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melisa1 [442]
1 year ago
8

USE THE FREQUENCY DISTRIBUTION TO DETERMINE THE FOLLOWING:A) THE TOTAL NUMBER OF OBSERVATIONS.B) THE WIDTH OF EACH CLASS.C) THE

MIDPOINT OF THE SECOND CLASS. D) THE MODAL CLASS.E) THE CLASS LIMITS OF THE NEXT CLASS IF AN ADDITIONAL CLASS WERE TO BE ADDEDSEE PHOTO BELOW

Mathematics
1 answer:
lutik1710 [3]1 year ago
4 0

A) The total number is the sum of all the frequencies

2+5+8+12+11+6=44

Answer: 44

B) The width is (upper limit - lower limit) / 2

(35-21) / 2 = 7

(50-36) / 2 = 7

(65-51) / 2 = 7

(80-66) / 2 = 7

(95-81) / 2 = 7

(110-96) / 2 = 7

Answer: the width is 7

C) The midpoint is the lower limit + the width

36+7=43

Answer: the midpoint is 43

D)The modal is the class with more frequency

In this case 66-80 with 12

Answer: class 66-80

E) We use the width formula

lower limit = 111

width = 7

upper = (width * 2) - lower

upper = (7 * 2) + 111

upper = 125

Answer: class 111-125

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How much does 4 5/7 kg of candy cost if 5/6 kg costs $7.50
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Answer:

It costs $42.43

Step-by-step explanation:

Lets explain how to solve the problem

- The cost of \frac{5}{6} kilogram of candy is $7.5

- We need to find the cost of 4\frac{5}{7} kilogram

We  can use the ratio to find the answer

======= Dollar   :    Kilogram

======= 7.5        :    \frac{5}{6}

=======  c           :    4\frac{5}{7}  

By using cross multiplication

∴ c × \frac{5}{6} = 7.5 × 4\frac{5}{7}

∵ 4\frac{5}{7} = \frac{4*7+5}{7}

∴ 4\frac{5}{7} = \frac{33}{7}

∴ c × \frac{5}{6} = 7.5 × \frac{33}{7}

∴ c × \frac{5}{6} = \frac{495}{14}

Divide both sides by \frac{5}{6}

∴ c = \frac{495}{14} ÷ \frac{5}{6}

Change the division sign to multiplication sign and reciprocal the

fraction after the division sign

∴ c = \frac{495}{14} × \frac{6}{5}

∴ c = \frac{297}{7} = 42.43 dollars

∵ c represents the cost of 4\frac{5}{7} kg

∴ It costs $42.43

7 0
4 years ago
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Step-by-step explanation:

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Step-by-step explanation:

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5 0
2 years ago
Read 2 more answers
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