(New page: 1) total outage is (.001)^k 2)(1-(.001^k))^365 gives you at least one connection every day of the year 3) doing 1 - (step 2) or 1-(1-(.001^k))^365 gives the probability of total outage a...) |
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4)plug in values of k until the number is less than .001 | 4)plug in values of k until the number is less than .001 | ||
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| + | //Comments | ||
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| + | for step 4, it is possible to solve the equation for k instead of plugging in values | ||
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| + | Hamad Al Shehhi | ||
Latest revision as of 03:38, 15 September 2008
1) total outage is (.001)^k
2)(1-(.001^k))^365 gives you at least one connection every day of the year
3) doing 1 - (step 2) or 1-(1-(.001^k))^365 gives the probability of total outage at least once
4)plug in values of k until the number is less than .001
//Comments
for step 4, it is possible to solve the equation for k instead of plugging in values
Hamad Al Shehhi
