Acatalog company promises to deliver online orders within 3 days of the order being placed. follow-up calls were made to randomly selected customers to see whether their orders arrived on time (within 3 days). a 95% confidence interval for percentage of on-time arrival is 88% +/- 6%. what does this mean? are these conclusions correct? explain.
a) 95% of all random samples of customers will show that 88% of orders arrive on time.
b) 95% of all random samples of customers will show that 82% to 94% of orders arrive on time.
c) we are 95% sure that between 82% and 94% of the orders placed by the sampled customers arrived on time.
d) on 95% of the days, between 82% and 94% of the orders will arrive on time.

c) 95% of all random samples of customers will show that 82% to 94% oders arrive on time

Step-by-step explanation:

From problem statement   we take 88 as a mean of a normal distribution, in that case we know, about relation between mean and standard deviation

μ  ± σ        ⇒ [ μ  -  σ ;   μ  +  σ ]   ⇒  [  88  - 6  ;  88  + 6 ]  ⇒ [  82  ; 94 ]

The above mentioned interval get 95.7 % of all values for a normal distribution, so we must be sure that a 95 % of all random samples of customers will show that 82% to 94% of orders arrive on time

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

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