Answer:
The answer would be 0.36
Step-by-step explanation:
Answer:
K is equal to 56
Step-by-step explanation:
Answer:
Well, these simulation are based on the statistics (lognormal-distributed PE, χ²-distributed s²). If you believe that only the ‘gold-standard’ of subject-simulations are valid, we can misuse the function sampleN.scABEL.sdsims() – only for the 3- and 4-period full replicates and the partial replicate:
# define a reg_const where all scaling conditions are ‘switched off’
abe <- reg_const("USER", r_const = NA, CVswitch = Inf,
CVcap = Inf, pe_constr = FALSE)
CV <- 0.4
2x2x4 0.05 0.4 0.4 0.95 0.8 1.25 34 0.819161 0.8
Since the sample sizes obtained by all simulations match the exact method, we can be confident that it is correct. As usual with a higher number of simulations power gets closer to the exact value.
Step-by-step explanation:
We have that
f²{k^6/[k^4*k^3]}^-3
step 1
f²{k^6/[k^(4+3)]}^-3---------> f²{k^6/[k^7]}^-3-------> step 1 is correct
step 2
f²{k^6/[k^7]}^-3------> f²{k^(6-7)]}^-3----> f²{k^(-1)}^-3----> step 2 is correct
step 3
f²{k^(-1)}^-3------> f²{k^[(-1)*(-3)]}-----> f²{k^(3)}
therefore
Martha made a mistake in the step 3
the answer is
step 3
f²k³
Y=24 because 60/20 is 3. 3*8 is 24
