1.3, 1 1/3, 1.34
The decimal version of 1 1/3 is 1.333 (so on)
so 1.3 is less than 1.333 and 1.333 is less than 1.34
1.3<1.333<1.34
It would go over the five, so im guessing the hundreds
Part I)
The module of vector AB is given by:
lABl = root ((- 3) ^ 2 + (4) ^ 2)
lABl = root (9 + 16)
lABl = root (25)
lABl = 5
Part (ii)
The module of the EF vector is given by:
lEFl = root ((5) ^ 2 + (e) ^ 2)
We have to:
lEFl = 3lABl
Thus:
root ((5) ^ 2 + (e) ^ 2) = 3 * (5)
root ((5) ^ 2 + (e) ^ 2) = 15
Clearing e have:
(5) ^ 2 + (e) ^ 2 = 15 ^ 2
(e) ^ 2 = 15 ^ 2 - 5 ^ 2
e = root (200)
e = root (2 * 100)
e = 10 * root (2)
The general way to work this out is to solve the general expression for
the remaining quantity versus half-life, using logarithms. But that's not
necessary with these numbers.
Look at the numbers:
-- 3 mg is 1/4 of 12 mg.
-- 1/4 is the product of (1/2) x (1/2).
-- So the 3 mg is what's left of 12 mg after 2 half-lives.
The 26 minutes must be two half-lives.
-- The half-life of that substance is 26/2 = <em>13 minutes</em>.
Go Maggie !
Total number of marbles in the bag =20
Number of blue marbles =4
Now we just want two blue marbles out of four.
Probability of taking out first marble which is blue =4/20
Now after we took out one blue marble , number of blue marbles left are 3.
Probability of taking out second marble which is blue =3/20
Total probability = 4/20 *3/20 = 12/400
So probability of taking out two blue marbles is 0.03 or 12/400.
