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shutvik [7]
3 years ago
14

We know that a class interval must be greater than what the I formula solution is; however... What is the solution for class int

erval (for the formula) for a frequency distribution if the data ranges from 100 to 220 with 50 observations?
Mathematics
1 answer:
stich3 [128]3 years ago
3 0

Answer:

Therefore the answer is 20.

Step-by-step explanation:

We know that

class interval = range / number of classes

But here number of classes is not given , so we use the formula

class interval = range / ( 1+ 3.322 log N)

where , range =maximum - minimum = 220-100 = 120

N= number of observations = 50

class interval = 120 / ( 1+ 3.322 * log 50) = 18.06

Rounding up to a convinient number

Thus , class intervai = 20

Therefore the answer is 20.

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What is another way to write the expression t. (14--5) ?
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Answer: Second option.

Step-by-step explanation:

It is important to remember the Distributive Property in order to solve this exercise.

The Distributive property states that:

a(b+c)=ab+ac\\\\\\a(b-c)=ab-ac

In this case you have the following expression provided in the exercise:

t(14-5)

Then, in order to write this expression in another way, you can apply the Distributive property. Multiply each number inside the parentheses by "t".

Applying this procedure, you get:

t(14-5)=(t14)-(t5)

Notice that this expression matches with the one shown in the the second option.

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