Using it's concept, it is found that the mean absolute deviation of 2.8 means that the heights differ from the mean by an average of 2.8 inches.
<h3>What is the mean absolute deviation of a data-set?</h3>
- The mean of a data-set is given by the sum of all observations divided by the number of observations.
- The mean absolute deviation of a data-set is the sum of the absolute value of the difference between each observation and the mean, divided by the number of observations.
- The mean absolute deviation represents the average by which the values differ from the mean.
In this problem, the mean is given by:
M = (65 + 58 + 64 + 61 + 67)/5 = 63.
Hence the mean absolute deviation is given by:
MAD = (|65-63| + |58-63| + |64-63| + |61-63| + |67-63|)/5 = 14/5 = 2.8.
The mean absolute deviation of 2.8 means that the heights differ from the mean by an average of 2.8 inches.
More can be learned about mean absolute deviation at brainly.com/question/3250070
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Answer:
D. 25 ft.
Step-by-step explanation:
To find the missing length of any right triangle, you can use the Pythagorean Theorem (A^2+B^2=C^2) to solve. A and B represent the legs (which we do have) and C represents the hypotenuse (which we are trying to find). Insert our known values into the theorem, and we will get 24^2+7^2=C^2. Now we just do the basic math for the numbers that we have (24^2 is 576<em>;</em><em> </em>7^2 is 49), and that added up would equal to 625. Now, since C is being squared on the other side, we now have to root both sides. On the side where C is, the radical and the square will cancel out each other, leaving us with just C. But, on the side where 625 is, it should root to 25. Since C is by itself now, the number on the other side (25) will be our answer. Therefore, the missing side is 25 feet long.
Answer:
which of the following is not a possible answer to a probability question 2 by 3 that is not a probability questions answer because always a probability will come in the probability of an event ok so 232 want to come
W-69 = 69 less than a number w
