A cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds
2 answers:
Answer:
Your friend is not correct.
It takes the cheetah Approximately 21.67 second running at 90 ft/s to reach the moving antelope.
Step-by-step explanation:
Consider the provided information.
A cheetah can run at top speed for only about 20 seconds and the top speed of cheetah is 90 feet per second
That means in 20 second cheetah will cover a distance of:
20 × 90 = 1800 feet
In 20 seconds cheetah will cover a distance of 1800 feet.
It is given that the antelope runs 60 feet per sec
That means in 20 second antelope will cover a distance of:
20 × 60 = 1200 feet
In 20 second antelope will cover a distance of 1200 feet.
Now add the distance ran initially, because the antelope was ahead of the cheetah from 650 feet.
1200 feet + 650 feet = 1850 feet
But the maximum distance cheetah can cover is 1850 feet. Thus the antelope is still 50 ft ahead of the cheetah.
Therefore, the antelope is probably safe.
Hence, Your friend is not correct.
The speed of cheetah is 30 feet per second more that antelope.
Cheetah need to cover 50 feet to catch antelope.
20 seconds + 1.67 second = 21.67 seconds
Thus, it takes the cheetah Approximately 21.67 second running at 90 ft/s to reach the moving antelope.
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Answer:
(a) The mean is 52 and the median is 55.
(b) The first quartile is 44 and the third quartile is 60.
(c) The value of range is 33 and the inter-quartile range is 16.
(d) The variance is 100.143 and the standard deviation is 10.01.
(e) There are no outliers in the data set.
(f) Yes
Step-by-step explanation:
The data provided is:
S = {55, 56, 44, 43, 44, 56, 60, 62, 57, 45, 36, 38, 50, 69, 65}
(a)
Compute the mean of the data as follows:
Thus, the mean is 52.
The median for odd set of values is the computed using the formula:
Arrange the data set in ascending order as follows:
36, 38, 43, 44, 44, 45, 50, 55, 56, 56, 57, 60, 62, 65, 69
There are 15 values in the set.
Compute the median value as follows:
The 8th observation is, 55.
Thus, the median is 55.
(b)
The first quartile is the middle value of the upper-half of the data set.
The upper-half of the data set is:
36, 38, 43, 44, 44, 45, 50
The middle value of the data set is 44.
Thus, the first quartile is 44.
The third quartile is the middle value of the lower-half of the data set.
The upper-half of the data set is:
56, 56, 57, 60, 62, 65, 69
The middle value of the data set is 60.
Thus, the third quartile is 60.
(c)
The range of a data set is the difference between the maximum and minimum value.
Maximum = 69
Minimum = 36
Compute the value of Range as follows:
Thus, the value of range is 33.
The inter-quartile range is the difference between the first and third quartile value.
Compute the value of IQR as follows:
Thus, the inter-quartile range is 16.
(d)
Compute the variance of the data set as follows:
Thus, the variance is 100.143.
Compute the value of standard deviation as follows:
Thus, the standard deviation is 10.01.
(e)
An outlier is a data value that is different from the remaining values.
An outlier is a value that lies below 1.5 IQR of the first quartile or above 1.5 IQR of the third quartile.
Compute the value of Q₁ - 1.5 IQR as follows:
Compute the value of Q₃ + 1.5 IQR as follows:
The minimum value is 36 and the maximum is 69.
None of the values is less than 20 or more than 80.
Thus, there are no outliers in the data set.
(f)
Yes, the data provided indicates that the Walmart is meeting its goal for reducing the number of hourly employees