Answer:
The Answer is: There are 8 small boxes and 9 large boxes. See explanation below for variables and variable definitions.
Step-by-step explanation:
Let s = small boxes. Let b = large boxes.
s + b = 17
You can solve for s:
s = 17 - b
You can solve for b:
b = 17 - s
10 times the number of small boxes plus 24 times the number of large boxes is equal to 296 granola bars.
10s + 24b = 296
Substitute:
10(17 - b) + 24b = 296
170 - 10b + 24b = 296
14b = 296 - 170
14b = 126
b = 126 / 14 = 9 large boxes
Find the number of small boxes, s:
s = 17 - b = 17 - 9 = 8 small boxes
There are 8 small boxes and 9 large boxes.
Proof:
10(8) + 24(9) = 296
80 + 216 = 296
296 = 296
Answer:
-16
Step-by-step explanation:
m^2 + n^2
-5^2 + 3^2 = -16
Answer:
12. The second one
13. The first one
14. The last one
15. The first one
16. 30 degrees
Step-by-step explanation:
Complete Question
A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?
Answer:
The value required is
Step-by-step explanation:
From the question we are told that
The upper specification is 
The lower specification is
The sample mean is
The standard deviation is 
Generally the capability index in mathematically represented as
![Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%20%5Cfrac%7BUSL%20-%20%20%5Cmu%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%20%2C%20%20%5Cfrac%7B%5Cmu%20-%20LSL%20%7D%7B%203%20%2A%20%20%5Csigma%20%7D%20%5D)
Now what min means is that the value of CPk is the minimum between the value is the bracket
substituting value given in the question
![Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%20%5Cfrac%7B1.68%20-%20%201.6%20%7D%7B%203%20%2A%20%200.03%20%7D%20%20%2C%20%20%5Cfrac%7B1.60%20-%20%201.52%20%7D%7B%203%20%2A%20%200.03%7D%20%5D)
=> ![Cpk = min[ 0.88 , 0.88 ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%200.88%20%2C%200.88%20%20%5D)
So

Now from the question we are asked to evaluated the value of standard deviation that will produce a capability index of 2
Now let assuming that

So

=> 
=> 
So

=> 
Hence
![Cpk = min[ 2, 2 ]](https://tex.z-dn.net/?f=Cpk%20%20%3D%20%20min%5B%202%2C%202%20%5D)
So

So
is the value of standard deviation required