Absolute value because for any number given the y or answer will be no less than 0 it will.always be positive
Answer:
B and D
Step-by-step explanation:
B:
Anything to a negative power means that it is 1/that to the positive power.
E.g. x^-1 = 1/x^1
In other words, anything to the power of a negative switches sides of a fraction (i.e. if in numerator moves to denominator and vice versa.)
1/x^-1 = 1/1/x^1 which is just equal to x, because there are x number of 1/xs in one (1/x * x =1) Therefore Option B is equal to just x.
D: (assuming the first given term is x^1/3 and not X1/3 (?) Correct me if I'm wrong).
x^1/3 * x^1/3 * x^1/3 is also equal to just x.
This is because when multiplying together terms with the same base (x in this case) the exponents just add together, so:
x^1/3 * x^1/3 * x^1/3 = x^(1/3 +1/3 +1/3) = x^1 = x.
Therefore B and D are equivalent because they both equal x.
Hope this helped!
54981.35 is the square root
Answer:
Probability of stopping the machine when
is 0.0002
Probability of stopping the machine when
is 0.0013
Probability of stopping the machine when
is 0.0082
Probability of stopping the machine when
is 0.0399
Step-by-step explanation:
There is a random binomial variable
that represents the number of units come off the line within product specifications in a review of
Bernoulli-type trials with probability of success
. Therefore, the model is
. So:
![P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%209%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%209%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%209%7D%20%280.91%29%5E%7B9%7D%280.09%29%5E%7B6%7D%2B...%2B%7B%2015%20%5Cchoose%2015%7D%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0002%20)
![P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2010%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%2010%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2010%7D%280.91%29%5E%7B10%7D%280.09%29%5E%7B5%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0013%20)
![P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2011%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%2011%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2011%7D%280.91%29%5E%7B11%7D%280.09%29%5E%7B4%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0082)
![P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2012%29%20%3D%201-%20P%20%28X%20%5Cgeq%2012%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2012%7D%280.91%29%5E%7B12%7D%280.09%29%5E%7B3%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0399%20)
Probability of stopping the machine when
is 0.0002
Probability of stopping the machine when
is 0.0013
Probability of stopping the machine when
is 0.0082
Probability of stopping the machine when
is 0.0399