2 equations, 2 unknowns. Call height H and base B.

From the first sentence:
HB = 216

From the second sentence:
H = ⅔ B

Plug the second into the first and solve for B:
(⅔ B)B = 216
B^2 = 324
B = 18

Plug that back into either of the original 2 equations and solve for H:
H(18) = 216
H = 12

3 0

3 years ago

Someone please help me with this ASAP I have only 4 minutes!

Firlakuza [10]

Answer:

Step-by-step explanation:

3 0

3 years ago

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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:

Step:-(1)

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

Step :-2

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

Alternative hypothesis :H₁: 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.

3 0

3 years ago