I solved this
I will use @ instead of that sign
@4 = 2x4^2 + 2 = 32 + 2 = 34
The ans I got is 34 not @34
In this I considered cases like for [x] I took 4 to be the greatest no as compared to the odd nos 1 & 3 but how do I incorporate the next logic
For this I used the formula: Sn = n/2 (2a + (n-1)d)
The approach is right right? Or there is another approach
For Amy I considered
rate initiallly was 8% later I divided by 2 and made it 4 as 1 year comprises 2 6 months
(1000) x (1.04)^2
For Bob
rate was initially 8% Then i calculated the no of months in one quarter that 12/4 = 3
so then i divide the rate by 3
later I will calculate = 1000 x (8/3)^4 as 4 quarters make a year
Is my logic right
I am unable to comprehend it
what i followed is
H - Halflife I- isotope C-Carbon
H(I)= 50/100 * X (X is a sample)
H(C) = 5730
Shud I equate the two
I get 32
for the first question, you can simplify options to get the correct answer
simplify? I got the ans as @34
@Leaderboard
Is my approach right?
- Notice that A and C are nested: show that \lceil{-1} = 4. Hence \lceil{\lceil{-1}} = \lceil{4}.
- -2 is even, so greatest integer less than -2 is -3. Conversely, the least integer greater than 5 is 6. (-3 - 6) = -9
- Your approach is fine.
- You are only looking for 6 months; this means that both Amy and Bob would have completed one and two compounding periods respectively (not 2 and 4).
- It’s a decay problem. 50% of the isotope will decay in 5730 years. 50% of that (75% of total sample) will decay in 14600 years. Let the number of times this happens be n. Then the total number of years would be 5730 \times n.
So what you want is to find this fraction. That’s (15500/16000) = 96.875%. Then the equation is
\frac{15500}{16000} = \left(1 - \frac{1}{2^n}\right)
Solve for n. We get n = 5. This means that it takes 5730 \times 5 = 28650 years in total.
