Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 60 manufacturing companies located in the Southwest. Estimate the mean and the standard deviation of advertising expenses. (Round squared deviations to nearest whole number and final answers to 2 decimal places.) Advertising Expenditure ($ millions) Number of Companies 25 up to 35 5 35 up to 45 10 45 up to 55 21 55 up to 65 16 65 up to 75 8 Total 60 Mean
Standard deviation

Step-by-step explanation:

Given

Solving (a): The mean

First, we calculate the class midpoint by calculating the average of the intervals.

For instance:

The midpoint of 25 -35 is:

So, the table becomes:

The mean is then calculated as:

Solving (b): The standard deviation

This is calculated using:

So, we have:

step-by-step explanation:

. your answer will contain x - and y -terms. (1 point) x ( ¿¿ 2 − 7 ) 2 +( y 2 − 9 ) 2 √ ¿ 12. what is the distance from the directrix (the crocodile river) to the point ( x , y )? write this equation. your answer will contain a y -term.

first, we define the variables:

x: number of years after 1950

f (x): amount of vinyl sold.

then, with the variables defined, we have:

68594 vinyl records were sold in 1958 > f (8) = 68594

91299 vinyl records were sold in 1961 > f (11) = 91299

38720 vinyl records were sold in 1952 > f (2) = 38720

161743 vinyl records were sold in 1967 > f (17) = 161743

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step-by-step explanation: