Answer:
Using the concept of speed-distance-time, we find that the diameter of the circular path is 12 inches.
Step-by-step explanation:
Given that the object completes one revolution in 8 seconds. So, the circular speed of the object is 2πr/8 = πr/4 inches per seconds, where r is the radius of the circular path. The speed can be written as:
πr/4 inches per seconds = 60πr/4 inches per minute = 15πr inches per minute.
But, it is given that the circular speed of the object is 90π inches per minute. So,
90π = 15πr
r = 6 inches
Since, diameter is the double of radius, d = 2r = 12 inches.
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Answer:
20/9 or 2 2/9 or 2.2 repeating
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
Given the data : 29, 2, 28, 30, 26, 31
Outlier ;
Lower :Q1 - (1.5 * IQR)
Upper : Q3 + (1.5 * IQR)
Q1 = Lower quartile ; Q3 = upper quartile ; IQR = Interquartile range
Using calculator :
Q1 = 26
Q3 = 30
IQR = (Q3 - Q1) = 30 - 26 = 4
Lower : 26 - (1.5 * 4) = 20
Upper : 30 + (1.5 * 4) = 36
Hence, the number in the given data which falls outside the range is 2
3 x + 7 = 2
3 x = 2 - 7
3 x = - 5
x = - 5/3
3 ( -5/3) + 7 = - 15/3 + 7 = - 5 + 7 = 2
Answer:
6x + 7y +84
Step-by-step explanation:
