Home
Class 12
MATHS
1.01^365=37.8 , 0.99^365=0.03...

`1.01^365=37.8` , `0.99^365=0.03`

Text Solution

Verified by Experts

(i) Let `t = 1.01^365`
`:. log_10t = 365log_10(1.01)`
`=> log_10t = 365(log_10(101) - log_10(100))`
`=> log_10t = 365(2.00432 - 2)`.... (As `log_10(101) = 2.00432`)
`=> log_10t = 365(.00432)`
`=> log_10t = 1.5773`
`=> t = 10^(1.57773)`
`=> t = 37.78`
...
Promotional Banner

Similar Questions

Explore conceptually related problems

Calculate mole fraction of KI if the density of 20% (mass/mass) aqueous KI solution is 1.202g mL^-1 (a)0.0263 , (b)0.569 , (c)0.365 , (d)0.987

If 3.65 x 0.5 = 1.825 then, find the value of 365 x 0.5:

Calculate mole fraction of KI if the density of 20% (mass/mass) aqueous KI solution is 1.202gmL^(-1) (a) 0.0263 (b) 0.569 (c) 0.365 (d) 0.987

Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = {omega_(1), omega_(2), omega_(3), omega_(4), omega_(5), omega_(6), omega_(7)} Assignment omega_(1)" "omega_(2)" "omega_(3)" "omega_(4)" "omega_(5)" "omega_(6) (a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6 (b) 1/7 1/7 1/7 1/7 1/7 1/7 1/7 (c) 0.1 0.2 0.3 0.4 0.5 - 0.6 - 0.7 (d) -0.1 0.2 0.3 0.4 -0.2 0.1 0.3 (e) 1/14 2/14 3/14 4/14 5/14 6/14 15/14

Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = {omega_(1), omega_(2), omega_(3), omega_(4), omega_(5), omega_(6), omega_(7)} Assignment omega_(1)" "omega_(2)" "omega_(3)" "omega_(4)" "omega_(5)" "omega_(6) (a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6 (b) 1/7 1/7 1/7 1/7 1/7 1/7 1/7 (c) 0.1 0.2 0.3 0.4 0.5 - 0.6 - 0.7 (d) -0.1 0.2 0.3 0.4 -0.2 0.1 0.3 (e) 1/14 2/14 3/14 4/14 5/14 6/14 15/14