A Food Marketing Institute found that 32% of households spend more than $125 a week on groceries. Assume the population proportion is 0.32 and a simple random sample of 402 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.31

1/6

step-by-step explanation:

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2

< 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)

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