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Vladimir [108]
3 years ago
11

The inhabitants of the island of jumble use the standard kobish alphabet ($20$ letters, a through t). each word in their languag

e is $4$ letters or less, and for some reason, they insist that all words contain the letter a at least once. how many words are possible?
Mathematics
1 answer:
rjkz [21]3 years ago
7 0
We know that

<span>1. The number of all 4-letter words written using the alphabet of 20 symbols (from A to T) is 
20</span><span>^4=160,000
</span><span>
 2. The number of all 4-letter words written using the alphabet of 19 symbols (from B to T) is 
19</span><span>^4=130,321
</span><span>
 3. The difference represents exactly the number of all words of the inhabitants of the island of Jumble
difference=</span>(160,000-130,321)= 29679.

the answer is
29679
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Calculating the value from the standard normal table we have,

1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%

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P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z

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d) P( calls last between 4 and 6 minutes)

P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z

P(4 \leq x \leq 6) = 91.45\%

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P( X > x) = P( z > \displaystyle\frac{x - 4.7}{0.50})=0.03  

= 1 -P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.03  

=P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.997  

Calculation the value from standard normal z table, we have,  

P(z < 2.75) = 0.997

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Hence, the call lengths must be 6.08 minutes or greater for them to lie in the highest 3%.

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