Problem 1.(32 points) A random sample of 100 Uber rides in Chicago has on average 5.4miles per ride. Assume the distribution of the distance of Uber rides to be approximatelynormal with a standard deviation of 1.3 miles.(a) (5 points) Construct a 99% confidence interval for the average distance (in miles) ofan Uber ride. (round to one decimal place)(b) (4 points) What can we assert with 99% confidence about the possible size of our errorif we estimate the average distance of an Uber ride to be 5.4 miles

a) CI = ( 5,1 ; 5,7 )

b) SE =  0,1  

Step-by-step explanation:

a) Sample random     n  = 100

Mean  =  μ = 5,4

Standard deviation   s = 1,3

CI = 99 %    α = 1 %   α = 0,01     α/2 =  0,005

z(c) for  0,005 is from z-table    z(c)  = 2,575

z(c)  =  ( X - μ ) /s/√n          CI  =  μ ± z(c) * s/√n

CI =  5,4 ± 2,575* 1,3/10

CI = 5,4 ± 0,334

CI = ( 5,1 ; 5,7 )

b) SE = Standard deviation / √n

SE = 1,3 /10       SE =  0,1  

We can support that with 99 % of probability our random variable will be in the CI.

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.