Answers to Sample WASL Questions
Answer to #1
This is a rather odd solution. We need to explore all possibilities of shapes for the fenced are. First lets explore a square fenced area. With 40 feet of fencing, we could have 10 feet on each side that would create an area of 100 square feet. This doesn't seem large enough to be a maximum area. How about a circle? Circumferance is pi times diameter so we can set up the equation 40=pi x d solve for d and find that d = 12.7 and r would then = 6.4. We can now determine area of the circular area. Area = pi times r squared. Area = 128 square feet. This is the maximum area.
Answer to question #2
6ab + 4b - 3a - 2ab + b(3a - 4)+ 9
6ab + 4b - 3a - 2ab + 3ab - 4b + 9 Distributing
6ab - 2ab + 3ab + 4b - 4b - 3a + 9 Grouping like terms
7ab - 3a + 9 Combining like terms
Answer to question #3
If we have 50 tickets numbered 1 through 50, we have 25 even numbers, and 25 odd numbers. So the probability of drawing an odd numbered ticket is 25 out of 50 or .5 But we are looking for the probability of drawing 3 odd tickets in a row without replacing the tickets we have drawn. If the first drawn is odd them there are 24 remaining odd tickets out of the 49 remaining tickets, which is a probability of .49 If the second drawn is also an odd there are 23 remaining odd tickets of the 48 remaining tickets which is a probability of .48 to determine the probability of drawing all three we need to multiply all these probabilities together. Our answer is a probability of .12
Answer to question #4
To determine the percentage of bulbs that passed on Saturday, we also need to know how many bulbs passed on Saturday. To datermine the percentage of bulbs that passed on Sunday, we also need to know how many failed on Sunday.
Keep Checking, there are more questions and answers coming!!!
Click here to return to the home page.