<br><br>and<br><br>47×5.92
eimsori [14]
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3 0
3 years ago
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for a recipe Peter uses 2 cups of sugar for every 3 cups of flour which table shows equivalent ratios for this table
Novosadov [1.4K]

Answer:  

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Step-by-step explanation:

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4 0
3 years ago
Scores on a college entrance exam are normally distributed with a mean of 550 and a standard deviation of 100. Find the value th
Alinara [238K]

Answer:

The value that represents the 90th percentile of scores is 678.

Step-by-step explanation:

Problems of normally distributed samples are 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 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 = 550, \sigma = 100

Find the value that represents the 90th percentile of scores.

This is the value of X when Z has a pvalue of 0.9. So X when Z = 1.28.

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

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The value that represents the 90th percentile of scores is 678.

4 0
2 years ago
Place the grouping symbols to make this equation true. 84-48/8-4=72
vlabodo [156]
84-6-4=72
48/4=12. 84-12= 72
84 - (48/(8-4)) = 84 - (48/4) -4 = 72
84 - 12 = 72
4 0
3 years ago
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