Use the normal distribution to find a confidence interval for a difference in proportions p1-p2 given the relevant sample results. Assume the results come from random samples. A 99% confidence interval for p1-p2 given that p^1=0.20 with n1=40 and p^2=0.35 with n2=80

step-by-step explanation:

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you never say the total so we cant find the percentage used for petrol

step-by-step explanation:

i use my brain for prombles like this maybe you could try it

step-by-step explanation:

< 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)