All categories

3 years ago

It takes 54 minutes for 4 people to paint 6 walls. How many minutes does it take 6 people to paint 7 walls?

Mathematics

2 answers:

Simora [160]3 years ago

8 0

Answer:

It will take 42 minutes.

Step-by-step explanation:

In order to solve this we first have to calulate how long will it take one person to paint one wall:

54/6=9minutes

That is between 4 people it will take 9 minutes to paint a wall, one person alone supposing they all work at the same rate will take 36 minutes.

Now we just multiply that by the number of walls to be painted:

36*7=252

And we divide that by the number of people painting:

252/6=42

So it will take 42 minutes to paint 7 walls with 6 people painting.

Nady [450]3 years ago

6 0

It takes 35 mins for 6 people to paint 7 walls if you would like this answer described to you let me know please

You might be interested in

obesity: The National Center for Health Statistics conducted the National Health Interview Survey (NHIS) for 27,787 U.S. civilia

WARRIOR [948]

Answer:

False.

Step-by-step explanation:

Hello!

29.9% is the proportion of adults aged 20 and over that were obese in January - September 2014.

This percentage was obtained from a sample of 27 787 adults, it's not a parameter but an estimation of the population proportion obtained from this sample.

I hope it helps!

5 0

3 years ago

Dionte needs to make at least $140 over winter break to pay back his parents. If he makes $15 per sidewalk that he shovels for,

lbvjy [14]

15 dollars from 1 sidewalk

140 dollars from ? sidewalks

140/15 =9.333(bar)

Dionte needs to shovel 10 sidewalks

3 0

3 years ago

Using the graph f(x)=x^2 as a guide, describe the transformations, and then graph each function.

Margarita [4]

Down refers to the movement along the y axis and left and right refer to movement along the x axis. 20. down 2, no y movement 21. 5 to the left, no y movement 22. to the right 1, no y movement 23. 4 to the left, down 3 24. to the left 2 and up 2 25. 4 to the right and down 9 26. opens up a bit wider, no x or y movement (picture it opening like a flower blooming) 27. opens downward and is very very close to the y axis 28. opened up SUPER wide, like a large shallow bowl.

6 0

3 years ago

Solve for n. (7^2)^4 = n^8

enyata [817]

7^2x4 = n^8
7^8 = n^8
n = 7

4 0

3 years ago

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

3 years ago