Answer:
0.006 is the probability that the record will be broken in next year.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 62 seconds
Standard Deviation, σ = 0.8 seconds
We are given that the distribution of face time is a bell shaped distribution that is a normal distribution.
Formula:
The record will be broken if the time is less than 60 seconds.
P(time is less than 60 seconds)
P(x < 60)
Calculation the value from standard normal z table, we have,

Thus, 0.006 is the probability that the record will be broken in next year.
20 mitiplay by 18 thats the answer
Answer:
The number N of bacteria in a refrigerated food is given by N(T) = 10T^2 - 20T + 600, 2 ≤ T ≤ 20 where T is the temperature of the food in degrees Celsius. When the food is removed from refrigeration, the temperature of the food is given by T(t) = 3t + 2, 0 ≤ t ≤ 6 where t is the time in hours.
Step-by-step explanation:
Answer:
a. P(X = 0) = 0.02586
b. 
c. 
Step-by-step explanation:
From the given information:
a. If the manufacturer stocks 120 components, what is the probability that the 120 orders can be filled without reordering components?


P(X = 0) = 1 × 1 ( 0.97)¹²⁰ ⁻ ⁰
P(X = 0) = 0.02586
b. ) If the manufacturer stocks 122 components, what is the probability that the 120 orders can be filled without reordering components?
![P(X \leq 2 ) = [ P(X=0) + P(X =1) + P(X = 2) ]](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%20%29%20%3D%20%5B%20P%28X%3D0%29%20%2B%20P%28X%20%3D1%29%20%2B%20P%28X%20%3D%202%29%20%5D)
![P(X \leq 2 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2}]](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%20%29%20%3D%20%5B%28%5E%7B122%7D_%7B0%7D%29%280.03%29%5E0%20%280.97%29%5E%7B122-0%7D%2B%28%5E%7B122%7D_%7B1%7D%29%280.03%29%5E1%20%20%280.97%29%5E%7B122-1%7D%2B%28%5E%7B122%7D_%7B2%7D%29%280.03%29%5E2%20%280.97%29%5E%7B122-2%7D%5D)
![P(X \leq 2 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120}]](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%20%29%20%3D%20%5B%5Cdfrac%7B122%21%7D%7B0%21%28122-0%29%21%20%7D%20%5Ctimes%201%20%5Ctimes%20%20%280.97%29%5E%7B122%7D%2B%5Cdfrac%7B122%21%7D%7B1%21%28122-1%29%21%20%7D%20%5Ctimes%20%280.03%29%20%20%280.97%29%5E%7B121%7D%2B%5Cdfrac%7B122%21%7D%7B2%21%28122-2%29%21%20%7D%20%5Ctimes%209%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%20%280.97%29%5E%7B120%7D%5D)
![P(X \leq 2 ) = [(1 \times 1 \times 0.02433 )+(122 \times (0.03) \times 0.025083)+(7381 \times 9 \times 10^{-4} \times 0.02586)]](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%20%29%20%3D%20%5B%281%20%5Ctimes%20%201%20%5Ctimes%20%200.02433%20%29%2B%28122%20%5Ctimes%20%280.03%29%20%20%5Ctimes%200.025083%29%2B%287381%20%5Ctimes%209%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%200.02586%29%5D)

(c) If the manufacturer stocks 125 components, what is the probability that the 120 orders can be filled without reordering components?
![P(X \leq 5 ) = [ P(X=0) + P(X =1) + P(X = 2) +P(X = 3)+P(X = 4)+ P(X = 5) ]](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%20%29%20%3D%20%5B%20P%28X%3D0%29%20%2B%20P%28X%20%3D1%29%20%2B%20P%28X%20%3D%202%29%20%20%2BP%28X%20%3D%203%29%2BP%28X%20%3D%204%29%2B%20P%28X%20%3D%205%29%20%20%20%20%5D)
![P(X \leq 5 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2} + (^{122}_{3})(0.03)^3 (0.97)^{122-3} + (^{122}_{4})(0.03)^4 (0.97)^{122-4}+ (^{122}_{5})(0.03)^5 (0.97)^{122-5}]](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%20%29%20%3D%20%5B%28%5E%7B122%7D_%7B0%7D%29%280.03%29%5E0%20%280.97%29%5E%7B122-0%7D%2B%28%5E%7B122%7D_%7B1%7D%29%280.03%29%5E1%20%20%280.97%29%5E%7B122-1%7D%2B%28%5E%7B122%7D_%7B2%7D%29%280.03%29%5E2%20%280.97%29%5E%7B122-2%7D%20%2B%20%28%5E%7B122%7D_%7B3%7D%29%280.03%29%5E3%20%280.97%29%5E%7B122-3%7D%20%2B%20%28%5E%7B122%7D_%7B4%7D%29%280.03%29%5E4%20%280.97%29%5E%7B122-4%7D%2B%20%28%5E%7B122%7D_%7B5%7D%29%280.03%29%5E5%20%280.97%29%5E%7B122-5%7D%5D)
![P(X \leq 5 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120} + \dfrac{122!}{3!(122-3)! }*(0.03)^3(0.97)^{122-3)}+ \dfrac{122!}{4!(122-4)! }*(0.03)^4(0.97)^{122-4)} +\dfrac{122!}{5!(122-5)! } *(0.03)^5(0.97)^{122-5)}]](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%20%29%20%3D%20%5B%5Cdfrac%7B122%21%7D%7B0%21%28122-0%29%21%20%7D%20%5Ctimes%201%20%5Ctimes%20%20%280.97%29%5E%7B122%7D%2B%5Cdfrac%7B122%21%7D%7B1%21%28122-1%29%21%20%7D%20%5Ctimes%20%280.03%29%20%20%280.97%29%5E%7B121%7D%2B%5Cdfrac%7B122%21%7D%7B2%21%28122-2%29%21%20%7D%20%5Ctimes%209%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%20%280.97%29%5E%7B120%7D%20%2B%20%5Cdfrac%7B122%21%7D%7B3%21%28122-3%29%21%20%7D%2A%280.03%29%5E3%280.97%29%5E%7B122-3%29%7D%2B%20%5Cdfrac%7B122%21%7D%7B4%21%28122-4%29%21%20%7D%2A%280.03%29%5E4%280.97%29%5E%7B122-4%29%7D%20%2B%5Cdfrac%7B122%21%7D%7B5%21%28122-5%29%21%20%7D%20%2A%280.03%29%5E5%280.97%29%5E%7B122-5%29%7D%5D)

Answer:
Exactly one solution.
Step-by-step explanation:
2x-4y=8
x+2y=4 /*2
<em>2x-4y=8</em>
<em><u>2x+4y=8 (+)</u></em>
4x=16
x=4
4+2y=4
y=0
(x;y)={(4; 0)}
Exactly one solution.