Answer:
f(x)=1200(1.04)^n
Step-by-step explanation:
Answer:
% Po lost = 100[1 - e^(-0.005t)] %; 73.0 g
Step-by-step explanation:
p(t) = 100e^(-0.005t)
Initial amount: p(0) = 100
Amount remaining: p(t) = 100e^(-0.005t)
Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]
% of Po lost = amount lost/initial amount × 100 %
= [1 - e^(-0.005t)] × 100 % = 100[1 - e^(-0.005t)] %
p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g
The mass of polonium remaining after 63 days is 73 g.
Answer: We should reject the null if the test statistic is greater than <u>1.895</u>.
Step-by-step explanation:
We assume the population to be normally distributed.
Given: Sample size : 
, which is asmall sample (n<30), so we use t-test.
We always reject the null hypothesis if the absolute t-value is greater than critical value.
Therefore, We should reject the null if the test statistic is greater than <u>1.895</u>.
The answer would be iii: 10-15%
the way to find this is to divide 43 by 381 to get 11.2% which falls in the rage of 10-15%
