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Natasha2012 [34]
3 years ago
13

Find the LCMof75, 5,3​

Mathematics
2 answers:
frozen [14]3 years ago
6 0

Answer:

LCM = 75

Step-by-step explanation:

1: Multiply the factor by the greatest number

Description:

The least common  multiple for 75,5,3 is 75.

LCM= Least common Multiple

Please mark brainliest

<em><u>Hope this helps.</u></em>

Sholpan [36]3 years ago
5 0

Answer:

75

Step-by-step explanation:

Break each number into prime factors

75 = 25*3 = 5*5*3

5 = 5*1

3 = 3*1

Multiply by the greatest number of each factor

3 = 1 time

5 = =2 times

The least common multiple = 3 * 5*5 = 75

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Step-by-step explanation:

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3 years ago
A fast food restaurant executive wishes to know how many fast food meals teenagers eat each week. They want to construct a 99% c
Lana71 [14]

Answer:

The minimum sample size required to create the specified confidence interval is 2229.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

What is the minimum sample size required to create the specified confidence interval

This is n when M = 0.06, \sigma = 1.1

0.06 = 2.575*\frac{1.1}{\sqrt{n}}

0.06\sqrt{n} = 2.575*1.1

\sqrt{n} = \frac{2.575*1.1}{0.06}

(\sqrt{n})^{2} = (\frac{2.575*1.1}{0.06})^{2}

n = 2228.6

Rounding up

The minimum sample size required to create the specified confidence interval is 2229.

7 0
3 years ago
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