Answer:top left corner
Step-by-step explanation:
we conclude that at 5:00 p.m. there are 8 more inches of snow than at 8:00 a.m.
<h3>How many more inches of snow were on the ground at 5:00 p.m. than at 8:00 a.m.?</h3>
We know that at 8:00 a.m. there were t inches of snow in the ground.
At 5:00 p.m. there were 3t inches of snow in the ground.
Then the difference between the heights of the snow is:
3t- t = 2t
And we know that at 5:00 p.m. there were 12 inches of snow then we can solve the linear equation for t:
3t = 12in
t = (12in)/3 = 4 in
Replacing that in the difference of heights:
2t = 2*4in = 8in
From this, we conclude that at 5:00 p.m. there are 8 more inches of snow than at 8:00 a.m.
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Answer:
Its about 36 feet
Step-by-step explanation:
I took this question before.
The answers are given below:-
- If the data is symmetrical, then the mean is the best measure of central tendency to use, and the standard deviation is the best spread to use.
- If the data is unsymmetrical, the median is the best measure of central tendency to use, and the inter-quarterly range is the best spread to use.
<h3>What are symmetrical and asymmetrical data?</h3>
A histogram for symmetrical data will give a symmetrical shape, and the mean, median and mode will all be the same value. Therefore, the best measure of the central tendency to use is the mean. The standard deviation shows how far away the values in a given data set are from the mean, and since the mean is used as the measure of central tendency in this case, the standard deviation should be used as the spread.
A histogram for a an asymmetric data set will give an asymmetric shape, and the mean is not always equal to the median. Therefore, the best measure of central tendency to use is the median. The inter-quarterly range shows the range of the middle 50% of a certain data, which is considered from the median value. Since the median is used as the measure of central tendency in this case, it is wise to use the inter-quarterly range as the measure of spread.
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