The average can be given by the following formula:
Average = (a1 + a2 + a3 + a4 + a5) / (5).
Equivalently.
Average = (ai) / (i).
Where,
i = 1,2,3,4, ..., n
Substituting
Average = (90 + 85 + 80 + 75 + 75) / (5).
Average = 81
answer:
Average = (a1 + a2 + a3 + a4 + a5) / (5).
or
Average = (ai) / (i).
Where,
i = 1,2,3,4, ..., n
Answer:
what is this? math or physics?
Step-by-step explanation:
Answer:
can l
Step-by-step explanation:
can l? can l? can l~
Answer:
Prof B and Prof D
Step-by-step explanation:
Step-by-step explanation:
From the question we are told that
Sample size 100
Instructor Number of Failures
Prof. A 13
Prof. B 0
Prof. C 11
Prof. D 16
Confidence level= 0.95
From Z table
Z=1.96
Generally proportion for failure is mathematically given as
![Proportion\ of\ Failure\ P=\frac{Number\ of\ Failure\ N}{Sample\ size\ S}](https://tex.z-dn.net/?f=Proportion%5C%20of%5C%20Failure%5C%20P%3D%5Cfrac%7BNumber%5C%20of%5C%20Failure%5C%20N%7D%7BSample%5C%20size%5C%20S%7D)
![P=\frac{N}{S}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BN%7D%7BS%7D)
For Prof A
![P_A=\frac{N_A}{S}](https://tex.z-dn.net/?f=P_A%3D%5Cfrac%7BN_A%7D%7BS%7D)
![P_A=\frac{13}{100}](https://tex.z-dn.net/?f=P_A%3D%5Cfrac%7B13%7D%7B100%7D)
![P_A=0.13](https://tex.z-dn.net/?f=P_A%3D0.13)
For Prof B
![P_B=\frac{N_B}{S}](https://tex.z-dn.net/?f=P_B%3D%5Cfrac%7BN_B%7D%7BS%7D)
![P_B=\frac{0}{100}](https://tex.z-dn.net/?f=P_B%3D%5Cfrac%7B0%7D%7B100%7D)
![P_B=0](https://tex.z-dn.net/?f=P_B%3D0)
For Prof C
![P_C=\frac{N_C}{S}](https://tex.z-dn.net/?f=P_C%3D%5Cfrac%7BN_C%7D%7BS%7D)
![P_C=\frac{11}{100}](https://tex.z-dn.net/?f=P_C%3D%5Cfrac%7B11%7D%7B100%7D)
![P_C=0.11](https://tex.z-dn.net/?f=P_C%3D0.11)
For Prof D
![P_D=\frac{N_D}{S}](https://tex.z-dn.net/?f=P_D%3D%5Cfrac%7BN_D%7D%7BS%7D)
![P_D=\frac{16}{100}](https://tex.z-dn.net/?f=P_D%3D%5Cfrac%7B16%7D%7B100%7D)
![P_D=0.16](https://tex.z-dn.net/?f=P_D%3D0.16)
Generally the average proportional failure is mathematically given as
![P_a_v_g=\frac{0.13+0+0.11+0.16}{4}](https://tex.z-dn.net/?f=P_a_v_g%3D%5Cfrac%7B0.13%2B0%2B0.11%2B0.16%7D%7B4%7D)
![P_a_v_g=0.10](https://tex.z-dn.net/?f=P_a_v_g%3D0.10)
Therefore having this we calculate for the control limits
Generally the upper control limit UCl is mathematically given as
Upper control limits:
![UCL =P_a_v_g +Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }](https://tex.z-dn.net/?f=UCL%20%3DP_a_v_g%20%2BZ%5Csqrt%7B%5Cfrac%7BP_a_v_g%281-P_a_v_g%29%20%7D%7BS%7D%20%7D)
![UCL =0.1 +1.96\sqrt{\frac{0.1(1-0.1) }{100}}](https://tex.z-dn.net/?f=UCL%20%3D0.1%20%2B1.96%5Csqrt%7B%5Cfrac%7B0.1%281-0.1%29%20%7D%7B100%7D%7D)
![UCL =0.1 +1.96*0.03](https://tex.z-dn.net/?f=UCL%20%3D0.1%20%2B1.96%2A0.03)
![UCL =0.1 +0.0588](https://tex.z-dn.net/?f=UCL%20%3D0.1%20%2B0.0588)
![UCL =0.1588](https://tex.z-dn.net/?f=UCL%20%3D0.1588)
Generally the upper control limit UCl is mathematically given as
Lower limit control limits:
![UCL =P_a_v_g _Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }](https://tex.z-dn.net/?f=UCL%20%3DP_a_v_g%20_Z%5Csqrt%7B%5Cfrac%7BP_a_v_g%281-P_a_v_g%29%20%7D%7BS%7D%20%7D)
![UCL =0.1 -1.96\sqrt{\frac{0.1(1-0.1) }{100}}](https://tex.z-dn.net/?f=UCL%20%3D0.1%20-1.96%5Csqrt%7B%5Cfrac%7B0.1%281-0.1%29%20%7D%7B100%7D%7D)
![UCL =0.1 -1.96*0.03](https://tex.z-dn.net/?f=UCL%20%3D0.1%20-1.96%2A0.03)
![UCL =0.1 -0.0588](https://tex.z-dn.net/?f=UCL%20%3D0.1%20-0.0588)
![UCL =0.0412](https://tex.z-dn.net/?f=UCL%20%3D0.0412)
Therefore the Range is 0.0412 to 0.1588
Given the range 0.0412 to 0.1588 Prof B and Prof D are outside the Range making them the correct options
-5=-5
-12+7 = x+3
-5 = x+3
-5-3 =x
x = -8
-12+7 = -8 +3
-5 = -5