Hi everybody, I've failed the math portion of the CBEST twice now, but passed the reading and writing. Does anyone have any advice on passing the math portion of the exam? Thanks!

Okay: 1. Estimate the product of 2.42 • 1.67, and show (or at least tell me) your work. 2. Because you've been asked to contribute a gallon of soda for a teachers' meeting, you picked up two six-packs of Hansom Cab's lime-and-cactus-fruit soda and stashed them in the back of the teachers' lounge fridge. If each can contains 12 ounces of soda, did you buy enough soda? Again, show (or at least tell me) your work.

1. You round, then multiply 2.4 (1.7)=4.08 2. 1 gallon= 128 ounces 6 cans (12 ounces)= 72 ounces times 2= 144 ounces. (That should be right) So yes, there is enough soda

1. Not quite. You used a calculator or pencil and paper for the multiplication, yes? The point of estimating is to produce inputs that you can process in your head. Round to the first significant digit: 2.42 rounds to 2, 1.67 rounds to 2, and 2 times 2 is 4. 2. Good: so you know liquid measurement. 3. A quilted wall hanging measures 46 inches long by 22 inches wide. If a one-inch border is added to all four sides, what is the perimeter in yards of the finished hanging? 4. What's 1,234.567 in scientific notation? 5. If the circumference of a table is 9 m, what is a good estimate of its diameter? 6. An isosceles triangle has a perimeter of 8 ft. If its shortest side is 2 ft, what is the length of one of the other sides?

7. Find the mean, median, and mode of this series of numbers: 5, 7, 3, 9, 5 8. Define the following in your own words: (a) stanine; (b) grade-equivalent score.

I'll add that improving in one of the other areas of CBEST math will also help you pass: what matters is not that you collect a certain number of correct answers per area but that you pull together enough correct answers across the whole thing to achieve the magic 41 scaled points. So chances are that it's more than estimation, measurement, and statistical principles that's pulling you down.

3. 46(2)+ 22(2)+ 4 additional inches 92+44+4= 140 inches 4. 1,234.567= 1.234567×10³ (because you move the decimal 3 places) 5. Circumference= 9 meters Diameter is half of circumference, so the diameter would be about 4.5 meters 6. The other side of the triangle is 4 ft (2x4=8) 7. Mode: 5 (it repeats more than once) Median: 3 (its between all of the listed values Mean: (5+7+3+9+5)= 29÷5 =5.8 8. A. Stanine- a score between 1-9 that determines if someone's performance is below average, average or above average. B. Grade equivalent score- the score that most people achieve at a certain grade level.

Thank you for all your help. I'll definitely keep that in mind and work on everything until I retest again.

3. You're missing four inches. Following your method, what you wanted was 46(2) in + 4 in + 22(2) in + 4 in; the reason is that while, yes, you're adding two inches to EACH of the sides. It may be easier to get this right, and to see what I mean, if we add the inches to each side FIRST: that would be (46 + 2)2 in + (22 + 2)2 in, which gives 144 in. That, however, is still not the right answer: I asked for an answer in yards. 4. Yes, you move the decimal point seven places - but then you have to truncate the answer so it has no more decimal places than the original number did. So the answer is 1.235 ×10³. (And now I know how to get superscripts into A to Z: that is sooo cool!) 5. Diameter is half of circumference? Nope. Pick a circle, any circle; measure its circumference (C) and its diameter (d); divide C by d; and if you've measured carefully enough, your answer will be somewhere in the neighborhood of 3.14159..., which is to say pi (, and in case your computer doesn't like my font, I'm going to keep calling the darned thing "pi), and it's pi no matter how big the circle is. It follows that C divided by pi gives diameter; since pi is a bit over 3, the diameter must be a bit under 3 meters. (The standard circumference formula is either C=pi(d) or C=pi(2r), where r = radius.) 6. Good. 7. Pretty good on mode and mean, but you're struggling on median. Line the numbers up in order: 3, 5, 5, 7, 9. Find the number that's in the middle (like a road median, see?): that's the median. Since we have an odd number of numbers, the median is the 5 that's right in the middle. If we had an even number of numbers - try adding a 6, which would give us 3, 5, 5, 6, 7, 9 - find the middle two numbers and take their mean: (5+6)/2 = 5.5.

8a. Stanine is short for "standard nine": it tells us how typical or how odd a given score is in a set of scores, and for teacher-test purposes it does so by dividing a stereotypical bell curve into nine unequal chunks. The middle 20% (that is, from the 40th percentile to the 60th percentile, with the 50% percentile point right in the middle) is the fifth stanine: it contains scores that are absolutely average, and it's the biggest chunk. The fourth and sixth stanines each account for 17%, so the fourth stanine goes down to the 23rd percentile and can be thought of as "low normal", while the sixth stanine goes up to the 67th percentile and is called "high normal". The third and seventh stanines account for the next 12% below and above (down to the 11th percentile, up to the 89th percentile), and are "slightly below normal" and "slightly above normal", respectively. The second and eighth stanines account for the next 7% below and above (down to the fourth percentile, up to the 96th percentile) and are "below normal" and "above normal", respectively. The first and ninth stanines account for the remaining 4% above and below and are "significantly below normal" and "significantly above normal", respectively: this is where one finds the outliers. 8b. The concept of "grade-equivalent score" is easier to explain with a made-up example. Suppose that a test company has developed a test for fourth graders. To determine grade-equivalent scores, the company rounds up groups of absolutely typical students (fifth stanine students, if you will) in a range of grades and months within grade and administers the test to each group. Suppose that the mean score achieved by the group of absolutely average kids in grade 2, month 3 is 12. Well, from then on the score 12 on that fourth-grade test is associated with the grade equivalent score 2.3 (second grade, month 3); the fourth grader who scores a grade-equivalent 2.3 is worryingly behind what we expect fourth graders to be able to do. Similarly, if the mean score by the group of average seventh graders in month 9 is 49, then we call that the grade equivalent score 7.9, and the fourth grader who scores like that is significantly ahead of grade. A grade equivalent score that is above a student's actual grade level correlates with a higher stanine and percentile; a grade equivalent score that is below grade level correlates with a lower stanine and percentile.