13. In engineering notation, the number 0.000452 is expressed as
1 answer:
You might be interested in
Answer:
A.The mean would increase.
Step-by-step explanation:
Outliers are numerical values in a data set that are very different from the other values. These values are either too large or too small compared to the others.
Presence of outliers effect the measures of central tendency.
The measures of central tendency are mean, median and mode.
The mean of a data set is a a single numerical value that describes the data set. The median is a numerical values that is the mid-value of the data set. The mode of a data set is the value with the highest frequency.
Effect of outliers on mean, median and mode:
- Mean: If the outlier is a very large value then the mean of the data increases and if it is a small value then the mean decreases.
- Median: The presence of outliers in a data set has a very mild effect on the median of the data.
- Mode: The presence of outliers does not have any effect on the mode.
The mean of the test scores without the outlier is:
*Here <em>n</em> is the number of observations.
So, with the outlier the mean is 86 and without the outlier the mean is 86.9333.
The mean increased.
Since the median cannot be computed without the actual data, no conclusion can be drawn about the median.
Conclusion:
After removing the outlier value of 72 the mean of the test scores increased from 86 to 86.9333.
Thus, the the truer statement will be that when the outlier is removed the mean of the data set increases.
We can set it up like this, where <em>s </em>is the speed of the canoeist:
To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
![s(s-5)[\frac{18}{s} + \frac{4}{s-5} = 3] \\ 18(s-5)+4s=3s(s-5) \\ 18s - 90+4s=3 s^{2} -15s](https://tex.z-dn.net/?f=s%28s-5%29%5B%5Cfrac%7B18%7D%7Bs%7D%20%2B%20%5Cfrac%7B4%7D%7Bs-5%7D%20%3D%203%5D%20%5C%5C%2018%28s-5%29%2B4s%3D3s%28s-5%29%20%5C%5C%2018s%20-%2090%2B4s%3D3%20s%5E%7B2%7D%20-15s)
If we rearrange this, we can turn it into a quadratic equation and factor:
Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.
The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.
To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
If we rearrange this, we can turn it into a quadratic equation and factor:
Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.
The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.