Find the volumen of the semisphere and subtract the volumes of the two cylinders
1) Volume of the semisphere:
[1/2] (4/3)π(r^3) =[ 2π(1.5m)^3 ]/3 = 7.0686 m^3
2) Volumen of the cylinders:π(r^2)h
a) π(0.75/2m)^2 (1.75m) = 0.7731 m^3
b) π(1/2m)^2 (1.25m) = 0.9818 m^3
3) 7.0686 m^3 - 0.7731 m^3 - 0.9818m^3 = 5.3137 m^3
Answer: 5.3 m^3
The group rent 11 river rent tubes and 4 cooler tubes.
Step-by-step explanation:
Rent for river rent tube = $20
Rent for cooler tubes = $12.50
Total spent = $270
Total tubes rented = 15
Let,
x be the number of river rent tube.
y be the number of cooler tubes.
According to given statement;
x+y=15 Eqn 1
20x+12.50y=270 Eqn 2
Multiplying Eqn 1 by 20
Subtracting Eqn 2 from Eqn 3
Dividing both sides by 7.5
Putting y=4 in Eqn 1
The group rent 11 river rent tubes and 4 cooler tubes.
Keywords: linear equations, subtraction
Learn more about linear equations at:
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Answer:
13.8%
Step-by-step explanation:
P = nCr pʳ qⁿ⁻ʳ
P = ₉C₅ (0.37)⁵ (1−0.37)⁹⁻⁵
P ≈ 0.138
D. 1/3
A cube root can be written as an exponent: 1/3
![\sqrt[3]{x} = x^{\frac{1}{3} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D)
<h3>
Answer: 0.5</h3>
This is equivalent to the fraction 1/2
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Explanation:
The distance from A to B is 3 units. We can count out the spaces, or subtract the x coordinates of the two points and apply absolute value.
|A-B| = |-5-(-8)| = |-5+8| = |3| = 3
or
|B-A| = |-8-(-5)| = |-8+5| = |-3| = 3
We can say that segment AB is 3 units long.
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The distance from A' to B' is 1.5 units because...
|A'-B'| = |-2.5-(-4)| = |-2.5+4| = |1.5| = 1.5
or
|B'-A'| = |-4-(-2.5)| = |-4+2.5| = |-1.5| = 1.5
The absolute values ensure the distance is never negative.
We can say A'B' = 1.5
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Now divide the lengths of A'B' over AB to get the scale factor k
k = (A'B')/(AB)
k = (1.5)/(3)
k = 0.5
0.5 converts to the fraction 1/2.
The smaller rectangle A'B'C'D' has side lengths that are exactly 1/2 as long compared to the side lengths of ABCD.
