how many groups of 3 are there in 18?
there are 6 groups of 3 in 18. 3*6=18.
there are 6 groups of 3 in 18. 3*6=18.
18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units 318 since, atomic number of an atom = the number of protons.hence, the number of protons is 17.now, the number of protons = the number of electrons.∴ the number of electrons is also 17.given: mass number = 35we know that mass number = number of protons + number of neutrons.∴ number of protons + number of neutrons = 35⇒ 17 + number of neutrons = 35⇒ number of neutrons = 35 – 17⇒ number of neutrons = 18 let x denote the number of sixes. p(getting a six when a die is thrown) = p = `(1)/(6)` ∴ q = 1 – p = `1 - (1)/(6) = (5)/(6)` given, n = 6∴ x ∼ b`(6, 1/6)`the p.m.f. of x is given by p(x = x) = `""^6"c"\_x (1/6)^x (5/6)^(6 - x), x` = 0, 1,...,6p(getting at most 2 sixes) = p(x ≤ 2) = p(x = 0 or x = 1 or x = 2) = p(x = 0) + p(x = 1) + p(x = 2) = `""^6"c"\_0(1/6)^0 (5/6)^6 + ""^6"c"\_1(1/6)(5/6)5 + ""^6"c"\_2(1/6)^2 (5/4)^4` = `(5/6)^4 + 6(1/6)(5/6)^5 + (6!)/(2! xx 4!)(1/6)^2 (5/6)^4` = `(5/6)^4 [(5/6)^2 + (5/6) + (6 xx 5 xx 4!)/(2 xx 1 xx 4) xx (1/6)^2]` = `(5/6)^4 [25/6^2 + 30/6^ + 15/6^2]` = `(5/6)^4 (70/36)` = `(5/6)^4 (35/18)` = `(5/4)^4 ((5 xx 7)/(6 xx 3))` = `(7)/(3)(5/6)^5`. Total no. of possible outcomes = 30 {1, 2, 3, … 30}  

  

E ⟶ event of getting no. divisible by 3.  

  

No. of favourable outcomes = 10 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}  

  

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)  

  

P(E) = 10/30 = 1/3  

  

Bar E ⟶ event of getting no. not divisible by 3.  

  

P(Bar E) = 1 – P(E)  

  

= 1 - 1/3 = 2/3 switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4 we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12 of three numbers, second is twice the first and is also thrice the third. if the average of the three numbers is 44, the largest number is:

option 1) 24                              option 2) 56option 3) 72               option 4) 108option 5)    36 you can pick 1 senior in 5 ways, 1 junior in 4 ways, 1 sophomore in 3 ways and i freshman in 2 ways.
in all you can make these selections i 5 * 4 * 3 * 2 ways = 120 ways.
 
please note: selecting a committee is a "combination" problem, not a "permutation". [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2]