IQR= Q3-Q1
Airport 1
Q3 is 8
Q1 is 2
8-2= 6
Airport 1 IQR is 6
Airport 2
Q3 is 8
Q1 is 3
8-3= 5
Airport 2 IQR Is 5
Distance=speed times time
if they take the same time, we cal the time , t
b=bike speed
w=walking speed
bikedistnace=17=bt
walkdistance=9=wt
b is 4 more than w
b=4+w
we have
17=bt
9=wt
b=4w
ok so
multiply first equation by 9 and 2nd by 17
153=9bt
153=17wt
set equal
9bt=17wt
divide both sides by t
9b=17w
sub b=4+w for b
9(4+w)=17w
distribute
36+9w=17w
minus 9w both sides
36=8w
divide both sides by 8
4.5=w
he walks 4.5kph
Answer:
Please Find the answer below
Step-by-step explanation:
Domain : It these to values of x , for which we have some value of y on the graph. Hence in order to determine the Domain from the graph, we have to determine , if there is any value / values for which we do not have any y coordinate. If there are some, then we delete them from the set of Real numbers and that would be our Domain.
Range : It these to values of y , which are as mapped to some value of x in the graph. Hence in order to determine the Range from the graph, we have to determine , if there is any value / values on y axis for which we do not have any x coordinate mapped to it. If there are some, then we delete them from the set of Real numbers and that would be our Range .
Answer:
A.
Step-by-step explanation:
Answer:

Given series is convergence by using Leibnitz's rule
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given series is an alternating series
∑
Let 
By using Leibnitz's rule


Uₙ-Uₙ₋₁ <0
<u><em>Step(ii):-</em></u>


= 

=0

∴ Given series is converges