5. Past experience shows that the standard deviation of the distances traveled by consumers to patronize Chinese restaurants is 2 km. How large a sample is needed to estimate the population mean distance traveled to within 0.4 km

96

Step-by-step explanation:

Given that :

Standard deviation (s) = 2km

Error in estimation is within 0.4 km ; Error (E) = 0.4 km

Using the relation :

Sample size = ((Zα/2 * s) /E)^2

Using α = 0.05

The Zcritical at 0.05 = Z0.025 = 1.96

Sample size = ((1.96 * 2) / 0.4)^2

Sample size = (3.92 / 0.4)^2

Sample size = 9.8²

Sample size = 96.04

Required sample size = 96

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)