Find the reciprocal of 2/5.
Use the / as the fraction bar.​

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

13.46

step-by-step explanation:

16.80 x .20 = 13.46

0 and 5 minutes

step-by-step explanation:

given that one morning eight buses arrive late by the number of minutes given below:

0 8 2 6 9 2 0 7 8

write these in ascending order

0   0 2 6 7 8 8 9

in the afternoon two more buses arrive at the bus stop say by x and y minutes late.

total number of buses = 10

median = average of 4th and 5th after arranging in ascending order

since 4th entry is 6, the values of x and y should be below 6

also given 0 is the mode so either x or y or both should be 0 because then only 0 gets the maximum frequency and hence mode

thus we have

0 0 0 2 6 7 8 8 9 and another unknown entry which is below 6.

since median is 3.5, we have the unknown entry to be 5 only because then only median =( 2+5)/2

thus the after noon buses were late by 0 and 5 minutes.