Answer- 8:5:7
Explanation:
Find the greatest common factor for all of them. Which is 6.
Then you just divide each number by 6
48/6= 8
30/6= 5
42/6= 7
Making it 8:5:7
Answer:
The value is 
Step-by-step explanation:
From the question we are told that
The amount charge per year is 
The number of members it will have at this amount is 
The amount amount increase that will lead to the loss of a single member is 
Generally the total amount the club would obtain from its members is mathematically represented as

Now let x denote the number of member lost
Hence

=> 
=> 
Thus the number of members that be removed to give the maximum income from dues is obtained by differentiating the above equation and equating it to zero

=> 
So from
we have


So the amount the club should charge is


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
I think the answer is 4
If my memory serves me right then you have to do Q3-Q1
This is 6-2
The answer would be 4
Hope this is right and it helps :)
Answer:
Step-by-step explanation:
I can’t really see it what it say