The area of a rectangular deck is 680 square feet. The deck's width is 17 feet. What is its perimeter?
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Answer:
It will take 39.27 gallons to repaint the barn.
Step-by-step explanation:
The image of the setup is obtained from online and attached to this solution.
The rectangular prism is a cuboid of
Length = 84 ft
Breadth = 38 ft
Height = 20 ft
And the triangular prism on top of the rectangular prism has a vertical height of 12 ft.
The surface area of the composite structure that is available for painting include the 4 sides of the top triangular prism and the four sides of rectangular prism
Two of the faces of the triangular prism are triangles with base of 38 ft and height of 12 ft.
The other two faces of the triangular prism are rectangles with length 84 ft and breadth of the hypoteneuse of the right angled triangle on top.
B² = 12² + 19²
B = 22.47 ft
Total Surface area of the faces of the rectangular prism is then
2×(84×20) + 2×(38×20) = 4880 ft²
Total surface area of the faces of the triangular prism is then
2×(0.5×12×38) + 2×(84×22.47) = 4,230.96 ft²
Total surface area = 4880 + 4230.96 = 9110.96 ft²
1 gallon of paint = 232 ft²
x gallons of paint = 9110.96 ft²
x = (1×9110.96/232) = 39.27 gallons.
Hope this Helps!!!
Answer:
Step-by-step explanation:
Hello!
The objective is to test if rising the process temperature reduces the thickness of the plastic film that coats a substrate material To do so, two samples of substrates are coated at different temperatures:
Sample 1
X₁: Thickness of the plastic film after the substrate is coated at 125F
n₁=11
X[bar]₁= 101.28
S₁= 5.08
Sample 2
X₂: Thickness of the plastic film after the substrate is coated at 150F
n₂= 13
X[bar]₂= 101.70
S₂= 20.15
Does the data support this claim? Use the P-value approach and assume that the two population standard deviations are not equal.
Now if the higher the heat, the thinner the thickness of the plastic coating, then the average thickness of the coating done at 150F should be less than the average thickness of the coating done at 125F, symbolically: μ₂ < μ₁
Then the hypotheses are:
H₀: μ₂ ≥ μ₁
H₁: μ₂ < μ₁
α:0.05 (there is no α level stated so I've chosen the most common one)
Assuming that both variables have a normal distribution since the population standard deviations are not equal, the statistic to use is the Welch's t-test:
This test is one-tailed to the left, meaning that you'll reject the null hypothesis at small values of t. The p-value is also one-tailed and has the same direction as the test. To calculate it you have to first calculate the degrees of freedom of the Welch's t:
The distribution is a Student's t with 13 degrees of freedom, then you can calculate the p-value as:
P(t₁₃≤0.072)= 0.4718
Using the p-value approach, the decision rule is:
If the p-value ≤ α, the decision is to reject the null hypothesis.
If the p-value > α, the decision is to not reject the null hypothesis.
The p-value is greater than the significance level, so the decision is to nor reject the null hypothesis.
Using a significance level of 5%, there is no significant evidence to support the claim that the average thickness pf the plastic coat processed at 150F is less than the average thickness pf the plastic coat processed at 125F.
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