Step-by-step explanation:
[tex]6.\\\\\dfrac{\tan x}{\cot x}\qquad\text{use}\ \cot x=\dfrac{1}{\tan x}\\\\=\dfrac{\tan x}{\frac{1}{\tan x}}=\tan x\cdot\dfrac{\tan x}{1}=\tan x\cdot\tan x=\left(\tan x\right)^2\qquad\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\=\left(\dfrac{\sin x}{\cos x}\right)^2\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\=\dfrac{\sin^2x}{\cos^2x}\qquad\text{use}\ \sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\=\dfrac{\sin^2x}{1-\sin^2x}[/tex]
[tex]7.\\\\\cos x\cdot\cot x+\sin x\qquad\text{use}\ \cot x=\dfrac{\cos x}{\sin x}\\\\=\cos x\cdot\dfrac{\cos x}{\sin x}+\sin x=\dfrac{\cos^2x}{\sin x}+\dfrac{\sin^2x}{\sin x}=\dfrac{\cos^2x+\sin^2x}{\sin x}=\dfrac{1}{\sin x}[/tex]
A population has a mean mu μ equals = 87 and a standard deviation σ = 24. Find the mean and standard deviation of a sampling distribution of sample means with sample size n equals = 36 mu μx equals = nothing (Simplify your answer.) sigma Subscript x overbar σx equals = nothing (Simplify your answer.
Answer:
The mean of a sampling distribution of sample means is 87
The standard deviation of a sampling distribution of sample = 4
Step-by-step explanation:
* Lets revise some definition to solve the problem
- The mean of the distribution of sample means is called μx
- It is equal to the population mean μ
- The standard deviation of the distribution of sample means is
called σx
- The rule of σx = σ/√n , where σ is the standard deviation and n
is the size of the sample
* lets solve the problem
- A population has a mean (μ) is 87
∴ μ = 87
- A standard deviation of 24
∴ σ = 24
- A sampling distribution of sample means with sample size n = 36
∴ n = 36
∵ The mean of the distribution of sample means μx = μ
∵ μ = 87
∴ μx = 87
* The mean of a sampling distribution of sample means is 87
∵ The standard deviation of a sampling distribution of sample
means σx = σ/√n
∵ σ = 24 and n = 36
∴ σx = 24/√36 = 24/6 = 4
* The standard deviation of a sampling distribution of sample = 4
The mean of the sampling distribution of sample means is 87, identical to the population mean. The standard deviation of this sampling distribution is 4, which is the population standard deviation (24) divided by the square root of the sample size (36).
Explanation:When working with populations and sampling distributions, the Central Limit Theorem is critical for understanding the behavior of sample means. Given a population with mean (μ) and standard deviation (σ), the mean of the sampling distribution of sample means will be the same as the population mean, and the standard deviation of the sampling distribution (σx) is equal to the population standard deviation divided by the square root of the sample size (n).
In this instance, the population has a mean (μ) of 87 and a standard deviation (σ) of 24. For a sample size (n) of 36, the mean of the sampling distribution of sample means (μx) is equal to the population mean:
μx = μ = 87
The standard deviation of the sampling distribution of sample means (σx) is calculated as follows:
σx = σ / √ n = 24 / √ 36 = 24 / 6 = 4
Therefore, the mean of the sampling distribution is 87, and standard deviation is 4.
The rectangular coordinates of a point are (5.00, y) and the polar coordinates of this point are ( r, 67.4°). What is the value of the polar coordinate r in this case?
Answer:
r ≈ 13.01
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you that ...
Cos = Adjacent/Hypotenuse
cos(67.4°) = 5.00/r . . . . . . filling in the given values
Solving for r gives ...
r = 5.00/cos(67.4°) ≈ 13.01
_____
Check your requirements for rounding. We rounded to 2 decimal places because the x-coordinate, 5.00, was expressed using 2 decimal places.
Determine if the frame can support a load of P = 20 kN if the factor of safety with respect to buckling of member AB is F.S. = 3. Assume that AB is made of steel and is pinned at its ends for x-x axis buckling and fixed at its ends for y-y axis buckling. ????????????????????????=200????????????????????????, ????????????????=360???????????? ???? .
Final answer:
To assess if the frame can support a 20 kN load given the factor of safety for buckling of member AB is 3, the critical buckling loads for both axes, considering the end conditions, should be calculated and compared after adjusting for the safety factor.
Explanation:
To determine if the frame can support a load of P = 20 kN with a factor of safety with respect to buckling of member AB being F.S. = 3, we need to calculate the critical buckling load for member AB. The member's critical buckling load depends on its end conditions, which are pinned for the x-x axis and fixed for the y-y axis buckling. The Euler Buckling Formula, which is Pcr = (π2EI) / (KL2), where E is the modulus of elasticity, I is the moment of inertia, K is the column effective length factor, and L is the actual length of the column, can be applied separately for each axis. For pin-ended columns, K=1, and for fixed-ended columns, K=0.5.
To ensure safety, the critical load calculated for both axes should be divided by the factor of safety, F.S. = 3. If the adjusted critical load for either axis is greater than the applied load (P = 20 kN), the frame can support the load. However, specific values for E, I, and L are required to perform these calculations, which are not provided in the question. Generally, for steel, E=200 GPa is a typical value, and the values for I and L need to be determined based on the cross-sectional and length dimensions of member AB. Without these specifics, an exact answer cannot be provided.
A rectangular pyramid has a volume of 90 cubic feet. What is the volume of a rectangular prism with the same size base and same height?
Answer:
30 cubic feet
Step-by-step explanation:
Here we are given the volume of rectangular pyramid as 90 cubic feet as we are required to find the volume of rectangular prism.
For that we need to use the theorem which says that
the volume prism is always one third of the volume of the pyramid . Whether it is rectangular of triangular base. Hence in this case also the volume of the rectangular prism will be one third of the volume of the rectangular pyramid.
Volume of Rectangular prism = [tex]\frac{1}{3}[/tex] * Volume of rectangular pyramid
= [tex]\frac{1}{3}[/tex] * 90
= 30
Answer:
270 is the answer just took test.
Step-by-step explanation:
Which of the following is most likely the next step in the series?
Answer:
D
Step-by-step explanation:
It's a repeating pattern
Answer:
D
Reasoning:
the pattern keeps the previous color and alternates between blue and red. The pattern is also increasing in number by 1. Therefore there would be no reason for the fourth step to change previous color or numbers. That eliminates A and B. For C, since the pattern is adding a different color clockwise in each step, it is more likely that the pattern will continue to alternate, therefore D is most likely.
Factor the higher degree polynomial
5y^4 +11y^2 +2
Answer:
[tex]\large\boxed{5y^4+11y^2+2=(y^2+2)(5y^2+1)}[/tex]
Step-by-step explanation:
[tex]5y^4+11y^2+2=5y^4+10y^2+y^2+2=5y^2(y^2+2)+1(y^2+2)\\\\=(y^2+2)(5y^2+1)[/tex]
a. The following events are mutually exclusive: Living in California and watching American Idol. True or False b. The number of patients seen by an outpatient practice is an example of a discrete random variable. True or False
c.The law of large numbers states that as the number of times an event experiment is conducted increases, the likelihood of the actual probability of an event approaching the theoretical probability decreases. True or False
d. Measuring the time it takes for patients to enter the operating room is an example of a continuous random variable. True or False
Answer:
a) False. Because you can live in California AND watch American Idol at the same time
b) True. Because the number of patients is a whole number, like 1, 2 or 3. There is no 1.5 patient
c) False. The actual probability should become closer to the theoretical probablity
d) True
Factor the Higher degree polynomial
5y^4 + 11y^2 + 2
[tex]\bf 5y^4+11y^2+2\implies 5(y^2)^2+11y^2+2\implies (5y^2+1)(y^2+2)[/tex]
For this case we must factor the following polynomial:
[tex]5y ^ 4 + 11y ^ 2 + 2[/tex]
We rewrite [tex]y ^ 4[/tex]as [tex](y^ 2) ^ 2[/tex]:
[tex]5 (y ^ 2) ^ 2 + 11y ^ 2 + 2[/tex]
We make a change of variable:
[tex]u = y ^ 2[/tex]
We replace:
[tex]5u ^ + 11u + 2[/tex]
we rewrite the middle term as a sum of two terms whose product of 5 * 2 = 10 and the sum of 11.
So:
[tex]5u ^ 2 + (1 + 10) u + 2[/tex]
We apply distributive property:
[tex]5u ^ 2 + u + 10u + 2[/tex]
We factor the highest common denominator of each group.
[tex](5u ^ 2 + u) + (10u + 2)\\u (5u + 1) +2 (5u + 1)[/tex]
We factor again:
[tex](u + 2) (5u + 1)[/tex]
Returning the change:
[tex](y ^ 2 + 2) (5y ^ 2 + 1)[/tex]
ANswer:
[tex](y ^ 2 + 2) (5y ^ 2 + 1)[/tex]
c. Using a standard deck of 52 cards, the probability of selecting a 4 of diamonds or a 4 of hearts is an example of a mutually exclusive event. True of False
Answer:
True
Step-by-step explanation:
If two events X and Y are mutually exclusive,
Then,
P(X∪Y) = P(X) + P(Y)
Let A represents the event of a diamond card and B represent the event of a heart card,
We know that,
In a deck of 52 cards there are 4 suit ( 13 Club cards, 13 heart cards, 13 diamond cards and 13 Spade cards )
That is, those cards which are heart can not be diamond card,
⇒ A ∩ B = ∅
⇒ P(A∩B) = 0
Since, P(A∪B) = P(A) + P(B) - P(A∩B)
⇒ P(A∪B) = P(A) + P(B)
By the above statement,
Events A and B are mutually exclusive,
Hence, the probability of selecting a 4 of diamonds or a 4 of hearts is an example of a mutually exclusive event is a true statement.
(a) (8%) Compute the probability of an even integer among the 100 integers 1!, 2!, 3!, .., until 100! (here n! is n factorial or n*(n-1)*(n-2) *… 1) (b) (16%) Compute the probability of an even integer among the 100 integers: 1, 1+2, 1+2+3, 1+2+3+4, …., 1+2+3+… + 99, and 1+2+3+… + 100
Answer:
(a) 99%
(b) 50%
Step-by-step explanation:
(a) All factorials after 1! have 2 as a factor, so are even. Thus 99 of the 100 factorials are even, for a probability of 99%.
__
(b) The first two sums are odd; the next two sums are even. The pattern repeats every four sums. There are 25 repeats of that pattern in 100 sums, so 2/4 = 50% of sums are even.
To the nearest degree, what is the measure of the central angle for faucets? 37° 24° 48° 43°
Answer:
[tex]\large\boxed{43^o}[/tex]
Step-by-step explanation:
[tex]\text{Faucets}\to12\%\\\\p\%=\dfrac{p}{100}\to12\%=\dfrac{12}{100}=0.12\\\\12\%\ \text{of}\ 360^o\to0.12\cdot360^o=43.2^o\approx43^o[/tex]
Answer: [tex]43^{\circ}[/tex]
Step-by-step explanation:
From the given pie-chart, the percentage for faucets = 12 %
We know that every circle has angle of [tex]360^{\circ}[/tex].
Now, the central angle for faucets is given by :-
[tex]\text{Central angle}=\dfrac{\text{Percent of Faucets}}{\text{100}}\times360^{\circ}\\\\\Rightarrow\ \text{Central angle}=\dfrac{12}{100}\times360^{\circ}=43.2^{\circ}\approx43^{\circ}[/tex]
Hence, the measure of the central angle for faucets [tex]\approx43^{\circ}[/tex]
b7
__
b6
Multiply or divide as indicated.
For this case we have the following expression:
[tex]\frac {b ^ 7} {b ^ 6}[/tex]
By definition of division of powers of the same base, we have to place the same base and subtract the exponents, that is:
[tex]\frac {a ^ m} {a ^ n} = a ^ {n-m}[/tex]
So:
[tex]\frac {b ^ 7} {b ^ 6} = b ^ {7-6} = b ^ 1 = b[/tex]
Answer:
b
Answer: [tex]b[/tex]
Step-by-step explanation:
You need to remember a property called "Quotient of powers property". This property states the following:
[tex]\frac{a^m}{a^n}=a^{(m-n)}[/tex]
You can observe that the bases of the expression [tex]\frac{b^7}{b^6}[/tex] are equal, then you can apply the property mentioned before.
Therefore, you can make the division indicated in the exercise.
Then you get this result:
[tex]\frac{b^7}{b^6}=b^{(7-6)}=b[/tex]
In the example we modeled the world population in the second half of the 20th century by the equation P(t) = 2560e^(0.017185t). Use this equation to estimate the average world population during the time period of 1950 to 2000. (Round your answer to the nearest million.)
Answer: There are 6045 millions world population during the period of 1950 to 2000.
Step-by-step explanation:
Since we have given that
The world population in the second half of the 20 the century by the equation:
[tex]P(t)=2560e^{0.017185t}[/tex]
We need to find the average world population during the period of 1950 to 2000.
So, there are 50 years between 1950 to 2000.
So, t = 50 years.
Therefore, the average population would be
[tex]P(50)=2560e^{0.017185\times 50}\\\\P(50)=6045.15\\\\P(50)\approx 6045\ millions[/tex]
Hence, there are 6045 millions world population during the period of 1950 to 2000.
Consider kite WXYZ. What are the values of a and b? a = 4; b = 10. a = 4; b = 40, a = 8; b = 10, a = 8; b = 40
Check the picture below.
let's recall that a kite is a quadrilateral, and thus is a polygon with 4 sides
sum of all interior angles in a polygon
180(n - 2) n = number of sides
so for a quadrilateral that'd be 180( 4 - 2 ) = 360, thus
[tex]\bf 3b+70+50+3b=360\implies 6b+120=360\implies 6b=240 \\\\\\ b=\cfrac{240}{6}\implies b=40 \\\\[-0.35em] ~\dotfill\\\\ \overline{XY}=\overline{YZ}\implies 3a-5=a+11\implies 2a-5=11 \\\\\\ 2a=16\implies a=\cfrac{16}{2}\implies a=8[/tex]
Answer:
D.)a = 8; b = 40
Step-by-step explanation:
it is on edgeinuity
Biologists estimate that the number of animal species of a certain body length is inversely proportional to the square of the body length.1 Write a formula for the number of animal species, N, of a certain body length as a function of the length, L. Use k as the constant of proportionality.
Answer:
[tex]N(L)=\frac{k}{L^2}[/tex]
Step-by-step explanation:
Here, N represents the number of animal species and L represents a certain body length,
According to the question,
[tex]N\propto \frac{1}{L^2}[/tex]
[tex]\implies N=\frac{k}{L^2}[/tex]
Where, k is the constant of proportionality,
Since, with increasing the value of L the value of N is decreasing,
So, we can say that, N is dependent on L, or we can write N(L) in the place of N,
Hence, the required function formula is,
[tex]N(L)=\frac{k}{L^2}[/tex]
Certainly! When we say that the number of animal species \( N \) is inversely proportional to the square of the body length \( L \), what we mean mathematically is that as the body length increases, the number of species decreases at a rate that is the square of the increase in length. This can be represented by the following formula:
\[ N = \frac{k}{L^2} \]
Here \( N \) is the number of species, \( L \) is the body length, and \( k \) is the constant of proportionality. This constant \( k \) represents the number of species at the unit body length (when \( L = 1 \)). The constant of proportionality is determined by the specific biological context, based on empirical data or theoretical considerations.
In this formula, \( L^2 \) denotes the body length squared, and the fraction represents the inverse relationship.
In summary, to find the number of species \( N \) for a given body length \( L \), we use the inverse square relationship with the constant of proportionality \( k \).
Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval. (Round your answer to four decimal places. Enter your answers as a comma-separated list.) f(x) = 8 x , [4, 9]
Consider two algorithms that perform the same function, that run in n/4 and log2(n), respectively, where n ∈ N (i.e. natural numbers).(a) Plot these runtimes on the same graph with the values n ∈ [1, 50] (don’t forget labels). Provide the set of intervals over N, where n/4 is the strictly better algorithm to use (think greater than, not greater than or equal).
Answer:
[2, 15]
Step-by-step explanation:
The graph shows the n/4 algorithm to be better (smaller run time) for n in the range 2 to 15.
evaluate the function at the fiven values of the variables:
f(x)= 5x^2 + 5x *+ 3
a f (-3)
b f (-9)
Answer:
f(-3)=5x^2+5x*3
-f*3=5x^2+5*3x
-3f=5x^2+15x
f=-5x(x+3)/3
f(-9)=5x^2+5x*3
-f*9=5x^2+5*3x
-9f=5x^2+15x
f=-5x(x+3)/9
Step-by-step explanation:
hope it helps you?
s f(-x)= x^2 -1 odd, even or neither
Answer:
f(x) = f(-x) = x^2 -1 is an even function
Step-by-step explanation:
When f(x) = f(-x), the function is symmetrical about the y-axis. That is the definition of an even function.
___
An odd function is symmetrical about the origin: f(x) = -f(-x).
Solve |x| > 5
{-5, 5}
{x|-5 < x < 5}
{x|x < -5 or x > 5}
Answer: Last option.
Step-by-step explanation:
Given the inequality [tex]|x| > 5[/tex] you need to set up two posibilities. These posibilities are the following:
- FIRST POSIBILITY :
[tex]x>5[/tex]
- SECOND POSIBILTY:
[tex]x<-5[/tex]
Then you get that:
[tex]x<-5\ or\ x>5 [/tex]
Therefore, through this procedure, you get that the solution set of the inequality [tex]|x| > 5[/tex] is this:
{[tex]x|x<-5\ or\ x>5[/tex]}
Answer:
C. {x|x < -5 or x > 5}
Step-by-step explanation:
Suppose r(t) = cos t i + sin t j + 3tk represents the position of a particle on a helix, where z is the height of the particle above the ground. (a) Is the particle ever moving downward? When? (If the particle is never moving downward, enter DNE.) t = (b) When does the particle reach a point 15 units above the ground? t = (c) What is the velocity of the particle when it is 15 units above the ground? (Round each component to three decimal places.) v = (d) When it is 15 units above the ground, the particle leaves the helix and moves along the tangent line. Find parametric equations for this tangent line. (Round each component to three decimal places.)
The particle has position function
[tex]\vec r(t)=\cos t\,\vec\imath+\sin t\,\vec\jmath+3t\,\vec k[/tex]
Taking the derivative gives its velocity at time [tex]t[/tex]:
[tex]\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k[/tex]
a. The particle never moves downward because its velocity in the [tex]z[/tex] direction is always positive, meaning it is always moving away from the origin in the upward direction. DNE
b. The particle is situated 15 units above the ground when the [tex]z[/tex] component of its posiiton is equal to 15:
[tex]3t=15\implies\boxed{t=5}[/tex]
c. At this time, its velocity is
[tex]\vec v(5)=-\sin 5\,\vec\imath+\cos5\,\vec\jmath+3\,\vec k\approx\boxed{0.959\,\vec\imath+0.284\,\vec\jmath+3\,\vec k}[/tex]
d. The tangent to [tex]\vec r(t)[/tex] at [tex]t=5[/tex] points in the same direction as [tex]\vec v(5)[/tex], so that the parametric equation for this new path is
[tex]\vec r(5)+\vec v(5)t\approx\boxed{(0.284+0.959t)\,\vec\imath+(-0.959+0.284t)\,\vec\jmath+(15+3t)\,\vec k}[/tex]
where [tex]0\le t<\infty[/tex].
Upper A 4-ft-tall fence runs parallel to a wall of a house at a distance of 24 ft. (a) Find the length of the shortest ladder that extends from the ground to the house without touching the fence. Assume the vertical wall of the house and the horizontal ground have infinite extent. (a) Let L be the length of the ladder, x be the distance from the base of the ladder to the fence, d be the distance from the fence to the house and b be the distance from the ground to the point the ladder touches the house. What is the objective function, in terms of x?
Answer:
(a) shortest ladder length ≈ 35.7 ft (rounded to tenth)
(b) L = (d/x +1)√(16+x²) . . . . where 16 is fence height squared
Step-by-step explanation:
It works well to solve the second part of the problem first, then put in the specific numbers.
We have not been asked anything about "b", so we can basically ignore it. Using the Pythagorean theorem, we find the length GH in the attached drawing to be ...
GH = √(4²+x²) = √(16+x²)
Then using similar triangles, we can find the ladder length L to be that which satisfies ...
L/(d+x) = GH/x
L = (d +x)/x·√(16 +x²)
The derivative with respect to x, L', is ...
L' = (d+x)/√(16+x²) +√(16+x²)/x - (d+x)√(16+x²)/x²
Simplifying gives ...
L' = (x³ -16d)/(x²√(16+x²))
Our objective is to minimize L by making L' zero. (Of course, only the numerator needs to be considered.)
___
(a) For d=24, we want ...
0 = x³ -24·16
x = 4·cuberoot(6) ≈ 7.268 . . . . . feet
Then L is
L = (24 +7.268)/7.268·√(16 +7.268²) ≈ 35.691 . . . feet
__
(b) The objective function is the length of the ladder, L. We want to minimize it.
L = (d/x +1)√(16+x²)
The objective function for the length of the ladder in terms of x is L = sqrt((x+b)^2 + d^2).
Explanation:To find the length of the shortest ladder that extends from the ground to the house without touching the fence, we can use the concepts of similar triangles. Let L be the length of the ladder, x be the distance from the base of the ladder to the fence, d be the distance from the fence to the house, and b be the distance from the ground to the point the ladder touches the house. In terms of x, the objective function for the length of the ladder is:
L = sqrt((x+b)^2 + d^2)
Learn more about Objective function for ladder length here:https://brainly.com/question/35413844
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Suppose you are to throw a dart at a circular dart board with radius 2 inches. Let (X, Y ) denote the point that you hit on the board (you can assume the board is centered at the origin (0, 0), and that the dart hits somewhere on the board uniformly at random). (a) Find the joint PDF of X and Y . (b) Find the marginal PDFs of X and Y . (c) Find the conditional PDFs fX|Y and fY |X. (d) Suppose that the “bulls eye” on the target consists of a small circle centered at the origin with radius 0.25. Explain how you would use one of these PDFs to compute the probability of a dart hitting the bulls eye. Find this probability an even easier way. (e) What is the probability that X > 1? (f) If you know that X = 1, what is the PDF of Y conditioned on this fact? What’s the probability that Y > 0.3 conditioned on this fact?
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.
It is estimated that only 68% of drivers wear their safety belt. How many people should a police officer expect pull over until she finds a driver not wearing a safety belt
A.1
B 2
C 4
D 3
E 5
Answer:
D. 4
Step-by-step explanation:
Percent of drivers who wear seat belts = 68
Percent of drivers who do not wear seat belts = 100 - 68 = 32
Now, we know that for every 100 pull overs, 32 drivers will not be wearing belt.
32 drivers without seat-belt = 100 pullovers
1 driver without seat-belt = 100/32 pullovers
1 driver without seat-belt = 3.125 pullovers (4 pullovers)
So, a police officer should expect 4 pullovers until she finds a driver not wearing a seat-belt.
The police officer should expect to pull over approximately 4 drivers until finding one not wearing a safety belt.
To determine how many people a police officer should expect to pull over until she finds a driver not wearing a safety belt, given that only 68% of drivers wear their safety belt, follow these steps:
Step 1:
Calculate the probability of a driver not wearing a safety belt.
Since 68% of drivers wear their safety belt, the probability of a driver not wearing a safety belt is [tex]\( 1 - 0.68 = 0.32 \).[/tex]
Step 2:
Calculate the expected number of people to pull over.
The expected number of people to pull over is the reciprocal of the probability of a driver not wearing a safety belt. So, it's [tex]\( \frac{1}{0.32} \).[/tex]
Step 3:
Perform the calculation.
[tex]\[ \frac{1}{0.32} \approx 3.125 \][/tex]
Step 4:
Interpret the result.
Since we can't pull over a fraction of a person, rounding up to the nearest whole number, the police officer should expect to pull over 4 drivers until finding one not wearing a safety belt.
So, the correct answer is option C: 4.
f p(x) and q(x) are arbitrary polynomials of degreeat most 2, then the mapping< p,q >= p(-2)q(-2)+ p(0)q(0)+ p(2)q(2)defines an inner product in P3.Use this inner product to find < p,q >, llpll, llqll, and the angletetha, between p(x) and q(x) forp(x) = 2x^2+6x+1 and q(x) = 3x^2-5x-6.< p;q >= ?llpll = ?llqll= ?
We're given an inner product defined by
[tex]\langle p,q\rangle=p(-2)q(-2)+p(0)q(0)+p(2)q(2)[/tex]
That is, we multiply the values of [tex]p(x)[/tex] and [tex]q(x)[/tex] at [tex]x=-2,0,2[/tex] and add those products together.
[tex]p(x)=2x^2+6x+1[/tex]
[tex]q(x)=3x^2-5x-6[/tex]
The inner product is
[tex]\langle p,q\rangle=-3\cdot16+1\cdot(-6)+21\cdot(-4)=-138[/tex]
To find the norms [tex]\|p\|[/tex] and [tex]\|q\|[/tex], recall that the dot product of a vector with itself is equal to the square of that vector's norm:
[tex]\langle p,p\rangle=\|p\|^2[/tex]
So we have
[tex]\|p\|=\sqrt{\langle p,p\rangle}=\sqrt{(-3)^2+1^2+21^2}=\sqrt{451}[/tex]
[tex]\|q\|=\sqrt{\langle q,q\rangle}=\sqrt{16^2+(-6)^2+(-4)^2}=2\sqrt{77}[/tex]
Finally, the angle [tex]\theta[/tex] between [tex]p[/tex] and [tex]q[/tex] can be found using the relation
[tex]\langle p,q\rangle=\|p\|\|q\|\cos\theta[/tex]
[tex]\implies\cos\theta=\dfrac{-138}{22\sqrt{287}}\implies\theta\approx1.95\,\mathrm{rad}\approx111.73^\circ[/tex]
The question is about calculating an inner product, norms, and the angle between two polynomials in a two-dimensional space, represented by their coefficients. We use the provided polynomials and input values from the question to calculate these values.
Explanation:The mathematics problem given refers to calculating the inner product and magnitude of two polynomials, as well as the angle between them, if they are represented in a two-dimensional space.
To find the values asked in this problem, we will use the given polynomials p(x) = 2x^2+6x+1 and q(x) = 3x^2-5x-6. The inner product < p,q > is calculated as follows: p(-2)q(-2) + p(0)q(0) + p(2)q(2).
To find the magnitudes, or norms, ||p|| and ||q||, we need to solve the expression for < p,p > and < q,q > respectively, and then take the square root of the result. The angle between the vectors, denoted as Theta (θ), can be computed using the formula cos θ = < p,q > / (||p||*||q||). Because cos θ is the cosine of the angle, θ = arccos(< p,q > / (||p||*||q||)).
Note that the exact value of the inner product, norms, and angle depends on the input values of -2, 0, and 2 for each polynomial.
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True or False? For any integer m, 6m(2m + 10) is divisible by 4. Explain
Answer:
True
Step-by-step explanation:
6m (2m + 10)
= (6m)(2m) + (6m)(10)
= 12m² + 60m
= 4(3m² + 15m)
Because we factored out a "4" in the expression, by observation, we can see that regardless of what value of integer m is chosen, the entire expression is divisible by the "4" which has been factored out of the parentheses. Hence it is True.
What is the divisibility test of 4?
If a number's last two digits are divisible by 4, the number is a multiple of 4 and totally divisible by 4.
Solving the problem.6m(2m+10) = 6m cannot be said to be always divisible by 4 but is divisible by 2 as 6m = 2*3m
(2m+10) cannot say if it is divisible by but can say is divisible by 2 as we can take 2 commons from both the integer and rewrite it as 2*(m+5).
Hence multiplying both the numbers we get = 2*3m*2(m+5) and now we group 2's together we get 4*3m*(m+5), as the whole number is multiplied by 4 it is at least divisible by 4 once.
Hence proved that 6m(2m+10) is divisible by 4 and the answer is true.
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Use the power rules for exponents to simplify the expression. (Type as a fraction, use exponential form)
Answer: [tex]\bold{\dfrac{h^9}{g^9}}[/tex]
Step-by-step explanation:
The power rule is "multiply the exponents".
You must understand that the exponent of both h and g is 1.
Multiply 1 times 9 for both variables.
[tex]\bigg(\dfrac{h}{g}\bigg)^9=\bigg(\dfrac{h^1}{g^1}\bigg)^9=\dfrac{h^{1\times 9}}{g^{1\times 9}}=\large\boxed{\dfrac{h^9}{g^9}}[/tex]
Trucks in a delivery fleet travel a mean of 80 miles per day with a standard deviation of 30 miles per day. The mileage per day is distributed normally. Find the probability that a truck drives between 97 and 107 miles in a day. (Round your answer to 4 decimal places)
Answer: 0.1037
Step-by-step explanation:
Given : Mean : [tex]\mu=80\text{ miles per day}[/tex]
Standard deviation : [tex]\sigma = 30\text{ miles per day}[/tex]
The formula to calculate the z-score is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = 97 miles per day ,
[tex]z=\dfrac{97-80}{30}\approx0.57[/tex]
For x = 107 miles per day ,
[tex]z=\dfrac{97-80}{30}=0.9[/tex]
The P-value =[tex]P(0.57<z<0.9)=P(z<0.9)-P(z<0.57)[/tex]
[tex]=0.8159398-0.7122603=0.1036795\approx0.1037[/tex]
Hence, the probability that a truck drives between 97 and 107 miles in a day = 0.1037
The probability that a truck drives between 97 and 107 miles in a day is approximately 0.101. We calculated this using the Z-scores for 97 and 107 miles, the standard normal distribution, and the properties of the normal distribution.
Explanation:To find the probability that a truck drives between 97 and 107 miles in a day, we'll use the normal distribution. This probability is the equivalent of finding the area under the curve of the normal distribution between 97 and 107 miles.
First, we calculate the Z-scores for 97 and 107, where Z = (X - μ)/σ. Here, μ is the mean (80 miles), and σ is the standard deviation (30 miles).
Z1 = (97 - 80)/30 ~ 0.567Z2 = (107 - 80)/30 ~ 0.9Now, we look up these Z-scores in the standard normal distribution table or use a calculator with this function. Suppose the table gives us P(Z < 0.567) = 0.715 and P(Z < 0.9) = 0.816.
Then, the probability that a truck drives between 97 to 107 miles is P(0.567 < Z < 0.9) = P(Z < 0.9) - P(Z < 0.567) = 0.816 - 0.715 = 0.101.
Please note that results can vary slightly depending on the Z-table or calculator you use. All of the numbers after the decimal point are rounded to 4 decimal places.
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a. The sample space consists of the results of a single probability experiment. True or False
Answer:
True
Step-by-step explanation:
The sample space consists of any possible outcome
Pretend you're playing a carnival game and you've won the lottery, sort of. You have the opportunity to select five bills from a money bag, while blindfolded. The bill values are $1, $2, $5, $10, $20, $50, and $100. How many different possible ways can you choose the five bills? (Order doesn't matter, and there are at least five of each type of bill.)
The total number of ways are:
462
Step-by-step explanation:When we are asked to select r items from a set of n items that the rule that is used to solve the problem is:
Method of combination.
Here the total number of bills of different values are: 7
i.e. n=7
( $1, $2, $5, $10, $20, $50, and $100 )
and there are atleast five of each type of bill.
Also, we have to choose 5 bills i.e. r=5
The repetition is allowed while choosing bills.
Hence, the formula is given by:
[tex]C(n+r-1,r)[/tex]
Hence, we get:
[tex]C(7+5-1,5)\\\\i.e.\\\\C(11,5)=\dfrac{11!}{5!\times (11-5)!}\\\\C(11,5)=\dfrac{11!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7\times 6!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2}\\\\\\C(11,5)=462[/tex]
Hence, the answer is:
462
There are 462 different possible ways to choose five bills from a set of seven types with repetition allowed.
To determine the number of different ways to choose five bills from a set of bills with values $1, $2, $5, $10, $20, $50, and $100, we can use the combinatorial concept known as "combinations with repetition."
Given:
- Bill values: [tex]\( \{1, 2, 5, 10, 20, 50, 100\} \)[/tex]
- Total number of types of bills: ( n = 7 )
- Number of bills to choose: ( r = 5 )
The formula for combinations with repetition (also known as "stars and bars") is:
\[
\binom{n + r - 1}{r}
\]
Substituting the values ( n = 7 ) and ( r = 5 ):
[tex]\[\binom{7 + 5 - 1}{5} = \binom{11}{5}\][/tex]
Now, calculate [tex]\( \binom{11}{5} \):[/tex]
[tex]\[\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!}\][/tex]
First, compute the factorials:
[tex]\[11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]\[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\][/tex]
Next, compute [tex]\( \frac{11!}{6!} \):[/tex]
[tex]\[\frac{11!}{6!} = 11 \times 10 \times 9 \times 8 \times 7\]\[= 11 \times 10 = 110\]\[110 \times 9 = 990\]\[990 \times 8 = 7920\]\[7920 \times 7 = 55440\][/tex]
Now, compute:
[tex]\[\frac{55440}{120} = 462\][/tex]
Therefore, the number of different possible ways to choose the five bills is:
[tex]\[\boxed{462}\][/tex]