a. 
× 
A
b. 
× 
A
c. 
A
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
<h3>How to determine the current</h3>
The formula for finding current
I = V/R
Where v = voltage
R = resistance
A. V = 12V
R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series
× A
B. V = 9V
R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series
× A
C. V= 12V
1/R = 
= 
×
R = 
= 310. 56 Ω
A
It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
Learn more about Ohms law here:
brainly.com/question/14296509
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Answer:
0.1032
Step-by-step explanation:
F(18)=(1,2,3,6,9,10) f(27) =(1,3,9,27) GCF =9 GCF =250, numbers are 2 (850) and 3 (850) or 1700 and 2550
Solution :
The sample proportion 
= 0.024
<u>90% confidence interval </u>
The standard deviation = 
= 
= 0.00484
z = 1.645 for 90% CI
Upper band = 0.024 + (0.00484 x 1.645 ) = 0.03196
Lower band = 0.024 - (0.00484 x 1.645 ) = 0.01603
Therefore, the 90% CI is (0.016, 0.032)
<u />
<u>99% confidence interval </u>
z = 2.576 for 99% CI
Upper band = 0.024 + (2.576 x 0.00484 ) = 0.0365
Lower band = 0.024 - (2.576 x 0.00484 ) = -0.0115
Therefore, the 99% CI is (0.0115, 0.0365)
Given:
z = (x - μ)/σ = z value
x = independent variable = 11
μ = mean = 8
σ = standard deviation = 3
Solution:
X ~ N (8, 3)
For day 1, n = 5
P (X < = 11) = P (Z < = 11 – 8 / 3/Sqrt5) = P (Z <
=2.23) = 0.987126
For day 1, n = 6
P (X < = 11) = P (Z < = 11 – 8 / 3/sqrt6) = P (Z <
=2.45) = 0.992857
For both days:
P (X <= 11) = P (X <= 11) = (0.987126) (0.992857) =
0.9801
