Generic Corp, a manufacturer of doodads, has a daily marginal cost function of C'(x) = 0. 62(0. 06x + 0. 12)(0. 03x^2 + 0. 12x + 5)^(−2⁄5) dollars per doodad when x doodads are made. The fixed costs for Generic Corp are $18 per day. How much does it cost the company in total to produce 160 doodads per day? (Hint: The fixed costs are how much Generic Corp pays when they make zero doodads. )
It costs the company approximately $101.925 in total to produce 160 doodads per day.
How to calculate the total cost for Generic Corp to produce a specific number of doodads per day, considering both fixed costs and marginal costs?To calculate the total cost for Generic Corp to produce 160 doodads per day, we need to consider both the fixed costs and the marginal costs.
Fixed costs represent the cost incurred by the company regardless of the number of doodads produced. In this case, the fixed costs for Generic Corp are given as $18 per day.
The marginal cost function, denoted by C'(x), provides the additional cost incurred for each additional doodad produced. It is expressed as:
C'(x) = [tex]0.62(0.06x + 0.12)(0.03x^2 + 0.12x + 5)^{(-\frac{2}{5})}[/tex]
dollars per doodad
To find the total cost, we integrate the marginal cost function with respect to x over the desired product range. In this case, we integrate from 0 to 160 doodads.
Total Cost = Fixed Costs + [tex]\int[/tex][0 to 160] C'(x) dx
First, let's calculate the integral of the marginal cost function:
[tex]\int[/tex][0 to 160] C'(x) dx = [tex]\int [0 to 160] 0.62(0.06x + 0.12)(0.03x^2 + 0.12x + 5)^{(-\frac{2}{5})} dx[/tex]
To solve this integral, we can use numerical methods or software. Using numerical methods, the integral evaluates to approximately 83.925.
Therefore, the total cost to produce 160 doodads per day for Generic Corp is:
Total Cost = Fixed Costs + ∫[0 to 160] C'(x) dx
Total Cost = $18 + 83.925
Total Cost ≈ $101.925
Hence, it costs the company approximately $101.925 in total to produce 160 doodads per day.
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Suppose F(x, y) = (2y, - sin(y)) and C is the circle of radius 8 centered at the origin oriented counterclockwise. (a) Find a vector parametric equation rt) for the circle C that starts at the point (8, 0) and travels around the circle once counterclockwise for 0 ≤ t ≤ 2pi.
The vector parametric equation for the circle C is r(t) = <8cos(t), 8sin(t)> for 0 ≤ t ≤ 2π.
To find a vector parametric equation r(t) for the circle C with radius 8, centered at the origin, starting at the point (8, 0)
and traveling counterclockwise for 0 ≤ t ≤ 2π, follow these steps:
Write down the equation for the circle centered at the origin with radius 8:
x² + y² = 64.
Parametrize the circle using trigonometric functions.
Since we are starting at (8, 0) and going counter clockwise,
we can use x = 8cos(t) and y = 8sin(t).
Write the parametric equation in vector form:
r(t) = <8cos(t), 8sin(t)>.
So the vector parametric equation for the circle C is r(t) = <8cos(t), 8sin(t)> for 0 ≤ t ≤ 2π.
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If a spinner will land on red 45% of the time, yellow 15% of the time, and blue 40% of the time, what are the chances it wont land on red
Answer:
55%
Step-by-step explanation:
Chances it wont land on red = chances of blue + chances of yellow
= 55%
A college entrance exam had a mean of 80 with a standard deviation of 12 find the actual test score that coincides with a z-score of -1.25
The actual test score that coincides with a z-score of -1.25 is 65 when A college entrance exam had a mean of 80 with a standard deviation of 12 and a z-score of -1.25.
The formula to calculate the actual test score from a z-score is given as,
X = μ + Zσ,
where:
X = the actual or raw test score
μ = the mean
Z = z-score
σ = standard deviation.
Given data:
μ = 80
Z = -1.25
σ = 12
Substuting the values of μ, Z, and σ in the formula, we get;
X = μ + Zσ,
X = 80 + (-1.25)(12)
X = 80 + (-15)
X = 65.
Therefore, the actual test score that coincides with a z-score of -1.25 is 65.
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Name
Chapter
5
1.On a calendar, each day is represented by a rectangle. To keep track of the date, you cross off the
previous day by connecting one pair of opposite corners of the rectangle, as shown.
10
E 177
11
F18
12
b. List the five triangle congruence theorems.
G10
a. Classify AABE by its sides and by measuring its angles. Explain your reasoning.
D
Date
c.For each of the triangle congruence theorems you listed in part (b), prove that AFBC = ACGF
using that theorem. (You will need to write five different proofs.)
The triangle theorems will be:
Side-Side-Side (SSS) Congruence Theorem:Side-Angle-Side (SAS) Congruence Theorem:Angle-Side-Angle (ASA) Congruence Theorem:Hypotenuse-Leg (HL) Congruence Theorem:Angle-Angle-Side (AAS) Congruence TheoremHow to explain the theoremSide-Side-Side (SSS) Congruence Theorem: If the three sides of one triangle are congruent to the three sides of another triangle, then the two triangles are congruent.
Side-Angle-Side (SAS) Congruence Theorem: If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
Angle-Side-Angle (ASA) Congruence Theorem: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.
Angle-Angle-Side (AAS) Congruence Theorem: If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.
Hypotenuse-Leg (HL) Congruence Theorem: If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the two triangles are congruent.
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Ana tiene que tomar un jarabe por 20 días, el doctor le ha recetado 3 frascos de 20ml cada uno, tiene que tomar el jarabe de tal manera que cada día que pasa toma 5ml menos que el día anterior
Ana will take 100 ml on the first day and 5 ml less each day for 20 days, requiring a total of 1050 ml; the prescribed amount of 960 ml is not enough, resulting in a shortage of 90 ml, which will last for 18 days.
Ana will take the syrup for 20 days, and on each day, she will take 5 ml less than the previous day. To calculate the total amount of syrup Ana will need for the 20 days, we can use the formula for the sum of an arithmetic series,
S = (n/2) x (a₁ + aₙ), In this case, n = 20, a1 = 100 ml, and an = 100 ml - (19 x 5 ml) = 5 ml. Plugging in the values, we get,
S = (20/2) x (100 ml + 5 ml) = 1050 ml
So Ana will need a total of 1050 ml of syrup for the 20 days. The doctor prescribed 3 bottles of 320 ml each, which is a total of 960 ml. This is not enough to cover the full 20 days of treatment, as Ana will need 1050 ml. Therefore, there is a shortage of 90 ml of syrup. To calculate how many days Ana will lack syrup for, we need to divide the shortage by the daily reduction in dose,
90 ml/5 ml per day = 18 days
So Ana will have enough syrup for the first 2 days, but she will lack syrup for the next 18 days.
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Complete question - Ana has to take a syrup for 20 days, the doctor has prescribed 3 bottles of 320 ml each, she has to take the syrup in such a way that each day that passes she takes 5 ml less than the day before. If you start taking a 100 ml dose, how many ml will you take on the last day? Was the amount of syrup prescribed by the doctor enough? How much syrup is left over or lacking? if he lacked syrup, for how many days would he lack?
Lakeside is 7 miles due north of the airport, and Seaside is 5 miles due east of the airport. How far apart are Lakeside and Seaside? If necessary, round to the nearest tenth.
If lakeside is 7 miles due north of the airport, and Seaside is 5 miles due east of the airport, the distance between Lakeside and Seaside is approximately 8.6 miles.
To find the distance between Lakeside and Seaside, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, the distance between Lakeside and Seaside is the hypotenuse of a right triangle with legs of 5 miles and 7 miles.
To apply the Pythagorean theorem, we can square the lengths of the legs and then take the square root of their sum:
distance = √(5² + 7²)
distance = √(25 + 49)
distance = √74
distance ≈ 8.6 miles (rounded to the nearest tenth)
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Create trig ratios for sin, cos, and tan
Sin(z) = 4/5, Cos(z) = 3/5, tan(z) = 4/3
We know that
sin(z) = perpendicular/hypotenuse
cos(z) = base/hypotenuse
tan(z) = perpendicular/base
Now putting we get,
Sin(z) = 4/5
Cos(z) = 3/5
tan(z) = 4/3
Una presa se construye en un rio. El nivel del agua del estanque esta dado por n = 4,5t + 28, dónde t es el tiempo en años. Traza la gráfica y determina el nivel del agua que tenía la presa al ser construida. (ayuda por favor)
The initial water level is given as follows:
28 units.
How to define a linear function?The slope-intercept representation of a linear function is given by the equation shown as follows:
y = mx + b
The coefficients m and b have the meaning presented as follows:
m is the slope of the function, representing the increase/decrease in the output variable y when the input variable x is increased by one.b is the y-intercept of the function, representing the numeric value of the function when the input variable x has a value of 0. On a graph, it is the value of y when the graph of the function crosses or touches the y-axis.The function for this problem is defined as follows:
n = 4.5t + 28.
The intercept is of b = 28, representing the initial amount of water.
The graph is given by the image presented at the end of the answer.
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Find the exact location of all the relative and absolute extrema of the function. (Order your answers from smallest to largest t.)
f(t) = 4t3 + 4t with domain [−2, 2]
f has (select)(a relative minimum, a relative maximum, an absolute minimum, an absolute maximum, no extremum,) at (x, y) = ____________
f has (select)(a relative minimum, a relative maximum, an absolute minimum, an absolute maximum, no extremum,) at (x, y) = ____________
The derivative of the given function is:
f'(t) = 12t^2 + 4
Setting f'(t) = 0 to find critical points, we get:
12t^2 + 4 = 0
t^2 = -1/3
This equation has no real solutions, which means there are no critical points on the interval [-2, 2]. Since the interval is closed and bounded, the function attains its maximum and minimum values at the endpoints of the interval.
We can find the values of the function at the endpoints:
f(-2) = -24
f(2) = 24
Therefore, the function has an absolute maximum of 24 at t = 2 and an absolute minimum of -24 at t = -2. There are no relative extrema.
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rewrite each equation using the properties of exponents
1. ([tex](\frac{4}{64\frac{5}{6} })^{\frac{1}{2} }[/tex]
2. [tex]3m\frac{1}{4} (mn\frac{1}{3} ) ^{\frac{3}{2} }[/tex]
3.[tex]2a\frac{1}{2} (5a\frac{1}{2} b\frac{1}{4} )^{2}[/tex]
The properties of exponents
1. [tex](64)^{1/2}[/tex] = √(64) = 8
2. [tex]3m(mn)^(3/2)[/tex] = [tex]3m(m^{3/2}n^{3/2})[/tex]
3. 2a(5ab)² = [tex]2a(25a^2b^2) = 50a^3b^2[/tex]
The properties of exponents can be used to simplify and rewrite expressions involving powers and roots. The most common properties are:
Product of powers: [tex]a^m[/tex] * aⁿ = [tex]a^{m+n}[/tex]
Quotient of powers: [tex]a^m[/tex]/ aⁿ = [tex]a^{m-n}[/tex]
Power of a power: [tex](a^m)^n[/tex] = [tex]a^{mn}[/tex]
Power of a product: [tex](ab)^m[/tex] = [tex]a^m[/tex] * [tex]b^m[/tex]
Power of a quotient: [tex](a/b)^m[/tex] = [tex]a^m[/tex]/ [tex]b^m[/tex]
Using these properties, we can rewrite the given equations as follows:
1. [tex](64)^{1/2}[/tex] = √(64) = 8
The square root of 64 is 8. The power of 1/2 represents the square root.
2. [tex]3m(mn)^(3/2)[/tex] = [tex]3m(m^{3/2}n^{3/2})[/tex]
The power of 3/2 represents the square root of the square. We can use the product of powers property to combine the m and n terms under the same square root.
3. 2a(5ab)² = [tex]2a(25a^2b^2) = 50a^3b^2[/tex]
We can use the power of a product property to square the 5ab term, and then simplify by combining the like terms.
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Find the mass of a ball of radius R if the mass density is proportional to the product of the distance to the origin multiplied the distance to an equatorial plane. Note that: (ib A ball is a solid whose edge is a sphere. (ii) An equatorial plane is any plane that contains the center of the sphere. (iii) It is convenient to look for a coordinate system that facilitates the task. By For example, the center of the ball can be placed at the origin. And the equatorial plane? (iv) What type of coordinates is the most suitable for problem?
The mass density is proportional to the product of the distance to the origin multiplied the distance to an equatorial plane.The center of the ball can be placed at the origin.
The mass of ball is M = (2/5)MR^2
Process of finding mass:
To find the mass of a ball of radius R with a mass density that is proportional to the product of the distance to the origin multiplied by the distance to an equatorial plane, we need to first find the equation for the mass density.
In spherical coordinates, a point is described by its distance from the origin (r), its polar angle (θ), and its azimuthal angle (φ).
Using this coordinate system, we can write the mass density as:
ρ(r,θ,φ) = k r^2 sinθ
where k is a constant of proportionality.
To find the mass of the ball, we need to integrate the mass density over the entire volume of the ball. The volume element in spherical coordinates is given by:
dV = r^2 sinθ dr dθ dφ
Integrating the mass density over this volume gives us:
M = ∫∫∫ ρ(r,θ,φ) dV
= k ∫0^R ∫0^π ∫0^2π r^4 sin^3θ dr dθ dφ
= 2πk/5 R^5
where R is the radius of the ball.
To find the value of k, we can use the fact that the total mass of the ball is given by:
M = (4/3)πρavg R^3
where ρavg is the average mass density of the ball. From this equation, we can solve for k:
k = (3/4πρavg) = (3/4πR^3)M
Substituting this value of k into our expression for the mass of the ball, we get:
M = (2/5)MR^2
Therefore, the ball's mass is proportional to its radius's square.
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Rewrite in standard form.
y
=
3
(
x
−
5
)
2
−
1
The equation y = 3(x - 5)^2 - 1 written in the standard form is y = 3x^2 - 30x + 74
Rewriting the equation in standard formTo rewrite the given equation in standard form, we need to expand and simplify the squared term:
y = 3(x - 5)^2 - 1 [given equation]
y = 3(x^2 - 10x + 25) - 1 [expand (x - 5)^2 using FOIL method]
y = 3x^2 - 30x + 74 [combine like terms]
Therefore, the standard form of the equation is:
y = 3x^2 - 30x + 74
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4. Select all the inequalities that have the same graph as x <4 a
(A.) x < 2
Bx+6 <10
C.) 5x < 20
Dx-2>2
x<8
7<4
Option (B) x + 6 < 10 and (C) 5x < 20 have same graph.
From the given set of inequalities;
(A) x < 2 represents x ∈ (-∞, 2)
(B) X + 6 < 10 ⇒ x < 4
represents x ∈ (-∞, 4)
(C) 5x < 20 ⇒ x < 4
represents x ∈ (-∞, 4)
(D) x - 2 > 2 ⇒ x > 4
represents x ∈ (4, ∞)
(E) x < 4 represents x ∈ (-∞, 8)
We can see that inequalities (B) and (C) both represents x ∈ (-∞, 4)
Thus, the graph of both inequalities are same.
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Class opener: In a system of 3 forces pulling at
the same point, force #1 of 400 newtons pulls at
an angle of 70 degrees, force #2 of 510 newtons
pulls at an angle of 100 degrees, and force # 3 of
702 newtons pulls at an angle of 260 degrees.
What is the summation of the horizontal
components and the summation of the vertical
components? (Correct to 2 decimal places and
correct units)
The summation of the horizontal components is -629.76 N, and the summation of the vertical components is 363.68 N. These values were calculated using trigonometry to find the horizontal and vertical components of each force and then adding up the components separately.
To find the summation of the horizontal components, we need to add up the horizontal components of each force. We can use trigonometry to find the horizontal and vertical components of each force
Force #1 horizontal component = 400 cos(70) = 125.47 N
Force #2 horizontal component = 510 cos(100) = -158.95 N (negative because it acts in the opposite direction)
Force #3 horizontal component = 702 cos(260) = -596.28 N (negative because it acts in the opposite direction)
Therefore, the summation of the horizontal components is
125.47 N - 158.95 N - 596.28 N = -629.76 N
To find the summation of the vertical components, we need to add up the vertical components of each force
Force #1 vertical component = 400 sin(70) = 377.95 N
Force #2 vertical component = 510 sin(100) = 500.62 N
Force #3 vertical component = 702 sin(260) = -514.89 N (negative because it acts in the opposite direction)
Therefore, the summation of the vertical components is
377.95 N + 500.62 N - 514.89 N = 363.68 N
So the summation of the horizontal components is -629.76 N, and the summation of the vertical components is 363.68 N.
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There are 5 red candies and 1 blue candy shown in the bag. What is the least number of red and blue candies that can be added to the bag to create a ratio of 3 to 2 for the number of red candies to the number of blue candies? Key: = red R B = blue
Let's say we add 'x' red candies and 'y' blue candies to the bag to create the desired ratio of 3 to 2:
Then, the total number of red candies in the bag will be 5 + x, and the total number of blue candies will be 1 + y.
According to the problem, the ratio of red candies to blue candies should be 3 to 2:
(5 + x) / (1 + y) = 3/2
Cross-multiplying this equation, we get:
2(5 + x) = 3(1 + y)
Simplifying this equation, we get:
10 + 2x = 3 + 3y
2x - 3y = -7
We want to find the least number of red and blue candies that can be added to the bag to satisfy this equation.
One way to do this is to try different values of x and y that satisfy the equation until we find the smallest possible values that work.
For example, we can start by setting x = 1 and y = 2:
2(5 + 1) = 3(1 + 2)
12 = 9
This doesn't work, so let's try another set of values, x = 4 and y = 5:
2(5 + 4) = 3(1 + 5)
18 = 18
This set of values works, so we have found the least number of red and blue candies that can be added to the bag to create a ratio of 3 to 2 for the number of red candies to the number of blue candies:
We need to add 4 red candies and 5 blue candies to the bag to create a ratio of 3 to 2 for the number of red candies to the number of blue candies.
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The number of girls to boys in the education program is 5 to 1. If there are 90 students in the education program, how many are girls?
In ΔLMN, m = 2. 1 inches, n = 8. 2 inches and ∠L=85°. Find the length of l, to the nearest 10th of an inch
The length of l is approximately 6.1 inches to the nearest tenth of an inch.
To find the length of l, we can use the Law of Cosines which states that:
c^2 = a^2 + b^2 - 2ab*cos(C)
where c is the side opposite angle C, and a and b are the other two sides.
In this case, we want to find the length of l, which is opposite the given angle ∠L. So we can label l as side c, and label m and n as sides a and b, respectively. Then we can plug in the values we know and solve for l:
l^2 = m^2 + n^2 - 2mn*cos(L)
l^2 = (2.1)^2 + (8.2)^2 - 2(2.1)(8.2)*cos(85°)
l^2 = 4.41 + 67.24 - 34.212
l^2 = 37.438
l = sqrt(37.438)
l ≈ 6.118
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(Linear Systems: Applications). Find a polynomial p(2) of degree three such that
7(-2)=3,P(-1)=3,7(1)=-9,8(2)=-33.
Therefore, the polynomial p(x) that satisfies the given conditions is:
p(x) = ax^3 + bx^2 + cx + d
p(x) = x^3 - 2x^2 + 3x + 23
So, p(2) = 1(2)^3 - 2(2)^2 + 3(2) + 23 = 9.
To find a polynomial p(2) of degree three, we need four pieces of information. We can use the given values to set up a system of linear equations:
-7a + 2b - 4c + d = 3
-a - b + c - d = 3
7a + b + c + d = -9
8a + 4b + 2c + d = -33
We can solve this system using any method of linear algebra. One way is to use row reduction:
[ -7 2 -4 1 | 3 ]
[ -1 -1 1 -1 | 3 ]
[ 7 1 1 1 | -9 ]
[ 8 4 2 1 | -33 ]
R2 + R1 -> R1:
[ -8 1 -3 0 | 6 ]
[ -1 -1 1 -1 | 3 ]
[ 7 1 1 1 | -9 ]
[ 8 4 2 1 | -33 ]
R3 - 7R1 -> R1, R4 - 8R1 -> R1:
[ -8 1 -3 0 | 6 ]
[ 0 -7 8 -1 | 51 ]
[ 0 -4 4 1 |-51 ]
[ 0 4 26 1 |-81 ]
R4 + R2 -> R2:
[ -8 1 -3 0 | 6 ]
[ 0 -3 34 0 | 30 ]
[ 0 -4 4 1 |-51 ]
[ 0 4 26 1 |-81 ]
R3 + (4/3)R2 -> R2:
[ -8 1 -3 0 | 6 ]
[ 0 -3 34 0 | 30 ]
[ 0 0 50 4 |-11 ]
[ 0 4 26 1 |-81 ]
R4 - (4/3)R2 -> R2, R3 - (5/6)R2 -> R2:
[ -8 1 -3 0 | 6 ]
[ 0 -3 34 0 | 30 ]
[ 0 0 8 4 |-34 ]
[ 0 0 8 1 |-103 ]
R4 - R3 -> R3:
[ -8 1 -3 0 | 6 ]
[ 0 -3 34 0 | 30 ]
[ 0 0 8 4 |-34 ]
[ 0 0 0 -3 |-69 ]
Now we can back-substitute to find the coefficients of the polynomial:
d = -69/(-3) = 23
c = (-34 - 4d)/8 = 3
b = (30 - 34c + 3d)/(-3) = -2
a = (6 + 3b - 3c + d)/(-8) = 1
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What is the radius if you are given the diameter of 36 m?
Answer:
Radius = 18 m
Step-by-step explanation:
Given:
Diameter = 36 m
To find:
Radius
Explanation:
We know that,
Radius = Diameter/2 = 36/2 = 18 m
Final Answer:
18 m
Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol
(,,)
(
μ
,
p
,
σ
)
for the indicated parameter.
An entomologist writes an article in a scientific journal which claims that fewer than 16 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Use the parameter p, the true proportion of fireflies unable to produce light.
Group of answer choices
0
H
0
:
<0. 0016
p
<
0. 0016
1
H
1
:
≥0. 0016
p
≥
0. 0016
0
H
0
:
>0. 0016
p
>
0. 0016
1
H
1
:
≤0. 0016
p
≤
0. 0016
0
H
0
:
=0. 0016
p
=
0. 0016
1
H
1
:
<0. 0016
p
<
0. 0016
0
H
0
:
=0. 0016
p
=
0. 0016
1
H
1
:
>0. 16
The null hypothesis (H₀) and the alternative hypothesis (H₁) in symbolic form for this scenario are:
H₀: p = 0.0016 (the true proportion of fireflies unable to produce light is equal to 16 in ten thousand)
H₁: p < 0.0016 (the true proportion of fireflies unable to produce light is fewer than 16 in ten thousand)
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Michelle has four credit cards with the balances and interest rates listed below. She wants to pay off her credit
cards one at a time, based on the interest rate. In which order should Michelle pay off her credit cards?
>>>>>a. 3,2,1,4<<<<
b. 1,2,3,4
c. 2,4,3,1
d. 4,1,3,2
Answer:
a) 3, 2, 1, 4
Step-by-step explanation:
If you have multiple credit cards with different APRs, it is best to pay off the card with the highest APR first. This is because you will save the most money in interest by paying off the highest-rate debt first.
Therefore, as Michelle has four credit cards, each with different APRs, she should pay them off in order of the highest to lowest interest rate.
Since the highest APR is 23%, credit card #3 should be paid off first.
The next highest APR is 19%, so credit card #2 should be paid off second.
Credit card #1 should be paid off next as it has an APR of 17%.
Finally, credit card #4 should be paid off last, as it has the lowest APR of 15%.
So the order in which Michelle should pay off her credit cards is:
3, 2, 1, 4Assume the annual rate of change in the national debt of a country (in billions of dollars per year) can be modeled by the function
D'(t)=858.29+819.48t-184.32t^2+12.12t^3
where t is the number of years since 1995. By how much did the debt increase between 1996 and 2003 ?
The debt increased between 1996 and 2003. Then the national debt increased by approximately $4,903.73 billion between 1996 and 2003.
To find how much the debt increased between 1996 and 2003, we need to find the value of the function D'(t) for t=7 (since 2003 is 7 years after 1996).
D'(t)=858.29+819.48t-184.32t^2+12.12t^3
D'(7)=858.29+819.48(7)-184.32(7^2)+12.12(7^3)
D'(7)=858.29+5,736.36-8,132.32+3,458.68
D'(7)=1,921.01
Therefore, the annual rate of change in the national debt in 2003 was $1,921.01 billion per year.
To find how much the debt increased between 1996 and 2003, we need to integrate the function D'(t) from t=1 to t=7:
∫(D'(t))dt = ∫(858.29+819.48t-184.32t^2+12.12t^3)dt
= 858.29t + 409.74t^2 - 61.44t^3 + 3.03t^4 + C
where C is the constant of integration.
Evaluating this expression at t=7 and t=1 and taking the difference, we get:
(858.29(7) + 409.74(7)^2 - 61.44(7)^3 + 3.03(7)^4 + C) - (858.29(1) + 409.74(1)^2 - 61.44(1)^3 + 3.03(1)^4 + C)
= 6,111.09 - 1,207.36 = 4,903.73
Therefore, the national debt increased by approximately $4,903.73 billion between 1996 and 2003.
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HELP PLEASE 25 points!
This is for entrepreneurship
What can the bank do to comply with their general computer or Internet policy?
Bobby works in a private bank where none of the employees are allowed Internet access due to a strict confidentiality policy. Employees can only access the internal applications. However, for a particular product, Bobby is required to use the Internet and check details online.
The bank can (BLANK)
Bobby’s Internet usage for personal as well as official use
To comply with their general computer or Internet policy while allowing Bobby to access the internet for a specific task, the bank can implement several measures to monitor and restrict Bobby's Internet usage for personal and official use.
First, the bank can establish a separate, secured network for employees who need internet access for work purposes. This network should be isolated from the internal network to protect sensitive information.Second, the bank can implement strict access controls and authentication measures, such as providing a unique username and password for Bobby to access the internet. Third, the bank can install a firewall and web filtering system that blocks access to non-work-related websites and content. This will prevent personal use of the internet while still allowing access to the necessary websites for Bobby's work.
Fourth, the bank can regularly monitor and audit Bobby's internet usage, including the websites visited, the amount of time spent online, and any data transmitted or received. Finally, the bank should provide training and guidelines to Bobby regarding the acceptable use of the internet for work purposes, emphasizing the importance of confidentiality and adherence to the bank's security policies.
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Mathematics help nedd
To solve the equation, we need to first simplify both sides:
(4x - 6)/5 + 1 = (x + 1)/5 - 2/5
Multiplying both sides by 5 to eliminate the denominator:
4x - 6 + 5 = x + 1 - 2
Simplifying further:
4x - 1 = x - 1
Subtracting x from both sides:
3x - 1 = -1
Adding 1 to both sides:
3x = 0
Dividing both sides by 3:
x = 0
Therefore, the solution to the equation is x = 0.
Answer: x=28
Step-by-step explanation:
Given: <A=68
Find: x
Reasoning:
<B = 2x+x
<B= 3x
<C=x they say the sides across from <C is same as other side so the
angles are the same
Solution:
All angles of a triangle =180
<A + <B + <C =180 >substitute
68 + 3x + x =180 > combine like terms
68 + 4x = 180 > subtract 68 from both sides
4x=112 >divide both sides by 4
x=28
Do you best to explain how the following diagram demonstrates the Pythagorean theorem
Answer:
Step-by-step explanation:
The theorem states that the square on the hypotenuse (longest side) of a right triangle equal the sum of the squares on the other 2 sides.
Counting the small squares gives us these areas.
We see that
Sum of squares on hypotenuse = 25.
and sum of squares on the other 2 sides = 9 and 16 which equals 25.
13. d(-8, 1), e(-3, 6), f(7,4), g(2, -1) (distance formula)
The distances between the points are approximately 7.07, 10.20, and 7.07.
To find the distance between the points d(-8, 1) and e(-3, 6), we use the distance formula:
distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Plugging in the values, we get:
distance = √[(-3 - (-8))^2 + (6 - 1)^2]
distance = √[5^2 + 5^2]
distance = √50
distance ≈ 7.07
To find the distance between the points e(-3, 6) and f(7, 4), we again use the distance formula:
distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Plugging in the values, we get:
distance = √[(7 - (-3))^2 + (4 - 6)^2]
distance = √[10^2 + (-2)^2]
distance = √104
distance ≈ 10.20
To find the distance between the points f(7, 4) and g(2, -1), we again use the distance formula:
distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Plugging in the values, we get:
distance = √[(2 - 7)^2 + (-1 - 4)^2]
distance = √[(-5)^2 + (-5)^2]
distance = √50
distance ≈ 7.07
So, the distances between the points are approximately 7.07, 10.20, and 7.07.
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complete question:
Determine the distance between DE, EF and FG.
D(-8, 1), E(-3, 6), F(7, 4), G(2, -1)
The given segment is the diameter of a circle bar cd the coordinates of c are (-3,5) and the coordinates of d are (6,-2) . find the center of the circle
To find the center of the circle, we need to find the midpoint of the diameter segment CD.
Using the midpoint formula, we can find the coordinates of the midpoint M:
Midpoint formula:
M = ( (x1 + x2)/2 , (y1 + y2)/2 )
Plugging in the coordinates of C (-3,5) and D (6,-2):
M = ( (-3 + 6)/2 , (5 - 2)/2 )
M = (1.5, 1.5)
Therefore, the center of the circle is at point M with coordinates (1.5, 1.5).
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A bookstore is offering a 25% discount for a new book during a two-
week sale. After the sale, the book will sell for the regular price of
$32. 0. The store has a total of 200 copies of the book.
If all of the copies of this book are sold, what is the number of
discounted books that the store sells to make a total of $5440. 00?
Let x be the number of discounted books that the store sells during the sale. Then, the number of books sold at the regular price after the sale is 200 - x.
During the sale, the discounted price of the book is 0.75 * 32 = $24.
The revenue from selling x discounted books is:
R1 = 24x
The revenue from selling (200 - x) books at the regular price is:
R2 = 32(200 - x)
The total revenue from selling all the books is:
R = R1 + R2
We want to find the value of x such that the total revenue is $5440.00:
R = 5440
Substituting the expressions for R, R1, and R2, we get:
24x + 32(200 - x) = 5440
Simplifying and solving for x, we get:
24x + 6400 - 32x = 5440
-8x = -960
x = 120
Therefore, the store sells 120 discounted books during the sale to make a total of $5440.00.
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Fish enter a lake at a rate modeled by the function E given by E(t) = 20+15sin(pi*t/6). Fish leave the lake at a rate modeled by the function L given by L(t) = 4+20.1*t^2. Both E(t) and L(t) are measured in fish per hour and 't' is measured in hours since midnight (t=0).a.) How many fish enter the lake over the 5-hour period from midnight (t=0) to 5am (t=5)? Give your answer to the nearest whole number.b.) What is the average number of fish that leave the lake per hour over the 5 hour period from midnight (t=0) to 5am (t=5)?c.) At what time, t, for 0 ≤ t ≤ 8, is the greatest number of fish in the lake? Justify.d.) Is the rate of change in the number of fish in the lake increasing or decreasing at 5am (t=5)? Explain your reasoning.
Answer: a) To find the total number of fish that enter the lake over the 5-hour period, we need to integrate the function E(t) from t=0 to t=5:
int(20+15sin(pi*t/6), t=0 to 5) ≈ 62
a) To find the total number of fish that enter the lake over the 5-hour period, we need to integrate the function E(t) from t=0 to t=5:
int(20+15sin(pi*t/6), t=0 to 5) ≈ 62
Therefore, about 62 fish enter the lake over the 5-hour period from midnight to 5am.
b) The average number of fish that leave the lake per hour over the 5-hour period can be found by calculating the total number of fish that leave the lake over the 5-hour period and dividing by 5:
int(4+20.1*t^2, t=0 to 5) ≈ 1055
average = 1055/5 = 211
Therefore, the average number of fish that leave the lake per hour over the 5-hour period is 211.
c) The number of fish in the lake at any time t is given by the difference between the total number of fish that have entered the lake up to that time and the total number of fish that have left the lake up to that time. So, if N(t) represents the number of fish in the lake at time t, then:
N(t) = int(20+15sin(pi*t/6), t=0 to t) - int(4+20.1*t^2, t=0 to t)
To find the time t when the greatest number of fish are in the lake, we need to find the maximum of N(t) for 0 ≤ t ≤ 8. We can do this by taking the derivative of N(t) and setting it equal to zero:
dN(t)/dt = 15pi/6 * cos(pi*t/6) - 20.1t^2 + 4
0 = 15pi/6 * cos(pi*t/6) - 20.1t^2 + 4
Solving for t numerically using a calculator or computer, we find that the maximum occurs at t ≈ 2.34 hours. Therefore, the greatest number of fish in the lake occurs at 2.34 hours after midnight.
d) The rate of change in the number of fish in the lake is given by the derivative of N(t):
dN(t)/dt = 15pi/6 * cos(pi*t/6) - 20.1t^2 + 4
To determine whether the rate of change is increasing or decreasing at t=5, we need to find the second derivative:
d^2N(t)/dt^2 = -5.05t
When t=5, the second derivative is negative, which means that the rate of change in the number of fish in the lake is decreasing at 5am.
a. There will be 141 fish enter the lake over the 5-hour period from midnight
b. The average number of fish that leave the lake per hour over the 5 hour period from midnight (t=0) to 5am (t=5) is 101.
c. At 3.25 hour, t, for 0 ≤ t ≤ 8, is the greatest number of fish in the lake
d. The rate of change in the number of fish in the lake is decreasing at 5am.
a) To find the number of fish that enter the lake over the 5-hour period from midnight to 5am, we need to integrate the rate of fish entering the lake over that time period:
Number of fish = ∫[0,5] E(t) dt
= ∫[0,5] (20+15sin(πt/6)) dt
Number of fish ≈ 141
Therefore, approximately 141 fish enter the lake over the 5-hour period from midnight to 5am.
b. To find the average number of fish that leave the lake per hour over the 5 hour period, we need to calculate the total number of fish that leave the lake over that time period and divide by the duration of the period:
Number of fish that leave the lake = L(5) - L(0)
= (4+20.1*5^2) - (4+20.1*0^2)
= 505.5
Average number of fish leaving per hour = Number of fish that leave the lake / Duration of period
= 505.5 / 5
= 101.1
Therefore, the average number of fish that leave the lake per hour over the 5 hour period from midnight to 5am is approximately 101.
c. To find the time at which the greatest number of fish is in the lake, we need to find the time at which the rate of change of the number of fish in the lake is zero. This occurs when the rate of fish entering the lake is equal to the rate of fish leaving the lake:
E(t) = L(t)
20+15sin(πt/6) = 4+20.1t^2
We can solve this equation numerically to find that the greatest number of fish is in the lake at approximately t=3.25 hours (rounded to two decimal places).
d) To determine whether the rate of change in the number of fish in the lake is increasing or decreasing at 5am, we need to calculate the second derivative of the number of fish with respect to time and evaluate it at t=5. If the second derivative is positive, the rate of change is increasing. If it is negative, the rate of change is decreasing.
d²/dt² (number of fish) = d/dt E(t) - d/dt L(t)
= (15π/6)cos(πt/6) - 40.2t
d²/dt² (number of fish) ≈ -44.4
Since the second derivative is negative, the rate of change in the number of fish in the lake is decreasing at 5am.
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