5(2x - 8) + 15 = -15
-15 -15 subtract 15 from each side
5(2x - 8) = -30
÷5 ÷5 divide both sides by 5
2x - 8 = -6
+8 +8 add 8 to each side
2x=2
÷2 ÷2 divide both sides by 3
x = 1
Checking:
5(2(1)-8) + 15 = -15
5(-6) + 15 = -15
-30 + 15 = -15
-15 = -15 Correct! x=1
Answer:
10 and 15, 20 and 25, and 35 and 45.
Step-by-step explanation:
Answer:
a. P(X = 0) = 0.02586
b. ![\mathbf{P(X \leq 2 ) =0.2879}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28X%20%5Cleq%202%20%29%20%3D0.2879%7D)
c. ![\mathbf{P(X \leq 5 ) =0.8387}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28X%20%5Cleq%205%20%29%20%3D0.8387%7D)
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)=(^{120}_{0}) (0.03)^0 (1-0.03)^{n-0}](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%3D%28%5E%7B120%7D_%7B0%7D%29%20%280.03%29%5E0%20%281-0.03%29%5E%7Bn-0%7D)
![P(X = 0)=\dfrac{120!}{0!(120-0)!} (0.03)^0 (1-0.03)^{n-0}](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%3D%5Cdfrac%7B120%21%7D%7B0%21%28120-0%29%21%7D%20%280.03%29%5E0%20%281-0.03%29%5E%7Bn-0%7D)
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)
![\mathbf{P(X \leq 2 ) =0.2879}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28X%20%5Cleq%202%20%29%20%3D0.2879%7D)
(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)
![\mathbf{P(X \leq 5 ) =0.8387}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28X%20%5Cleq%205%20%29%20%3D0.8387%7D)
Answer:
It equals 1.66666666666...
But you can round it to 1.7 if your rounding to the nearest tenth.