Suppose your marketing colleague used a known population mean and standard deviation to compute the standard error as 52.4 for samples of a particular size. You don't know the particular sample size but your colleague told you that the sample size is greater than 70. Your boss asks what the standard error would be if you quintuple (5x) the sample size. What is the standard error for the new sample size? Please round your answer to the nearest tenth. Note that the correct answer will be evaluated based on the full-precision result you would obtain using Excel.

answer:

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

explanation:

service only with 20% tip  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

in the opening of the question the word 'service' is associated only with the pretax value.

the closing part of the question asks: how much did she pay for the  

service

−−−−−−  

so technically the answer is  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

however, the intention of the question provider may have been that your answer is to be  

$

87.12

(

1

+

8

100

)

=

$

94.09

to 2 decimal places.

'

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

tony b

mar 21, 2017

answer:

trying to explain it another way

explanation:

service charge  

.

→

$

72.60

20% tip  

20

100

×

$

72.60

=

$

14.52

−−−−−−  

←

add

$

87.12

tax  

8

100

×

$

87.12

=

$

6.97

−−−−−  

←

add

$

94.09

related questions

hello, sorry being so rude after all that fighting i found it kind of funny, so as me saying sorry here is an answer!

since the table showed her that   2km = 2000m.

to determine, how many km in 2000m, she would divide by 1000.

= 2000 / 1000 = 2 km.

similarly, for 8000m = 8000 / 1000 = 8 km

cheers to savings

~

your pal papaguy