The probability that a two-digit number selected at random has a tens digit less than its units digit is 0.2667 (4/15).
There are 90 two-digit numbers (99-9). Of these, six numbers are divisible by 15 (15, 30, 45, 60, 75, 90). This is also divisible by 5. Therefore, the preferred case is 30-6 = 24. Therefore, the required probability is 24/90 = 4/15.
The probability of an event can be calculated by simply dividing the number of favorable results by the total number of possible results using a probabilistic expression. Whenever you are uncertain about the outcome of an event, you can talk about the probability of a particular outcome, that is, its potential.
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Answer:
answer : x = 3±√ 119 / 16
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1. 8c^2-26c+15= (4c-3) (2c-5). Break the expression into groups: =(8c^2-6c)+(-20c+15). Factor out 8c^2-6c: 2c(4c-3). Factor out -5 from -20c+ 15: -5(4c-3). Lastly factor out common term (4c-3) and thats how you'll get your answer (4c-3) (2c-5).
2. common factors for 270 and 360 is 90.To find this write the factors of each and find the largest one.270: 1, 270, 2, 135, 3, 90, 5, 54, 6, 45, 9, 30, 10, 27, 15, 18360: 1, 360, 2, 180, 3, 120, 4, 90, 5, 72, 6, 60, 8, 45, 9, 40, 10, 36, 12, 30, 15, 24, 18, 20
3. The factors for 8 a3b2 and 12 ab4 is 4. because 8: 1, 2, 4, 812: 1, 2, 3, 4, 6, 12.
4. 81a^2+36a+4= (9a+2)^2. Break down the expression into groups: (81a^2+18a)+(18a+4). Factor out 9a from 81a^2 +18a: 9a(9a+2). Factor out 2 from 18a+4: 2(9a+2). so the groups you got are now 9a(9a+2)+2(9a+2). Lastley factor out common term (9a+2) to get (9a+2) (9a+2). Finally you get the answer (9a+2)^2.
5. mn-15+3m-5n= (n+3)(m-5). factor out m from nm+3m: m(n+3). Factor out -5 from -5n-15: -5(n+3). And thats how you get the number (n+3)(m-5)
Hope this helped :) Have a great day
