a. All points on the board are equally likely to be hit with a probability of 1/(area of board), or
[tex]f_{X,Y}(x,y)=\begin{cases}\dfrac1{4\pi}&\text{for }x^2+y^2\le4\\\\0&\text{otherwise}\end{cases}[/tex]
b. To find the marginal distribution of [tex]X[/tex], integrate the joint distribution with respect to [tex]y[/tex], and vice versa. We can take advantage of symmetry here to compute the integral:
[tex]\displaystyle\int_y f_{X,Y}(x,y)\,\mathrm dy=2\int_0^{\sqrt{4-x^2}}\frac{\mathrm dy}{4\pi}=\frac{\sqrt{4-x^2}}{2\pi}[/tex]
[tex]f_X(x)=\begin{cases}\dfrac{\sqrt{4-x^2}}{2\pi}&\text{for }-2\le x\le2\\\\0&\text{otherwise}\end{cases}[/tex]
and by the same computation you would find that
[tex]f_Y(y)=\begin{cases}\dfrac{\sqrt{4-y^2}}{2\pi}&\text{for }-2\le y\le2\\\\0&\text{otherwise}\end{cases}[/tex]
c. We get the conditional distributions by dividing the joint distributions by the respective marginal distributions:
[tex]f_{X\mid Y=y}(x)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}[/tex]
[tex]f_{X\mid Y=y}(x)=\begin{cases}\dfrac1{2\sqrt{4-y^2}}&\text{for }-2\le y\le2\text{ and }x^2\le4-y^2\\\\0&\text{otherwise}\end{cases}[/tex]
and similarly,
[tex]f_{Y\mid X=x}(y)=\begin{cases}\dfrac1{2\sqrt{4-x^2}}&\text{for }-2\le x\le2\text{ and }y^2\le4-x^2\\\\0&\text{otherwise}\end{cases}[/tex]
d. You can compute this probability by integrating the joint distribution over a part of the circle (call it "B" for bullseye):
[tex]\displaystyle\iint_Bf_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^{0.25}\frac r{4\pi}\,\mathrm dr\,\mathrm d\theta=\frac1{64}[/tex]
(using polar coordinates) The easier method would be to compute the area of a circle with radius 0.25 instead, then divide that by the total area of the dartboard.
[tex]\dfrac{\pi\left(\frac14\right)^2}{\pi\cdot2^2}=\dfrac1{64}[/tex]
e. The event that [tex]X>1[/tex] is complementary to the event that [tex]X\le1[/tex], so
[tex]P(X>1)=1-P(X\le1)=1-F_X(1)[/tex]
where [tex]F_X(x)[/tex] is the marginal CDF for [tex]X[/tex]. We can compute this by integrate the marginal PDF for [tex]X[/tex]:
[tex]F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<-2\\\\\dfrac12+\dfrac1\pi\sin^{-1}\dfrac x2+\dfrac{x\sqrt{4-x^2}}{4\pi}&\text{for }-2\le x<2\\\\1&\text{for }x\ge2\end{cases}[/tex]
Then
[tex]P(X>1)=1-F_X(1)=\dfrac13-\dfrac{\sqrt3}{4\pi}\approx0.1955[/tex]
f. We found that either random variable conditioned on the other is a uniform distribution. In particular,
[tex]f_{Y\mid X=1}(y)=\begin{cases}\dfrac1{2\sqrt3}&\text{for }y^2\le3\\\\0&\text{otherwise}\end{cases}[/tex]
Then
[tex]P(Y>0.3\mid X=1)=1-P(Y\le0.3\mid X=1)=1-F_{Y\mid X=1}(0.3)[/tex]
where [tex]F_{Y\mid X=x}(y)[/tex] is the CDF of [tex]Y[/tex] conditioned on [tex]X=x[/tex]. This is easy to compute:
[tex]F_{Y\mid X=1}(y)=\displaystyle\int_{-\infty}^yf_{Y\mid X=1}(t)\,\mathrm dt=\begin{cases}0&\text{for }y<-\sqrt3\\\\\dfrac{y+\sqrt3}{2\sqrt3}&\text{for }-\sqrt3\le y<\sqrt3\\\\1&\text{for }y\ge\sqrt3\end{cases}[/tex]
and we end up with
[tex]P(Y>0.3\mid X=1)=\dfrac{10-\sqrt3}{20}\approx0.4134[/tex]
Final answer:
The joint PDF for a uniform distribution on a circular dart board is constant within the dart board and zero outside. Marginal and conditional PDFs are derived from the joint PDF. To compute probabilities such as the bulls eye hit or X > 1, integrate the corresponding PDF over the relevant range.
Explanation:
Finding the Joint and Marginal PDFs, and Conditional Probabilities
The joint probability density function (PDF) of X and Y for a uniform distribution over a circular dart board with radius 2 inches is constant within the circle and zero outside. First, we calculate the area of the circle, A = πr² = π(2)² = 4π square inches. The joint PDF f(x, y) will be 1/A for all points inside the dart board, and 0 otherwise.
The marginal PDFs are derived by integrating the joint PDF over the other variable. For instance, fX(x) is found by integrating f(x, y) over y, and fY(y) is found by integrating f(x, y) over x.
The conditional PDFs fX|Y and fY|X are derived from the joint PDF divided by the marginal PDF of the conditioned variable.
To find the probability of hitting the bulls eye, a circle of radius 0.25 inches, you'd integrate the joint PDF over the area of the bulls eye or simply calculate the area ratio of the bulls eye to that of the entire dart board.
The probability that X > 1 is found by integrating fX(x) from 1 to 2. If you know that X = 1, the conditional PDF of Y is fY|X(y|X=1). The probability that Y > 0.3 given X = 1 is calculated by integrating this conditional PDF from 0.3 to the upper limit set by the circle's boundary.
Find the dimensions of the box described. The length is twice as long as the width. The height is 4 inches greater than the width. The volume is 48 cubic inches. Find the length, width, height
Answer:
Width of box= 2inches
Length of box= 4inches
Height of box= 6inches.
Step-by-step explanation:
Let width of box=x inches
Length of box = twice of width=[tex]2\times x[/tex]=[tex]2x[/tex]
Height of box= 4 inches greater than width= [tex]x+4[/tex]
Volume of box= 48 cubic inches
We know that the formula of volume of cuboid= [tex] length\times breadth\times height[/tex]
Apply the formula
Volume of box= [tex]x\times 2x\times (x+4)[/tex]
Volume of cube = [tex]2x^2(x+4)[/tex]
[tex]2x^2(x+4)=48[/tex]
[tex]x^2(x+4)=24[/tex]
[tex]x^3+x^2-24[/tex]
Apply inspection method to solve the equation
Put [tex]x=0[/tex]
Then we get [tex]-24\neq0 [/tex]
Hence, x=0 is not the solution of x
Put x=1 in the equation then we get
[tex]-22\neq 0[/tex]
Hence x=1 is not the solution of equation.
Put x=2 then we get
[tex](2)^3+(4)^2-24[/tex]
8+16-24=0
Hence, x=2 is the solution of equation .
[tex] (x-2)(x^2+6x+12)[/tex]=0
Now substitute equation [tex]x^2+6x+12[/tex]=0
Sum roots =6
Product of roots=12
When sum of roots is greater than zero and product of roots is greater than zero then value of roots of equation is imaginary.
Hence, the roots of equation [tex]x^2+6x+12=0[/tex] are imaginary.
Lenght , widht and height are dimensions of box therefore, imaginary value are not possible.
Hence,[tex] x=2 [/tex] is the only real values of root of equation .Therefore, it is possible and other two imaginary value of roots are not possible .
Widht of box=2 inches
Length of box = [tex]2\times2[/tex]=4inches
Height of box=[tex]x+4[/tex]=2+4=6 inches
The dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.
Explanation:
Let's use the given information to solve for the dimensions of the box:
Let the width of the box be represented by x inches.
The length of the box is twice as long as the width, so the length is 2x inches.
The height is 4 inches greater than the width, so the height is (x + 4) inches.
The volume of a box can be calculated by multiplying the length, width, and height. Since the volume of the box is given as 48 cubic inches, we can set up the equation: 2x * x * (x + 4) = 48.
Simplifying the equation, we get 2x^3 + 8x^2 - 48 = 0.
Factoring the equation, we find that (x - 2)(x + 4)(x + 6) = 0.
The possible solutions are x = 2, x = -4, or x = -6.
Since we are dealing with dimensions, the width cannot be negative, so we can disregard the negative solutions. The width, therefore, is 2 inches.
The length is twice as long as the width, so the length is 2 * 2 = 4 inches.
The height is 4 inches greater than the width, so the height is 2 + 4 = 6 inches.
Therefore, the dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.
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What is the future value of $510 per year for 8 years compounded annually at 9 percent?
The future value of $510 per year for 8 years compounded annually at 9 percent is $1,016.21.
What is the future value?The investment's future value refers to the compounded value of the present cash flows in the future, using an interest rate.
The future value can be determined using the future value table or formula.
We can also determine the future value using an online finance calculator as below.
Data and Calculations:N (# of periods) = 8 years
I/Y (Interest per year) = 9%
PV (Present Value) = $510
PMT (Periodic Payment) = $0
Results:
FV = $1,016.21 ($510 + $506.21)
Total Interest = $506.21
Thus, the future value of $510 per year for 8 years compounded annually at 9 percent is $1,016.21.
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for the following right triangle find the side length x
Since there is a right angle, you can use Pythagoras' Theorem:
So x = √(24² + 7²) = 25
---------------------------------------------------------
Answer:
25
Suppose we wanted to differentiate the function h(x)= (5 - 2 x^6)^3 +1/(5 - 2 x^6) using the chain rule, writing the function h (x) as the composite function h(x)= f(g(x)). Identify the functions f (x) and g (x). f (x) = g (x) = Calculate the derivatives of these two functions f '(x) = g '(x) = Now calculate the derivative of h (x) using the chain rule
[tex]h(x)=(5-2x^6)^3+\dfrac1{5-2x^6}[/tex]
Let [tex]g(x)=5-2x^6[/tex] and [tex]f(x)=x^3+\dfrac1x[/tex]. Then [tex]h(x)=f(g(x))[/tex].
Set [tex]u=5-2x^6[/tex]. By the chain rule,
[tex]\dfrac{\mathrm dh}{\mathrm dx}=\dfrac{\mathrm dh}{\mathrm du}\cdot\dfrac{\mathrm du}{\mathrm dx}[/tex]
Since [tex]h(u)=u^3+\dfrac1u[/tex] and [tex]u(x)=5-2x^6[/tex], we have
[tex]\dfrac{\mathrm dh}{\mathrm du}=3u^2-\dfrac1{u^2}[/tex]
[tex]\dfrac{\mathrm du}{\mathrm dx}=-12x^5[/tex]
Then
[tex]\dfrac{\mathrm dh}{\mathrm dx}=\left(3u^2-\dfrac1{u^2}\right)(-12x^5)=\boxed{-12x^5\left(3(5-2x^6)^2-\dfrac1{(5-2x^6)^2}\right)}[/tex]
which we could rewrite slightly as
[tex]\dfrac{\mathrm dh}{\mathrm dx}=-\dfrac{12x^5(3(5-2x^6)^4-1)}{(5-2x^6)^2}[/tex]
To differentiate the given function using the chain rule, we need to identify the functions f(x) and g(x), then calculate their derivatives. Once we have the derivatives, we can apply the chain rule to find the derivative of the composite function h(x).
Explanation:Chain Rule
To differentiate the function h(x) = (5 - 2x^6)³ + 1/(5 - 2x^6) using the chain rule, we can write it as the composite function h(x) = f(g(x)).
Let's identify the functions f(x) and g(x):
f(x) = x³, g(x) = (5 - 2x^6)
Next, let's calculate the derivatives of f(x) and g(x):
f'(x) = 3x², g'(x) = -12x^5
Finally, we can apply the chain rule to differentiate h(x):
h'(x) = f'(g(x)) * g'(x) = (3(5 - 2x^6)²) * (-12x^5)
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A motorboat travels 180 km in 3 hours going upstream and 504 in 6 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?
Answer:
Speed of boat x = 84 km/hr
Speed of current = 12 km/hr
Step-by-step explanation:
Let 'x' be the speed of boat and 'y' be the speed of still water
Upstream speed = x - y and
Downstream speed = x + y
It is given that, A motorboat travels 180 km in 3 hours going upstream and 504 in 6 hours going downstream
Upstream speed = x - y = 180/3 = 60 km/hr
Downstream speed = x + y = 504/6 = 84 km/hr
To find the value of x and y
x + y = 84 ----(1)
x - y = 60 ----(2)
(1) + (2) ⇒
x + y = 84 ----(1)
x - y = 60 ----(2)
2x + 0 = 144
x = 144/2 = 72
x + y = 84
y = 84 - 72 = 12
Therefore speed of boat x = 84 km/hr
Speed of current = 12 km/hr
7(x - 2) = 3(x + 4)
Solve the following equation. Then enter your answer in the space provided using mixed number format.
Answer:
In mixed number format: 6 1/2
Step-by-step explanation:
To solve the following equation: 7(x - 2) = 3(x + 4), first we need to apply the distributive property:
7(x - 2) = 3(x + 4) → 7x -14 = 3x + 12
Solving for 'x' → 4x = 26 → x = 6.5
→ In mixed number format: 6 1/2
For this case we must solve the following equation:
[tex]7 (x-2) = 3 (x + 4)[/tex]
Applying distributive property to the terms within the parenthesis we have:
[tex]7x-14 = 3x + 12[/tex]
We subtract 3x on both sides of the equation:
[tex]7x-3x-14 = 12\\4x-14 = 12[/tex]
Adding 14 to both sides of the equation:
[tex]4x = 12 + 14\\4x = 26[/tex]
Dividing between 4 on both sides of the equation:
[tex]x = \frac {26} {4} = \frac {13} {2}[/tex]
ANswer:
[tex]x = \frac {13} {2}\\x = 6 \frac {1} {2}[/tex]
Four different prime numbers, each less than 20, are multiplied together. What is greatest possible result?
a. 21,879
b. 28,728
c. 40,755
d. 46,189
e. 49,172
Please show me how I can solve this!!
Answer:
46,189
Step-by-step explanation:
The prime numbers that are less than 20 are :
1,2,3,5,7,11,13,17,19
to get the greatest value, we multiply the four numbers with the largest values i.e
11 x 13 x 17 x 19 = 46,189
The greatest possible product of four different prime numbers each less than 20 is found by multiplying the four largest primes in that range: 19, 17, 13, and 11, which equals 46,189.
Explanation:To find the greatest possible product of four different prime numbers each less than 20, we should choose the four largest prime numbers in that range. The largest primes less than 20 are 19, 17, 13, and 11. Multiplying these together gives us:
19 \times 17 \times 13 \times 11 = 46,189.
Thus, the greatest possible result when multiplying four different prime numbers, each less than 20, is 46,189, which matches option 'd'.
Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48feet above the ground, the function h(t)=−16t2+32t+48 models the height, h, of the ball above the ground as a function of time, t. Find the times the ball will be 48feet above the ground.
[tex]\bf \stackrel{height}{h(t)}=-16t^2+32t+48\implies \stackrel{48~ft}{~~\begin{matrix} 48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=-16t^2+32t~~\begin{matrix} +48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \\\\\\ 0=-16t^2+32t\implies 16t^2-32t=0\implies 16t(t-2)=0\implies t= \begin{cases} 0\\ 2 \end{cases}[/tex]
t = 0 seconds, when the ball first took off, and t = 2, 2 seconds later.
Answer: [tex]t_1=0\\t_2=2[/tex]
Step-by-step explanation:
We know that the function [tex]h(t)=-16t^2+32t+48[/tex] models the height "h" of the ball above the ground as a function of time "t".
Then, to find the times in which the ball will be 48 feet above the ground, we need to substitute [tex]h=48[/tex] into the function and solve fot "t":
[tex]48=-16t^2+32t+48\\0=-16t^2+32t+48-48\\0=-16t^2+32t[/tex]
Factorizing, we get:
[tex]0=-16t(t-2)\\t_1=0\\t_2=2[/tex]
A house was valued at $100,000 in the year 1987. The value appreciated to $165,000 by the year 2002.
Use the compound interest form S=P(1+r)^t to answer the following questions.
A) What was the annual growth rate between 1987 and 2002? (Round to 4 decimal places.)
B) What is the correct answer to part A written in percentage form?
C) Assume that the house value continues to grow by the same percentage. What will the value equal in the year 2005? (Round to the nearest thousand dollars.)
Please help ASAP I need an answer by today!!! :(
Answer:
A) The annual multiplier was 1.0339; the annual increase was 0.0339 of the value.
B) 3.39% per year
C) $182,000
Step-by-step explanation:
A) Let's let t represent years since 1987. Then we can fill in the numbers and solve for r.
165000 = 100000(1 +r)^15
1.65^(1/15) = 1 +r . . . . . divide by 100,000; take the 15th root
1.03394855265 -1 = r ≈ 0.0339
The value was multiplied by about 1.0339 each year.
__
B) The value increased by about 3.39% per year.
__
C) S = $100,000(1.03394855265)^18 ≈ $182,000
Lines a and b are parallel. Line c is perpendicular to both line a and line b. Which statement about lines a, b, and c is NOT true?
Line a and line b have the same slope.
The sum of the slopes of line b and line c is 0.
The product of the slopes of line c and line b is −1.
The product of the slopes of line a and line c is −1.
m + (-1/m) ≠ 0
⇒ The sum of the slopes of line b and line c is 0.
⇒ False ⇒ NOT true
Answer:
The sum of the slopes of line b and line c is 0.
Step-by-step explanation:
Remember that the product of the slopes of two parallel lines is -1, so in order to be -1 you have to multiply M*-1/m=-1 so since to add them up you would do it like this m+(-1/m) taht wouldn´t get as result 0, so that would be the option that is not correct, remember that parallel lines have the same slope, so that also eliminates all of the other options.
The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?
Answer:
624
Step-by-step explanation:
The sequence is 49, 47, 45,...., 7, 5, 3. This is an arithmetic sequence, because the difference between terms is the same.
The sum of the first n terms of an arithmetic sequence is:
S = n/2 (a₁ + an)
where a₁ is the first term and an is the nth term.
Here, we know that a₁ = 49 and an = 3. But we need to find what n is. To do that, we use definition of an arithmetic sequence:
an = a₁ + (n-1) d
where d is the common difference (in this case, -2)
3 = 49 + (n-1) (-2)
2(n-1) = 46
n - 1 = 23
n = 24
So there are 24 terms in the sequence.
The sum is:
S = 24/2 (49 + 3)
S = 12 (52)
S = 624
There are 624 bricks in the wall.
The total number of bricks in the wall is 624. This is a math problem that involves arithmetic sequence, where each term is obtained from the previous one by subtracting a fixed number (2, in this case), and concepts from algebra (equations).
Explanation:The problem describes a scenario where each row of a brick wall has two fewer bricks than the row below it, which characterizes a sequence in mathematics. More specifically, this is an arithmetic sequence, which is characterized by a common difference between terms, in this case, the difference is -2.
To solve the problem, we need to find the sum of an arithmetic sequence. The formula of the sum is given by:
S = n/2 * (a1 + an)
Where S is the sum, n the number of terms, a1 the first term, and an the last term. Here, a1 is 49 and an is 3. To find n, we use the formula n = (a1 - an) / d + 1, with d being the common difference which is -2. Solving the equation we find that n = 24.
We now plug these values into the sum formula and find that the sum S, which represents the total number of bricks in the wall is
S = 24/2 * (49 + 3) = 12 * 52 = 624.
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x = Temperature (°C) 1100 1200 1300 1100 1500 1200 1300 y = Porosity (%) 30.8 19.2 6 13.5 11.4 7.7 3.6 (a) Fit the simple linear regression model using the method of least squares. Find the least squares estimates of the intercept and slope in the simple linear regression model.
Answer:
intercept: 55.6256slope: -0.0341585Step-by-step explanation:
This sort of problem is best worked by a tool such as a graphing calculator or spreadsheet.
Identify the parameters p and n in the following binomial distribution scenario. The probability of winning an arcade game is 0.718 and the probability of losing is 0.282. If you play the arcade game 20 times, we want to know the probability of winning more than 15 times. (Consider winning as a success in the binomial distribution.)
Answer:
p = 0.718 and n = 20
Step-by-step explanation:
p is the probability of success and n is the number of trials.
Here, p = 0.718 and n = 20.
Answer:
There is a 29.50% probability of winning more than 15 times.
Step-by-step explanation:
For each time you play the arcade game, there are only two possible outcomes. Either you win, or you lose. This means that we can solve this problem using the binomial probability distribution.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
The probability of winning a game is 0.718. So [tex]p = 0.718[/tex].
The game is going to be played 20 times, so [tex]n = 20[/tex].
If you play the arcade game 20 times, we want to know the probability of winning more than 15 times.
This is
[tex]P(X > 15) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.2950[/tex].
There is a 29.50% probability of winning more than 15 times.
What is the simplest form of
Answer:
The simplest form of [tex]\sqrt[3]{27a^{3}b^{7}}[/tex] is
3ab²(∛b)
Step-by-step explanation:
The given term is:
[tex]\sqrt[3]{27a^{3}b^{7}}[/tex]
To convert it into its simplest form, we will apply simple mathematical rules to simplify the power of individual terms.
[tex]\sqrt[3]{27a^{3}b^{7}}\\= \sqrt[3]{3^{3} a^{3}b^{7}}\\= \sqrt[3]{3^{3}a^{3}b^{6}b}\\= 3^{3/3} a^{3/3}b^{6/3}b^{1/3}}\\= 3ab^{2}(\sqrt[3]{b})[/tex]
While simplifying the term, we basically took the cube root of individual terms. The powers cancelled out cube root for some terms. In the end, we were left with the simplest form of the expression.
College algebra homework review... Having issues calculating this by hand and on TI-84 receiving errors like "8e12" when trying to calculate the actual quadratic equation it calls for in question B.... Please help
checking the vertex of this upside-down parabola, it has a vertex at (1000, 2000000), so that's the U-turn, when as the price "p" increases the revenue goes down.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{a}x^2\stackrel{\stackrel{b}{\downarrow }}{+b}x\stackrel{\stackrel{c}{\downarrow }}{+c} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
now, if we solve the quadratic using the value of 500000
[tex]\bf \stackrel{R(p)}{500000}=-2p^2+4000p\implies 250000=-p^2+2000p \\\\\\ p^2-2000p+250000=0[/tex]
and we run the quadratic formula on it, we get the values of x = 133.97 and x = 1866.03, one value is obviously when going upwards, the first one, and the other is when going downwards.
so we know that the R(p) is 500,000 at x = 133.97, and it keeps on going up, up to the vertex above at x = 1000, so we can say from x = [134, 1000] R(p) > 500000.
Explain why vertical lines are a special case in the definition of parallel lines.
Answer:
A vertical line has an infinite or undefined slope since the denominator is zero.
Step-by-step explanation:
Parallel lines by definition refers to lines that never intersect or meet since they have identical slopes. The slope of line is defined as;
(change in y)/(change in x)
For a vertical line, the y values are changing while the x values remain constant. The slope of this line will thus have a zero value in the denominator implying that its slope will not defined or will be infinity.
Answer:
A vertical line has an infinite or undefined slope since the denominator is zero.
Step-by-step explanation:
If y = e2x is a solution to y''- 5y' + ky = 0, what is the value of k?
Answer:
The value of k is 6
Step-by-step explanation:
we need to find the value of k
Given : - [tex]y=e^{2x}[/tex] is the solution [tex]y''-5y'+ky=0[/tex]
[tex]y=e^{2x}[/tex] ........(1)
differentiate [tex]y=e^{2x}[/tex] with respect to 'x'
[tex]\frac{dy}{dx}=\frac{d}{dx}e^{2x}[/tex]
Since, [tex]\frac{d}{dx}e^{x} =e^{x}\frac{d}{dx}(x)[/tex]
[tex]\frac{dy}{dx}=e^{2x}\frac{d}{dx}(2x)[/tex]
[tex]\frac{dy}{dx}=e^{2x}\times 2[/tex]
[tex]\frac{dy}{dx}=2e^{2x}[/tex]
so, [tex]y'=2e^{2x}[/tex] ..........(2)
Again differentiation above with respect to 'x'
[tex]\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}2e^{2x}[/tex]
[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\frac{d}{dx}(2x)[/tex]
[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\times 2[/tex]
[tex]\frac{d^{2}y}{dx^{2}}=4e^{2x}[/tex]
so, [tex]y''=4e^{2x}[/tex] ........(3)
Now, put the value of [tex]y\ ,y' \ \text{and} \ y''[/tex] in [tex]y''-5y'+ky=0[/tex]
[tex]4e^{2x}-5(2e^{2x})+(e^{2x})k=0[/tex]
[tex]4e^{2x}-10e^{2x}+e^{2x}k=0[/tex]
[tex]-6e^{2x}+e^{2x}k=0[/tex]
add both the sides by [tex]6e^{2x}[/tex]
[tex]e^{2x}k=6e^{2x}[/tex]
Cancel out the same terms from left and right sides
[tex]k=6[/tex]
Hence, the value of k is 6
when two dice are rolled, what is the probability the two numbers will have a sum of 10
A. 1/10
B.1/18
C.1/12
D.1/3
Answer:
The correct answer is option C. 1/12
Step-by-step explanation:
It is given that, two dies are rolled.
The outcomes of tossing two dies are,
(1,1), (1,2), (1,3), (1,4), (15), (1,6)
----- -------- ------ ------ ----- ----
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Number of possible outcomes = 36
To find the probability
The possible outcomes are getting sum 10 which are,
(4,6), (5, 5) and (6,4)
Number of possible outcomes = 3
Therefore probability of getting sum 10 = 3/36 = 1/12
The correct answer is option C. 1/12
Write 1.052 as a percent
1.052 is equal to 105.2 percent.
Given that a decimal number 1.052, we need to write 1.052 as a percent,
To express a decimal number as a percent, you need to multiply it by 100.
Let's calculate 1.052 as a percent:
1.052 x 100 = 105.2
To understand this, let's break it down:
The number 1.052 represents 105.2% because it is greater than 1 (100%). By multiplying it by 100, we shift the decimal point two places to the right, resulting in 105.2.
In percentage terms, 105.2% means that 1.052 is 105.2 parts out of 100. This can also be interpreted as 105.2 per hundred or simply 105.2 out of every 100 units.
Therefore, 1.052 can be written as 105.2%.
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How do I simply this radical expression?
Answer:
[tex]1000±100\sqrt{55}[/tex]
Step-by-step explanation:
To simplify that expression, first we need to find the largest common of the expression inside the radical, in this case: 2.200.000.
We know that 2.200.000 = 2 · 2 · 2 · 2 · 2 · 2 · 5 · 5 · 5 · 5 · 5 · 11 = [tex]2^{6}[/tex] ×[tex]5^{5}[/tex]× [tex]11[/tex]
Now, [tex]\sqrt{2^{6}5^{5}11} = 200\sqrt{55}[/tex].
Now we have: [tex]\frac{2000±200\sqrt{55}}{2}[/tex]
Dividing by 2: [tex]1000±100\sqrt{55}[/tex]
So the simplified expression is: [tex]1000±100\sqrt{55}[/tex]
Question 1: Factor out the Greatest
Common Factor
3t5s − 15t2s3
Question 1 options:
3(t5s − 5t2s3)
3t2(t3s − 5s3)
3t2s(t3 − 5s2)
-12t3s2
For this case we have that by definition, the GCF or (Greatest Common Factor) is given by the greatest common factor that divides both terms without leaving a residue.
15: 1,3,5,15
3: 1.3
Then we have the GCF of the expression is:
[tex]3t ^ 2s (t ^ 3-5s ^ 2)[/tex]
ANswer:
Option C
Find the derivative of the function using the definition of derivative. g(x) = 5 − x (1) Find g'(x) (2) State the domain of the function. (Enter your answer using interval notation.) (3) State the domain of its derivative. (Enter your answer using interval notation.)
Answer:
Answer is contained in the explanation
Step-by-step explanation:
[tex]g(x)=5-x\\g'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\\g'(x)=\lim_{h \rightarrow 0} \frac{[5-(x+h)]-[5-x]}{h}\\g'(x)=\lim_{h \rightarrow 0} \frac{5-x-h-5+x}{h}\\g'(x)=\lim_{h \rightarrow 0} \frac{-h}{h}\\g'(x)=\lim_{h \rightarrow 0} -1\\g'(x)=-1[/tex]
g(x)=5-x has domain all real numbers (you can plug an a number and always get a number back)
So in interval notation this is [tex](-\infty, \infty)[/tex]
g'(x)=-1 has domain all real numbers (the original function had domain issues... and no matter the number you plug in you do get a number, that number being -1)
So in interval notation this is [tex](-\infty, \infty)[/tex]
The derivative of given function g(x) is
g'(x)=-1
Domain of function g(x) is (-∞,∞)
Domain of derivative is (-∞,∞)
Given :
[tex]g(x) = 5 - x[/tex]
Lets find derivative using definition of derivative
[tex]\lim_{h \to 0} \frac{g(x+h)-g(x)}{h} \\g(x)=5-x\\g(x+h)=5-(x+h)\\g(x+h)=5-x-h\\\lim_{h \to 0} \frac{5-x-h-(5-x)}{h} \\\\\lim_{h \to 0} \frac{5-x-h-5+x}{h} \\\\\lim_{h \to 0} \frac{-h}{h} \\\\-1[/tex]
Derivative g'(x)=-1
g(x) is a linear function . for all linear function the domain is set of all real numbers
Domain of function g(x) is (-∞,∞)
Derivative function g'(x) =-1. For all values of x the value of y is -1
So domain is set of all real numbers
Domain of derivative is (-∞,∞)
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A town's January high temperatures average 36degreesF with a standard deviation of 8degrees, while in July the mean high temperature is 72degrees and the standard deviation is 9degrees. In which month is it more unusual to have a day with a high temperature of 57degrees? Explain.
Answer: July
Step-by-step explanation:
Formula of z score :
[tex]z=\dfrac{X-\mu}{\sigma}[/tex]
Given: The mean high temperature in January = [tex]\mu_1=36^{\circ} F[/tex]
Standard deviation : [tex]\sigma_1=8^{\circ}F[/tex]
For X = [tex]57^{\circ}F[/tex]
[tex]z=\dfrac{57-36}{8}=2.625[/tex]
The mean high temperature in July = [tex]\mu_1=72^{\circ} F[/tex]
Standard deviation : [tex]\sigma_1=9^{\circ}F[/tex]
[tex]z=\dfrac{57-72}{8}=-1.875[/tex]
⇒ 57° F is about 2.6 standard deviations above the mean of January high temperatures, and 57° F is about 1.9 standard deviations below the mean of July’s high temperatures.
A general rule says that z-scores lower than -1.96 or higher than 1.96 are considered unusual .
Hence, the 57˚F is more unusual in January.
Final answer:
A high temperature of 57 degrees is more unusual in January than in July, as it is 2.625 standard deviations above the January mean, compared to 1.667 standard deviations below the July mean.
Explanation:
To determine in which month it is more unusual to have a high temperature of 57 degrees Fahrenheit, we can calculate the z-score for each month. The z-score tells us how many standard deviations away from the mean a particular value is.
For January, the z-score is calculated as follows:
Z = (57 - 36) / 8 = 21 / 8 = 2.625
This means that a temperature of 57 degrees in January is 2.625 standard deviations above the January mean.
For July, the z-score is calculated as follows:
Z = (57 - 72) / 9 = -15 / 9 = -1.667
This means that a temperature of 57 degrees in July is 1.667 standard deviations below the July mean.
Since the absolute value of the January z-score (2.625) is higher than the absolute value of the July z-score (-1.667), a high temperature of 57 degrees is more unusual in January than in July.
Which expression is equivalent to
Answer:
The correct answer is second option
4a²b²c²∛b)
Step-by-step explanation:
It is given an expression, ∛(64a⁶b⁷c⁹)
Points to remember
Identities
ⁿ√x = x¹/ⁿ
To find the equivalent expression
We have, ∛(64a⁶b⁷c⁹)
∛(64a⁶b⁷c⁹) = (64a⁶b⁷c⁹)1/3
= (4³/³ a⁶/³ b⁷/³ c⁹/³) [Since 64 = 4³]
= 4a² b² b¹/³ c³
= 4a²b²c³(b¹/³)
= 4a²b²c³ (∛b)
Therefore the correct answer is second option
4a²b²c³(∛b)
Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²
The derivative of the function f(x) is:
[tex]f'(x)=-2x[/tex]
Step-by-step explanation:We are given a function f(x) as:
[tex]f(x)=5-x^2[/tex]
We have:
[tex]f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)[/tex]
( Since,
[tex](a+b)^2=a^2+b^2+2ab[/tex] )
Hence, we get:
[tex]f(x+h)=5-x^2-h^2-2xh[/tex]
Also, by using the definition of f'(x) i.e.
[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]
Hence, on putting the value in the formula:
[tex]f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x[/tex]
Hence, the derivative of the function f(x) is:
[tex]f'(x)=-2x[/tex]
Answer:
The derivative of given function is -2x.
Step-by-step explanation:
The first principle of differentiation is
[tex]f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]
The given function is
[tex]f(x)=5-x^2[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x+h)^2-(5-h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x^2+2xh+h^2)-5+h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-x^2-2xh-h^2-5+h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2-2xh}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-\frac{2xh}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-2x[/tex]
Apply limit.
[tex]f'(x)=\frac{-x^2}{0}-2x[/tex]
[tex]f'(x)=0-2x[/tex]
[tex]f'(x)=-2x[/tex]
Therefore, the derivative of given function is -2x.
Solve the system y = -x + 7 and y= 0.5(x - 3)2
Answer:
The solutions of the system of equations are (-1,8) and (5,2)
Step-by-step explanation:
[tex]y=-x+7[/tex] -------> equation A (equation of a line)
[tex]y=0.5(x-3)^{2}[/tex] ----> equation B (vertical parabola open upward)
Solve the system of equations by graphing
Remember that the solution is the intersection points both graphs
using a graphing tool
The intersection points are (-1,8) and (5,2)
see the attached figure
therefore
The solutions of the system of equations are (-1,8) and (5,2)
Answer: (-1,8) and (5,2)
Step-by-step explanation: The person above me is correct. Give him five stars and a thanks!
Federal Rent-a-Car is putting together a new fleet. It is considering package offers from three car manufacturers. Fred Motors is offering 5 small cars, 5 medium cars, and 10 large cars for $500,000. Admiral Motors is offering 5 small, 10 medium, and 5 large cars for $400,000. Chrysalis is offering 10 small, 5 medium, and 5 large cars for $300,000. Federal would like to buy at least 700 small cars, at least 600 medium cars, and at least 700 large cars. How many packages should it buy from each car maker to keep the total cost as small as possible?
Answer:
40 packages from Fred Motors20 packages from Admiral Motors40 packages from ChrysalisStep-by-step explanation:
I would formulate the problem like this. Let f, a, c represent the numbers of packages bought from Fred Motors, Admiral Motors, and Chrysalis, respectively. Then the function to minimize (in thousands) is …
objective = 500f +400a +300c
The constraints on the numbers of cars purchased are …
5f +5a +10c >= 700
5f +10a +5c >= 600
10f +5a +5c >= 700
Along with the usual f >=0, a>=0, c>=0. Of course, we want all these variables to be integers.
Any number of solvers are available in the Internet for systems like this. Shown in the attachments are the input and output of one of them.
The optimal purchase appears to be …
40 packages from Fred Motors20 packages from Admiral Motors40 packages from ChrysalisThe total cost of these is $40 million.
This is a linear programming problem that requires to minimize a cost function subject to several constraints about the total number of small, medium, and large cars in the fleet. It can be set up using the system of inequalities and then solved using methods like the Simplex one.
Explanation:This problem can be solved through linear programming, which involves creating a system of inequalities to represent the constraints of the problem, and then optimizing a linear function. To start, let's define the variables: x is the number of packages bought from Fred Motors, y is the number from Admiral Motors, and z is the number from Chrysalis.
The fleet requirements translate to the following constraints: 5x + 5y + 10z ≥ 700 (small cars), 5x + 10y + 5z ≥ 600 (medium cars), and 10x + 5y + 5z ≥ 700 (large cars).
Then, the cost to minimize is: $500,000x + $400,000y + $300,000z.
This is a linear programming problem and can be solved using various methods, such as the Simplex method or graphically. Exact solutions would require a more detailed analysis.
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Last year, Susan had 10,000 to invest. She invested some of it in an account that paid 6%
simple interest per year, and she invested the rest in an account that paid 5% simple interest per year. After one year, she received a total of %560 in interest. How much did she invest in each account?
Answer:
In the account that paid 6% Susan invest [tex]\$6,000[/tex]
In the account that paid 5% Susan invest [tex]\$4,000[/tex]
Step-by-step explanation:
we know that
The simple interest formula is equal to
[tex]I=P(rt)[/tex]
where
I is the Final Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
Part a) account that paid 6% simple interest per year
in this problem we have
[tex]t=1\ years\\ P=\$x\\r=0.06[/tex]
substitute in the formula above
[tex]I1=x(0.06*1)[/tex]
[tex]I1=0.06x[/tex]
Part b) account that paid 5% simple interest per year
in this problem we have
[tex]t=1\ years\\ P=\$10,000-\$x\\r=0.05[/tex]
substitute in the formula above
[tex]I2=(10,000-x)(0.05*1)[/tex]
[tex]I2=500-0.05x[/tex]
we know that
[tex]I1+I2=\$560[/tex]
substitute and solve for x
[tex]0.06x+500-0.05x=560[/tex]
[tex]0.01x=560-500[/tex]
[tex]0.01x=60[/tex]
[tex]x=\$6.000[/tex]
therefore
In the account that paid 6% Susan invest [tex]\$6,000[/tex]
In the account that paid 5% Susan invest [tex]\$4,000[/tex]
Susan invested $6,000 at 6% and the remainder, $4,000, at 5% interest.
Susan invested $10,000 in two different accounts, one with a 6% simple interest and the other with a 5% simple interest. After one year, she received a total of $560 in interest. We need to find out how much she invested in each account.
Let's denote x as the amount invested at 6% and (10,000 - x) as the amount invested at 5%. Using the formula for simple interest, interest = principal × rate × time, we can set up two equations based on the given information:
The interest from the account with 6% interest: 0.06 × x
The interest from the account with 5% interest: 0.05 × (10,000 - x)
The sum of these interests is $560, so the equation is:
0.06x + 0.05(10,000 - x) = 560
Now we solve for x:
0.06x + 500 - 0.05x = 560
0.01x = 60
x = 60 / 0.01
x = $6,000
Therefore, Susan invested $6,000 at 6% and the remainder, $4,000, at 5% interest.
The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a population mean of 60,000 miles and a population standard deviation of 4,000 miles. Suppose we select a sample of 40 tires and use a simulator to determine the tread life. What is the likelihood of finding that the sample mean is between 59,050 and 60,950?
Answer: 0.8664
Step-by-step explanation:
Given : Mean : [tex]\mu = 60,000\text{ miles}[/tex]
Standard deviation : [tex]\sigma = 4,000\text{ miles}[/tex]
Sample size : [tex]n=40[/tex]
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 59,050
[tex]z=\dfrac{59050-60000}{\dfrac{4000}{\sqrt{40}}}\approx-1.50[/tex]
For x= 60,950
[tex]z=\dfrac{60950-60000}{\dfrac{4000}{\sqrt{40}}}\approx1.50[/tex]
The P-value : [tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)[/tex]
[tex]=0.9331927-0.0668072=0.8663855\approx0.8664[/tex]
Hence, the likelihood of finding that the sample mean is between 59,050 and 60,950=0.8664
The likelihood of finding that the sample mean is between 59,050 and 60,950 miles, according to the given normal distribution, is approximately 86.64%.
Explanation:To solve this problem, we consider that the population mean is 60,000 and the standard deviation is 4,000. If we choose a sample of 40 tires, the standard deviation of the sample mean (standard error) is the standard deviation divided by the square root of the sample size (σ/√n).
This gives us 4,000/√40 = 633. The z-scores for the lower and upper bounds of our interval (59,050 and 60,950) are calculated by subtracting the population mean from these values, and dividing by the standard error. For 59,050: (59,050 - 60,000)/633 = -1.5 and for 60,950: (60,950 - 60,000)/633 = 1.5.
Using standard normal distribution tables, we know that the probability associated with a z-value of 1.5 is 0.9332. Since the normal distribution is symmetric, the probability associated with -1.5 is also 0.9332. Therefore, the probability that the sample mean lies between 59,050 and 60,950 is 0.9332 - (1 - 0.9332) = 0.8664 or approximately 86.64%.
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Question Help For the month of MarchMarch in a certain city, 5757% of the days are cloudycloudy. Also in the month of MarchMarch in the same city, 5555% of the days are cloudycloudy and foggyfoggy. What is the probability that a randomly selected day in MarchMarch will be foggyfoggy if it is cloudycloudy?
Answer: There is probability of 96.4% that a day in March will be foggy if it is a cloudy.
Step-by-step explanation:
Since we have given that
Probability of the days in March are cloudy = 57%
Probability of the cloudy days in March are foggy = 55%
Let A be the event of cloudy days in March.
Let B be the event of foggy days in March.
So, here,
P(A) = 0.57
P(A∩B) = 0.55
We need to find the probability that days are foggy given that it is cloudy.
We would use "Conditional probability":
[tex]P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{0.55}{0.57}=0.964=96.4\%[/tex]
Hence, There is probability of 96.4% that a day in March will be foggy if it is a cloudy.