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Vlada [557]
3 years ago
7

Solve the system of equations - 4x + 2y = -30 and – 2x + 2y = -10 by combining the equations.

Mathematics
1 answer:
I am Lyosha [343]3 years ago
3 0

Answer:

  (x, y) = (10, 5)

Step-by-step explanation:

We notice that the y-coefficient is the same in both equations, and that the x-coefficient is smaller in the first equation. Subtracting one equation from the other will eliminate the y-variable. Subtracting the first equation from the second will result in a positive coefficient for x.

Subtracting the first equation from the second, we have ...

  (-2x +2y) -(-4x +2y) = (-10) -(-30)

  2x = 20 . . . .simplify

  x = 10 . . . . . divide by 2

Substituting for x in the second equation gives ...

  -2(10) +2y = -10

  2y = 10 . . . . add 20

  y = 5 . . . . . . divide by 2

The solution is (x, y) = (10, 5).

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Complete Question

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.)  

t=8.169  

P-value= ?

Answer:

a)  P-value=0

b)  Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/  cellulose composite exceeds 48 MPa.

Step-by-step explanation:

From the question we are told that:

Sample size n=10

Mean \=x= 51.3

Standard deviation \sigma=1.2

Significance level is taken as \alpha=0.05

t test statistics

t=8.169

Therefore

P-Value=P(t>8.169)

Critical point

t_{\alpha,df}

\alpha=0.05

df=10-1=>9

Therefore

P-value from T distribution table

P-value=0

Conclusion

P-value (0)< \alpha(0.05)

We Reject the Null Hypothesis H_0

Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/  cellulose composite exceeds 48 MPa.

7 0
3 years ago
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