Egbert is making trail mix out of 18 bags of nuts and 9 bags of dried fruit. He wants each new portion of trail mixto be identic
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Answer:
A. 4.26 in^2
Step-by-step explanation:
Step 1: Find the area of the sector DBC. Here we have to use the formula.
Area of a sector = Central Angle/360 *π
The area of the sector DBC = (54/360)*3.14*
= 30.16
Step 2: Area of segment CFD = Area of the sector DBC - Area of the ΔCBD
= 30.17 - 25.9
Area of segment CFD = 4.27in^2
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
Given n₁=n₂=60.
On calculation, we get
Z =
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.