Three snack bars contain 1/5, 0.22, and 19% of their calories from fat. Which snack bar contains the least amount of calories fr
1 answer:
You might be interested in
Answer:
Third option:
Step-by-step explanation:
<h3> The correct exercise is attached.</h3>
The equation given is:
The steps to find the value of "x" are shown below:
1. Add 2 to both sides of the equation:
2. Descompose 9 and 27 into their prime factors:
3. Substitute them into the equation:
4. Knowing that If , then
, we get:
5. Apply Distributive property:
6. Add 2 to both sides:
7. Divide both sides of the equation by 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.