Is this supposed to be a question?
Answer: Windows 2000
Explanation: I researched it and multiple sites said that it was Windows 2000.
<u><em>Hope this helps!!! :)</em></u>
The balance exercises used for introducing balance training should initially involve little joint motion and improve the Reflexive (automatic) joint-stabilization contractions.
<h3>What is the main goal of balance training?</h3>
Balance training is known to be a kind of an exercise where a person that is doing the exercises works out so as to strengthen the muscles that aids them to keep to be upright, such as their legs and core.
Note that kinds of exercises are done so as to improve stability and help hinder falls.
Therefore, The balance exercises used for introducing balance training should initially involve little joint motion and improve the Reflexive (automatic) joint-stabilization contractions.
Learn more about balance exercises from
brainly.com/question/16218940
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Answer:
40, 5 and 3
See explaination for the details
Explanation:
A)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =4bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/4
= 160/4
= 40
b)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =32bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/32
= 160/32
= 5
c)
Consider the data.
Assume D-cache is empty
An integer size =4bytes
The cache block size =64bytes
Therefore, the number of D-cache misses for reading the first 40 integers is,
= (40×4)/64
= 160/64
= 3
Answer:
double the bandwidth assigned per channel to 40 MHz
Explanation:
The best way of doing this would be to double the bandwidth assigned per channel to 40 MHz. This will make sure that the capacity is more than sufficient. This is simply because the bandwidth of a channel represents how much information can pass through the channel at any given second, the larger the channel, the more information/data that can pass at the same time. Therefore, if 20 MHz is enough for the network, then doubling this bandwidth channel size would be more than sufficient capacity for the network to handle all of the data.
