assume that the variable under consideration has a density curve. it is given that 27.1% of all possible observations of the variable are less than 17. Determine the area under the density curve that lies to the right of 17.

72.9%

Step-by-step explanation:

Given that :

27.1% of all possible observation of a certain variable are less than 17 ;

That is ;

Let variable = y ;

Proportion of (y < 17) = 27.1% = (27.1 / 100) = 0.271

Proportion that lies to the right :

P(y > 17) = 1 - P(y < 17)

P(y > 17) = 1 - 0.271

P(y > 17) = 0.729 = 72.9%

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

~just a girl in love with shawn mendes