By using The half-angle formula
cos X/2 = ±√[(1 + cos X)/ 2]
But we have cos x=2/3
= ±√[(1 + 2/3)/ 2]
= ±√[(5/3 )/ 2]
= ±√[(5/5]
=±√1
Therefore answer is ±1
cos A/2 = ±√[(1 + cos A )/ 2]
Answer:
1/2
Step-by-step explanation:
Answer:
a) 0.3125 per hour
b) 2.225 hours
c) 8.9 hours
d) 12.1 hours
e) 80%
Step-by-step explanation:
Given that:
mean time = 3.2 hours, standard deviation (σ) = 2 hours
The mean service rate in jobs per hour (λ) = 2 jobs/ 8 hour = 0.25 job/hour
a) The average number of jobs waiting for service (μ)= 1/ mean time = 1/ 3.2 = 0.3125 per hour
b) The average time a job waits before the welder can begin working on it (L) is given by:
c) The average number of hours between when a job is received and when it is completed (Wq) is given as:
d) The average number of hours between when a job is received and when it is completed (W) is given as:
e) Percentage of the time is Gubser's welder busy (P) is given as:
Answer:
3 term
Step-by-step explanation:
Count the amount of numbers between operations
Answer:
a) f(16) = 42
b) f(16) = 54
c) f(16) = 162
d) f(16) = 30
Step-by-step explanation:
a) y = mx + b ∧ m = (f(8) - f(4))/(8-4) ⇒ m = (18 - 6)/(8 - 4) = 3
b = y - mx = 6 - 3(4) = 6 - 12 = - 6
f(16) = 3(16) - 6 = 42
b) y = kxⁿ ∧ f(4) = 6 = k4ⁿ ∧ f(8) = 18 = k8ⁿ ⇒ 18/6 = (k8ⁿ)/(k4ⁿ) ⇒ 3 = 2ⁿ
n = ㏑(3) / ㏑(2) ⇒ k = y/xⁿ ⇒ k = 6/4ⁿ = 2/3
f(16) = 2/3 × 16ⁿ = 54
c) y = aeᵇˣ ∧ f(4) = 6 = aeᵇ⁴ ∧ f(8) = 18 = aeᵇ⁸ ⇒ 18/6 = (aeᵇ⁸)/(aeᵇ⁴) ⇒ 3 = e⁴ᵇ
b = ㏑(3/4) ∧ a = y / eᵇˣ ⇒ a = 6 / e⁴ᵇ = 2
f(16) = 2eᵇ¹⁶ = 162
d) y = a㏑(bx) ∧ f(4) = 6 = a㏑(b4) ∧ f(8) = 18 = a㏑(b8)
⇒ 18 - 6 = a㏑(b8) - a㏑(b4) ⇒ 12 = a㏑(8b/4b) ⇒ a = 12 / ㏑(2)
f(4) = 6 = a㏑(4b) ⇒ b = (√2)/4
f(16) = a㏑(b16) = 30
