All categories

3 years ago

It takes 52 minutes for 5 people to paint 5 walls. How many minutes does it take 20 people to paint 20walls?

Mathematics

2 answers:

abruzzese [7]3 years ago

7 0

Answer:

I'm not sure if I'm correct or not but, it would still take 52 minutes as the ratio of people per wall does not change.

snow_tiger [21]3 years ago

5 0

Answer:

52 Minutes

Step-by-step explanation:

There is one person for every wall in the first part of the question. 5 people for 5 walls.

Therefore, the ratio of walls to people (walls:people) is 5:5.

5 to 5 is a one to one ratio.

5/5 = 1, which means it takes 1 person 52 minute to paint 1 wall.

Likewise, for the second part of the question, there are 20 people for 20 walls.

This is also a 1 to 1 ratio,

which means that it will also take 1 person 52 minutes to paint 1 wall.

Further Explanation:

People. Walls. Time

5. 5. 52

20. 20. 52

5. 20. 52 × 4 = 208

1. 20. 52 × 20 = 1040

20. 10. 52 / 2 = 26

You might be interested in

Paul bought a student discount card for the bus. The card allows him to buy daily bus passes for $1.60. After one month, Paul bo

Annette [7]

Paul spent $6 on the student discount card. 23 times 1.60 equals 36.80. Subtract that from 42.80 and you get 6

6 0

3 years ago

Easy math problems!!!

neonofarm [45]

Answer:

Rad 68

Step-by-step explanation:

84- 4^2=68

7 0

3 years ago

Read 2 more answers

A radiator contains 15 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure anti

Annette [7]

%Antifreeze=VAntifreezeVFluid VFluid=VAntifreeze%Antifreeze VAntifreeze=Vfluid∗%Antifreeze I want to find the amount of antifreeze in a 15 quart solution with 30% antifreeze VAntifreeze=15∗0.30 =18/4 quarts of antifreeze Similarly, I want to find the amount of antifreeze in a 15 quart solution with 35% antifreeze first. VAntifreeze=15∗0.35 = 21/4 quarts of antifreeze

the difference between 21/4 and 18/4 is 3/4 quarts, which is the amount of pure antifreeze I've added in.

SO the V_fluid I replaced with 3/4 quarts of antifreeze is (3/4)/ 0.35

5 0

4 years ago

Use the rules for multiplication by powers of 10 to calculate 3 x 100

Yakvenalex [24]

300

because 3 x 1 = 3 +00 = 300... make sense?

4 0

3 years ago

In a population of 10,000, there are 5000 nonsmokers, 2500 smokers of one pack or less per day, and 2500 smokers of more than on

Kazeer [188]

Answer:

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

Step-by-step explanation:

We have to write the transition matrix M for the population.

We have three states (nonsmokers, smokers of one pack and smokers of more than one pack), so we will have a 3x3 transition matrix.

We can write the transition matrix, in which the rows are the actual state and the columns are the future state.

- There is an 8% probability that a nonsmoker will begin smoking a pack or less per day, and a 2% probability that a nonsmoker will begin smoking more than a pack per day. Then, the probability of staying in the same state is 90%.

- For smokers who smoke a pack or less per day, there is a 10% probability of quitting and a 10% probability of increasing to more than a pack per day. Then, the probability of staying in the same state is 80%.

- For smokers who smoke more than a pack per day, there is an 8% probability of quitting and a 10% probability of dropping to a pack or less per day. Then, the probability of staying in the same state is 82%.

The transition matrix becomes:

$\begin{vmatrix} &NS&P1&PM\\NS& 0.90&0.08&0.02 \\ P1&0.10&0.80 &0.10 \\ PM& 0.08 &0.10&0.82 \end{vmatrix}$

The actual state matrix is

$\left[\begin{array}{ccc}5,000&2,500&2,500\end{array}\right]$

We can calculate the next month state by multupling the actual state matrix and the transition matrix:

$\left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4950&2650&2400\end{array}\right]$

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

To calculate the the state for the second month, we us the state of the first of the month and multiply it one time by the transition matrix:

$\left[\begin{array}{ccc}4950&2650&2400\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4912&2756&2332\end{array}\right]$

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

If we repeat this multiplication 12 times from the actual state (or 10 times from the two-months state), we will get the state a year from now:

$\left( \left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] \right)^{12} =\left[\begin{array}{ccc}4792.63&3005.44&2201.93\end{array}\right]$

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

3 0

4 years ago