A)0.5
B)2.5
C)-0.5
D)-2.5
Answers
Answer:
B)2.5
Step-by-step explanation:
When x = 3 the actual = 6
Residual = actual - predicted = 6 - 3.5 = 2.5
Related Questions
-(a-b)(a-bc) if a=3 b=-4 c=2
(7h + 7) + (7h + 8)........
Find the first partial derivatives of the function. f(x, t) = e−9t cos(πx)
. A coin is tossed three times, and the sequence of heads and tails is recorded.(a) Determine the sample space, Ω.(b) List the elements that make up the following events: i.A= exactly two tails, ii.B= at least twotails, iii.C= the last two tosses are heads(c) List the elements of the following events: i.A, ii.A∪B, iii.A∩B, iv.A∩C
Answers
Answer: 6 days
Step-by-step explanation:
take 22 off of the final price 46-22=24
now divid 24 by 4
and u have 6
How many additional wells should be drilled to obtain the maximum amount of oil per day?
Answers
Answer:
The additional wells for maximum amount of oil per day is 3 wells.
Step-by-step explanation:
Given;
initial number of wells, n = 6
total production, T = 1800
average production per well, = 1800/6 = 300 barrels per day
Let the additional well = y
total number of wells after optimization = 6 + y
new production per well = 300 - 25y
new total production = (6+y)(300-25y)
t = 1800 - 150y + 300y - 25y²
t = 1800 + 150y - 25y²
dt / dy = 150 -50y
for maximum value, dt/dy = 0
150 - 50y = 0
50y = 150
y = 150 / 50
y = 3
Therefore, the additional wells for maximum amount of oil per day is 3 wells.
Final answer:
By setting up the equation of the total daily oil production and finding its maximum, we learn that approximately 13 additional wells should be drilled to maximize the daily oil production.
Explanation:
To find out how many additional wells should be drilled to obtain the maximum
amount of oil
per day, we must firstly set up an equation to represent the situation. The total daily oil production is equal to the number of wells multiplied by the daily production per well. Given the conditions in the question, we can express this as:
Total daily oil production = (6 + x) * (1800 - 25x)
where x represents the number of additional wells that should be drilled. In order to find the maximum of this function, we would have to differentiate this equation and set the derivative equal to 0 then solve for x. This would be up to the individual's level of mathematical experience. However, one can use a financial calculator or a graphic calculator to find the maximum and get approximately 13 additional wells.
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Answers
Answer:
-5
Step-by-step explanation:
-4x +10 =5(x +11)
-4x +10 =5x +55
-4x - 5x =55 - 10
-9x =45
x=-45 :9
x=-5
Answers
Answer:
x = 22
y = 35
Step-by-step explanation:
Because its an isosceles triangle, you know the base angles are equal so:
Solving:
subtract 2x from both sides
add 11 to both sides
Next, find the angle measures of the two base angles:
To find y, you know that the interior angles of the whole triangle add up to 180, so you have
solving,
subtract 110 from both sides
divide both sides by 2
Answer:
x =22°
y =35°
Step-by-step explanation:
3x-11 = 2x+11
then x =22°
and 2y = 180-(55+55)
2y = 70
y = 45°
Answers
P(yellow or blue) = 15 + 20 = 35%
P(not green) = 100 - 20 = 80%
P(striped) = 0%
Answers
Final answer:
In damped harmonic motion, we calculate damping coefficient γ by comparing the periods of damped and undamped motion. For the given situation where the quasi-period is 90% greater than the undamped period, the damping coefficient is approximately 0.7416.
Explanation:
The subject of this question involves Damped Harmonic Motion, a concept in Physics, related to vibrations and waves. The equation given, u'' + γu' + u = 0, describes the motion where γ denotes the damping coefficient. Here, we have to calculate this damping coefficient when the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.
To solve this, we must use the relationship between damped and undamped periods. The quasi-period T' of a damped harmonic motion relates to the undamped period T as: T' = T/(sqrt(1 - (γ/2)^2)). Now, given that T' = 1.9T, we can but these two equations together:
1.9 = 1/(sqrt(1 - (γ/2)^2))
Solving this for γ, we get γ ≈ 0.7416. Hence, the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is approximately 0.7416.
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Final answer:
The value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the undamped motion is the one that satisfies γ=2*ω*0.9, where ω is the natural frequency of oscillation.
Explanation:
The given equation is for a damped harmonic oscillator, a physical system that oscillates under both a restoring force and a damping force proportional to the velocity of the system. The damping coefficient γ determines the behavior of the system and in this case, we need to find the value of γ such that the quasi period of the damped motion is 90% greater than the period of the undamped motion.
The period of the undamped motion, T₀, is calculated by the formula T₀=2π/sqrt(ω), where ω is the natural frequency of oscillation. The quasi period of the damped motion, Td, is increased by a factor of 1+η (in this case, 1.9 as the increase is 90%) and calculated by the formula Td=T₀(1+η) = T₀*1.9.
The damping ratio η is determined by the damping coefficient γ as η=γ/2ω. Therefore, by combining these expressions and rearranging the terms, we extract γ from these formulas as γ=2ω*η => γ=2*ω*(0.9). Thus, the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is the one which satisfies γ=2*ω*0.9.
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