Place value: hundredsFace value: three hundred.Place value: hundredsFace value: three hundred.Place value: hundredsFace value: three hundred.Place value: hundredsFace value: three hundred.
Find the value of the fraction.Find the value of the fraction.Find the value of the fraction.Find the value of the fraction.
the value of 6 is 60 000 the value of 7 is 7000 the value of 2 is 200 the value of 4 is 40 the value of 1 is 1
The value is 300.
The answer depends on the value of WHAT! The value of your degree education or the value of your student loan debt!
4c2 + 5c + 2c = 4c2 + 7c = c(4c + 7)
(b2 - 20c2)(b2 + 4c2)
4C2 * (-2CD2) = -8C3D2
24c3 + 375z3 = 3*(8c3 + 125z3) = 3*(2c + 5z)*(4c2 - 10cz + 25z2)
4C2(1/2)4 = 6/16 = 3/8
Let c represent cos x. Then, we have, 4c2 - 2 = 0; whence, 4c2 = 2, c2 = ½, and c = ±√½ = ±½√2; that is, cos x = ±½√2. From this, it is evident that, for 0 ≤ x < 360°, x = 45°, 135°, 225°, or 315°; or, if you prefer, for 0 ≤ x < 2π, x = ¼π, ¾π, 1¼π, or 1¾ π.
6 of them. 4C2 = 4!/(2!*2!) = 4*3/(2*1) = 6
5 cards out of 52 may be drawn in 52C5 = 2 598 960 ways. 2 aces out of 4 may be drawn in 4C2 = 6 ways. 3 non-aces out of 48 non-aces may be drawn in 48C3 = 17 296 ways P( 2 aces ) = (4C2)(48C3) / (52C5) = (6)(17 296)/( 2 598 960) = 0.0399
This one is much more straightforward. There are 5C2 = 10 ways to choose two parallel lines from the set of five. There are 4C2 = 6 ways to choose two parallelograms from a set of four. Any parallelogram is uniquely determined by one pair of lines from the five, and one pair of lines from the four. Thus, the number of possible parallelograms is(5C2)*(4C2) = (10)*(6) = 60
24c2 + 96c + 90 = 6*(4c2 + 16c + 15) = 6*(2c + 3)*(2c + 5)
let the pairs be named as A1 B1,A2 B2,A3 B3,A4 B4.LET US SELECT TWO FROM BOYS AND ONE FROM GIRL THIS WE CAN DO IN 4c2*2c1=12 NOW THE OTHER WAY i.e 2 GIRLS AND 1 BOY 4c2*2c1=12 OTHER WAY IS TO SELECT ALL THE THREE FROM GIRLS OR BOYS=2*4c3 TOTAL 30
it has three cases 1st E1- two balls are whiteE2- three balls are whiteE4-four balls are whiteE4/w= (p(E3)*4C2/4C2)/(P(E1)*2C2/4C2+P(E2)3C2/4C2+P(E3)*4C2/4C2)WHERE-P(E1)=P(E2)=P(E3)=1/3----------------------------------------------------------------------------------------------------2nd opinionLet's say we have 3 boxes with 4 balls each.Box A has 4 white balls.Box B has 3 white balls.Box C has 2 white balls.The probability of drawing 2 W balls from;Box A, P(2W│A)=(4/4)∙(3/3)=1Box B, P(2W│B)=(3/4)∙(2/3)=1/2Box C, P(2W│C)=(2/4)∙(1/3)=1/6Say the probability of picking any of the 3 boxes is the same, we have;P(A)=1/3P(B)=1/3P(C)=1/3Question is, given the event of drawing 2 W balls from a box taken blindlyfrom the 3 choices, what is the probability that the balls came from box A,P(A│2W).Recurring to Bayes Theorem:P(A│2W)=[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)]=[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)]=6/10=0.60=60%P(A│2W)=0.60=60%Read more:Solution_to_a_bag_contains_4_balls_Two_balls_are_drawn_at_random_and_are_found_to_be_white_What_is_the_probability_that_all_balls_are_white