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2 years ago

Describe each occupation

Engineering

1 answer:

vitfil [10]2 years ago

5 0

Answer:

Carpenter — A carpenter is someone who works with wood. They build houses and make cabinets etc.

Logging — Loggers are people who cut trees. They use strong chain saws to cut trees.

Shipwrights — Shipwrights build, design, and repair all sizes of boats.

Wood Machinist — Wood Machinists repair and cut timber or any kind of wood for construction projects. They also operate woodworking machines, as their name suggest.

Furniture finishers — Furniture finishers shape, decorate, and restore damaged and worn out furniture.

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Suppose an assembly requires five components from five different vendors. To guarantee starting the assembly on time with 90 per

spayn [35]

Answer:

The service level for each component must be 97.91%

Explanation:

If we want a 90% confidence of starting on time, that means we need

$P_{\mbox{starting on time}}=P_{\mbox{every component being ready on time}}=0.9\\$

As the probability of each component being ready is independent from the others, that means that the probability of the 5 components being ready is equal to multiply each probability:

$0.9=P_{\mbox{component 1 ready on time} } * P_{\mbox{component 2 ready on time} } *\\ P_{\mbox{component 3 ready on time} } * P_{\mbox{component 4 ready on time} } *\\P_{\mbox{component 5 ready on time} }$

The probability of being ready on time is equal to the service level (in fraction), and all 5 are equal so we can say:

$0.9=(\mbox{service level(in fraction)})^5\\\\\sqrt[5]{0.9} =\mbox{service level(in fraction)}=0.9791\\\mbox{In percentage}: \mbox{service level (in fraction)}*100 = 97.91\%$

6 0

3 years ago

Select the correct answer. Felix aspires to be an engineer working for the government. What credentials will Felix require to ap

Aneli [31]

Answer:

Explanation:

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2 years ago

How to plot 0.45 gradation chart for sieve analysis ?

LekaFEV [45]

Answer:

532235w3r35w3r

Explanation:

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2 years ago

A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

Part 2

For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

And we can quantify the decrease using the relative change:

$\% Change = \frac{5s}{55 s} *100 = 9.09\%$ of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

8 0

2 years ago

A construction crew lifts approximately 400 lb. of material several times during a day from a flatbed truck to a 25 ft. rooftop.

Irina18 [472]

Answer:

2ib

Explanation:

if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8

there goes your answer.

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1 year ago