While you cannot solve 946x+642y=911 for numerical values of x and y, you can indeed solve 946x+642y=911 first for x and then for y:
-911 - 642y
For x: 946x+642y=911 becomes 946x = 911 - 642y, or x = ------------------
946
911-946x
For y: 946x+642y=911 becomes 642y = 911-946x, or y = ---------------
642
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
Answer:
Answer: 19.2
Step-by-step explanation:
To calculate percentage, multiply the number 80 by 0.24, and the answer is 19.2.
Is there a picture or options to this ?