1)
.
2)
.
3) The particle is moving right when the velocity function is positive:
or
.
4) When
the particle is slowing down because the acceleration is close to zero
the particle is speeding up when acceleration is increasing away from zero:
.
5)
.
<u>Answer-</u>
The standard error of the confidence interval is 0.63%
<u>Solution-</u>
Given,
n = 2373 (sample size)
x = 255 (number of people who bought)
The mean of the sample M will be,
Then the standard error SE will be,
Therefore, the standard error of the confidence interval is 0.63%
Answer:
the rate is: 6 cups of flour per cup of water
Step-by-step explanation:
Recall that rate involve quotient of the two quantities in question:
Then cups of flour per cups of water is the quotient: 2 cups of flour divided by 1/3 cup of water:
this means 6 cups of flour per cup of water
To solve this problem you must apply the proccedure shown below:
1. You have that the hyperbola <span>has a vertex at (0,36) and a focus at (0,39).
2. Therefore, the equation of the directrices is:
a=36
a^2=1296
c=39
y=a^2/c
3. When you susbtitute the values of a^2 and c into </span>y=a^2/c, you obtain:
<span>
</span>y=a^2/c
<span> y=1296/13
4. When you simplify:
y=432/13
Therefore, the answer is: </span><span>y = ±432/13</span>
The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin.
<span>Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. </span>
<span>If F is the mean of f over the region 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>∴ F = 10935π/8 ÷ 81π/2 = 135/4</span>
