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lukranit [14]
4 years ago
7

One day for plumbers in five helpers earned $350. At the same rate of pay, another group of five plumbers and six helpers earned

$430 how much does a plumber and how much does a helper earn each day
Mathematics
1 answer:
schepotkina [342]4 years ago
3 0

Answer: a plumber earns $50 each day and a helper earns $30 each day.

Step-by-step explanation:

Let x represent the amount that a plumber earns in a day.

Let y represent the amount that a helper earns in a day.

One day, four plumbers and five helpers earned $350. This means that

4x + 5y = 350 - - - - - - - - - - -1

At the same rate of pay, another group of five plumbers and six helpers earned $430. This means that

5x + 6y = 430 - - - - - - - - - - - - 2

Multiplying equation 1 by 5 and equation 2 by 4, it becomes

20x + 25y = 1750

20x + 24y = 1720

Subtracting, it becomes

y = 30

Substituting y = 30 into equation 1, it becomes

4x + 5 × 30 = 350

4x + 150 = 350

4x = 350 - 150

4x = 200

x = 200/4 = 50

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pashok25 [27]

Answer:

a) y=\frac{28}{75}(x-1900)+47.3

b) 86.5 years

Step-by-step explanation:

All values are rounded to 2 decimal places

a) We can find line by using y =mx+b

y=mx+b\\m=\frac{y^{2}-y^{1} }{x^{2}-x^{1}}\\m=\frac{69.7-47.3}{1960-1900} \\m=\frac{22.4}{60} \\m=\frac{28}{75}\\c=y-intercept\\y-intercept=47.3\\y=\frac{28}{75} (x-1900)+47.3

Therefore the line of best fit is y=\frac{28}{75}(x-1900)+47.3\\

b) We can do this by using the formula of the best fit line to estimate the life expectancy of someone born in 2005

y=\frac{28}{75}(x-1900)+47.3\\y=\frac{28}{75}(2005-1900)+47.3\\y=\frac{28}{75}\times105+47.3\\y=86.5

Therefore the estimated life expectancy of someone born in 2005 is 86.5 years

PS. Please give brainliest answer this was a lot of working out

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3 years ago
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