You measure 32 turtles' weights, and find they have a mean weight of 73 ounces. Assume the population standard deviation is 7.9 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean turtle weight. Give your answer as a decimal, to two places

Margin of error = 2.74 ounces  

Step-by-step explanation:

We are given the following in the question:

Sample mean, = 73 ounces

Sample size, n = 32

Alpha, α = 0.05

Population standard deviation, σ = 7.9 ounces

Margin of error:

Putting the values, we get,

Thus, the maximum margin of error associated with a 95% confidence interval for the true population mean turtle weight is 2.74 ounces

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58