Ask question

Ask question

All categories

3 years ago

15

Jane works twice as fast as her daughter Anna. If it takes 15 minutes to clean the kitchen together, how long would it take Anna

to clean the kitchen by herself?

1 answer:

Sonbull [250]3 years ago

7 0

Here, Twice fast means, she takes half time in doing same amount of work.
Let, Anna takes = 2t & Jane takes = t

It would be: 1/t + 1/2t = 1/15
2t + t / 2t² = 1/15
15(3t) = 2t²
2t² = 45t
Divide both sides by t,
2t = 45
t = 45/2
t = 22.5
& 2t = 22.5(2) = 45

In short, Anna would take 45 minutes

Hope this helps!

You might be interested in

Evaluate the expression if m = − 4 , n = 1 , p = 2 , q = − 6 , r = 5 , and t = − 2 | 16 + 4 ( 3 q + p )

Arisa [49]

The evaluation of the expression, if m = − 4 , n = 1 , p = 2 , q = − 6 , r = 5 , and t = − 2 | 16 + 4 ( 3 q + p ) is 46.

<h3>How can the expression be simplified?</h3>

the given expression is t = − 2 | 16 + 4 ( 3 q + p )

Then since we are given m = − 4 , n = 1 , p = 2 , q = − 6 , r = 5

Then, we can substitute all the given values of the terms into the given expression as :

t = − 2 | 16 + 4 ( 3 q + p )

t= -2 I 16+4(-18+2)

t= -2 I 16+4(-16)

t= -2 I 16 -64

t= -2 I -48

t =46

Read more about expression at:

brainly.com/question/723406

#SPJ1

4 0

1 year ago

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$

Dropping the units and simplifying, we get the linear ODE

$A'=10+5\sin\dfrac t4-\dfrac A{10}$

$10A'+A=100+50\sin\dfrac t4$

Multiplying both sides by $e^{10t}$ allows us to identify the left side as a derivative of a product:

$10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}$

$\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}$

$e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt$

Integrate and divide both sides by $e^{10t}$ to get

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}$

The tanks starts off with 30 lb of salt, so $A(0)=30$ and we can solve for $C$ to get a particular solution of

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}$

6 0

3 years ago

Two consecutive odd integers have sum of 44. Find the integers

satela [25.4K]

Answer:

21, and 23

Step-by-step explanation:

divide 44 by 2 to find the integer that spits it up perfectly into two. Then just add 1 to one 22 and minus 1 to the other. Then its 21 and 23.

8 0

2 years ago

What is the end behavior of y=-2x^13+25x^8-3 as x approaches infinity?

RUDIKE [14]

Odd degree and negative leading coefient
means it goes from the top left to bottom right
goes infinitely down

as x aproaches infinity, y approaches neagtive infinity

6 0

3 years ago

Toll Brothers is a luxury home builder that would like to test the hypothesis that the average size of new homes exceeds 2,400 s

boyakko [2]

Answer:

Since the pvalue of the test is 0.09 > 0.01, we do not reject the null hypothesis that the average size of new homes is of 2400 square feet.

Step-by-step explanation:

Toll Brothers is a luxury home builder that would like to test the hypothesis that the average size of new homes exceeds 2,400 square feet. Test to see if the average size of new homes exceeds 2,400 square feet.

At the null hypothesis, we test if the average is of 2400 square feet, that is:

$H_o: \mu = 2400$

At the alternate hypothesis, we test if the average is greater than 2400 square feet, that is:

$H_a: \mu > 2400$

The test statistic is:

Since we have the standard deviation for the sample, we use the t-distribution.

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

2400 is tested at the null hypothesis:

This means that $\mu = 2400$

A random sample of 36 newly constructed homes had an average of 2,510 square feet with a sample standard deviation of 480 square feet.

This means that $n = 36, X = 2510, s = 480$

Test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{2510 - 2400}{\frac{480}{\sqrt{36}}}$

$t = 1.375$

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a sample mean of at least 2510, which is the pvalue of t = 1.375 with 36 - 1 = 35 degrees of freedom, using a one-tailed test.

With the help of a calculator, this pvalue is of 0.09

Since the pvalue of the test is 0.09 > 0.01, we do not reject the null hypothesis that the average size of new homes is of 2400 square feet.

6 0

2 years ago