Exponent notation is 10^2 square feet, and in words it is one hundred square feet.
The paraboloid intersects the x-y plane when x²+y²=9, defining a circle with a radius of 3, centered at the origin.
<span>Utilizing cylindrical coordinates (r,θ,z), the paraboloid transforms into z = 9−r² and f = 5r²z. </span>
<span>If F represents the average of f over the area R then F ∫ (R)dV = ∫ (R)fdV </span>
<span>∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz </span>
<span>= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 </span>
<span>∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz </span>
<span>= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ </span>
<span>= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ </span>
<span>= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 </span>
<span>Thus, F = 10935π/8 ÷ 81π/2 = 135/4</span>
Answer:
Indeed, the equation is solvable by factoring. By applying the given equation, you can take the square root of both sides. Since both 169 and 9 are perfect squares, the left-hand side simplifies to plus or minus 13/3, producing rational results. Adding 6 to 13/3 yields a rational number while subtracting it does too. Thus, a quadratic equation is factorable if its solutions are rational.
Subtracting seven from sixteen results in nine
The solution is nine
Answer:
The anticipated number of tests required to identify 680 acceptable circuits is 907.
Step-by-step explanation:
For any circuit, there are two potential results: it either passes the test or it fails. The likelihood of passing is independent between circuits. Therefore, we apply the binomial probability distribution to address this scenario.
Binomial probability distribution
This distribution calculates the chance of obtaining exactly x successes across n trials, where x has only two possible outcomes.
To find the expected number of trials to achieve r successes with a probability p, the formula is given by:
Circuits from a specific factory pass a certain quality evaluation with a probability of 0.75.
Thus, to determine the expected number of tests needed for 680 acceptable circuits, let’s denote this as E where r = 680.
The expected number of tests necessary to find 680 acceptable circuits is 907.
