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Answer:
a. 0.11
b. 110 students
c. 50 students
d. 0.46
e. 460 students
f. 540 students
g. 0.96
Explanation:
(See attachment below)
a. Probability that a student got an A
To get an A, the student needs to get 9 or 10 questions right.
That means we want P(X≥9);
P(X>9) = P(9)+P(10)
= 0.06+0.05=0.11
b. How many students got an A on the quiz
Total students = 1000
Probability of getting A = 0.11 ---- Calculated from (a)
Number of students = 0.11 * 1000
Number of students = 110 students
So,the number of students that got A is 110
c. How many students did not miss a single question
For a student not to miss a single question, then that student scores a total of 10 out of possible 10
P(10) = 0.05
Total Students = 1000
Number of Students = 0.05 * 1000
Number of Students = 50 students
We see that 5
d. Probability that a student pass the quiz
To pass, a student needed to get at least 6 questions right.
So we want P(X>=6);
P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)
=0.08+0.12+0.15+0.06+0.05=0.46
So, the probability of a student passing the quiz is 0.46
e. Number of students that pass the quiz
Total students = 1000
Probability of passing the quiz = 0.46 ----- Calculated from (d)
Number of students = 0.46 * 1000
Number of students = 460 students
So,the number of students that passed the test is 460
f. Number of students that failed the quiz
Total students = 1000
Total students that passed = 460 ----- Calculated from (e)
Number of students that failed = 1000 - 460
Number of students that failed = 540
So,the number of students that failed is 540
g. Probability that a student got at least one question right
This means that we want to solve for P(X>=1)
Using the complement rule,
P(X>=1) = 1 - P(X<1)
P(X>=1) = 1 - P(X=0)
P(X>=1) = 1 - 0.04
P(X>=1) = 0.96
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
