Can someone help me solve this problem?
2 answers:
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Answer:
Step-by-step explanation:
Given
Girls = 3
Boys = 5
Required
How many ways can 2 boys and girls be chosen?
The keyword in the question is chosen;
This implies combination and will be calculated as thus;
For Boys;
n = 5 and r = 2
For Girls;
n = 3 and r = 2
Total Selection is calculated as thus;
Well, keeping in mind there are 12 inches in 1 foot, 20 feet is 20 * 12, or 240 inches.
Answers:
- 10
- decrease
- increase
- does
- incorrect
So it would look like this:
The MAD of the data in the table is <u> 10 </u>.If the additional drawings are Amy's, the MAD of the date set will <u> decrease </u>.If they are Emily's, the MAD will <u> increase </u>.The MAD of the new data set <u> does </u> depend on whether it was Amy or Emily who turned in the additional drawings. So, Greg is <u> incorrect </u>.
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Explanation:
The MAD will have us find the mean first, which I'll call xbar
To find xbar, we add up the values and then divide by n which is the number of items in the set.
xbar = (sum of items)/n
xbar = (6+34+35+37+43)/5
xbar = 155/5
xbar = 31
Then we'll subtract this xbar from from each x value of the data set. Use absolute value bars to make sure the result isn't negative. This forms the third column in each table shown below. The value in yellow is the average of the stuff in the third column (ignoring the yellow value itself of course). So that's how we get a MAD of 10 for the original data set.
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When we consider case 1, which is where Amy made those 25 additional drawings, then her count goes from 6 to 6+25 = 31. We repeat the steps earlier and we get a MAD of 3.2, which is a decrease from 10 earlier.
Why is this? Well notice how Amy's count, before those 25 extra drawings were done, was far lower than the rest of the class. Her being an outlier will make the MAD fairly big. The larger the MAD, the more spread out the data is. If you condense the data set, then the MAD shrinks.
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Now onto case 2. We'll reset Amy's count back to 6 and instead add 25 to Emily's count to go from 43 to 43+25 = 68
This will increase the MAD because we're effectively spreading the data out more (since Emily's value is becoming more of an outlier). The same steps to compute the MAD will be done as earlier. This time we get 13.2 as the MAD.
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Referring back to the previous 2 sections, we can see that the MAD of the new set will depend on whether Amy or Emily submitted those 25 extra drawings. Therefore, Greg is incorrect in thinking that the MAD will stay the same.
The only time the MAD stays the same is if we can keep the spread of the data the exact same. That would mean we have to add the same number to each person to keep things balance. Think of it like a see-saw. If we add something to one side, then we have to do the same thing to the other side; otherwise, things will change.
