Answer:
d. There is a 98% chance that the true proportion of customers who click on ads on their smartphones is between 0.56 and 0.62.
Step-by-step explanation:
Confidence interval:
x% confidence
Of a sample
Between a and b.
Interpretation: We are x% sure(or there is a x% probability/chance) that the population mean is between a and b.
In this question:
I suppose(due to the options) there was a small typing mistake, and we have a 98% confidence interval between 0.56 and 0.62.
Interpreation: We are 98% sure, or there is a 98% chance, that the true population proportion of customers who click on ads on their smartphones is between 0.56 and 0.62. Option d.
Answer: see below
<u>Step-by-step explanation:</u>
If the figures are similar, then their corresponding sides are proportional.
ABCD ~ WXYZ
![\dfrac{AB}{WX} = \dfrac{BC}{XY} = \dfrac{CD}{YZ} = \dfrac{AD}{WZ}](https://tex.z-dn.net/?f=%5Cdfrac%7BAB%7D%7BWX%7D%20%3D%20%5Cdfrac%7BBC%7D%7BXY%7D%20%3D%20%5Cdfrac%7BCD%7D%7BYZ%7D%20%3D%20%5Cdfrac%7BAD%7D%7BWZ%7D)
Ok i tried and ig youre looking for what a and b equal for both systems..... it'll be 1) a=-2b+15 b=7.5-(1/2)a <-- fraction........2) a=-6+b b=6-a
Answer:
![\frac{x-2}{x+3}](https://tex.z-dn.net/?f=%5Cfrac%7Bx-2%7D%7Bx%2B3%7D)
Step-by-step explanation:
given ![\frac{(x-3)(x-2)}{(x+3)(x-3)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-3%29%28x-2%29%7D%7B%28x%2B3%29%28x-3%29%7D)
the factor (x - 3) can be cancelled from the numerator/ denominator leaving
![\frac{x-2}{x+3}](https://tex.z-dn.net/?f=%5Cfrac%7Bx-2%7D%7Bx%2B3%7D)
Since it's y/x, and x is what you input into the function (and y is what you get out), we get that the amount donated is based off of the lunch specials, so we get y(12)/x(60)=12/60=1/5Since it's y/x, and x is what you input into the function (and y is what you get out), we get that the amount donated is based off of the lunch specials, so we get y(12)/x(60)=12/60=1/5.