Answer:
-a - 4
Step-by-step explanation:
3(a-2)+4(3-a)-10
3a - 6 + 12 - 4a - 10
-a - 4
9514 1404 393
Answer:
- absolute error: 0.4 kg
- relative error: 6.1%
Step-by-step explanation:
The absolute error is the difference between the estimate and the actual weight:
7.0 kg -6.6 kg = 0.4 kg . . . . absolute error
__
The relative error is the ratio of the absolute error to the actual weight:
(0.4 kg)/(6.6 kg) × 100% = 6.060606...% ≈ 6.1% . . . . relative error
I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs
are:
cos(x)tan(x) – ½ = 0
→ π/6 and 5π/6
cos(π/6)tan(π/6) – ½ = 0
cos(5π/6)tan(5π/6) – ½ = 0
sec(x)cot(x) + 2 =
0 → 7π/6 and 11π/6
sec(7π/6)cot(7π/6) + 2 = 0
sec(11π/6)cot(11π/6) + 2 = 0
sin(x)cot(x) +
1/sqrt2 = 0 → 3π/4 and 5π/4
sin(3π/4)cot(3π/4) + 1/sqrt2 = 0
sin(5π/4)cot(5π/4) + 1/sqrt2 = 0
csc(x)tan(x) – 2 = 0 → π/3 and 5π/3
csc(π/3)tan(π/3) – 2 = 0
csc(5π/3)tan(5π/3) – 2 = 0
Answer:
(a) The probability of having exactly four arrivals during a particular hour is 0.1754.
(b) The probability that at least 3 people arriving during a particular hour is 0.7350.
(c) The expected arrivals in a 45 minute period (0.75 hours) is 3.75 arrivals.
Step-by-step explanation:
(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:
The probability of having exactly four arrivals during a particular hour is 0.1754.
(b) The probability that at least 3 people arriving during a particular hour can be written as
Using
We get
The probability that at least 3 people arriving during a particular hour is 0.7350.
(c) The expected arrivals in a 45 minute period (0.75 hours) is
X=3/4
hope this helps u out :D
