12 ÷ 75 = 0.16.
So she runs 0.16 miles per minute.
Hope it helps!
To solve this, we use the z test.
The formula:
z = (x – u) / s
where x is sample value = 20, u is the mean = 15, and s is
the standard deviation = 2.5
z = (20 – 15) / 2.5
z = 2
Since we are looking for values greater than 20, this is
right tailed test. We use the standard distribution tables to find for P.
P = 0.0228
Therefore:
number of students = 100 * 0.0228 = 2.28
<span>2 to 3 students will get greater than 20 measurement</span>
Part A: y-intercept = 12 miles
Part B: Find the average rate of change: 36 - 20/2 = 16/2 = 8 mph
This tells us that the bird can travel long distances.
Part C: 172 - 12 = 160 miles at a rate of 8 mph
160 mi ÷ 8mph = 20 hrs
Domain of function = 0 ≤ x ≤ 20
Average of 5 games = 13
Total of 5 games = 13 x 5 = 65
Average of 6 games = 17
Total of 6 games = 102
6th game = 102 - 65 = 37
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Answer: Her score for the 6th game as 37.
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.2 converted to a fraction would be 1/5
