Question
21 rotten bananas are accidentally mixed with 135 good
ones. It is not possible to just look at a banana and tell whether or not it is rotten. One banana is taken out at random from this lot. Determine the probability that the banana is taken out is a good one.Solution
Numbers of bananas = Numbers of rotten bananas + Numbers of good bananas ∴ Total number of bananas = 135 + 21 = 156 bananas P(E) = (Number of favourable outcomes) / (Total number of outcomes) P(picking a good banana) = 135/156 = 45/52Â
319.995 × 15.98 ÷ 4.002 - ? × 7.95 = 1679.89 ÷ 2.005
564.932 + 849.029 – 425.08 = 612.095 + ?
`root(3)(725.87)` `-:` `sqrt(81.033)` Â + 49.88% of 809.77 = ? - (14.78Â `xx` Â 52.2)
9.8 × 77.9 ÷ (2.3)2 = ?
What value should come in place of question mark (?) in the following question. (You need not to calcualte the exact value)
?/647 = 226/ ?
499.99 + 1999 ÷ 39.99 × 50.01 = ?
79.99% of (84.89 × 5.99) - (3.89)2 × 21.87 = ?
(112 × 4) ÷ 16 + 484 = (?)3
72.8% of (215.69 + 189.38) - 5.97² + (3.01 of 7.8) = ? of (64.02 - 38.95)
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