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Sav [38]
3 years ago
7

1.) 6x + 3x^2 - x - x^2

Mathematics
2 answers:
svlad2 [7]3 years ago
7 0

Answer:

the answer is x=1/3y+1 y=3x−3

but wat the question?

shutvik [7]3 years ago
5 0

\large \mathfrak{Solution : }

1. 6x + 3x² - x - x²

  • 3x² - x² + 6x - x

  • 2x² + 5x

2. 5 + 2x + y + 2x - 1

  • 2x + 2x + y + 5 - 1

  • 4x + y + 4

3. 2x + 2y + x² - x + x²

  • x² + x² + 2x - x + 2y

  • 2x² + x + 2y

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Ruby and her friends are going into a restaurant. How much would it cost to buy 3
Gnoma [55]

Answer:

Step-by-step explanation:

3x2.25=6.75

4x4=16

16+6.75=22.75

8 0
3 years ago
A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 6%
Georgia [21]

Answer:

a

\% E =  0.9 \%

b

\%E_1 = 10.1 \%

Step-by-step explanation:

From the question we are told that

The probability that an employees suffered lost-time accidents last year is P(e) =  0.06

The probability that an employees suffered lost-time accident during the current year is

P(c) =  0.05

The probability that an employee will suffer lost time during the current year given that the employee suffered lost time last year is

P(c | e) =  0.15

Generally the probability that an employee will experience lost time in both year is mathematically represented as

P(c \ n \ e) =  P(e) *  P(c \ |\ e)

=> P(c \ n \ e) =  0.06*   0.15

=> P(c \ n \ e) = 0.009

Generally the percentage of employees that will experience lost time in both year is mathematically represented as

\% E =  P(e \ n \ c ) * 100

=> \% E =  0.009 * 100

=> \% E =  0.9 \%

Generally the probability that an employee will experience at least one lost time accident over the two-year period is mathematically represented as

P(e \ u \ c) =  P(e) + P(c) - P(e \ n \  c)

=> P(e \ u \ c) =  0.06 + 0.05 - 0.009

=> P(e \ u \ c) =  0.101

Generally the percentage of the employees who will suffer at least one lost-time accident over the two-year period is mathematically represented as

\%E_1 = P(e \ u \ c) *  100

=> \%E_1 = 0.101*  100

=> \%E_1 = 10.1 \%

7 0
3 years ago
We have a bag that contains 100 balls, 50 of them red and 50 blue. Select 5 balls at random. What is the probability that 3 are
MakcuM [25]

Answer:

Likely, I think I havent been in this for a long time so i may be a little rusty. But anyways, I hope this helped!

Step-by-step explanation:

5 0
2 years ago
The mean and the median are the same.<br> 2,3,4,5
Ber [7]

Step-by-step explanation:

mean \: ( \bar  x) =  \frac{2 + 3 + 4 + 5}{4}  =  \frac{14}{4}  = 3.5 \\  \\ median \: =  \frac{3 + 4 }{2}  =  \frac{7}{4}  = 3.5 \\  \\

Thus, mean and median are same.

7 0
3 years ago
15 times as much as 1 fith of 12
Mila [183]
The answer is 36.

12 / 5 = 2.4

2.4 • 15 = 36

I hope this helps!
8 0
3 years ago
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