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lilavasa [31]
4 years ago
7

If the wages of 12 workers for 5 days are $7500, find the wages of 17 workers for 6 days.

Mathematics
2 answers:
a_sh-v [17]4 years ago
6 0
The correct answer would be 12,125
rewona [7]4 years ago
3 0
$7500 divided by 12 = $625

$625 divided by 5 = $125

each person gets $125 a day

$125 X 6 = $750

$750 X 17 = $12,750 -- Answer
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The sum of two numbers is -20. The difference of the two numbers is -48. What are the two numbers?
Lostsunrise [7]
34 and -14 as proven here:

Sum: 34 + -14 = 20
Difference: 34 - -14 = 48

just the opposite (when it says negative it gonna be positive)
3 0
3 years ago
Two angles are supplementary. The measure of one angle is 54 more than the measure of the other angle. What is the measure of th
il63 [147K]

Answer:

bigger angle is 117°

Step-by-step explanation:

x+x+54= 180 (when two angles are complementary it means that their sum is 180 degrees)

2x+54=180

2x= 180-54

2x=128

x=126/2

x= 63

x1= smaller angle

x1= 63

x2= bigger angle

x2= 63+54

x2=117

4 0
3 years ago
Use counting rules to find the following probabilities.
lions [1.4K]

Answer:

0.00144

Step-by-step explanation:

(13/1)(4/3)(12/1)(4/2)/(52/5)=(13)(4)(12)(6)/2598960=3744/2598960≈0.00144

8 0
4 years ago
The school that Scott goes to is selling tickets to the annual dance
zalisa [80]
Senior tickets (x)
Child tickets (y)

First day: 3x + 5y = 70

second day: 12x + 12y = 216

Solve the system of equations (use elimination)

Multiply first equation by -4 -4(3x + 5y = 70), which makes it

-12x - 20y = -280
(+)12x + 12y = 216 add to second equation

-8y = -64

divide by -8. y = 8

Plugin the y value to either equation ( I will choose first equation)

3x + 5(8) =70

3x+ 40 = 70

3x = 30

x = 10

Senior tickets are $10, child tickets are $8
4 0
3 years ago
A landscaper has enough cement to make a patio with an area of 150 Sq ft. The homeowner wants the length of the patio to be 6 ft
Natasha_Volkova [10]
A=lw=150\\l=w+6\\(w+6)w=150\\w^2+6w=150\\w^2+6w-150=0\\w=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\w=\frac{-(6)+-\sqrt{(6)^2-4(1)(-150)}}{2(1)}\\w=\frac{-6+-\sqrt{36+600}}{2}\\w=\frac{-6+-\sqrt{636}}{2}\\w=\frac{-6+-2\sqrt{159}}{2}\\w=-3+-\sqrt{159}>0\\w=-3+\sqrt{159}\\\\l=(-3+\sqrt{159})+6\\l=3+\sqrt{159}

Width = -3 + sqrt(159)
Length = 3 + sqrt(159)
4 0
3 years ago
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