Answer:
1. Divide each bonus by regular bonus apply this to all the data
2. In cell C11 write, "Average" press tab key on the keyboard and then select the range of the cells either by typing "C7:C10" or by selecting it through the mouse.
Explanation:
The average bonus multiplier can be found by dividing each bonus with the regular bonus applying this to all the data and then putting the average formula and applying it to the cells C7:C10.
After dividing the bonus with regular bonus, in cell C11 write, "Average" press tab key on the keyboard and then select the range of the cells either by typing "C7:C10" or by selecting it through the mouse.
Answer:
switchboard
Explanation:
The option that he should implement would be a switchboard interface design. These design structures for graphic user interfaces use a single main page, large icons/buttons, a fixed navigation menu, and all the necessary functionality right in front of the user. This design is made with simplicity in mind in order to make it as easy as possible for a new user to pick up and efficiently and intuitively navigate the user interface. Therefore, since Calvin needs a simple yet professional design, this would be the best implementation.
Answer:
scope because the scope will determine the Time for the employee to work
Answer:
Around seven billion people
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
