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Arada [10]
4 years ago
9

PLEASE HELP ASAP FIRST TO ANSWER CORRECTLY GETS BRAINLYIST (question in the pic)

Mathematics
1 answer:
Simora [160]4 years ago
5 0

C i believe because they both can break it up those ways even if they might not be the most effective

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An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were
vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

<u>Step:-(1)</u>

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

<u>Step :-2</u>

<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

<u>Alternative hypothesis :H₁: </u>there is  significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

                     Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} }  } }

Given  n₁=n₂=60.

                    Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60}  } }

On calculation, we get

                   Z =   \frac{4}{\sqrt{0.6833} }

                   z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

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3 years ago
What is the greatest common factor of 20 and 55?
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Answer: 5

Step-by-step explanation:

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