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If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumpe

lakkis [162]

If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumper-to-bumper

Answer: We are given there are 49 cars

Also these 49 cars are in a line stretched 528 feet.

Now the average length of the car is:

Average length = $\frac{(length-covered-by-49-cars)}{Number-of-cars}$

= $\frac{528}{49}=10.78$

Therefore, the average length of car is 10.78 feet

6 0

3 years ago

What is the value of 9/10÷1/5 ?

Anna35 [415]

The answer of 9/10 divided by 1/5 is 4.5

3 0

3 years ago

Read 2 more answers

Which graph represents the solution for the equation 5x-2=4

Jet001 [13]

You didnt list any graph?? Can you add it please :)

7 0

3 years ago

Read 2 more answers

two cars travel at same speed to different destinations. car A reaches its destination in 24 minutes. car B reaches its destinat

Harlamova29_29 [7]

Answer:

The speeds of the cars is: 0.625 miles/minute

Step-by-step explanation:

We use systems of equations in two variables to solve this problem.

Recall that the definition of speed (v) is the quotient of the distance traveled divided the time it took : $v=\frac{distance}{time}$ . Notice as well that the speed of both cars is the same, but their times are different because they covered different distances. So if we find the distances they covered, we can easily find what their speed was.

Writing the velocity equation for car A (which reached its destination in 24 minutes) is:

$v\,*\,24\,min=d_A$

Now we write a similar equation for car B which travels 5 miles further than car A and does it in 32 minutes:

$v\,*\,32\,min=d_A+5\,miles$

Now we solve for $d_A$ in this last equation and make the substitution in the equation for car A:

$v\,*\,32\,min=d_A+5\,miles\\d_A=v\,*\,32\,min-5\,miles\\\\v\,*\,24\,min=v\,*\,32\,min-5\,miles\\v\,(24\,min-32\,min)=-5\,miles\\v\,(-8\,min)=-5\, miles\\v=\frac{-5}{-8} \frac{miles}{min} \\v=0.625\,\frac{miles}{min}$