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suter [353]
4 years ago
8

If z is a positive integer, does 4 + 3(2z-5) represent a number that is greater than,less than ,or equal to 2(3z-4)? Please show

me how you got the answer.
Mathematics
2 answers:
EastWind [94]4 years ago
5 0

Answer:

The number will be greater than 2(3z-4)

Step-by-step explanation:

Thinking process:

Let the expressions be:

4 + 3(2z-5)            ...1

and

2(3z-4)                 ...2

Now, let z be a positive number, say 2

Substituting gives:

4 + 3(2*2-5) = 7     ...1

and

2(3*2-4)        = 4     ...2

As it can be seen from the solutions, the expression 4 + 3(2z-5)  will be greater than  2(3z-4) for any value of z.

That is,  4 + 3(2z-5) > 2(3z-4) for any real value of z.

DochEvi [55]4 years ago
4 0
4 + 3( 2(4) - 5)
= 4 + 24-5
= 28-5
= 23

2(3(4) -4)
= 2(12) -4
= 24 -4 = 20



4+3(2z-5) > 2(3z-4)
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Quadrilateral JKLM has vertices J(-4,-1), K(-1,2), L(6,2). For what coordinates of point M is JKLM a parallelogram?
Burka [1]
Answer: B

Look At the pictures Please. :)

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7 0
3 years ago
Complete the equivalent equation for -7x-60 =x^2 +10x
alex41 [277]

Answer:

(x+12)(x+5)

Step-by-step explanation:

Formula use: a²+bx+c

  • Make one side equal to zero:

Original:

-7x-60 =x² +10x

New:

x² + 17x + 60

  • Multiply A with C

New:

(1)x x 60 = 60

  • Find factors of 60 that when added, equal to 17.

New:

10 × 6, 60 × 1, 20 × 3, <u>5 × 12</u>, 4 × 15

5 times 12 equal 60, but when added equal to 17.

  • Replace the 17 with 5 and 12

New:

x² + 5x + 12x + 60

  • Break them off into two equations

New:

x² + 5x  l  12x + 60

  • Divide each equation into it's simpilest form. Make sure the numbers in the ( ) are the same.

New:

x(x + 5)  l  +12(x+5)

  • Answer: (x+12)(x+5)

6 0
3 years ago
How do you do 69% as a fraction? Please show work.​
MakcuM [25]
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7 0
3 years ago
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7 0
3 years ago
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A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod
tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1                2              3

495      525            470

500         515           480

505        505            460

<u>500         515             470        </u>

<u>∑2000     2060         1880</u>

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4=  470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using  μ ± σ , μ±2 σ  or μ ± 3 σ.

In this question the limits are determined by using  μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4=  470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other  two means are within the given UCL and LCL.

This can be shown by the diagram.

8 0
2 years ago
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