I think that it gets smaller. (not entirely sure)
The decimal you put would be “two ten-millionths” if read out loud. If the allowable amount is 10 parts per billion, it would be 1 part per 100 million. Because 2 parts per 10 million is also 20 parts per 100 million, it is not not an acceptable amount
Answer:
a) ![P(X = 1) = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%200.38742)
b) ![P(X = 3) = 0.05740](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%200.05740)
c) ![P(X = 9) = 0.00000](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%200.00000)
d) ![P(X \geq 5) = 0.00163](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%200.00163)
Step-by-step explanation:
For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.
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 combinatios 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.
In this problem
There are 10 containers, so
.
A food-packaging apparatus underfills 10% of the containers, so
.
a) This is P(X = 1)
![P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B10%2C1%7D.%280.1%29%5E%7B1%7D.%280.9%29%5E%7B9%7D%20%3D%200.38742)
b) This is P(X = 3)
![P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B10%2C3%7D.%280.1%29%5E%7B3%7D.%280.9%29%5E%7B7%7D%20%3D%200.05740)
c) This is P(X = 9)
![P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.1%29%5E%7B9%7D.%280.9%29%5E%7B1%7D%20%3D%200.00000)
d) This is
.
Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:
![P(X < 5) + P(X \geq 5) = 1](https://tex.z-dn.net/?f=P%28X%20%3C%205%29%20%2B%20P%28X%20%5Cgeq%205%29%20%3D%201)
![P(X \geq 5) = 1 - P(X < 5)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29)
In which
![P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%28X%20%3C%205%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%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%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_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B10%2C0%7D.%280.1%29%5E%7B0%7D.%280.9%29%5E%7B10%7D%20%3D%200.34868)
![P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B10%2C1%7D.%280.1%29%5E%7B1%7D.%280.9%29%5E%7B9%7D%20%3D%200.38742)
![P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B10%2C2%7D.%280.1%29%5E%7B2%7D.%280.9%29%5E%7B8%7D%20%3D%200.1937)
![P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B10%2C3%7D.%280.1%29%5E%7B3%7D.%280.9%29%5E%7B7%7D%20%3D%200.05740)
![P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B10%2C4%7D.%280.1%29%5E%7B1%7D.%280.9%29%5E%7B9%7D%20%3D%200.38742)
So
![P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837](https://tex.z-dn.net/?f=P%28X%20%3C%205%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%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%3D%200.34868%20%2B%200.38742%20%2B%200.19371%20%2B%200.05740%20%2B%200.01116%20%3D%200.99837)
Finally
![P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29%20%3D%201%20-%200.99837%20%3D%200.00163)
Answer:
-0.25
Step-by-step explanation:
-8= 2/x
multiply both sides by x -> -8x =2
divide both sides by -8 -> x=2/-8
Ans= -1/4 = -0.25
Answer: i am not sure but thanks for the points have a good day
Step-by-step explanation: