Answer: The required answer is
[tex]f(x)=e^x+\cos x+\sin x.[/tex]
Step-by-step explanation: We are given to find the inverse Laplace transform of the following function as a function of x :
[tex]F(s)=\dfrac{2s^2}{(s-1)(s^2+1)}.[/tex]
We will be using the following formulas of inverse Laplace transform :
[tex](i)~L^{-1}\{\dfrac{1}{s-a}\}=e^{ax},\\\\\\(ii)~L^{-1}\{\dfrac{s}{s^2+a^2}\}=\cos ax,\\\\\\(iii)~L^{-1}\{\dfrac{1}{s^2+a^2}\}=\dfrac{1}{a}\sin ax.[/tex]
By partial fractions, we have
[tex]\dfrac{s^2}{(s-1)(s^2+1)}=\dfrac{A}{s-1}+\dfrac{Bs+C}{s^2+1},[/tex]
where A, B and C are constants.
Multiplying both sides of the above equation by the denominator of the left hand side, we get
[tex]2s^2=A(s^2+1)+(Bs+C)(s-1).[/tex]
If s = 1, we get
[tex]2\times 1=A(1+1)\\\\\Rightarrow A=1.[/tex]
Also,
[tex]2s^2=A(s^2+1)+(Bs^2-Bs+Cs-C)\\\\\Rightarrow 2s^2=(A+B)s^2+(-B+C)s+(A-C).[/tex]
Comparing the coefficients of x² and 1, we get
[tex]A+B=2\\\\\Rightarrow B=2-1=1,\\\\\\A-C=0\\\\\Rightarrow C=A=1.[/tex]
So, we can write
[tex]\dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s+1}{s^2+1}\\\\\\\Rightarrow \dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}.[/tex]
Taking inverse Laplace transform on both sides of the above, we get
[tex]L^{-1}\{\dfrac{2s^2}{(s-1)(s^2+1)}\}=L^{-1}\{\dfrac{1}{s-1}\}+L^{-1}\{\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}\}\\\\\\\Rightarrow f(x)=e^{1\times x}+\cos (1\times x)+\dfrac{1}{1}\sin(1\times x)\\\\\\\Rightarrow f(x)=e^x+\cos x+\sin x.[/tex]
Thus, the required answer is
[tex]f(x)=e^x+\cos x+\sin x.[/tex]
If the probability of losing is 7/10, what is the probability of winning?
1/10
7/10
3/10
3/3
Answer:
3/10
Step-by-step explanation:
If the probability of losing is 7/10, then the probability of winning is 3/10. To find out the probability of winning we just have to subtract the probability of losing from the total which is 10:
10/10 = total probability
7/10 = probability of losing
Probability of winning is 3/10.
Solve the ODE numerically using Euler's method, 10 steps. Graph the computed values and the solution curve on the same axes. y'= -0.25y, y(0) = 6, h=1
Answer:
0=6
Step-by-step explanation:
-0.25yxyx0=6
To determine the order that cars line up for the Daytona 500, drivers must complete qualifying laps around the Daytona International Speedway. The fastest ever qualifying lap was in 1987 by Bill Elliott, with a time of 42.782 seconds.That was 210.364 miles per hour (or 338.548 kilometers per hour).
A. What was the frequency of the car (measured in laps per second?)
B. How many laps were completed in 60 seconds (1 minute)?
Answer:
The frequency of the car is 0.0233743 laps/sec and 1.4024 laps were completed in 60 seconds
Step-by-step explanation:
We are given that he fastest ever qualifying lap was in 1987 by Bill Elliott, with a time of 42.782 seconds.
[tex]Frequency = \frac{1}{Time}[/tex]
[tex]Frequency = \frac{1}{42.782}[/tex]
[tex]Frequency = 0.0233743[/tex]
No. of laps completed in 42.782 seconds. = 1
No. of laps completed in 60 seconds. = [tex]\frac{1}{42.782} \times 60[/tex]
= [tex]1.4024[/tex]
Hence the frequency of the car is 0.0233743 laps/sec and 1.4024 laps were completed in 60 seconds
A ball with mass m kg is thrown upward with initial velocity 28 m/s from the roof of a building 17 m high. Neglect air resistance Use g = 9.8 m/s. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. Fend Click If you would like to Show Work for this question: Open Show Work LINK TO TEXT
Answer:
The ball will take 6.3 seconds to reach the maximum height and hit the ground.
Step-by-step explanation:
a). When a ball was thrown upwards with an initial velocity u then maximum height achieved h will be represented by the equation
v² = u² - 2gh
where v = final velocity at the maximum height h
and g = gravitational force
Now we plug in the values in the equation
At maximum height final velocity v = 0
0 = (28)² - 2×(9.8)h
19.6h = (28)²
h = [tex]\frac{(28)^{2}}{19.6}[/tex]
= [tex]\frac{784}{19.6}[/tex]
= 40 meter
B). If the ball misses the building and hits the ground then we have to find the time after which the ball hits the ground that will be
= Time to reach the maximum height + time to hit the ground from the maximum height
Time taken by the ball to reach the maximum height.
Equation to find the time will be v = u - gt
Now we plug in the values in the equation
0 = 28 - 9.8t
t = [tex]\frac{28}{9.8}[/tex]
= 2.86 seconds
Now time taken by the ball to hit the ground from its maximum height.
H = ut + [tex]\frac{1}{2}\times g\times (t)^{2}[/tex]
(17 + 40) = 0 + [tex]\frac{1}{2}\times g\times (t)^{2}[/tex]
57 = 4.9(t)²
t² = [tex]\frac{57}{4.9}[/tex]
t² = 11.63
t = √(11.63)
= 3.41 seconds
Now total time taken by the ball = 2.86 + 3.41
= 6.27 seconds
≈ 6.3 seconds
Therefore, the ball will take 6.3 seconds to reach the maximum height and hit the ground.
Sketch the Cartesian product on the x-y plane R^2: Zx Z.
Answer:
[tex]\mathbb{Z}\times \mathbb{Z}=\{(a,b)\lvert a, b\in \mathbb{Z}\}[/tex]
Step-by-step explanation:
In general, the Cartesian product of two sets [tex]A,B[/tex] is a new set defined by
[tex]A\times B=\{(a,b)\lvert a\in A,b\in B\}[/tex]
The pair [tex](a,b)[/tex] is ordered pair because the order is important, that is to say, in general [tex](a,b)\neq (b,a)[/tex].
One of the most important Cartesian products in mathematics is [tex]\mathbb{R}\times \mathbb{R}=\{(x,y) \lvert x,y \in \mathbb{R}\}[/tex] which is precisely the Cartesian Plane xy. The set [tex]\mathbb{Z}\times \mathbb{Z}[/tex] is a subset of [tex]\mathbb{R}\times \mathbb{R}[/tex] which is the set of all the points in the Cartesian plane whose coordinates are integers numbers. So, sketching the set [tex]\mathbb{Z}\times \mathbb{Z}[/tex] we have a picture as the shown below.
0 / 2 pts 5000 kg = _______short tons Round UP to nearest 100th
Answer:
5000kg = 5.5 short tons
Step-by-step explanation:
This can be solved as a rule of three problem.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.
Unit conversion problems, like this one, is an example of a direct relationship between measures.
Each kg has 0.0011 short tons. How many short tons are there in 5000 kg. So?
1kg = 0.0011 short tons
5000kg - x short tons
[tex]x = 5000*0.0011[/tex]
[tex]x = 5.5[/tex] short tons.
5000kg = 5.5 short tons
To convert 5000 kg to short tons, multiply by the conversion factor and round up.
Explanation:The conversion factor between kilograms and short tons is 0.00110231. To convert 5000 kg to short tons, we multiply by the conversion factor:
5000 kg * 0.00110231 = 5.51155 short tons
Rounding up to the nearest 100th gives us 5.52 short tons.
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If the earth travels around the sun one time each year and Jupiter travels around the sun one time every 12 years and they met at a point in 2002 when would they meet at the point again?
Answer:
They will meet again in 2014
Step-by-step explanation:
Once every year the Earth will go back through the point where it met with Jupiter. Let's write those years:
Earth = {2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016...}
Once every 12 years Jupiter will go back through the point where it met with the Earth. Let's write it down:
Jupiter= {2002, 2002+12=2014, 2014+12=2026, 2026+12=2038...}
The Earth and Jupiter will meet again for the first time (after 2002) the year they both go trought that point.
Earth = {2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016...}
Jupiter= {2002, 2014, 2026, 2038...}
As we can see, 2014 is the first year the'll meet again.
determine whether the set is closed for the given operation.
explain your reasoning. -- -the set of integers for division
-the set of positive integers for subtraction.
Closure Property for the set of integers for Subtraction
Take two integers,
A=6
B=9
A-B
=6-9
= -3
Which is also an Integer.Hence Closure Property is satisfied by Integers with respect to Subtraction.
⇒Closure Property for the set of integers for Division
Take two integers,
A=6
B=9
[tex]\rightarrow \frac{A}{B}\\\\=\frac{6}{9}\\\\=\frac{2}{3}-----\text{Not an Integer}[/tex]
Which is not an Integer.Hence Closure Property is not satisfied by Integers with respect to Division.
Set P = {x, 4, 6, 10, -2, x, 6, 1, x} In set P, if the value of x is 4 greater than the largest numerical value present, what is the range of the set?
Answer:
16.
Step-by-step explanation:
The largest numerical value present in the set P is 10, then x is 4+10 = 14. Then the set P is
P = {14, 4, 6, 10, -2, 14, 6, 1, 14}.
Now, the range is the difference between the greatest and the smallest element in the set. The greatest number is 14 and the smallest is -2, then the range is
14-(-2) = 14+2 = 16.
Then the range of the set P is 16.
Baytown Village Stone Creations is making a custom stone bench. the recommended height for the bench is 18 in. the depth of the stone bench is 3 3/8 in. Each of the two supporting legs is made up of three stacked stones. Two of the stones measure 3 1/2in. and 5 1/4 in. how much must the third stone measure?
Answer:
x = tex]5\frac{\textup{7}}{\textup{8}}\ in[/tex]
Step-by-step explanation:
Given:
Total height of the bench = 18 in
Depth of the stone bench = [tex]3\frac{\textup{3}}{\textup{8}}\ in[/tex] = [tex]\frac{\textup{27}}{\textup{8}}\ in[/tex] = 3.375 in
Measure of stones = [tex]3\frac{\textup{1}}{\textup{2}}\ in[/tex] = [tex]\frac{\textup{7}}{\textup{2}}\ in[/tex] = 3.5 in
measure of another stone = [tex]5\frac{\textup{1}}{\textup{4}}\ in[/tex] = [tex]\frac{\textup{21}}{\textup{4}}\ in[/tex] = 5.25 in
let the height of the third stone be 'x'
Now,
The total height of the bench = depth of bench + Measure of two stones + x
18 = 3.375 + 3.5 + 5.25 + x
or
x = 18 - 12.125
or
x = 5.875 in
or
x = tex]5\frac{\textup{7}}{\textup{8}}\ in[/tex]
Assume that for a particular tire brand, the probability of wearing out before 30,000 miles is 0.25. For someone who buys a set of iour of these tires, what is the probability that all will last at least 30,000 miles? You be a suitable model in this case.
Answer:
The probability that all will last at least 30,000 miles is 0.3164.
Step-by-step explanation:
Consider the provided information.
For a particular tire brand, the probability of wearing out before 30,000 miles is 0.25. Someone who buys a set of four of these tires,
That means the probability of not wearing out is 1-0.25 = 0.75
Now use the binomial distribution formula.
[tex]^nC_r (p^{n-r})(q^r)[/tex]
Substitute p = 0.25, q=0.75, r=4 and n=4
[tex]^4C_4 (0.25)^{4-4}(0.75)^4[/tex]
[tex]1 \times 0.75^4[/tex]
[tex]0.3164[/tex]
The probability that all will last at least 30,000 miles is 0.3164.
The square matrix A is called orthogonal provided
thatAT=A-1. Show that the determinant of such
amatrix must be either +1 or -1.
Answer: The proof is done below.
Step-by-step explanation: Given that the square matrix A is called orthogonal provided that [tex]A^T=A^{-1}.[/tex]
We are to show that the determinant of such a matrix is either +1 or -1.
We will be using the following result :
[tex]|A^{-1}|=\dfrac{1}{|A|}.[/tex]
Given that, for matrix A,
[tex]A^T=A^{-1}.[/tex]
Taking determinant of the matrices on both sides of the above equation, we get
[tex]|A^T|=|A^{-1}|\\\\\Rightarrow |A|=\dfrac{1}{|A|}~~~~~~~~~~~~~~~~~~~~[\textup{since A and its transpose have same determinant}]\\\\\\\Rightarrow |A|^2=1\\\\\Rightarrow |A|=\pm1~~~~~~~~~~~~~~~~~~~~[\textup{taking square root on both sides}][/tex]
Thus, the determinant of matrix A is either +1 or -1.
Hence showed.
Estimate the product. Round the first factor to the nearest whole number, round the second factor to the nearest hundred, and then multiply.
5 4/7 * 598
The product is approximately blank
Answer:
The product is approximately 3600.
Step-by-step explanation:
The whole number nearest the first factor is 6.
The hundred nearest the second factor is 600.
The product of these is 6×600 = 3600, your estimated product.
Susie has planned a trip to a city 60 miles away. She wishes to have an average speed of 60 miles/hour for the trip. Due to a traffic jam, however, she only has an average speed of 30 miles/hour for the first 30 miles. How fast does she need to go for the remaining 30 miles so that her average speed is 60 miles/hour for the whole trip?
Answer:
90 mi/h
Step-by-step explanation:
Given,
For first 30 miles, her speed is 30 miles per hour,
Let x be her speed in miles per hour for another 30 miles,
Since, here the distance are equal in each interval,
So, the average speed of the entire journey
[tex]=\frac{\text{Average speed for first 30 miles + Average speed for another 30 miles}}{2}[/tex]
[tex]=\frac{30+x}{2}[/tex]
According to the question,
[tex]\frac{30+x}{2}=60[/tex]
[tex]30+x=120[/tex]
[tex]\implies x = 90[/tex]
Hence, she needs to go 90 miles per hour for remaining 30 miles.
Linear Algebra Question
The general equation for a circle is
a(x2+y2)+bx+cy+d=0.
There is exactly one circle passing through the points (2,3),(−3,2), and (0,0). Find an equation for this circle.
To find the specific circle passing through points (2,3), (-3,2), and (0,0), we started by substituting these coordinates in the general circle equation. This gave us a system of equations, which when solved, resulted in the coefficients for the circle equation. The equation of the circle we were seeking is (x^2 + y^2)/13 - (3x/32) + (15y/32) = 0.
Explanation:In order to find the equation for a specific circle we need to plugin the coordinates for (x, y) in the general equation a(x^2+ y^2) + bx + cy + d = 0 which represent three points on the circle. Let's start to solve this system using the points (2,3), (−3,2), and (0,0).
First, let's plug in (2,3) in general equation: 13a + 2b + 3c + d = 0Second, we will plugin (−3,2): 13a - 3b + 2c + d = 0Lastly, we will use the point (0,0): 0 = dNow, we have a system of three equations with three variables that we can solve. By subtracting the second from the first to eliminate d and simplifying, we obtain 5b + c = 0. Substituting d = 0 into the first and second equations, we get 13a + 2b + 3c = 0 and 13a - 3b + 2c = 0. Solving this system gives a = 1/13, b = -3/32 and c = 15/32.
Therefore, the equation of the circle is (x^2 + y^2)/13 - (3x/32) + (15y/32) = 0.
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If a child accidentally swallowed 1.6 fluidounces of FEOSOL Elixir, containing 2/3 gr of ferrous sulfate per 5 mL, how many milligrams (to the nearest hundreth) of ferrous sulfate did the child ingest? (Hint: use the smaller available conversion factor] Your Answer: Answer units
The child ingested approximately 20.18 milligrams of ferrous sulfate, which is determined by unit conversion.
Given data:
1 fluid ounce (oz) = 29.5735 milliliters (ml)FEOSOL Elixir concentration: 2/3 gr (grains) of ferrous sulfate per 5 mlFirst, let's convert the 1.6 fluid ounces to millilitres:
[tex]1.6 \text{ fluid ounces} \times 29.5735 \text{ ml/oz} \approx 47.3176 \text{ ml}[/tex]
Now, calculate the amount of ferrous sulfate in milligrams:
[tex]\text{Concentration of ferrous sulfate} = \frac{2}{3} \text{ gr per 5 ml} \\= \frac{2}{3} \times \frac{1}{5} \text{ gr/ml}[/tex]
Since 1 grain (gr) is approximately equal to 64.79891 milligrams (mg), we'll convert the concentration to milligrams per millilitre (mg/ml):
[tex]\text{Concentration of ferrous sulfate} = \frac{2}{3} \times \frac{1}{5} \times 64.79891 \text{ mg/ml}[/tex]
To calculate the total amount of ferrous sulfate ingested:
[tex]\text{Amount of ferrous sulfate} = 47.3176 \text{ ml} \times \frac{2}{3} \times \frac{1}{5} \times 64.79891 \text{ mg/ml}[/tex]
Calculating this gives:
[tex]\text{Amount of ferrous sulfate} \approx 20.1817 \text{ mg}[/tex]
Therefore, the child ingested approximately 20.18 milligrams of ferrous sulfate.
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an airpline flies 165 miles from point A in the direction 130degrees and then travels in the direction 245 degrees for 80 miles. approximately how far is the airplane from point A?
Answer:
205.4 miles
Step-by-step explanation:
As per the question,
We have been provided that an airplane started flies 165 miles from point A in the direction of 130 degrees and after that, it changes its direction to 245 degrees and travel to 80 miles again, which is drawn in the figure as below:
From the figure,
Let XY be the plane.
Now,
By using the cosine rule in triangle APR, we get
AP² = AR² + PR² - 2 × AR × PR × cos 115°
Now put the value from figure:
AP² = 165² + 80² - 2 × 165 × 80 × (-0.325)
AP² = 42205
AP = 205.4 miles.
Hence, the airplane is 205.4 miles far from point A.
mathematical induction of 3k-1 ≥ 4k ( 3k = k power of 3 )
For making mathematical induction, we need:
a base caseAn [tex]n_0[/tex] for which the relation holds true
the induction stepif its true for [tex]n_i[/tex], then, is true for [tex]n_{i+1}[/tex]
base casethe relationship is not true for 1 or 2
[tex]1^3-1 = 0 < 4*1[/tex]
[tex]2^3-1 = 8 -1 = 7 < 4*2 = 8[/tex]
but, is true for 3
[tex]3^3-1 = 27 -1 = 26 > 4*3 = 12[/tex]
induction steplets say that the relationship is true for n, this is
[tex]n^3 -1 \ge 4 n[/tex]
lets add 4 on each side, this is
[tex]n^3 -1 + 4 \ge 4 n + 4[/tex]
[tex]n^3 + 3 \ge 4 (n + 1)[/tex]
now
[tex](n+1)^3 = n^3 +3 n^2 + 3 n + 1[/tex]
[tex](n+1)^3 \ge n^3 + 3 n [/tex]
if [tex]n \ge 1[/tex] then [tex]3 n \ge 3[/tex] , so
[tex](n+1)^3 \ge n^3 + 3 n \ge n^3 + 3 [/tex]
[tex](n+1)^3 \ge n^3 + 3 \ge 4 (n + 1) [/tex]
[tex](n+1)^3 \ge 4 (n + 1) [/tex]
and this is what we were looking for!
So, for any natural equal or greater than 3, the relationship is true.
75 is ___% of 725.50
Answer:
75 is 10.34% of 725.50.
Step-by-step explanation:
Let 75 is x % of 725.50.
It can be written as mathematical equation.
x % of 725.50 = 15
[tex]\frac{x}{100}\times 725.50=75[/tex]
Multiply both sides by 100.
[tex]\frac{x}{100}\times 725.50\times 100=75\times 100[/tex]
[tex]x\times 725.50=7500[/tex]
Divide both sides by 725.50.
[tex]\frac{x\times 725.50}{725.50}=\frac{7500}{725.50}[/tex]
[tex]x=\frac{7500}{725.50}[/tex]
[tex]x=10.3376981392[/tex]
[tex]x\approx 10.34[/tex]
Therefore, 75 is 10.34% of 725.50.
The differential equation xy' = y(in x – Iny) is neither separable, nor linear. By making the substitution y(x) = xv(x), show that the new equation for v(x) equation is separable. N.B. you do not have to actually solve the ODE.
Answer:
We can place everything with v(x) on one side of the equality, everything with x on the other side. This is done on the step-by-step explanation, and shows that the new equation is separable.
Step-by-step explanation:
We have the following differential equation:
[tex]xy' = y(ln x - ln y)[/tex]
We are going to apply the following substitution:
[tex]y = xv(x)[/tex]
The derivative of y is the derivative of a product of two functions, so
[tex]y' = (x)'v(x) + x(v(x))'[/tex]
[tex]y' = v(x) + xv'(x)[/tex]
Replacing in the differential equation, we have
[tex]xy' = y(ln x - ln y)[/tex]
[tex]x(v(x) + xv'(x)) = xv(x)(ln x - ln xv(x))[/tex]
Simplifying by x:
[tex]v(x) + xv'(x) = v(x)(ln x - ln xv(x))[/tex]
[tex]xv'(x) = v(x)(ln x - ln xv(x)) - v(x)[/tex]
[tex]xv'(x) = v(x)((ln x - ln xv(x) - 1)[/tex]
Here, we have to apply the following ln property:
[tex]ln a - ln b = ln \frac{a}{b}[/tex]
So
[tex]xv'(x) = v(x)((ln \frac{x}{xv(x)} - 1)[/tex]
Simplifying by x,we have:
[tex]xv'(x) = v(x)((ln \frac{1}{v(x)} - 1)[/tex]
Now, we can apply the above ln property in the other way:
[tex]xv'(x) = v(x)(ln 1 - ln v(x) -1)[/tex]
But [tex]ln 1 = 0[/tex]
So:
[tex]xv'(x) = v(x)(- ln v(x) -1)[/tex]
We can place everything that has v on one side of the equality, everything that has x on the other side, so:
[tex]\frac{v'(x)}{v(x)(- ln v(x) -1)} = \frac{1}{x}[/tex]
This means that the equation is separable.
Ecology: Wetlands Government agencies carefully monitor water quality and its effect on wetlands (Reference: Environmental Protection Agency Wetland Report EPA 832-R-93-005). Of particular concern is the concentration of nitrogen in water draining from fertilized lands. Too much nitrogent can kill fish and wildlife. Twenty-eight samples of water were taken at random from a lake. The nitrogen concentration (milliagrams of nitrogen per liter of water) was determined for each sample.a) Identify the variable.b) is the variable quantitatie or qualitative?c) What is the implied population?
Answer:
a) The variable of the study is: milligrams of nitrogen per liter of water.
This is the amount that needs to be measured and analyzed to reach conclusions in the study.
b) The variable is quantitative. The quantitative variables are those that represent quantities. This variables can be measured on a continuous or discrete scale. Then, all the variables that you can measure or count are quantitative variables(height of trees, number of passengers per car, wind speed, milligrams of nitrogen per liter, etc). On the other hand, qualitative variables are those that can’ t be measured, and they represent attributes, like apple colors (red, green), size of trousers (small, medium, large) and so on.
c) The population under study is the milligrams of nitrogen per liter of water that are in the entire lake. You can estimate the parameters of the population by taking samples (In the example, 28 samples are taken).
Eli walked 12 feet down the hall of his house to get to the door. He conti in a straight line out the door and across the yard to the mailbox, a distan 32 feet. He came straight back across the yard 14 feet and stopped to pet hil dog. a. Draw a diagram of Eli's walking pattern. b. How far has he walked? c. How far from the house is he now?
Answer:
Eli walked 12 feet down the hall of his house to get to the door. He continued in a straight line out the door and across the yard to the mailbox, a distance of 32 feet.
He came straight back across the yard 14 feet and stopped to pet his dog.
Part A : find the image attached.
Part B : [tex]12+32+14=58[/tex] feet
Part C : [tex]32-14=18[/tex] feet from his house.
dy/dx = y/x , y(1) = −2
Answer with Step-by-step explanation:
The given differential equation is variable separable in nature and hence will be solved accordingly as follows:
[tex]\frac{dy}{dx}=\frac{y}{x}\\\\=\frac{dy}{y}=\frac{dx}{x}\\\\\int \frac{dy}{y}=\int \frac{dx}{x}\\\\ln(y)=ln(x)+ln(c)\\\\ln(y)=ln(cx)\\\\(\because ln(ab)=ln(a)+ln(b))\\\\\therefore y=cx[/tex]
where 'c' is constant of integration whose value shall be obtained using the given condition [tex]y(1)=-2[/tex]
Thus we have
[tex]-2=c\times 1\\\\\therefore c=-2\\[/tex]
Thus solution becomes
[tex]y=-2x[/tex]
Consider the following statement: The square of a prime number is not prime. (a) Write this as an if-then statement, using careful mathematical language and notation. (b) Prove or disprove the statement.
The square of a prime number is not prime.
a) let x ∈ R, If x ∈ {prime numbers}, then [tex]x^{2}[/tex]∉{prime numbers}
there says that if x is a real and x is in the set of the prime numbers, then the square of x isn't in the set of prime numbers.
b) Prove or disprove the statement.
ok, if x is a prime number, then x only can be divided by himself. Now is easy to see that [tex]x^{2}[/tex] = x*x can be divided by himself and x, then x*x is not a prime number, because can be divided by another number different than himself
Final answer:
Rephrased as an if-then statement, 'If p is a prime number, then p² is not prime.' It is proved through contradiction that the square of a prime number cannot be prime, as it would have a factor other than 1 and itself.
Explanation:
The question asks us to consider the statement: "The square of a prime number is not prime." Let's tackle this in two parts:
(a) If-Then Statement
An if-then statement using mathematical language and notation for the given statement might look like this: "If p is a prime number, then p² is not a prime number." Written symbolically, if we let p represent a prime number, then we can express the statement as: "If p is a prime, then p² is not prime," or more formally, "∀p ∈ Primes, (¬ Prime(p²))."
(b) Proof
We can prove this statement by contradiction. Assume the contrary: that there exists a prime number p such that p² is also prime. Since p is prime, p ≥ 2. So p² will be greater than p and have p as a factor, which contradicts the definition of a prime number (a prime number has no positive divisors other than 1 and itself). Hence, the square of a prime number cannot be prime, confirming the original statement.
Convert the following to rational numbers: (a) 0.12345, (b) 5.4321.
Answer:
A) [tex]\frac{2469}{20000}[/tex]
B) [tex]\frac{54321}{100000}[/tex]
Step-by-step explanation:
To answer this question first we define what are rational numbers.
A rational number is a number that can be expressed in the form of a fraction [tex]\frac{x}{y}[/tex], where x and y are integers and y ≠ 0.
a) 0.12345
This can be written in fraction form as
[tex]\frac{12345}{100000}[/tex]
Now, we simplify this fraction to lowest form. Simplified form of the above fraction is [tex]\frac{2469}{20000}[/tex]. This simplified form was obtained by dividing both numerator an denominator by 5.
b) This number can be written in fraction form as [tex]\frac{54321}{100000}[/tex]
Now, we simplify this fraction.
The simplified form of the above fraction is [tex]\frac{54321}{100000}[/tex] as the numerator and denominator have no common factor.
A frequency polygon having a peak in the center of its distribution is:
symmetric
skewed
kurtosis
a histogram
Which of the following is a true statement regarding the mode?
The mean, median, and mode may be the same value.
The mode is equal to the second quartile, Q2.
A dataset cannot have more than one mode.
The mode is the tallest bar in a histogram.
A measure of spread which magnifies the effect of values far from the mean is the:
median
interquartile range
variance
mode
In Chap4 Practice Exercise Prob #5, the histogram interval width is:
20 units
400
6 classes
Answer:
(a) symmetric
(b) The mean, median, and mode may be the same value.
and, The mode is the tallest bar in a histogram.
(c) variance
Step-by-step explanation:
Data is said to be symmetric if it is not skewed. i.e. if it has peaked at the center of the distribution.
Skewness gives us an idea about the shape of the curve.
Kurtosis enables us to have an idea about the flatness and peakedness of the frequency.
Mode represents the highest repetition of the observation.
The mean, median, and mode can have the same value.
The histogram represents the frequency of observation by the length of the bar this is the same as Mode. Thus, the mode is the tallest bar in a histogram.
Variance is the square of standard deviation and it is defined as, the sum of the square of the distance of an observation from the mean.
So, Variance is the correct option.
Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a CD player. Let A1 be the event that the receiver functions properly throughout the warranty period, A2 be the event that the speakers function properly throughout the warranty period, and A3 be the event that the CD player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with P(A1) 5 .95, P(A2) 5 .98, and P(A3) 5 .80.a. What is the probability that all three components function properly throughout the warranty period?b. What is the probability that at least one component needs service during the warranty period?c. What is the probability that all three components need service during the warranty period?d. What is the probability that only the receiver needs service during the warranty period?e. What is the probability that exactly one of the three components needs service during the warranty period?f. What is the probability that all three components function properly throughout the warranty period but that at least one fails within a month after the warranty expires?
Answer:
2.063X10^-4
0.999
0.833
3.26X10^-3
9.85x10^-3
Step-by-step explanation:
B="At least one component needs service during the warranty period"
C="All three components need service during the warranty period"
D="Only the receiver needs service during the warranty period"
E="Exactly one of the three components needs service during the warranty period"
P(A1)=0.0595
P(A2)=0.0598
P(A3)=0.058
a) P(A1∩A2∩A3)=P(A1)P(A2)P(A3)=0.0595*0.0598*0.058=2.063X10^-4
b) P(B)=1-P(A1∩A2∩A3)=1-2.063X10^-4=0.999
c) P(C)=P(A1'∩A2'∩A3')=P(A1')P(A2')P(A3')=0.9405*0.9402*0.942=0.833
d) P(D)=P(A1'∩A2∩A3)=P(A1')P(A2)P(A3)=0.9405*0.0598*0.058=3.26X10^-3
e) P(E)=P(A1'∩A2∩A3)+P(A1∩A2'∩A3)+P(A1∩A2∩A3')=3.26X10^-3 + 0.0595*0.9402*0.058+0.0595*0.0598*0.942=9.85x10^-3
Convert 375 minutes to hours?
Answer:
6.25 hours
Step-by-step explanation:
As we know that,
1 hour = 60 minutes
⇒ [tex]1 \ minute = \frac{1}{60} \ hour[/tex]
⇒ [tex]375 \ minutes =375\times \frac{1}{60} \ hours[/tex]
⇒ 375 mintes = 6.25 hours
Thus, 375 mintes = 6.25 hours
If you roll one die and flip one coin, does the occurrence of a particular outcome on the die affect the probability of a particular outcome on the coin? Why or why not?
Answer:
No, because these events are independent.
Step-by-step explanation:
When we roll one die (a fair die), we will have outcomes in the numbers from 1 to 6.
And when we flip a coin, we will have either a heads or a tails.
Now these two events are independent events as the occurrence of any result in one event is not affecting the result of second event,
Hence, these two are independent events. So, the answer is no, the occurrence of a particular outcome on the die will not affect the probability of a particular outcome on the coin.
Final answer:
The outcomes of a die roll and a coin toss are independent events, meaning the result of the die does not affect the result of the coin toss.
Explanation:
The occurrence of a particular outcome on the die does not affect the probability of a particular outcome on the coin because each event is independent of the other.
For example, if you roll a die and flip a coin, the result of the die does not influence the result of the coin toss as they are separate events.
Additionally, the concept of independence in probability states that the outcomes of the coin toss and die roll do not impact each other.
Write a short statement that expresses a possible relationship between the variables. (size of file, time to download the file)
Answer:
So you may know that that the time for downloading a file of A (MB) is B (seconds), where A and B are numbers.
Then suppose if some file size is C, which is 10 times more than A, and you assume that the download speed sees this as downloading 10 times the A (MB) file. Thus the time for downloading this second file will be 10*B (seconds).
As, if [tex]C\ (MB) file=10*A\ (MB) file[/tex]
then [tex]C\ (MB)file =10*B\ (seconds)[/tex]
B(seconds) is the time required to download A (MB) file.