amount of data that can be transferred in a given time

What is the maximum number of data refreshes per day?

Which term describes the amount of data that can be moved from one point on a network to another point in a given time?*1 pointa) Bandwidthb) Speedc) Flowd) Bits per second

The volume of data that can be sent from one point to another in a given time period is calledA. broadbandB. byteC. baudD. bandwidth

The amount of time required by a storage device to retrieve data and programs is called its

Video ExampleEXAMPLE 5 The graph shows data traffic on the link from the United States to SWITCH, the Swiss academic and research network, on February 10, 1998. D(t) is the data throughput, measured in megabits per second (Mb/s). Use Simpson's rule to estimate the total amount of data transmitted on the link up to noon on that day.SOLUTION Because we want the units to be consistent and D(t) is measured in megabits per second, we convert the units for t from hours to seconds. If we let A(t) be the amount of data (in megabits) transmitted by time t, where t is measured in seconds, then A'(t) = D(t). So, by the Net Change Theorem, the total amount of data transmitted by noon (when t = 12 × 602 = 43,200) isA(43,200) = 43,2000D(t) dt.We estimate the values of D(t) at hourly intervals from the graph and compile them in the table.t (hours) t (seconds) D(t) t (hours) t (seconds) D(t)0 0 3.2 7 25,200 1.31 3,600 2.7 8 28,800 2.82 7,200 1.9 9 32,400 5.73 10,800 1.7 10 36,000 7.14 14,400 1.3 11 39,600 7.75 18,000 1.0 12 43,200 7.96 21,600 1.1 Then we use Simpson's Rule with n = 12 and Δt = 3600 to estimate the integral.43,2000A(t) dt ≈ Δt3[D(0) + 4D(3600) + 2D(7200) + ... + 4D(39,600) + D(43,200)]≈ 360033.2 + 4(2.7) + 2(1.9) + 4(1.7) + 2(1.3) + 4(1.0) + 2(1.1) +  4(1.3) + 2(2.8) + 4(5.7) + 2(7.1) + 4(7.7) + 7.9  =  Thus the total amount of data transmitted up to noon is about megabits.