Answer:
the company must buy 22 gallons to paint this entire area
Step-by-step explanation:
The circumference of the tank is given and is C = 2(pi)r, where r is the area.
118 ft
Here the circumference is C = 2(pi)(r) = 118 ft, which leads to r = ------------ ≈
18.79 ft ≈ r 2(pi)
The area of the sides is A = (circumference)(height), or approximately
(118 ft)(50 ft) = 5900 ft², and the area of the top is A = πr², which here comes to (π)(18.79 ft)² ≈ 1109 ft². Combining these two sub-areas, we get:
A(total) = 1109 ft² + 5900 ft² ≈ 7009 ft²
To determine how many gallons of paint will be needed to paint only the top and sides, we divide 7009 ft² by the coverage rate, which is
320 ft²
-----------
1 gallon
which results in:
7009 ft²
---------------------- ≈ 21.9 gallons
320 ft² / gallon
Since the paint comes only in full gallon cans, the company must buy 22 gallons to paint this entire area.
Answer:
(12.1409, 14.0591
Step-by-step explanation:
Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.
Group Group One Group Two
Mean 87.600 74.500
SD 1.000 1.500
SEM 0.316 0.433
N 10 12
The mean of Group One minus Group Two equals 13.100
standard error of difference = 0.556
90% confidence interval of this difference:
t = 23.5520
df = 20
Into? Is hard to answer without the rest of the problem
3(2)-18=-12
6-18 = -12
-12 = -12
Answer:
Area=190.091 cm^2
Step-by-step explanation:
Area = 1/2(Pi x r^2) one-half because it's a semi-circle
Area=1/2(3.14 x 11^2)
11^2=121 so, Area=1/2(3.14 x 121)
Area=1/2(379.94)
Area=189.97 cm^2
adjustment:
Area=1/2(3.142 x 11^2)
Area=1/2(3.142 x 121)
Area=1/2(380.182)
Area=190.091
