Answer:
2.6 seconds
Explanation:
We first start by calculating the speed up
The formula is given as:
n/1+(n-1)F
We have n = 3 which is the number of processors
F = 20% = percentage of algorithm
When we put values into the formula
3/1+(3-1)0.20
= 3/1+2*0.20
= 3/1+0.4
= 3/1.4
Speed up = 2.14
From here we calculate the expected time
T/speedups
= 5.6/2.14
= 2.6
Therefore the expected time is 2.6 seconds
In a database it allows you to filter out the key info which you are looking for so its simpler for you.
Hope this helped :P
<u>2 logs is </u>the virus removal credit awarded to a properly operated conventional treatment plant.
<h3>What is the log removal for viruses?</h3>
A log removal value (LRV) is a measurement of the ability of treatment processes to remove pathogenic microorganisms. LRVs are specified by taking the logarithm of the ratio of pathogen concentration in the influent and effluent water of a restorative process (shown in Equation 1).
<h3>What is log inactivation?</h3>
“Log inactivation” is a suitable way to express the number or percent of microorganisms inactivated (killed or unable to replicate) through the disinfection process. For example, a 3 log inactivation value means that 99.9% of microorganisms of claim have been inactivated.
To learn more about log removal value, refer
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