AI in product?

[cycloneturf]

:usflag:How do you figure the total AI in pounds on a fifty pound bag?
Example: 24-5-11 with Mach 2 1.33%

**** 1.33 % / 50lbs = .026 lbs ?????? ***** This does not sound right

Thanx for the help!
:usflag:

[ICT Bill]

If the number was 1% and you had 50 of anything it would be 0.50, yours would be somewhere around 0.65

[eagle1243]

Multiply the % AI times total weight of product. Then divide by 100.

1.33 % x 50 lbs = 66.5 lbs / 100 = .665 lbs

[JDUtah]

Multiply the % AI times total weight of product. Then divide by 100.

1.33 % x 50 lbs = 66.5 lbs / 100 = .665 lbs

:clapping:

[RigglePLC]

Buy two bags. You have a hundred pounds. 1.33 percent then equals 1.33 pounds.

[cycloneturf]

So with that said... a 19-0-6 Dimension .1 % would have .05 lbs of AI per 50 lb bag of fertilizer. :confused:

[Runner]

That is correct. Seems strange, doesn't it? That is like Crosscheck (bifenthrin)...It is like 3 lbs. that does an acre, I believe.