The highest rated of the four European cities under consideration: This can be done by multiplying factor and importance and summing for each city. A: 8050: Highest rating B: 6450 C: 7150 D: 7950

The question is not complete, as the table containing the data is missing, but I found a matching table that can be used to answer the question.

The Question is:

Which is the highest rated, of the four European cities under consideration, using the table.

The correct answer is: City A is the highest rated European city.

Step-by-step explanation:

The highest rated European city can be found by multiplying the factor and the importance of the factors, and summing up their final values. the cty with the highest number is the one with the highest rated city. Having this in mind, let us calculate the ratings for each of the cities as follows:

City A:

(70 × 20) + (80 × 20) + (100 × 20) + (80 × 10) + (90 × 10) + (65 × 10) + (70 × 10) = 1400 + 1600 + 2000 + 800 + 900 + 650 + 700 = 8050

City B:

(70 × 20) + (60 × 20) + (50 × 20) + (90 × 10) + (60 × 10) + (75 × 10) + (60 × 10) = 1400 + 1200 + 1000 + 900 + 600 + 750 + 600 = 6450

City C:

(60 × 20) + (90 × 20) + (75 × 20) + (65 × 10) + (50 × 10) + (85 × 10) + (65 × 10) = 1200 + 1800 + 1500 + 650 + 500 + 850 + 650 = 7150

City D:

(90 × 20) + (75 × 20) + (90 × 20) + (65 × 10) + (70 × 10) + (70 × 10) + (80 × 10)   = 1800 + 1500 + 1800 + 650 + 700 + 700 + 800 =7950

Therefore, from the ratings computed above, City A with a rating of 8050, is the highest rated, while City B with a rating of 6450, is the lowest rated.

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.