<span>A major software corporation has decided that its operating system is too difficult for the average user. Who should they hire to make the system easier to understand?
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The main difference between transnet and perfect competitor is that a transnet is a mode of transportation whereas a perfect competitor is a person who sells an identical product with others, they can garner a small share of their income, and that the cannot control the price of the market of their product.
Answer:
Hello. You did not put the answer options, but the role of moss in the transition to stage III is the creation of protonemes where the gametes that will act in stage III will be formed.
Explanation:
Stage III consists of the fertilization of the female gamete of the moss with the male gamete of another moss, thus creating a zygote that will give rise to other mosses in the future.
Moss is very important for this phase, because it will be responsible for creating a structure called protonemas. In the protonemas are the gametophytes that will have a structure called gametangium that will allow the formation of gametes.
Answer:
- <u><em>D. No, because $100,000 is much greater than the values used in the experiment.</em></u>
<u><em></em></u>
Explanation:
<em>Correlations</em>, when have strong correlation coefficients, which r = 0.9 is, may be good predictors within the limits of the range of the data.
Trying to extrapolate the <em>linear relationship</em> between the variables, <em>x = advertising spending and y= product sales</em>, way beyond the limits of the data used for the study, is too risky, because the data may be linear just for some stages (ranges) but behave very different in other ranges.
As, the option D. states, <em>$100,000 is much greater that the values used in the experiment</em>; hence, the correlation would likely would not be a good predictor for that input.
Answer:
The standard deviation is 4.83 inches.
Explanation:
We are given that the average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches.
The snowfall in Chillyville exceeds 60 inches in 15% of the years, we have to find the standard deviation.
Let X = <u><em>the average yearly snowfall in Chillyville</em></u>.
The z-score probability distribution for the normal distribution is given by;
Z = 
~ N(0,1)
where, 
= mean amount of rainfall = 55 inches
= standard deviation
Now, it is stated that the snowfall in Chillyville exceeds 60 inches in 15% of the years, that means;
P(X > 60 inches) = 0.15
P( > 
) = 0.15
P(Z > ) = 0.15
In the z table, the critical value of z that represents the top 15% of the area is given as 1.0364, that means;
= 4.83 inches
Hence, the standard deviation is 4.83 inches.
