Answer: D)0.205
Step-by-step explanation:
In binomial distribution, the probability of getting success in x trials is given by :-
![P(X=x)=^nC_xp^x(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%5EnC_xp%5Ex%281-p%29%5E%7Bn-x%7D)
, where n is the total number of trials , p is the probability of getting success in each trial .
Given : The probability that a component will fail is 0.2.
Let x be the number of components will stop working.
i.e. p=0.2
n=12
Since The machine will stop working if more than three components fail.
Then, the probability that the machine will stop working will be :-
![P(X>3)=1-P(X\leq3)\\\\=1-[P(0)+P(1)+P(2)+P(3)]](https://tex.z-dn.net/?f=P%28X%3E3%29%3D1-P%28X%5Cleq3%29%5C%5C%5C%5C%3D1-%5BP%280%29%2BP%281%29%2BP%282%29%2BP%283%29%5D)
![=1-[^{12}C_0(0.2)^0(0.8)^{12}+^{12}C_1(0.2)^1(0.8)^{11}+^{12}C_2(0.2)^2(0.8)^{10}+^{12}C_3(0.2)^3(0.8)^{9}]](https://tex.z-dn.net/?f=%3D1-%5B%5E%7B12%7DC_0%280.2%29%5E0%280.8%29%5E%7B12%7D%2B%5E%7B12%7DC_1%280.2%29%5E1%280.8%29%5E%7B11%7D%2B%5E%7B12%7DC_2%280.2%29%5E2%280.8%29%5E%7B10%7D%2B%5E%7B12%7DC_3%280.2%29%5E3%280.8%29%5E%7B9%7D%5D)
![=1-[(1)(0.8)^{12}+(12)(0.2)(0.8)^{11}+\dfrac{12!}{2!10!}(0.2)^2(0.8)^{10}+\dfrac{12!}{3!9!}(0.2)^3(0.8)^{9}]\\\\=1-(0.068719476736+0.206158430208+0.283467841536+0.23622320128 )\\\\=1-0.79456894976\\\\=0.20543105024\approx0.205](https://tex.z-dn.net/?f=%3D1-%5B%281%29%280.8%29%5E%7B12%7D%2B%2812%29%280.2%29%280.8%29%5E%7B11%7D%2B%5Cdfrac%7B12%21%7D%7B2%2110%21%7D%280.2%29%5E2%280.8%29%5E%7B10%7D%2B%5Cdfrac%7B12%21%7D%7B3%219%21%7D%280.2%29%5E3%280.8%29%5E%7B9%7D%5D%5C%5C%5C%5C%3D1-%280.068719476736%2B0.206158430208%2B0.283467841536%2B0.23622320128%20%29%5C%5C%5C%5C%3D1-0.79456894976%5C%5C%5C%5C%3D0.20543105024%5Capprox0.205)
Hence, the probability that the machine will stop working is 0.205.
Thus , the correct answer is D)0.205
Answer:
126
Step-by-step explanation:
A group of store managers must assemble 280 displays for an upcoming sale.
Displays assembled in first hour = 25 percent of the displays = ![25\% \times 280](https://tex.z-dn.net/?f=25%5C%25%20%5Ctimes%20280)
Displays assembled in first hour= ![\frac{25}{100} \times 280 =70](https://tex.z-dn.net/?f=%5Cfrac%7B25%7D%7B100%7D%20%5Ctimes%20280%20%3D70)
Remaining displays =280-70= 210
We are given that 40 percent of the remaining displays during the second hour
Displays assembled in second hour= ![\frac{40}{100} \times 210 =84](https://tex.z-dn.net/?f=%5Cfrac%7B40%7D%7B100%7D%20%5Ctimes%20210%20%3D84)
Displays assembled in first hour and second hour = 70+84=154
Remaining displays = 280-154 = 126
Hence 126 of the displays will not have been assembled by the end of the second hour
Area of a trapezoid is height times the average of the bases,
A= 1/2 h (a + b)
Here we have height h=12 and bases a=30, b=12 so area
A = (1/2) (12) (30 + 12) = 6(42)= 252
Answer: B
Answer:Answer is in a photo. I couldn't attach it here, but I uploaded it to a file hosting. link below! Good Luck!