The ratio of medium size cars to compacts is 7 to 5 and the ratio of compacts to full-size cars is 8 to 9. Since compact cars are there in both ratios , we need to make the number in the ratio for compact cars to be the same.

$\frac{medium}{compact} = \frac{7}{5}$

Multiply by 8 on numerator and denominator

$\frac{medium}{compact} = \frac{56}{40}$

$\frac{compact}{full size} = \frac{8}{9}$

Multiply numerator and denominator by 5.

$\frac{compact}{full size} = \frac{40}{45}$

Ratio of medium, compact and full size are in ratio 56,40,45.

Total number of cars =2820.

(56 + 40 + 45)x = 2820

x = $\frac{2820}{141}$

x = 20

Number of full size cars = $20\times 45$ =900

3 0

3 years ago

For the function given, state the middle line; f(t) = 40cos (80t + 20)

Digiron [165]

In the standard form of the equation

$\\ \ f(t)=Acos[b(t\pm c)]+k\\ \\$

The middle line =k

For our given problem

f(t) = 40cos (80t + 20)

On comparison we get k=0

Hence middle line=0

8 0

3 years ago