Answer:
0.762 = 76.2% probability that this shipment is accepted
Step-by-step explanation:
For each pen, there are only two possible outcomes. Either it is defective, or it is not. The probability of a pen being defective is independent from other pens. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
17 randomly selected pens
This means that ![n = 17](https://tex.z-dn.net/?f=n%20%3D%2017)
(a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)
This is
when
. So
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29)
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{17,0}.(0.1)^{0}.(0.9)^{17} = 0.167](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B17%2C0%7D.%280.1%29%5E%7B0%7D.%280.9%29%5E%7B17%7D%20%3D%200.167)
![P(X = 1) = C_{17,1}.(0.1)^{1}.(0.9)^{16} = 0.315](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B17%2C1%7D.%280.1%29%5E%7B1%7D.%280.9%29%5E%7B16%7D%20%3D%200.315)
![P(X = 2) = C_{17,2}.(0.1)^{2}.(0.9)^{15} = 0.280](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B17%2C2%7D.%280.1%29%5E%7B2%7D.%280.9%29%5E%7B15%7D%20%3D%200.280)
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.167 + 0.315 + 0.280 = 0.762](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%3D%200.167%20%2B%200.315%20%2B%200.280%20%3D%200.762)
0.762 = 76.2% probability that this shipment is accepted
The question only supplied us with the mass of the new baby which is:
![1.5\cdot10^3\operatorname{kg}\Rightarrow1500\operatorname{kg}](https://tex.z-dn.net/?f=1.5%5Ccdot10%5E3%5Coperatorname%7Bkg%7D%5CRightarrow1500%5Coperatorname%7Bkg%7D)
This mass is too bizzarre & unrealistic for
Answer:
1/3
Step-by-step explanation:
hope this helps :D
brainliest pls?
1.BD=AC
AC^2=(9^2)+(6^2)-2(9)(6)COS115
AC^2=117-108COS115
AC=√71.358
AC=8.447//
Are you kidding me, bro? lol
276-115=131
131 divided by 276 = 0.47
0.47% increase