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3 years ago

8

One painter can finish a certain job 5 hours before another painter who is doing the same job. When they work together they can

finish this job in 6 hours. How long does it take for each painter to do the job by himself?

1 answer:

Whitepunk [10]3 years ago

5 0

Answer:

A (painter 1) = 10 hours

B (painter 2) = 15 hours

Step-by-step explanation:

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a parking meter has 32 coins in nickels and quarters. the coins are worth $3.20. how many quarters and how many nickels are in t

BigorU [14]

Answer:

n = 24

q = 8

Step-by-step explanation:

Let:

n = nickels

q = quarters

5n + 25q = 320

n + q = 32

n = 32 - q

Substitute (to solve for q):

5(32 - q) + 25q = 320

160 - 5q + 25q = 320

160 + 20q = 320

20q = 320 - 160

20q = 160

q = 160 ÷ 20

q = 8

Substitute (to solve for n):

n = 32 - q

n = 32 - (8)

n = 24

8 0

3 years ago

a 16 inch wire is to be cut. one piece is to be bent into the shape of a square, whereas the other piece is to be bent into the

Lyrx [107]

1. Let x be a length of square side, then the square perimeter is 4x (you need to cut 4x from 16 in. to make a square with side x in.). Then 16-4x is remained length of wire and you have to make form this piece a rectangle with sides one of which is twice bigger than another. If y is length of the smaller side, then 2y is length of the bigger side and rectangle perimeter is y+2y+y+2y=6y. You have 16-4x in. of wire left, so 6y=16-4x and $y= \frac{16-4x}{6}$ .

2.
$A_{square}=x^2 \\ A_{rectangle}=y\cdot 2y=\frac{16-4x}{6}\cdot2\cdot\frac{16-4x}{6}= \frac{(16-4x)^2}{18} \\ A(x)=A_{square}+A_{rectangle}=x^2+\frac{(16-4x)^2}{18}$ .

3. Find the derivative of the function A(x):
$A'(x)=2x+ \dfrac{2(16-4x)\cdot(-4)}{18} =2x- \dfrac{4(16-4x )}{9}$ .

4. Solve the equation A'(x)=0:
$2x- \dfrac{4(16-4x )}{9}=0 \\ 18x-64+16x=0 \\ 34x=64 \\ x= \frac{32}{17}$

5. Since $y= \frac{16-4x}{6}$ you have

$y= \dfrac{16-4 \frac{32}{17} }{6} = \dfrac{16\cdot17-4\cdot32}{6\cdot 17} = \dfrac{24}{17}$ .

Answer: <span>the width of the rectangle is $\frac{24}{17}$ </span>

6 0

4 years ago

Find the maximum rate of change of f at the given point and the direction in which it occurs. f(p, q) = 8qe−p + 4pe−q, (0, 0) ma

Pachacha [2.7K]

Answer:

Step-by-step explanation:

$Given \ f (p,q) = 8qe^{-p} + 4pe^{-q} \ where \ P_o(0,0) \\ \text{thus the gradient of f is: } \\ \\ \bigtriangledown f(p,q) = \Big(\dfrac{\partial f}{\partial p}, \dfrac{\partial f}{\partial q} \Big) \\ \\ \dfrac{\partial f}{\partial p} = -8qe^{-p} + 4pe^{-q} \\ \\ \dfrac{\partial f}{\partial p} = 8qe^{-p} - 4pe^{-q} \\ \\ Then: \bigtriangledown f(p.q) = (-4qe^{-p}+ 8qe^{-q}, 4qe^{-p}- 8qe^{-q}) \\ \\ f(0,0) = (-4*(0)e^{-0}+ 8*(0)e^{-(0)}, 4*(0)e^{-(0)}- 8*(0)e^{-(0)}) \\ \\ = (0+8,4-0) = (8.4)$

$\Big| \Big | \bigtriangledown f(0,0) = \sqrt{(8)^2+4^2} \\ \\ = \sqrt{64+16} \\ \\ = \sqrt{80}$

$\mathbf{the \ direction \ of \ maximum \ change \ is }= \mathbf{\sqrt{80}} \\ \\ \mathbf{direction }(8,4)$

7 0

3 years ago

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.

8 0

3 years ago

Solve for f.<br><br> f+0.2=−3 HELP THIS IS URGENT

Mumz [18]

F= -3 -.2
F= -3.2
The final answer is -3.2

4 0

3 years ago

Read 2 more answers