Question 25: [1 + 2 + 2 + 2 + 2 + 1= 10 points]
Given student of classified as belonging to three colleges and gender males and Females in the following table.
Eng (E) Life (L) Sci (S) sum F 60 40 30 130
M 40 20 10 70 sum 100 60 40 200
a) [1 points] Find the probabilities whole events. () , (), (), () , and ()
b) [2 points] Find the probabilities of Intersection (AND) Events:
( ∩ ) = ( ∩ ) ( ∩ ) = ( ∩ ) ( ∩ ) = ( ∩ )
( ∩ ) = ( ∩ ) ( ∩ ) = ( ∩ ) ( ∩ ) = ( ∩ )
c) [2 points] Find the probabilities of Union (OR) disjoint Events:
( ∪ ) = ( ∪ ) ( ∪ ) = ( ∪ )
( ∪ ∪ ) = ( ∪ ∪ ) ( ∪ ) = ( ∪ )
d) [2 points] Find the probabilities of Union (OR) joint events: ( ∪ )
( ∪ )
e) [2 points] Find the probabilities of Conditional Probabilities
(/) (/)
(/) (/)
(/) (/)
(/) (/)
f) [1 points] Use The multiplication low for dependent events to find the Conditional probability. ( ∩ ) = () (/)
See below for the values of the probabilities
How to determine the probabilities?The probabilities of the whole events
For event A, the probability of the whole event is calculated using
P(A) = n(A)/Total
Using the table of values, we have:
P(F) = 130/200 = 0.65
P(M) = 70/200 = 0.35
P(L) = 60/200 = 0.30
P(S) = 40/200 = 0.20
The probabilities of the intersection events
For events A and B, the probability of the intersection events is calculated using
P(A n B) = n(A n B)/Total
Using the table of values, we have:
P(F n E) = P(E n F) = 60/200 = 0.30
P(F n L) = P(L n F) = 40/200 = 0.20
P(F n S) = P(S n F) = 30/200 = 0.15
P(M n E) = P(E n M) = 40/200 = 0.20
P(M n L) = P(L n M) = 20/200 = 0.10
P(M n S) = P(S n M) = 10/200 = 0.05
The probabilities of the Union (OR) disjoint events
For events A and B, the probability of the union (OR) disjoint events is calculated using
P(A u B) = P(A) + P(B)
Using the table of values, we have:
P(E u S) = P(E) + P(S) = 100/200 + 40/200 = 0.70
P(E u L) = P(E) + P(L) = 100/200 + 60/200 = 0.80
P(E u L u S) = P(E) + P(L) + P(S) = 100/200 + 60/200 + 40/100 = 1
P(F u M) = P(F) + P(M) = 130/200 + 70/200 = 1
The probabilities of the Union (OR) joint events
For events A and B, the probability of the union (OR) joint events is calculated using
P(A u B) = P(A) + P(B) - P(A n B)
Using the table of values, we have:
P(E u F) = P(E) + P(F) - P(E n F) = 100/200 + 130/200 - 60/100 = 0.85
P(L u M) = P(L) + P(M) - P(L n M) = 60/200 + 70/200 - 20/100 = 0.55
The probabilities of the conditional probabilities
For events A and B, the conditional probability is calculated using
P(A/B) = P(A n B)/P(B)
Using the table of values, we have:
P(F/S) = P(F n S)/P(S) = 0.15/0.20 = 0.75
P(F/E) = P(F n E)/P(E) = 0.30/0.50 = 0.60
P(M/S) = P(M n S)/P(S) = 0.05/0.20 = 0.25
P(M/L) = P(M n L)/P(L) = 0.10/0.30 = 0.33
P(S/F) = P(F n S)/P(F) = 0.15/0.65 = 0.23
P(E/F) = P(F n E)/P(F) = 0.30/0.65 = 0.46
P(S/M) = P(M n S)/P(M) = 0.05/0.35 = 0.14
P(L/M) = P(M n L)/P(M) = 0.10/0.35 = 0.29
The multiplication of dependent events
For events A and B, the conditional probability is calculated using
P(A n B) = P(A) * P(B/A)
Using the table of values, we have:
P(F n L) = P(F) * P(L/F)
This gives
P(F n L) = 0.65 * (0.20/0.30)
Evaluate
P(F n L) = 0.43
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A company's sales increased 20% this year, to $6740. What were their sales last year?
The sales last year of the company is 5617
How to determine the sales last year?The given parameters are:
Company sales = 6740
Percentage increase = 20%
The sales of the company last year (x) to date is calculated as:
Company sales = (1 + Proportion) * Last year sales
Substitute the known values in the above equation
6740 = (1 + 20%) * x
Evaluate the sum
6740 = 1.2 * x
Divide both sides by 1.2
x = 5617
Hence, the sales last year is 5617
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5. Write the following inequality in slope-intercept form. −8x + 4y ≥ −52 y ≤ 2x − 13 y ≥ 2x − 13 y ≤ 2x + 13 y ≥ 2x + 13
Answer:35443
Step-by-step explanation:
a dust mite is 400 pm long under a microscope, it looks 100,000 pm long. what magnification scale was used
Answer: 250:1
Step-by-step explanation:
Divide the measurement of appearance by the actual measurement.
[tex]100,000/400=250[/tex]
The magnification scale is 250:1. The scale is used by a 250X magnifying lense.
Instructions: Find the missing side of the triangle.
X= Answer
Answer: 15
Step-by-step explanation: Using the pythagorean theorem A^2 + B^2 = C^2 where C is the hypotenuse of the triangle and any A and B are any sides of the triangle that are not the hypotenuse; we can rewrite this as A^2 = C^2 - B^2 to find out the missing side. I picked A to be the unknown side but you could also choose B. We know the hypotenuse is 39, and side B is 36. 39^2 - 36^2, or 1521 - 1296, is 225.
Now we have A^2 = 225. To get rid of the square, we have to take the square root of both sides. The A^2 cancels to be A, and sqrt(225) is 15. Thus, side A, or the missing side is 15.
Hope this helped!
Find the area of the sector formed by the 60 degree central angle.
503π in2503π in2
103π in2103π in2
100π in2100π in2
None of the Above
The area of the sector of the circle is: A. 50/3π in.².
What is the Area of a Sector of a Circle?The area of a sector that is bounded by two radii of a circle is calculated using the formula, ∅/360 × πr², where we have the following parameters:
r = radius of the circle∅ = central angle formed by the sector.Thus, we are given the following regarding the sector of the circle:
Central angle (∅) = 60 degrees
Radius (r) = 10 inches.
Plug in the values into ∅/360 × πr²:
Area of sector = 60/360 × π(10²)
Area of sector = 1/6 × π(100)
Area of sector = 100/6 × π
Area of sector = 50/3 × π
Area of sector = 50/3π in.²
Thus, the area of the sector of the circle is: A. 50/3π in.².
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(4x³-12x +11) + (2x - 2)
6n−8(n+4)= what? Please help me
Answer:
-2n -32
Step-by-step explanation:
Acc. to BODMAS
we multiply first
⇒-6n - -8 x [(+4) (n)]
⇒ 6n - 8n x -32
[tex]\huge\text{Hey there!}[/tex]
[tex]\huge\textbf{Equation:}[/tex]
[tex]\mathsf{6n - 8(n + 4)}[/tex]
[tex]\huge\textbf{Solve: }[/tex]
[tex]\mathsf{6n - 8(n + 4)}[/tex]
[tex]\huge\textbf{Distribute \boxed{\bf -8} within the parenthesis:}[/tex]
[tex]\mathsf{= 6n - 8(n) - 8(4)}[/tex]
[tex]\mathsf{= 6n - 8n - 32}[/tex]
[tex]\huge\textbf{Combine the like terms: }[/tex]
[tex]\mathsf{= (6n - 8n) - (32)}[/tex]
[tex]\mathsf{= 6n - 8n - 32}[/tex]
[tex]\mathsf{= -2n - 32}[/tex]
[tex]\huge\textbf{Therefore, your answer should be:}[/tex]
[tex]\huge\boxed{\frak{{ -2n - 32}}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}}[/tex]
consider the function y=-2-cos(x-pi). What effect does pi have on the basic graph?
Using translation concepts, it is found that pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.
What is a translation?A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.
In this problem, the change is given as follows:
x -> x - pi
It means that the change is in the domain, in which pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.
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When pricing out a menu, we know it is important to use the Edible a portion Price. What would the effect on your profit margin be, if we only used the as purchased price to determine our cost and selling price?
The effect of Edible a portion Price on your profit margin , if we only use the as purchased price to determine our cost and selling price is that it will maximize the profit because it will account for every part of the production.
What is edible portion cost?
The portion cost can be calculated by multiplying the cost of a usable kg with the portion size.
This can be represented as : portion cost = (portion size x cost of usable kg)
It should be noted that Edible portion (EP) serves as the portion of food which will be given top the customer after the preparation.
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Answer:
Step-by-step explanation:
HOW MANY DISTINCT ARRANGEMENTS CAN BE MADE WITH THE LETTERS IN THE
WORD CONNECTION?
There are 151200 distinct ways to arrange the letters of the word CONNECTION
How to determine the number of arrangements?The word is given as:
CONNECTION
In the above word, we have the following parameters:
Total number of characters, n = 10
The repeated letters are:
C's = 2
O's = 2
N's = 3
The number of arrangements of the letters is then calculated as:
Arrangements = n!/(C! * O! * N!)
Substitute the known values in the above equation
Arrangements = 10!/(2! * 2! * 3!)
Evaluate the expression
Arrangements = 151200
Hence, there are 151200 distinct ways to arrange the letters of the word CONNECTION
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HOW MANY DIFFERENT COMMITTEES CAN BE FORMED FROM 4 TEACHERS AND 20
STUDENTS IF THE COMMITTEE IS TO CONSIST OF TWO TEACHERS AND THREE
STUDENTS?
Answer:
6840
Step-by-step explanation:
multiply combinations
n!/(r!(n − r)!)
n choose r
n (objects) = |n
r (sample) = r
(4 teachers, CHOOSE 2)*(20 students, CHOOSE 3)
C(4,2) times C(20,3) =
C(n,r)=?
C(n,r) = C(4,2)
4! / (2!(4-2)!)
4! /2! x 2!
= 6
C(n,r)=?
C(n,r) = C(20,3)
20! / (3!(20-3)!)
20! / 3! x 17!
= 1140
6 * 1140 = 6840
alegbracom Edwin McCravy
PLEASE HELP URGENT
Which equation could be solved using this application of the quadratic formula?
Answer:
Step-by-step explanation:
D because the quadratic formula is basically [tex]\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]±
so if we place the equation D into the formula
a = 1
b = +3
c = +24
we can clearly say that D eq is correctly used and can be.
A _____ is a solid consisting of a suite a point not in the same plane as the square and all points between them
Answer:
Square Pyramid
Step-by-step explanation:
Thats what a Square Pyramid is
Help me asappp w this question
[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]
The straight angles are angles that form a straight line, and their measure = 180°
In the given figure, the Straight angles is :
GEB[tex] \qquad \large \sf {Conclusion} : [/tex]
Correct choice is DFind a potential function for the vector field
(a) We want to find a scalar function [tex]f(x,y,z)[/tex] such that [tex]\mathbf F = \nabla f[/tex]. This means
[tex]\dfrac{\partial f}{\partial x} = 2xy + 24[/tex]
[tex]\dfrac{\partial f}{\partial y} = x^2 + 16[/tex]
Looking at the first equation, integrating both sides with respect to [tex]x[/tex] gives
[tex]f(x,y) = x^2y + 24x + g(y)[/tex]
Differentiating both sides of this with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y} = x^2 + 16 = x^2 + \dfrac{dg}{dy} \implies \dfrac{dg}{dy} = 16 \implies g(y) = 16y + C[/tex]
Then the potential function is
[tex]f(x,y) = \boxed{x^2y + 24x + 16y + C}[/tex]
(b) By the FTCoLI, we have
[tex]\displaystyle \int_{(1,1)}^{(-1,2)} \mathbf F \cdot d\mathbf r = f(-1,2) - f(1,1) = 10-41 = \boxed{-31}[/tex]
[tex]\displaystyle \int_{(-1,2)}^{(0,4)} \mathbf F \cdot d\mathbf r = f(0,4) - f(-1,2) = 64 - 41 = \boxed{23}[/tex]
[tex]\displaystyle \int_{(0,4)}^{(2,3)} \mathbf F \cdot d\mathbf r = f(2,3) - f(0,4) = 108 - 64 = \boxed{44}[/tex]
What is the domain of the relation graphed below?
The domain of the relation shown in the graph is -4 <= x <= 4
How to determine the domain of the relation shown in the graph?The relation on the graph is an ellipse function.
The domain is the set of x values the function can take
From the graph, we have the following x values
This means that the domain of x in the graph is -4 <= x <= 4
Hence, the domain of the relation shown in the graph is -4 <= x <= 4
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the figure at the right shows the dimensions of the garden in Marissa's back yard. What is the area of the garden?
The area of the garden is 74 square feet
How to determine the area of the garden?The complete question is added as an attachment
From the attached figure, we have the following shapes and dimensions:
Rectangle: 10 by 5 feetTrapezoid: Bases = 10 and 6; Height = 3The rectangular area is
A1 = 10 * 5
Evaluate
A1 = 50
The area of the trapezoid is
A2 = 0.5 * Sum of parallel bases * height
This gives
A2 = 0.5 * (10 + 6) * 3
Evaluate
A2 = 24
The total area is
Total = A1 + A2
This gives
Total = 50 + 24
Evaluate
Total = 74
Hence, the area of the garden is 74 square feet
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Does this set of ordered pairs represent a function? {(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)} A. The relation is a function. Each input value is paired with more than one output value. B. The relation is a function. Each input value is paired with one output value. C. The relation is not a function. Each input value is paired with only one output value. D. The relation is not a function. Each input value is paired with more than one output value.
The correct option regarding whether the relation is a function is:
B. The relation is a function. Each input value is paired with one output value.
When does a relation represent a function?A relation represent a function if each value of the input is paired with one value of the output.
In this problem, when the input - output mappings are given by:
{(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)}.
Which means that yes, each input value is paired with one output value, hence the relation is a function and option B is correct.
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Evaluate the limits
[tex]x > \ln(x)[/tex] for all [tex]x[/tex], so
[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}[/tex]
Similarly, [tex]\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}[/tex]
We can of course see the limits are identical by replacing [tex]x\mapsto e^x[/tex], so that
[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)[/tex]
You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,
[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)[/tex]
Using the rule, the limit here is
[tex]\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0[/tex]
so the overall limit is
[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right) = \ln(0) = \lim_{x\to0^+} \ln(x) = -\infty[/tex]
s−3(s+6)= please help
Answer:
-2s-18 is the simplified answer
Step-by-step explanation:
s-3(s+6)
distributive property
s-3s-18
-2s-18
TIME REMAINING
44:36
Which point is an x-intercept of the quadratic function f(x) = (x + 6)(x – 3)?
(0,6)
(0,–6)
(6,0)
(–6,0)
Answer:
d. (-6, 0)
Step-by-step explanation:
no explanation bc u need this quick
PLEASE HELP ME WITH THIS QUESTION I NEED HELP
Answer:
True
Step-by-step explanation:
This is true in an equilateral triangle.
please help!!!
maths functions
Answer:
Point A: (2, 10)
Point B: (-3, 0)
Point C: (-5, -4)
Point D: (-5, -32)
Step-by-step explanation:
Part (a)Points A and B are the points of intersection between the two graphs.
Therefore, to find the x-values of the points of intersection, substitute one equation into the other and solve for x:
[tex]\implies 2x+6=-2x^2+18[/tex]
[tex]\implies 2x^2+2x-12=0[/tex]
[tex]\implies 2(x^2+x-6)=0[/tex]
[tex]\implies x^2+x-6=0[/tex]
[tex]\implies x^2+3x-2x-6=0[/tex]
[tex]\implies x(x+3)-2(x+3)=0[/tex]
[tex]\implies (x-2)(x+3)=0[/tex]
[tex]\implies x=2, -3[/tex]
From inspection of the graph:
The x-value of point A is positive ⇒ x = 2The x-value of point B is negative ⇒ x = -3To find the y-values, substitute the found x-values into either of the equations:
[tex]\begin{aligned} \textsf{Point A}: \quad 2x+6 & =y\\2(2)+6 & =10\\ \implies & (2, 10)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point B}: \quad -2x^2+18 & =y\\-2(-3)^2+18 & =0\\ \implies & (-3,0)\end{aligned}[/tex]
Therefore, point A is (2, 10) and point B is (-3, 0).
Part (b)If the distance between points C and D is 28 units, the y-value of point D will be 28 less than the y-value of point C. The x-values of the two points are the same.
Therefore:
[tex]\textsf{Equation 1}: \quad y=2x+6[/tex]
[tex]\textsf{Equation 2}: \quad y-28=-2x^2+18[/tex]
As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:
[tex]\implies 2x+6-28=-2x^2+18[/tex]
[tex]\implies 2x^2+2x-40=0[/tex]
[tex]\implies 2(x^2+x-20)=0[/tex]
[tex]\implies x^2+x-20=0[/tex]
[tex]\implies x^2+5x-4x-20=0[/tex]
[tex]\implies x(x+5)-4(x+5)=0[/tex]
[tex]\implies (x-4)(x+5)=0[/tex]
[tex]\implies x=4,-5[/tex]
From inspection of the given graph, the x-value of points C and D is negative, therefore x = -5.
To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:
[tex]\begin{aligned} \textsf{Point C}: \quad 2x+6 & =y\\2(-5)+6 & =-4\\ \implies & (-5,-4)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point D}: \quad -2x^2+18 & = y \\ -2(-5)^2+18 & =-32\\ \implies & (-5, -32)\end{aligned}[/tex]
Therefore, point C is (-5, -4) and point D is (-5, -32).
Answer:
a) A = (2, 10) and B = (-3, 0)
b) C = (-5, -4) and D = (-5, -32)
Explanation:
a) To determine the coordinates of A and B, find the intersection points of the line "y = 2x + 6" and curve "y = -2x² + 18".
Solve the equation's simultaneously:
y = y
⇒ 2x + 6 = -2x² + 18
⇒ 2x² + 2x + 6 - 18 = 0
⇒ 2x² + 2x - 12= 0
⇒ 2x² + 6x - 4x - 12 = 0
⇒ 2x(x + 3) - 4(x + 3) = 0
⇒ (2x - 4)(x + 3) = 0
⇒ 2x - 4 = 0, x + 3 = 0
⇒ x = 2, x = -3
Then find value of y at this x points,
at x = 2, y = 2(2) + 6 = 10
at x = -3, y = 2(-3) + 6 = 0
Intersection points: A(2, 10) and B(-3, 0)
b) Given that CD = 28 units. Also stated parallel to y axis so x coordinates for both will be same but differ in y coordinate.
[tex]2x + 6 = -2x^2 + 18 + 28[/tex]
[tex]-2x^2 + 18 + 28-2x - 6 = 0[/tex]
[tex]-2x^2-2x+40=0[/tex]
[tex]-2x^2-10x+8x+40=0[/tex]
[tex]-2(x+5)+8(x+5)=0[/tex]
[tex](-2x+8)(x+5)=0[/tex]
[tex]x = -5, 4[/tex]
[tex]\leftrightarrow \sf C(-5, y_2), \ D(-5, y_2)[/tex]
Find y value for Point C : 2x + 6 = 2(-5) + 6 = -4
Find y value for Point D : -2x² + 18 = -2(-5)² + 18 = -32
[tex]\sf \rightarrow Point \ C = (-5, -4)\\ \\\rightarrow Point \ D = (-5, -32)[/tex]
Consider the ordinary differential equation (answer questions in picture)
a. Given the 2nd order ODE
[tex]y''(x) = 4y(x) + 4[/tex]
if we substitute [tex]z(x)=y'(x)+2y(x)[/tex] and its derivative, [tex]z'(x)=y''(x)+2y'(x)[/tex], we can eliminate [tex]y(x)[/tex] and [tex]y''(x)[/tex] to end up with the ODE
[tex]z'(x) - 2y'(x) = 4\left(\dfrac{z(x)-y'(x)}2\right) + 4[/tex]
[tex]z'(x) - 2y'(x) = 2z(x) - 2y'(x) + 4[/tex]
[tex]\boxed{z'(x) = 2z(x) + 4}[/tex]
and since [tex]y(0)=y'(0)=1[/tex], it follows that [tex]z(0)=y'(0)+2y(0)=3[/tex].
b. I'll solve with an integrating factor.
[tex]z'(x) = 2z(x) + 4[/tex]
[tex]z'(x) - 2z(x) = 4[/tex]
[tex]e^{-2x} z'(x) - 2 e^{-2x} z(x) = 4e^{-2x}[/tex]
[tex]\left(e^{-2x} z(x)\right)' = 4e^{-2x}[/tex]
[tex]e^{-2x} z(x) = -2e^{-2x} + C[/tex]
[tex]z(x) = -2 + Ce^{2x}[/tex]
Since [tex]z(0)=3[/tex], we find
[tex]3 = -2 + Ce^0 \implies C=5[/tex]
so the particular solution for [tex]z(x)[/tex] is
[tex]\boxed{z(x) = 5e^{-2x} - 2}[/tex]
c. Knowing [tex]z(x)[/tex], we recover a 1st order ODE for [tex]y(x)[/tex],
[tex]z(x) = y'(x) + 2y(x) \implies y'(x) + 2y(x) = 5e^{-2x} - 2[/tex]
Using an integrating factor again, we have
[tex]e^{2x} y'(x) + 2e^{2x} y(x) = 5 - 2e^{2x}[/tex]
[tex]\left(e^{2x} y(x)\right)' = 5 - 2e^{2x}[/tex]
[tex]e^{2x} y(x) = 5x - e^{2x} + C[/tex]
[tex]y(x) = 5xe^{-2x} - 1 + Ce^{-2x}[/tex]
Since [tex]y(0)=1[/tex], we find
[tex]1 = 0 - 1 + Ce^0 \implies C=2[/tex]
so that
[tex]\boxed{y(x) = (5x+2)e^{-2x} - 1}[/tex]
6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.
6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].
6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].
The given ordinary differential equation is y''(x) = 4y(x) + 4, y(0) = y'(0) = 1 ... (d).
We are also given a substitution function, z(x) = y'(x) + 2y(x) ... (z).
Putting x = 0, we get:
z(0) = y'(0) + 2y(0),
or, z(0) = 1 + 2*1 = 3.
Rearranging (z), we can write it as:
z(x) = y'(x) + 2y(x),
or, y'(x) = z(x) - 2y(x) ... (i).
Differentiating (z) with respect to x, we get:
z'(x) = y''(x) + 2y'(x),
or, y''(x) = z'(x) - 2y'(x) ... (ii).
Substituting the value of y''(x) from (ii) in (d) we get:
y''(x) = 4y(x) + 4,
or, z'(x) - 2y'(x) = 4y(x) + 4.
Substituting the value of y'(x) from (i) we get:
z'(x) - 2y'(x) = 4y(x) + 4,
or, z'(x) - 2(z(x) - 2y(x)) = 4y(x) + 4,
or, z'(x) - 2z(x) + 4y(x) = 4y(x) + 4,
or, z'(x) = 2z(x) + 4y(x) - 4y(x) + 4,
or, z'(x) = 2z(x) + 4.
The initial value of z(0) was calculated to be 3.
6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.
Transforming z(x) = dz/dx, and z = z(x), we get:
dz/dx = 2z + 4,
or, dz/(2z + 4) = dx.
Integrating both sides, we get:
∫dz/(2z + 4) = ∫dx,
or, {ln (z + 2)}/2 = x + C,
or, [tex]\sqrt{z+2} = e^{x + C}[/tex],
or, [tex]z =Ce^{2x}-2[/tex] ... (iii).
Substituting z = 3, and x = 0, we get:
[tex]3 = Ce^{2*0} - 2\\\Rightarrow C - 2 = 3\\\Rightarrow C = 5.[/tex]
Substituting C = 5, in (iii), we get:
[tex]z = 5e^{2x} - 2[/tex].
6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].
Substituting the value of z(x) in (z), we get:
z(x) = y'(x) + 2y(x),
or, 5e²ˣ - 2 = y'(x) + 2y(x),
which gives us:
[tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex] for the initial condition y(x) = 0.
6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].
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Jake goes to the grocery store and buys 3 apples, 2 cans of soup, and 1 box of cereal. The apples cost $0.89 each; the soup costs $2.98 per can; and the box of cereal costs $4.99. Write an equation that represents the total cost c of Jake’s purchases.
The equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.
How to Write the Equation of Total Cost?To write the equation that represents the total cost of a given scenario like the one given, you can use variables to represent each component that makes up the total cost in the given situation.
Thus, let:
a = apple = 3a
s = can of soup = 2s
b = box of cereal = b
c = total cost
We know that unit price of each items are:
Apple = $0.89
Can of soup = $2.98
Box of cereal = $4.99
Equation would be:
c = 3a + 2s + b
Plug in the values
c = 3(0.89) + 2(2.98) + 4.99
Therefore, the equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.
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HOW MANY DIFFERENT ARRANGEMENTS CAN BE MADE WITH THE NUMBERS
28535852
Using the arrangements formula, it is found that 1680 arrangements can be made with these numbers.
What is the arrangements formula?The number of possible arrangements of n elements is given by the factorial of n, that is:
[tex]A_n = n![/tex]
When there are repeated elements, repeating [tex]n_1, n_2, \cdots, n_n[/tex] times, the number of arrangements is given by:
[tex]A_n^{n_1, n_2, \cdots, n_n} = \frac{n!}{n_1!n_2! \cdots n_n!}[/tex]
For the number 28535852, we have that:
There are 8 numbers.5 repeats 3 times.2 repeats two times.8 repeats two times.Hence the number of arrangements is:
[tex]A_8^{3,2,2} = \frac{8!}{3!2!2!} = 1680[/tex]
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one of the acute angles of a right triangle is 36° find the other acute angles
Step-by-step explanation:
An acute angle is an angle less or equal to 90 degrees.
let the other angle be x.
x + 36 = 90
grouping like terms
X = 90 - 36
X = 54 degrees
Answer:
54
Step-by-step explanation:
= ∠A + ∠B + ∠C = 180o … [∵sum of the angles of a triangle is 180]
or 180 = 36 + 90 + ∠C
C=54
The width of a rectangle measures
(7h+3) centimeters, and its length measures
(8h−4) centimeters. Which expression represents the perimeter, in centimeters, of the rectangle?
The expression that represents the perimeter of the rectangle is 30h - 2
How to determine the perimeter expression?The dimensions of the rectangle are given as:
Length = 7h + 3
Width = 8h - 4
The perimeter of the rectangle is given as:
P = 2 * (Length + Width)
Substitute the known values in the above equation
P =2 * (7h + 3 + 8h - 4)
Evaluate the like terms
P = 2 * (15h - 1)
Evaluate the product
P = 30h - 2
Hence, the expression that represents the perimeter of the rectangle is 30h - 2
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15. Find the young modulus of a brass rod of diameter 25mm and of length 250mm which is
subjected to a tensile load of 50KN when the extension of the rod is equal to 0.3mm.
The Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².
How to find the young modulus of a brass rod?
Given: Physical Properties of the Brass rod exist
Length, L = 250 mm
Radius, R = Diameter / 2 = 25/2 mm
Load of 50 kN hanging on the brass rod which results in elongation of the rod to 0.3 mm.
To estimate the Young Modulus of the brass rod we will require Stress on the rod, Ratio of elongation to the total length.
To find the stress on the rod,
Stress(Pressure) = Force / Area
⇒ Stress = 50,000 / (Area of cross-section)
The rod exists shaped like a cylinder then the area of its cross-section will be πR² ( R = radius)
⇒ Stress = 50000 / (π (25/2 mm)² )
The force exists given in newton, then we must convert the unit of the area into m² to obtain the solution in pascal.
1 mm = 0.001 m
⇒ Stress = 50,000 / 22/7 ( 25/2 × 10⁻³ m)²
take, π = 22/7
⇒ Stress = 50000 / ( 22/7 × 625 / 4 × 10⁻⁶ )
⇒ Stress = 50000 / ( 11/14 × 625 × 10⁻6 )
⇒ Stress = 50000 × 14 × 10⁶ / 625×11
⇒ Stress = 80 × 14 × 10⁶ / 11
⇒ Stress = 1120/11 × 10⁶ Pa
Now, that we maintain Stressed, to estimate Strain (ratio of elongation to original length),
⇒ Strain = (Elongation) / (Original length)
⇒ Strain = 0.3 / 250
⇒ Strain = 3 / 2500
Therefore, Young Modulus, Y = Stress / Strain
⇒ Y = 1120/11 × 10⁶ / 3/2500
⇒ Y = 1120/11 × 2500/3 × 10⁶
⇒ Y = 101.8 × 833.3 × 10⁶
⇒ Y = 84829.94 × 10⁶
⇒ Y ≈ 8.5 × 10¹⁰ Pa
Therefore, the Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².
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