Answer:
oof that's gotta suck bro
2+1=3
15/3=5
split the 3 5's in ratio 2:1 and you get 10:5
Answer:
B. load-distance model
Step-by-step explanation:
A. trial and error
Trial and error is "a fundamental method of problem solving. It is characterised by repeated, varied attempts which are continued until success". But this method is not the best in order to compare effectiveness of layouts
B. load-distance model
The load-distance method is a "mathematical model used to evaluate locations based on proximity factors. The objective is to select a location that minimizes the total weighted loads moving into and out of the facility. The distance between two points is expressed by assigning the points to grid coordinates on a map". And that's the correct option since we are trying to measure the effectiveness of layouts quantitatively.
C. exponential smoothing
This is "a rule of thumb technique for smoothing time series data using the exponential window function". Wheighting observations using the exponential function. But this is a techinique used to smooth s time series not to compare effectiveness of layouts.
D.process control charts
The Control Chart is a "graph used to study how a process changes over time with data plotted in time order". But we don't want to see how the process changes the objective is quantitatively compare the effectiveness of layouts, and this one is not the best option for this.
E. mean absolute deviation (MAD)
The median absolute deviation(MAD) is "a robust measure of how spread out a set of data is. The variance and standard deviation are also measures of spread, but they are more affected by extremely high or extremely low values and non normality". But again is just a measure of spread and not allow to compare effectiveness of layouts.
Answer:
Step-by-step explanation:
For question #1:
For question #2:
Answer:
Yes
Step-by-step explanation:
First, suppose that nothing has changed, and possibility p is still 0.56. It's our null hypothesis. Now, we've got Bernoulli distribution, but 30 is big enough to consider Gaussian distribution instead.
It has mean μ= np = 30×0.56=16.8
standard deviation s = √npq
sqrt(30×0.56×(1-0.56)) = 2.71
So 21 is (21-16.8)/2.71 = 1.5494 standard deviations above the mean. So the level increased with a ˜ 0.005 level of significance, and there is sufficient evidence.
