Answer:
In response to the above problems, the predecessors have made many efforts. ese include collaborative computing between terminal devices and cloud servers [6][7][8], model compression and parameter pruning [9][10][11][12], or customized mobile implementation [13][14][15]. Despite all these efforts made by the predecessors, on the premise of ensuring the accuracy of the model required by the user, the service latency is minimized, and the user's hardware configuration and system status can be sensed to implement automatic model pruning and partition. ...
Step-by-step explanation:
I hope it will help you
One JOB = 1 and 2 hours = 120 min
Tyler Rate per minute: 1/120 (in 1 minute he performes 1/120 of the job)
Dakota<span> Rate per minute: 1/90 (in 1 minute he performes 1/90 of the job)
Tyler's + Dakota's rate per 1 minute = 1/120 + 1/90 = 7/360 (Job/minutes)
7/360 of the job was performed in 1 minute
</span>a complete JOB =1 to be performed in x minutes (Rule of three)
x = 1x1/(7/360) that equals to 360/7 and x (time of both) = 51 min 42
Basically all the information in this problem is useless. They included all those numbers to confuse you. If you read throughly you can see that all you need to know is that Adam's sister can only have 70 grams of fat each day and that her dinner will have 48 grams of fat.
To figure this out I use this formula all the time to figure out percentages.
is/of=x/100
Your trying to figure out what is 48 percent of 70 because that's how much Adam's sister is going to eat at dinner. So your is would be 48 and your of is 70. When you substitute those numbers your formula becomes 48/70=x/100. Now you cross multiply and get 70x=4800.All that's left is to divide 70 on both sides. Which gives you 68.57.The final answer is 69 percent.
<span>Look at your table for a Z value of 1.55. The numbers on the far left column are your z values. See the 1.5 row, then move over to the 0.05 column to make it 1.55.
You'll see 0.9394.
That's the area under the normal curve from 1.55 to negative infinity.
But you wanted the area under the curve greater than 1.55.
Take 1-0.9394=0.0606.
You subtract from 1 because you know that the area under the whole curve is 1, so it gives you the area you need.</span>
