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Mr. Brust needs 427 cans to cover the wall of dimensions 20 feet long and 9 feet tall.
<h3>What is area of rectangle?</h3>Area of rectangle is the product of its sides
i.e., Area of rectangle = length x breadth
<h3>What is area of circle?</h3>The area of a circle is pi times the radius squared (A = π r²).
<h3>What is circumference of circle?</h3>The circumference of a circle is defined as the linear distance around it.
( C= 2πr)
The can of red bull is in shape of cylinder.
If we open the cylinder it will have two circular surfaces( at bottom and top) and one rectangular sheet.
Like the given figure given.
- The can is 6.5 inches tall and has a diameter of about 2.5 inches.
It shows that the two circle will have the diameter 2.5 inches.
d= 2.5 inch
= 0.208 ft
radius, r=
r=
r= 0.104 ft
- To get the length of the rectangle , we have to calculate the circumference of the circle.
Circumference of the circle be,
= 2πr
= 2 x 3.14 x 0.104
= 0.653 ft
The rectangular sheet will have,
length= 0.653 ft, breadth= 6.5 inches = 0.541 ft
- Now, area of rectangular sheet be,
= length x breadth
= 0.653 x 0.541
=
- Area of 2 circle( top and bottom of can) be,
= 2 x π
= 2 x 3.14 x 0.104 x 0.104
= 6.28 x 0.010816
= 0.0679
- Total area of can be,
= Area of rectangular sheet + Area of 2 circle( top and bottom of can)
= 0.3532 + 0.0679
= 0.4211
Now, we have the wall of dimensions
length= 9 ft, breadth= 20 ft
- Area of wall be,
= length x breadth
=9 x 20
= 180
- Number of cans required to cover the whole wall be,
=
=
= 427.45 cans
≈ 428 cans
Learn more about the concept of area here:
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Answer:
= 1.56
Step-by-step explanation:
Hello!
You have two independent samples of fertilizers from distributor A and B, and need to test if the average content of nitrogen from fertilizer distributed by A is greater than the average content of nitrogen from fertilizer distributed by B.
Sample 1
X₁: Content of nitrogen of a fertilizer batch distributed by A
n₁= 4 batches
Sample mean X₁[bar]= 23pound/batch
σ₁= 4 poundes/batch
Sample 2
X₂: Content of nitrogen of a fertilizer batch distributed by B
n₂= 4 batches
Sample mean X₂[bar]= 18pound/batch
σ₂= 5 pounds/batch
Both variables have a normal distribution. Now since you have the information on the variables distribution and the values of the population standard deviations, you could use a pooled Z-test.
Your hypothesis are:
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
The statistic to use is:
Z= <u> X₁[bar] - X₂[bar] - (μ₁ - μ₂) </u>~N(0;1)
√(δ²₁/n₁ + δ²₂/n₂)
=<u> </u>(<u>23 - 18) - 0 </u> = 1.56
√(16/4 + 25/4)
I hope this helps!