I have a question. I got a question that is : b_x =x^2+7x+50 ; b_8? I was really confused by this since its impossible for b_8 to appear out of nowhere
Just wanted to give thanks for this video. I appreciate it! I recently enrolled in college, but there was roughly a seven-year period between graduating high school and enrolling. That being said, Maths was my weakest subject anyway, so getting a brief refresher on how to work the calculator is proving very helpful!
THANK YOU SO SO VERY MUCH. THIS HAS HELP WITH MY STATISTICS SUMMER CLASS. I DID NOT UNDERSTAND AT ALL HOW TO USE GEOGEBRA. MAY YOU HAVE A LOVELY AN AMAZING DAY. ❤
Interesting. In the Maya numeric system numeric system we use same concept of 10s, 100s, 1000s but base 20. So it's 20, 400, 8000 with the exception that they are implicit on each ascending row so we don't have to write the 20,400,8000. I'm very surprised, it's almost identical.
Guess it has its' uses, but till then I stay in Casio wonderland. :) sqrt(684) gives me 6sqrt(19) on screen without any hassle. But I can do the same way you did it too on my CG-50 though. But could someone please explain to me way TI are so clunky to use, and yes I own one Ti-83+ of which I still use at work next to my Casio f-991ex.
Nice explanation. A short cut if u will ... Once you know the first binomial by inspection you can find the 2nd... In your 2nd example 20x^2 - 47x + 24, once you know ( 4x - 3 ) the only other way to get 20x^2 and +24 are multiplication of first term by 5x and 2nd term by -8 hence ( 5x - 8 ) ... a quick FOIL check proves your answer
The midpoint of a class is the mean of the upper and lower values of the same class. If you subtract two consecutive lower limits, you will find the Class Width.
@@mathemporia In my stat class we are to find midpoints using the mean of two consecutive lower class limits (1-2, 3-4, (1+3)/2 midpoint= 2) but online the midpoint is made by the mean of the lower and upper class limit (1-2, (1+2)/2 midpoint=1.5) It is a small difference but I'm seeing the two different ways of finding the midpoint, they are slightly different...and has led me to get some wrong (but close) homework answers...
@@user-dg4it7wv5p If you are looking for the midpoint of a class, the midpoint is the average of the lower and upper limits of the interval (or class). In the example you show, you find "2" which is actually the upper limit of the first class , so it can't be the midpoint. There are times where you find the class boundaries which involves finding the average of the upper limit of one class and the lower value of the next class (or interval). Check out my explanation of that at around minute 4 of this video. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-1j0bv8yTO3Y.html
hello, is there a way to have the calculator give you the answer that has the correct significant figures? Without us having to know what answer to put for the correct sig figs?
Thank you so much for this video. The frequency magic you id in the table around 2 minutes ended my 2 hour suffering. It took some attempts to make it work as it always said 'SPILL!' but now it is finally done...
I knew there had to be a way to convert between improper and mixed. Coming from the middle school mathprint calcs, I was just struggling to find it on the 36X Pro. Thank you.
