Information or signals entered into a computer system is

Logs.

Like data logs. Sometimes people make these logs to keep tabs on other people or to get important information put down somewhere that way it is saved and can be looked back upon later. Anytime someone makes an action on the computer, it makes a TMP file representing a log of what you want it to do before the computer quickly get's rid of the file.

-Ɽ3₮Ɽ0 Ⱬ3Ɽ0

A) Most points are between these limits, so the process appears to be in control with respect to variability.

D) The value of S on the 5th day lies above the UCL, so an out-of-control signal is generated.

Step-by-step explanation:

We have the data of 30 days of daily samples, of size n=200. The data is expressed in "number of non-conforming chips". To graph the data in a p-chart, we have to calculate the proportion, so we have to divide the number of non-conforming chips in each sample by the sample size in order to get the proportion.

The data of non-conforming chips es:

[916241836196268253017161915201322]

The proportion of non-conforming chips is calculated as:

Then, the proportions of non-conforming chips are:

[0.0450.080.120.090.180.0950.030.130.040.1250.150.0850.080.0950.0750.10.0650.110.0450.1150.090.0750.0750.1250.160.0950.060.1150.0750.13

]

For the p-chart we calculate the average proportion:

Then, we calculate the control limits as:

With these values, we can graph the data in the p-chart.

We can see that most points are within these limits, as it is expected (we are calculating the control limits with the same data we are charting). There are 5 out of 30 (one sixth of the samples) that are outside the control limits but there are not more than one consecutive points outside the contorl limits, so we can conclude that the process is in control.

The value of the 5th day (p=0.180) is over the UCL (UCL=0.149), so an out-of-control signal is generated. As the 6th day sample is in control (p=0.095), we could conclude that a special condition makes the 5th day sample lies outside the control limits.

A) Most points are between these limits, so the process appears to be in control with respect to variability.

D) The value of S on the 5th day lies above the UCL, so an out-of-control signal is generated.

Step-by-step explanation:

We have the data of 30 days of daily samples, of size n=200. The data is expressed in "number of non-conforming chips". To graph the data in a p-chart, we have to calculate the proportion, so we have to divide the number of non-conforming chips in each sample by the sample size in order to get the proportion.

The data of non-conforming chips es:

[916241836196268253017161915201322]

The proportion of non-conforming chips is calculated as:

Then, the proportions of non-conforming chips are:

[0.0450.080.120.090.180.0950.030.130.040.1250.150.0850.080.0950.0750.10.0650.110.0450.1150.090.0750.0750.1250.160.0950.060.1150.0750.13

]

For the p-chart we calculate the average proportion:

Then, we calculate the control limits as:

With these values, we can graph the data in the p-chart.

We can see that most points are within these limits, as it is expected (we are calculating the control limits with the same data we are charting). There are 5 out of 30 (one sixth of the samples) that are outside the control limits but there are not more than one consecutive points outside the contorl limits, so we can conclude that the process is in control.

The value of the 5th day (p=0.180) is over the UCL (UCL=0.149), so an out-of-control signal is generated. As the 6th day sample is in control (p=0.095), we could conclude that a special condition makes the 5th day sample lies outside the control limits.

"Test Phase " is the correct choice.

Explanation:

NTQ

Explanation:

The given sequence is

BHE : FLI : JPM

If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.

B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.

Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.

H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.

Alphabets on third places are E, I, M. Difference between their place vales is 4.

E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.

Therefore, the next term is NTQ.

Energy equation from this week’s notes, your answer from #5, and Plank’s constant (6.63E-34) to find the approximate energy of this photon

Explanation:

1.The amount of energy in those photons is calculated by this equation, E = hf, where E is the energy of the photon in Joules; h is Planck's constant, which is always 6.63 * 10^-34 Joule seconds; and f is the frequency of the light in hertz

2.The first is Planck's equation, which was proposed by Max Planck to describe how energy is transferred in quanta or packets. Planck's equation makes it possible to understand blackbody radiation and the photoelectric effect. The equation is:

E = hν

where

E = energy

h = Planck's constant = 6.626 x 10-34 J·s

ν = frequency

x=float(input("Enter a number: "))

sub=(x-int(x))

print(sub)

Explanation:

Got it Right

B ). When cleaning a vessel, use regular  household detergents and cleaners.

Explanation:

Boat cleaning or vessel cleaning is a tedious job. It takes a lot of hard work to clean a vessel. We have to be careful while maintaining the vessel.

The products that we use to clean and maintain the vessels needs to chosen carefully. Many boat cleaning products have adverse effects on the marine life and aquatic animals. it also tends to pollute the water bodies.

These products may be corrosive in nature, toxic or poisonous.

So we must use the regular household detergents and cleaners to clean our vessels. The regular detergents do not cause any harm to the nature and the water bodies.

We can use phosphate free no detergent soaps which may include citrus based soap or vegetable based soaps.

Therefore, while cleaning or maintaining vessels , use the regular household detergents and cleaners.  

Thus option (B) is correct.

Learn more :

link

understanding others first before starting to explain your own point

Explanation:

Stephen covey believes the key to effective communication is understanding others first before starting to explain your own point. Communication is an important skill in life, in which one must have a good character in order to communicate effectively.