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gavmur [86]
3 years ago
13

Find the range of data: 52,48,57,47,49,60,59,51

Mathematics
2 answers:
ElenaW [278]3 years ago
6 0
15 20 54 64 67 637 637 636 5636 635 6635 6363
insens350 [35]3 years ago
6 0

Answer:

The range of #1 is 13

The range of #2 is 9

The range of #3 is 20

The range of #4 is 59

Step-by-step explanation:

I had a math teacher in 6th grade who taught me a corny, but helpful rhyme.

It is

Hey diddle-diddle

The median's the middle (organize the numbers from least to greatest, then find the number in the middle)

You add (all the numbers), then divide (by how many numbers there are) for the mean

The mode is the one that appears the most (which number do you see more than once?)

The range is the difference between (find the smallest number and the largest number, then subtract the small one from the large one)

See? Corny huh? I hope this helps bro!

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Lf y 1/4 · when x=5, find y when x=7. given that y varies directly with x.<br><br>y=​
egoroff_w [7]

Answer:

y= 0.35

Step-by-step explanation:

y 1/4

when x=5,

find y=? when x=7.

given that y varies directly with x.

y = kx

Where k is the constant of proportionality

k= y/x

k= (1/4)÷ 5

k= 0.05

y= kx

y= (0.05)(7)

y= 0.35

I Hope this will helpful...

5 0
3 years ago
Ian invests $13,670 in a savings account at his local bank which gives 1.9% simple annual interest. He also invests $6,040 in an
IceJOKER [234]
The formula says
I=prt
I Interest earned
P principle
R interest rate
T time

The interest earned of local account
13,670×0.019×9
=2,337.57

The interest earned of online account
6,040×0.045×9
=2,446.2

The difference
2,446.2−2,337.57
=109

So the answer is d
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25  Is the least number of tables
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Holding the price level constant, we would expect that if the multiplier is 1.5, the cumulative effect of this change in M on AD
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A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
3 years ago
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