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5.94. A student s grades in the laboratory, lecture, and recitation parts of a physics course were 71, 78, and 89, respectively, (a) If the weights accorded these grades are 2, 4, and 5, respectively, what is an appropriate average grade (b) What is the average grade if equal weights are used 5.95. Three teachers of economics reported mean examination grades of 79, 74, and 82 in their classes, which consisted of 32, 25, and 17 students, respectively. Determine the mean grade for all the classes. 5.96. The mean annual salary paid to all employees in a company was $5000. The mean annual salaries paid to male and female employees of the company were $5200 and $4200, respectively. Determine the percentages of males and females employed by the company. 5.97. Table 5-18 shows the distribution of the maximum loads in short tons (1 short ton 2000 lb) supported by certain cables produced by a company. Determine the mean maximum loading using (a) the long method, (b) the coding method.

Solving,

Table 5-18

From (9c) and (9d) we have !h !l a The bandwidth is de ned by !h !l a The quality factor Q is de ned by Q !0 = p b=a 10c 10b and !h !l b !2 0 10a

Maximum Load (short tons) 9 9.7 9.8 10.2 10.3 10.7 10.8 11.2 11.3 11.7 11.8 12.2 12.3 12.7 12.8 13.2 TOTAL Number of Cables 2 5 12 17 14 6 3 1 60 x f 462 98 480 75 498 56 516 42

Table 5-19

The quality factor measures the sharpness of the frequency response around the center frequency. This behavior is also called resonance (see Sections 12.11 to 12.15). When the quality factor is high, !l and !h may be approximated by !0 =2 and !0 =2, respectively.

Estimated Completion Date 1/15/01 2/15/01 2/25/01 4/15/01 6/1/01 7/1/01

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534 30 552 21 570 15 588 11 606 6 624 2

EXAMPLE 12.7 Consider the network function H s 10s= s2 300s 106 . and upper half-power frequencies, the bandwidth, and the quality factor. Since !2 106 , the center frequency !0 1000 rad/s. 0 The lower and upper half-power frequencies are, respectively, Find the center frequency, lower

5.98. Find x for the data in Table 5-19 using (a) the long method (b) the coding method. # 5.99. Table 5-20 shows the distribution of the diameters of the heads of rivets manufactured by a company. Compute the mean diameter.

Table 5-20

FREQUENCY RESPONSE, FILTERS, AND RESONANCE q q a2 =4 b a=2 3002 =4 106 300=2 861:2 rad=s q q !h a2 =4 b a=2 3002 =4 106 300=2 1161:2 rad=s !l The bandwidth !h !l 1161:2 861:2 300 rad/s. The quality factor Q 1000=300 3:3.

Diameter (inches) 0.7247 0.7249 0.7250 0.7252 0.7253 0.7255 0.7256 0.7258 0.7259 0.7261 0.7262 0.7264 0.7265 0.7267 0.7268 0.7270 0.7271 0.7273 0.7274 0.7276 0.7277 0.7279 0.7280 0.7282 TOTAL Frequency 2 6 8 15 42 68 49 25 18 12 4 1 250

Perform dc and ac analysis on the circuit in Fig. 15-28. Find the complex magnitude of V2 for f varying from 100 Hz to 10 kHz in 100 steps.

5.100. Compute the mean for the data in Table 5-21.

Actual Date Completed 2/1/01 2/15/01 3/1/01 4/1/01

Fig. 15-28 The source le is DC and AC Vs Is R1 R2 C .AC .PROBE .END analysis of Fig. 15-28 1 0 AC 0 2 DC 1 2 1k 2 0 2k 2 0 1 uF LIN 100 Vm(2) Vp(2) 10 1 mA 0

Table 5-21

Class 10 under 15 15 under 20 20 under 25 25 under 30 30 under 35 35 under 40 40 under 45 TOTAL Frequency 3 7 16 12 9 5 2 54

The output le contains the following results: SMALL SIGNAL BIAS SOLUTION NODE VOLTAGE NODE VOLTAGE (1) 0.0000 (2) .6667 VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 6:667E 04 TOTAL POWER DISSIPATION 0:00E 00 WATTS

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