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Dennis_Churaev [7]

2 years ago

7

Kate and Jean paint a house together in 6 days. If they work alone, Kate takes 5 fewer days than Jean would take to do the same

job. Which equation represents the time it takes Kate to paint the house by herself?
A. x^2-7x-30=0
B. x^2-5x-30=0
C. 2x-1=0
D. x^2-5x over 6 = 0

1 answer:

frutty [35]2 years ago

3 0

Number of jobs=1

Time:
<span>Since ‘x’ is given as a variable in the choices, we should reserve it for Kate and use ‘y’ for Jean as follows: </span>
<span>
Kate: => x </span>
<span>Jean: => y </span>

<span>But y = x - 5 </span>

<span>Combined rate: </span>
<span>1/x+ 1/y </span>
<span>= (x + y)/(xy) </span>

<span>Number of jobs done: </span>
<span>
Number of days X combined rate = number of jobs </span>
<span>6 [(x + y)/(xy)] = 1 </span>
<span>6x + 6y = xy </span>
<span>6x + 6(x - 5) = x(x - 5) </span>
<span>6x + 6x - 30 = x² - 5x </span>
<span>12x - 30 = x² - 5x </span>
<span>x² - 17x + 30 = 0
</span>
Therefore, the answer would be A.

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Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the

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Answer:

No, at the 0.05 significance level, the number of units produced on the night shift is not larger.

Step-by-step explanation:

We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.

Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.

Let $\mu_1$ = population mean number of units produced on the day shift

$\mu_2$ = population mean number of units produced on the night shift

So, <u>Null Hypothesis</u>, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1\geq\mu_2$ {means that the mean number of units produced on the night shift is same or lesser on the day shift}

<u>Alternate Hypothesis,</u> $H_A$ : $\mu_1-\mu_2$ or $\mu_1$ {means that the mean number of units produced on the night shift is larger}

The test statistics that will be used here is <u>Two-sample z test statistics</u> as we know about population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$ ~ N(0,1)

where, $\bar X_1$ = sample mean number of units produced by a sample of 54 day-shift workers = 345

$\bar X_2$ = sample mean number of units produced by a sample of 60 night-shift workers = 351

$\sigma_1$ = population standard deviation of the number of units produced on the day shift = 21

$\sigma_2$ = population standard deviation of the number of units produced on the day shift = 28

$n_1$ = sample of day-shift workers = 54

$n_2$ = sample of night-shift workers = 60

So, <em><u>test statistics</u></em> = $\frac{(345-351)-(0)}{\sqrt{\frac{21^{2} }{54}+\frac{28^{2} }{60} } }$

= -1.302

Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.

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2 years ago

A baby otter was born 3/4 of a month early. At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average

Assoli18 [71]

Answer:

The average weight of new born otter was, $1\frac{31}{40}$

Step-by-step explanation:

Let average weight of new born otter be x.

As per the given statement: At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average weight of a new born otter in the aquarium

" $\frac{9}{10}$ kg less than average weight of a new born otter" means $x-\frac{9}{10}$

As per the given information, we have;

$\frac{7}{8} = x -\frac{9}{10}$

Add $\frac{9}{10}$ both sides, we have;

$\frac{7}{8} + \frac{9}{10} = x$

Take LCM of 8 and 10 is, 40

⇒ $\frac{35+36}{40}=x$

Simplify:

$x = \frac{71}{40} = 1\frac{31}{40}$

Therefore, the average weight of new born otter was, $1\frac{31}{40}$

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2 years ago

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The number of bees that visit a plant is 500 times the number of years the plant is alive, where t represents the number of year

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here we have given that number of bees visit a plant id 500 times the number of years that the plant is alive.

we know that t is the number of years that plant is alive.

we know that $500*t$ is expression representing the number of bees that will visit the plant in its life time.

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The exchange rate is 13.25 Mexican pesos for each U.S. dollar. So 1 U.S. dollar = 13.25 Mexican pesos.
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2 years ago