Would appreciate if someone would help:) thanks!

OLEGan [10]

Tangent (R) = opposite / adjacent = 5 / 12 = 0.416666
arc tangent (.416666) = angle R = 22.62 Degrees

7 0

3 years ago

Can someone please explain how to find x and y I don’t understand this at all

Sloan [31]

3y is the supplement of 141 so that means 3y+140=180 so y is 13.33. Since they give you 68, 68+13.33+x=180. So 81.33+x=180 and x=98.66.

4 0

3 years ago

g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa

adoni [48]

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 $\sigma = 0.0133$

Step-by-step explanation:

From the question we are told that

The upper specification is $USL = 1.68 \ mm$

The lower specification is $LSL = 1.52 \ mm$

The sample mean is $\mu = 1.6 \ mm$

The standard deviation is $\sigma = 0.03 \ mm$

Generally the capability index in mathematically represented as

$Cpk = min[ \frac{USL - \mu }{ 3 * \sigma } , \frac{\mu - LSL }{ 3 * \sigma } ]$

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

$Cpk = min[ \frac{1.68 - 1.6 }{ 3 * 0.03 } , \frac{1.60 - 1.52 }{ 3 * 0.03} ]$

=> $Cpk = min[ 0.88 , 0.88 ]$

So

$Cpk = 0.88$

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

$\frac{\mu - LSL }{ 3 * \sigma } = 2$

So

$\frac{ 1.60 - 1.52 }{ 3 * \sigma } = 2$

=> $0.08 = 6 \sigma$

=> $\sigma = 0.0133$

So

$\frac{ 1.68 - 1.60 }{ 3 * 0.0133 }$

=> $2$

Hence

$Cpk = min[ 2, 2 ]$

So

$Cpk = 2$

So $\sigma = 0.0133$ is the value of standard deviation required

3 0

3 years ago

A circuit board manufacturer can produce 2,000 boards per hour. How many days will it take to produce 16,500 if the company has

mrs_skeptik [129]

Hi Mrsdiaz1jd1977owwfvt!
OK so 2,000 boards per hour in 10 hours is 2,000*10=20,000 boards in 10 hours. So it can produce 16,500 boards in less than a day.
I may be wrong but I hope this helps!

5 0

3 years ago

Read 2 more answers

Simplify (6^5/7^3)^2

o-na [289]

Simplify the following:
(6^5/7^3)^2

7^3 = 7×7^2:
(6^5/(7×7^2))^2

7^2 = 49:
(6^5/(7×49))^2

7×49 = 343:
(6^5/343)^2

6^5 = 6×6^4 = 6 (6^2)^2:
((6 (6^2)^2)/343)^2

6^2 = 36:
((6×36^2)/343)^2

| | 3 | 6
× | | 3 | 6
| 2 | 1 | 6
1 | 0 | 8 | 0
1 | 2 | 9 | 6:
((6×1296)/343)^2

6×1296 = 7776:
(7776/343)^2

(7776/343)^2 = 7776^2/343^2:
7776^2/343^2

| | | | 7 | 7 | 7 | 6
× | | | | 7 | 7 | 7 | 6
| | | 4 | 6 | 6 | 5 | 6
| | 5 | 4 | 4 | 3 | 2 | 0
| 5 | 4 | 4 | 3 | 2 | 0 | 0
5 | 4 | 4 | 3 | 2 | 0 | 0 | 0
6 | 0 | 4 | 6 | 6 | 1 | 7 | 6:
60466176/343^2

| | | 3 | 4 | 3
× | | | 3 | 4 | 3
| | 1 | 0 | 2 | 9
| 1 | 3 | 7 | 2 | 0
1 | 0 | 2 | 9 | 0 | 0
1 | 1 | 7 | 6 | 4 | 9:

Answer: 60466176/117649

6 0

3 years ago