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alina1380 [7]
2 years ago
9

Skylar is arranging 8 objects on a shelf. How many different ways can she arrange them? How can I figure this out using the form

ula for combinations?
Mathematics
1 answer:
Aneli [31]2 years ago
4 0

Personally, I never bothered learning the formula. 
When I run into a problem like this, I look at it this way: 
(I learned this method from my high school math teacher,
Mr. H. Carlisle Taylor, in 1956.  It not only works, but I can
even understand it !)

The first object can be any one of the 8 .  For each of those ...
The 2nd object can be any one of the other 7.  For each of those ...
The 3rd object can be any one of the other 6 .  For each of those ...
The 4th object can be any one of the other 5 .  For each of those ...
The 5th object can be any one of the other 4 .  For each of those ...
The 6th object can be any one of the other 3 .  For each of those ...
The 7th object can be any one of the other 2 .  For each of those ...
The 8th object has to be the 1 that's left.

Total number of possible ways to line them up is

                 (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) =  40,320 ways .

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What is the least common multiple of 2,6 and 15?
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The mean preparation fee H&amp;R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t
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b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)=\frac{ME*\sqrt{N}}{s } where

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  • s is the standard deviation of the population
  • N is the sample size

a.

z(p)=\frac{8*\sqrt{30}}{50 } ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)=\frac{8*\sqrt{50}}{50 } ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)=\frac{8*\sqrt{100}}{50 } ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥(\frac{z*s}{ME} )^2 where

  • N is the sample size
  • z is the corresponding z-score in 95% probability (1.96)
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  • ME is the margin of error (8)

then N≥(\frac{1.96*50}{8} )^2 ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

7 0
3 years ago
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