You earn $35 for washing 7 cars. How much do you earn for washing 4 cars?

Mathematics

1 answer:

Gre4nikov [31]2 years ago

5 0

Answer:

The earning by washing four cars is $20

Step-by-step explanation:

You earn $35 for washing 7 cars.

The following steps can be used in order to determine the earning by washing four cars:

Step 1 - The unitary method can be used in order to determine the earning by washing four cars.

Step 2 - According to the given data, the earning by washing seven cars is $35.

Step 3 - So, the earning for washing four cars is given by:

4/7 x 35

Step 4 - Simplify the above expression.

140/7

= $20

So, the earning by washing four cars is $20.

You might be interested in

The number of applicants to a university last year was 11,450. This year, the number of applicants grew by 2%. How many applican

Vitek1552 [10]

Answer:

11679

Step-by-step explanation:

Increased by 229

First you need to find the 2%

So to find the 2%

2/100 * 11450=229

Now add 11450+229= 11679

4 0

3 years ago

Read 2 more answers

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$