I've gone through C's 343-page state test prep book - that's 343 pages sans answers - which was sent home with the kids with instructions to "use the book."
We're using the book.
I figure C should do 3 lessons a day over break (test starts March 12). In theory the answer key will arrive on Wednesday, so that means I need to work the problems in 12 lessons.
Which I am doing.
Ed and I had a spat about this yesterday. I had planned to refuse to purchase the test prep book unless the school gave me the answer key, too.
Ed managed not to notice that I'd been plotting this move for weeks, and sent in the $11 with Christopher when asked.
So now we have the book, and the principal can just say no when I ask for the answers. (He already had said no when I brought it up a couple of weeks ago, shortly before he also said no to moving Christopher into the Phase 3 class for the remainder of the year and then back into Phase 4 come fall. "No" is the word!)
Yesterday Ed told me he "completely disagreed" with my plan to refuse to buy a test prep book with no answers; that would mean the principal would simply not sell me the book; etc.
I said I have books; I have at least $200-worth of books, all of them with answers in the back.
He said, Those books aren't targeted to the material being tested.
etc.
The fact is, it would make my life much easier to have a test prep book specifically constructed to provide review and practice of New York state math standards for grade 7.
It does not make my life easier to have such a book without the answers.
Ed's attitude was that I could just work the problems myself.
So.
Today he's sorry.
When he left 20 minutes ago to go to Stew Leonard's and Costco I was still working on state test prep problems.
I pointed this out (I'm going to be pointing it out for days) and Ed said, "Are you doing all the problems in the book?"
No.
I'm doing the first 12 lessons in the book.
This is how long it takes to do the first 12 lessons in the book.
etc.
Who's sorry now?!
His parting shot was an offer to do some of the problems himself.
I ought to take him up on that.
The reason I ought to take him up on that is that the first lessons in the book, the ones I'm doing, are on probability and he has no idea how to do problems on probability. I bet he doesn't even remember how to construct a proper tree diagram!
Which brings me to the reason I'm writing this post.
I can't do this problem:
Raymond has a deck of 5 cards, numbered 3-7. How many different ways can the cards be lined up?
help!
update 10-9-2007
Turns out all the answers are belong to us.
Spoke to COOG, at length.
Test prep answers are subject to FOIL.
woo hoo!
_____________
state test coming right up (2006)
throwing money at the problem
more stuff only teachers can buy
help desk 1
state test coming right up (2007)
help desk 2
my life and welcome to it
inflammatory
canadianteacher.com
progress report
despair
28 out of 30
I feel your pain.
ReplyDeleteMy husband wanted me to teach algebra out of a book with no solution guide or answer key and we've ended up in a similar situation. "If it's so easy then YOU write up the solutions." I'm keeping an update on the right column of my blog on which problem he's got up to.
The answer to your particular problem, I believe, is that you have five different possibilities for the first choice. Five cards = five possibilities. Once you choose the first card you'll be left with four possible cards to choose from (5 X 4) then three cards will be left..( 5 X 4 X 3) and then down to one. This is where the "factorial" comes in handy. 5!
And you read that very loudly, like this FIVE !
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletehmmm....
ReplyDeleteI'll try that.
Thanks!
Of course the annoying thing here is that each lesson is focused on just ONE aspect of probability, and factorials aren't in THIS LESSON.
I get 120, which IS one of the possibilities.
ReplyDeleteMUST HAVE ANSWER KEY
Thanks!
ReplyDeleteOK, I'm going to go do Lesson 13 Probability of Dependent Events."
I wonder if I'll ever get to go outdoors again?
For a walk, say?
The way to arrange 'N' distinct items (no duplicates) is N!
ReplyDeleteSo ... 5 cards numbered 3,4,5,6,7 are distinct. 5! = 120.
Sigh ... you have a long 4 or 5 days ahead.
-Mark Roulo
You can call these folks on Monday. I'm pretty sure you can order just one student book ($7.95). The teacher guide is $9.95 and can be purchased online.
ReplyDeleteA suggestion anyway. It's Grade 7 test prep just for New York.
Okay, maybe Tuesday.
ReplyDeleteThey also have last year's test here, which would give you a better snapshot of the performance indicators that are actually assessed. (Only about 60%-70% of objectives are tested.)
ReplyDeleteFactorial 57 of Math on Call.
ReplyDelete5! = 5 x 4 x 3 x 2 x 1 = 120
Mr. Person - thanks!
ReplyDeleteI'm going to have to call the Coach folks back (that's the book I have) and find out who they said they were working for.
I thought they said NYC, but I don't see why NYC would have two companies.
Does anyone know what "SAMIS" stands for?