Answer:
(a) 0.12924
(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Explanation:
(a)
n=200 for fair coin getting head, p= 0.5
Expectation = np =200*0.5=100
Variance = np(1 - p) = 100(1-0.5)=100*0.5=50
Standard deviation, 
Z value for 108, 
P( x ≥108) = P( z >1.13)= 0.12924
(b)
Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Answer:
Benefits that last a lifetime. The Army offers you money for education, comprehensive health care, generous vacation time, family services and support groups, special pay and cash allowances to cover the cost of living.
Answer:
A) Wet bulb temperature of #1 is less than that of #2
Explanation:
This can be gotten from pinpointing the states of the two containers on a psychometric chart.
Answer:
I have solved the problem below. I hope it will let you clear the concept.
For any inquiries ask me in the comments.
Explanation:
Answer:
V₂ = 20 V
Vt = 20 V
V₁ = 20 V
V₃ = 20 V
I₁ = 10 mA
I₃ = 3.33 mA
It = 18.33 mA
Rt = 1090.91 Ω
Pt = 0.367 W
P₁ = 0.2 W
P₂ = 0.1 W
P₃ = 0.067 W
Explanation:
Part of the picture is cut off. I assume there is a voltage source Vt there?
First, use Ohm's law to find V₂.
V = IR
V₂ = (0.005 A) (4000 Ω)
V₂ = 20 V
R₁ and R₃ are in parallel with R₂ and the voltage source Vt. That means V₁ = V₂ = V₃ = Vt.
V₁ = 20 V
V₃ = 20 V
Vt = 20 V
Now we can use Ohm's law again to find I₁ and I₃.
V = IR
I = V/R
I₁ = (20 V) / (2000 Ω)
I₁ = 0.01 A = 10 mA
I₃ = (20 V) / (6000 Ω)
I₃ = 0.00333 A = 3.33 mA
The current It passing through Vt is the sum of the currents in each branch.
It = I₁ + I₂ + I₃
It = 10 mA + 5 mA + 3.33 mA
It = 18.33 mA
The total resistance is the resistance of the parallel resistors:
1/Rt = 1/R₁ + 1/R₂ + 1/R₃
1/Rt = 1/2000 + 1/4000 + 1/6000
Rt = 1090.91 Ω
Finally, the power is simply each voltage times the corresponding current.
P = IV
Pt = (0.01833 A) (20 V)
Pt = 0.367 W
P₁ = (0.010 A) (20 V)
P₁ = 0.2 W
P₂ = (0.005 A) (20 V)
P₂ = 0.1 W
P₃ = (0.00333 A) (20 V)
P₃ = 0.067 W
