(a) The region Y₁ < 1/2 and Y₂ > 1/4 corresponds to the rectangle,
{(y₁, y₂) : 0 ≤ y₁ < 1/2 and 1/4 < y₂ ≤ 1}
Integrate the joint density over this region:
[tex]P\left(Y_1<\dfrac12,Y_2>\dfrac14\right) = \displaystyle\int_0^{\frac12}\int_{\frac14}^1 (y_1+y_2)\,\mathrm dy_2\,\mathrm dy_1 = \boxed{\dfrac{21}{64}}[/tex]
(b) The line Y₁ + Y₂ = 1 cuts the support in half into a triangular region,
{(y₁, y₂) : 0 ≤ y₁ < 1 and 0 < y₂ ≤ 1 - y₁}
Integrate to get the probability:
[tex]P(Y_1+Y_2\le1) = \displaystyle\int_0^1\int_0^{1-y_1}(y_1+y_2)\,\mathrm dy_2\,\mathrm dy_1 = \boxed{\dfrac13}[/tex]
Y₁ and Y₂ are not independent because
P(Y₁ = y₁, Y₂ = y₂) ≠ P(Y₁ = y₁) P(Y₂ = y₂)
To see this, compute the marginal densities of Y₁ and Y₂.
[tex]P(Y_1=y_1) = \displaystyle\int_0^1 f(y_1,y_2)\,\mathrm dy_2 = \begin{cases}\frac{2y_1+1}2&\text{if }0\le y_1\le1\\0&\text{otherwise}\end{cases}[/tex]
[tex]P(Y_2=y_2) = \displaystyle\int_0^1 f(y_1,y_2)\,\mathrm dy_1 = \begin{cases}\frac{2y_2+1}2&\text{if }0\le y_2\le1\\0&\text{otherwise}\end{cases}[/tex]
[tex]\implies P(Y_1=y_1)P(Y_2=y_2) = \begin{cases}\frac{(2y_1+1)(2y_2_1)}4&\text{if }0\le y_1\le1,0\ley_2\le1\\0&\text{otherwise}\end{cases}[/tex]
but this clearly does not match the joint density.
PLEASE HELPPP ASAP!!! I tried all sorts of equations but no correct answer! Not sure how to approach this problem.
Answer:
[tex]44[/tex]
Step-by-step explanation:
The dimensions of the garden is 12 by 8. If we have a walkway that surrounds the garden, the dimensions of the walkway is 2. Since it surrounds the rectangle all sides add 2 to each of the dimensions so now the dimensions of the garden and walkway is 14×10.
The area of the garden is 96 square ft.
The area of the garden and walkway is 140 so let subtract the area of the garden from the total area of both the garden and walkway.
[tex]140 - 96 = 44[/tex]
The area is 44.
Answer:
120 square feet
Step-by-step explanation:
(8+2*2)
(18+2*2) - 8*18 = 120 square feet.
1=5
2=12
3=39
4=148
5=?
What is the volume of a cone with a height of 27 cm
and a radius of 13 cm? Round your answer to the
nearest tenth.
Use the button on your calculator to complete this
problem.
V=
cm3
Answer:4778.3 cm^3
Step-by-step explanation: The formula for volume of a cone is V=1/3h pi r^2. By plugging in the height and the radius we get our answer.
Answer:
4778.4 :)
Step-by-step explanation:
please help me solve this math
Answer:
d
Step-by-step explanation:
I NEED HELP FAST!!!!!!
Answer:
6.
Step-by-step explanation:
.
Answer:
[tex]C)\:8[/tex]
8 units tiles must be added
--------------------------------------
~HOPE IT HELPS~
~HAVE A GREAT DAY!!~
An investor puts $1,200 into an account that pays 5.5% interest compounded annually. The total amount A in the account after t years is given by which function below?
A = 1200(1.55) t
A = 1200(1.055) t
A = 1200(1.055) t
A = 1200 + (1.055) t
Let's see
[tex]\\ \tt\leadsto A=P(1+r/n)^{nt}[/tex]
n=1[tex]\\ \tt\leadsto A=1200(1+0.055)^t[/tex]
[tex]\\ \tt\leadsto A=1200(1.055)^t[/tex]
Answer:
[tex]\sf A=1200(1.055)^t[/tex]
Step-by-step explanation:
Annual Compound Interest Formula
[tex]\large \text{$ \sf A=P\left(1+r\right)^{t} $}[/tex]
where:
A = final amountP = principal amountr = interest rate (in decimal form)t = time (in years)Given:
P = $1,200r = 5.5% = 0.055t = t yearsSubstitute the given values into the equation:
[tex]\implies \sf A=1200(1+0.055)^t[/tex]
[tex]\implies \sf A=1200(1.055)^t[/tex]
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You are a 60 year old male. You want $1,000,000.00 in term life insurance. It will cost you $13.22 per $1,000.
Calculate the annual premium.
A $11,220.00
B $12,220.00
C $13,220.00
D $14,220.00
Answer:
C $13,220
Step-by-step explanation:
if cotA=3/4 find sinA and cosA
Answer:
sin A = 4/5
cos A = 3/5
Step-by-step explanation:
SOHCAHTOA
cot A = 1/tan A
tan A = opp/adj
cot A = 1/tan A = adj/opp
cot A = 3/4
adj = 3; opp = 4
adj^2 + opp^2 = hyp^2
3^2 + 4^2 = hyp^2
9 + 16 = hyp^2
hyp = 5
sin A = opp/hyp
sin A = 4/5
cos A = adj/hyp
cos A = 3/5
Answer:
sin A = 4/5
cos A = 3/5
Which expression is equivalent to…
Answer:
D
Step-by-step explanation:
An irrigation system (sprinkler) has a parabolic pattern. The height, in feet, of the spray of water is given by the equation h(x) = -x^2+10x+7.5 where x is the number of feet away from the sprinkler head (along the ground) the spray is.
The irrigation system is positioned____ feet above the ground to start.
The spray reaches a maximum height of ____feet at a horizontal distance of feet away from the sprinkler head.
The spray reaches all the way to the ground at about_____ feet away
9514 1404 393
Answer:
7.5 ft32.5 ft, 5 ft10.7 ftStep-by-step explanation:
a) The starting height is h(0) = 7.5 feet, the constant in the quadratic function.
The irrigation system is positioned 7.5 feet above the ground
__
b) The axis of symmetry for quadratic ax^2 +bx +c is x = -b/(2a). For this quadratic, that is x=-10/(2(-1)) = 5. This is the horizontal distance to the point of maximum height. The maximum height is ...
h(5) = (-5 +10)(5) +7.5 = 32.5 . . . feet
The spray reaches a maximum height of 32.5 feet at a horizontal distance of 5 feet from the sprinkler head.
__
c) The maximum distance will be √32.5 + 5 ≈ 10.7 ft.
The spray reaches the ground at about 10.7 feet away.
Answer:
7.5
32.5
5
maximum
10.7
Step-by-step explanation:
2 triangles are shown. The first triangle has side lengths 35, 20, and 20. The second triangle has side lengths x, 44, 44.
What value of x will make the triangles similar by the SSS similarity theorem?
15.9
59
77
96.8
Answer:
[tex]x = 77[/tex]
Step-by-step explanation:
Given
[tex]First \to Second[/tex]
[tex]35 \to x[/tex]
[tex]20 \to 44[/tex]
[tex]20 \to 44[/tex]
Required
Find x by SSS
Represent the triangle sides as a ratio
[tex]35 : x = 20 : 44[/tex]
Express as fraction
[tex]\frac{x}{35} = \frac{44}{20}[/tex]
Multiply by 35
[tex]x = \frac{44}{20} * 35[/tex]
[tex]x = \frac{44 * 35}{20}[/tex]
[tex]x = \frac{1540}{20}[/tex]
[tex]x = 77[/tex]
Answer:
ccccccccccccccccccccccccccc
Step-by-step explanation:
1. Find the exact value of sin( a−B), given that sin a=−4/5 and cos B=12/13, with a in quadrant III and B in quadrant IV.
2. Find all real numbers in the interval [0,2pi) that satisfy the equation.
3sec^2 x tan x =4tan x
3. Simplify the following trigonometric expressions, using identities as needed:
sin(x)/1−cos(x) + 1−cos(x)/sin(x)
(1) Recall that
sin(x - y) = sin(x) cos(y) - cos(x) sin(y)
sin²(x) + cos²(x) = 1
Given that α lies in the third quadrant, and β lies in the fourth quadrant, we expect to have
• sin(α) < 0 and cos(α) < 0
• sin(β) < 0 and cos(β) > 0
Solve for cos(α) and sin(β) :
cos(α) = -√(1 - sin²(α)) = -3/5
sin(β) = -√(1 - cos²(β)) = -5/13
Then
sin(α - β) = sin(α) cos(β) - cos(α) sin(β) = (-4/5) (12/13) - (-3/5) (-5/13)
==> sin(α - β) = -63/65
(2) In the second identity listed above, multiplying through both sides by 1/cos²(x) gives another identity,
sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)
==> tan²(x) + 1 = sec²(x)
Rewrite the equation as
3 sec²(x) tan(x) = 4 tan(x)
3 (tan²(x) + 1) tan(x) = 4 tan(x)
3 tan³(x) + 3 tan(x) = 4 tan(x)
3 tan³(x) - tan(x) = 0
tan(x) (3 tan²(x) - 1) = 0
Solve for x :
tan(x) = 0 or 3 tan²(x) - 1 = 0
tan(x) = 0 or tan²(x) = 1/3
tan(x) = 0 or tan(x) = ±√(1/3)
x = arctan(0) + nπ or x = arctan(1/√3) + nπ or x = arctan(-1/√3) + nπ
x = nπ or x = π/6 + nπ or x = -π/6 + nπ
where n is any integer. In the interval [0, 2π), we get the solutions
x = 0, π/6, 5π/6, π, 7π/6, 11π/6
(3) You only need to rewrite the first term:
[tex]\dfrac{\sin(x)}{1-\cos(x)} \times \dfrac{1+\cos(x)}{1+\cos(x)} = \dfrac{\sin(x)(1+\cos(x))}{1-\cos^2(x)} = \dfrac{\sin(x)(1+\cos(x)}{\sin^2(x)} = \dfrac{1+\cos(x)}{\sin(x)}[/tex]
Then
[tex]\dfrac{\sin(x)}{1-\cos(x)}+\dfrac{1-\cos(x)}{\sin(x)} = \dfrac{1+\cos(x)+1-\cos(x)}{\sin(x)}=\dfrac2{\sin(x)}[/tex]
Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 217
Using a linear approximation, the estimated cube root of 217 is 6.00925.
Given that the number is,
The cube root of 217
Now, for the cube root of 217 using a linear approximation, use differentials.
So, the derivative of the function [tex]f(x) = x^{(1/3)[/tex] at a known point.
Taking the derivative of [tex]f(x) = x^{(1/3)[/tex], we get:
[tex]f'(x) = (\dfrac{1}{3} )x^{-2/3[/tex]
Now, we can choose a point near 217 to evaluate the linear approximation.
Let's use x = 216, which is a perfect cube.
Substituting x = 216 into the derivative, we get:
[tex]f'(216) = (\dfrac{1}{3} )(216)^{-2/3[/tex]
[tex]= 0.00925[/tex]
Next, use the linear approximation formula:
Δy ≈ f'(a)Δx
Since our known point is a = 216 and we want to estimate the cube root of 217,
since 217 - 216 = 1
Hence, Δx = 1
Δy ≈ f'(216)
Δx ≈ 0.00925 × 1
≈ 0.00925
Finally, add this linear approximation to the known value at the known point to get our estimate:
Estimated cube root of 217 ;
f(216) + Δy = 6 + 0.00925
= 6.00925
Therefore, the estimated cube root of 217 is 6.00925.
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A plane flying horizontally at an altitude of 2 miles and a speed of 410 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 5 miles away from the station.
Answer:
[tex]82\sqrt{21}\text{ or approximately 375.77 miles per hour}[/tex]
Step-by-step explanation:
Please refer to the diagram below. R is the radar station and x is the distance from the station to the plane.
We are given that the plane is flying horizontally at an altitude of two miles and at a speed of 410 mph. And we want to find the rate at which the distance from the plane to the station is increasing when it is five miles away from the station.
In other words, given da/dt = 410 and x = 5, find dx/dt.
From the Pythagorean Theorem:
[tex]a^2+4=x^2[/tex]
Implicitly differentiate both sides with respect to time t. Both a and x are functions of t. Hence:
[tex]\displaystyle 2a\frac{da}{dt}=2x\frac{dx}{dt}[/tex]
Simplify:
[tex]\displaystyle a\frac{da}{dt}=x\frac{dx}{dt}[/tex]
Find a when x = 5:
[tex]a=\sqrt{5^2-2^2}=\sqrt{21}[/tex]
Therefore, dx/dt when da/dt = 410, x = 5, and a = √(21) is:
[tex]\displaystyle \frac{dx}{dt}=\frac{(\sqrt{21})(410)}{5}=82\sqrt{21}\approx 375.77\text{ mph}[/tex]
The rate at which is distance from the plane to the radar station is increasing at a rate of approximately 375.77 miles per hour.
What is the value of x?
Answer:
22
Step-by-step explanation:
3x-14= 4(x-9)
3×-14= 4x-36
4x-36-3x+14=0
×-22÷0
x=22
What is the place value of the 4 in 4.09?
Choose 1 answer:
(Choice A)
Tens
(Choice B)
Ones
(Choice C)
Tenths
(Choice D)
Hundredths
Answer:
B: Ones.
Step-by-step explanation:
Because this number is 4.09, and the decimal is right next to the 4, that means that it is in the ones place. Decimals are always adjacent on the right to the ones place.
Plzz prove this tomorrow is my test plzz help me
Step-by-step explanation:
this is the correct answer for the question
What is the cost, in dollars, of 16 onions if 3 onions weigh 1.5 lb and the price of onions is 33 cents per kilogram
Answer:
The cost of 16 onions is $ 1.20.
Step-by-step explanation:
To determine what is the cost, in dollars, of 16 onions if 3 onions weigh 1.5 lb and the price of onions is 33 cents per kilogram, the following calculation must be performed:
1.5 pounds = 0.68 kilos
0.68 / 3 = 0.22666 kilos each onion
16 x 0.22666 = 3.626 kilos
0.33 x 3.626 = 1.20
Therefore, the cost of 16 onions is $ 1.20.
the volume of pyramid a is the volume of pyramid b. if the heigh of pyramid b increases to twice that of pyramid a the new volume of pyramid b the volume of pyramid a
Answer:
12.259-12.25 890654321
Can someone help me with this problem?
5.11.
A manufacturing process produces 500 parts per hour. A sample part is selected about every half hour, and after five parts are obtained, the average of these five measurements is plotted on an x control chart.
(a) Is this an appropriate sampling scheme if the assignable cause in the process results in an instantaneous upward shift in the mean that is of very short duration?
(b) If your answer is no, propose an alternative procedure. If your answer is yes, justify.
5.12.
Consider the sampling scheme proposed in Exercise 5.11. Is this scheme appropriate if the assignable cause results in a slow, prolonged upward drift in the mean? If your answer is no, propose an alternative procedure.
Answer:
Following are the response to the given points:
Step-by-step explanation:
For question 5.11:
For point a:
For all the particular circumstances, it was not an appropriate sampling strategy as each normal distribution acquired is at a minimum of 30(5) = 150 or 2.5 hours for a time. Its point is not absolutely fair if it exhibits any spike change for roughly 10 minutes.
For point b:
The problem would be that the process can transition to an in the state in less than half an hour and return to in the state. Thus, each subgroup is a biased selection of the whole element created over the last [tex]2 \frac{1}{2}[/tex] hours. Another sampling approach is a group.
For question 5.12:
This production method creates 500 pieces each day. A sampling section is selected every half an hour, and the average of five dimensions can be seen in a [tex]\bar{x}[/tex]line graph when 5 parts were achieved.
This is not an appropriate sampling method if the assigned reason leads to a sluggish, prolonged uplift. The difficulty would be that gradual or longer upward drift in the procedure takes or less half an hour then returns to a controlled state. Suppose that a shift of both the detectable size will last hours [tex]2 \frac{1}{2}[/tex] . An alternative type of analysis should be a random sample of five consecutive pieces created every [tex]2 \frac{1}{2}[/tex] hour.
indicate the following pairs of lines are coinciding, parallel ,perpendicular or neither
Two lamps marked 100 W - 110 V and 100 W - 220 V are connected i
series across a 220 V line. What power is consumed in each lamp?
The power consumed in the lamp marked 100W - 110V is 15.68W
The power consumed in the lamp marked 100W - 220V is 62.73W
Step-by-step explanation:
Given:
First lamp rating
Power (P) = 100W
Voltage (V) = 110V
Second lamp rating
Power (P) = 100W
Voltage (V) = 220V
Source
Voltage = 220V
i. Get the resistance of each lamp.
Remember that power (P) of each of the lamps is given by the quotient of the square of their voltage ratings (V) and their resistances (R). i.e
P = [tex]\frac{V^2}{R}[/tex]
Make R subject of the formula
⇒ R = [tex]\frac{V^2}{P}[/tex] ------------------(i)
For first lamp, let the resistance be R₁. Now substitute R = R₁, V = 110V and P = 100W into equation (i)
R₁ = [tex]\frac{110^2}{100}[/tex]
R₁ = 121Ω
For second lamp, let the resistance be R₂. Now substitute R = R₂, V = 220V and P = 100W into equation (i)
R₂ = [tex]\frac{220^2}{100}[/tex]
R₂ = 484Ω
ii. Get the equivalent resistance of the resistances of the lamps.
Since the lamps are connected in series, their equivalent resistance (R) is the sum of their individual resistances. i.e
R = R₁ + R₂
R = 121 + 484
R = 605Ω
iii. Get the current flowing through each of the lamps.
Since the lamps are connected in series, then the same current flows through them. This current (I) is produced by the source voltage (V = 220V) of the line and their equivalent resistance (R = 605Ω). i.e
V = IR [From Ohm's law]
I = [tex]\frac{V}{R}[/tex]
I = [tex]\frac{220}{605}[/tex]
I = 0.36A
iv. Get the power consumed by each lamp.
From Ohm's law, the power consumed is given by;
P = I²R
Where;
I = current flowing through the lamp
R = resistance of the lamp.
For the first lamp, power consumed is given by;
P = I²R [Where I = 0.36 and R = 121Ω]
P = (0.36)² x 121
P = 15.68W
For the second lamp, power consumed is given by;
P = I²R [Where I = 0.36 and R = 484Ω]
P = (0.36)² x 484
P = 62.73W
Therefore;
The power consumed in the lamp marked 100W - 110V is 15.68W
The power consumed in the lamp marked 100W - 220V is 62.73W
What is the slope of (-0,-1) and (3,1)
Answer:
Step-by-step explanation:
(0, -1) & (3 ,1)
[tex]Slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{1-[-1]}{3-0}\\\\=\frac{1+1}{3}\\\\=\frac{2}{3}[/tex]
The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 54 and a standard deviation of 3. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 54 and 63?
Answer:
The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
Approximately 68% of the measures are within 1 standard deviation of the mean.
Approximately 95% of the measures are within 2 standard deviations of the mean.
Approximately 99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 54, standard deviation = 3.
What is the approximate percentage of lightbulb replacement requests numbering between 54 and 63?
63 = 54 + 3*3
So between the mean and 3 standard deviations above the mean.
The normal distribution is symmetric, which means that 50% of the values are below the mean and 50% are above.
Of those 50% above, 99.7% are below 63. So
0.5*0.997 = 0.4985
0.4985*100% = 49.85%
The approximate percentage of lightbulb replacement requests numbering between 54 and 63 is of 49.85%.
Order these numbers from least to greatest.
5.772 , 11/2, 5 6/11, 5.77
Answer:
6/11, 11/2, 5.77, 5.772
Step-by-step explanation:
A ball is thrown vertically upward with an initial velocity of 19 m/s. Its height, h(t)metres after t seconds, is given by the equation h(t) = -3t2 + 20t + 2.0.
The time taken by the ball to reach the maximum height is ________ seconds. Round your answer to the nearest tenth.
Answer:
Step-by-step explanation:
There are 2 different ways to do this: calculus and by completing the square. In this particular instance, calculus is WAY easier, and since I don't know for what class you are doing this, I'll do both ways. First the calculus way. We know the position equation, and the first derivative of the position is velocity. We also know that when the velocity is equal to 0 is when the object is at its max height. So we'll find the derivative first, then solve it for t:
If [tex]s(t)=-3t^2+20t+2[/tex] then the first derivative is
v(t) = -6t + 20 Solving for t requires that we set the velocity equal to 0 (again, this is where the object is at its max height), so
0 = -6t + 20 and
-20 = -6t so
t = 3.3 seconds. Now that we know that at 3.3 seconds the object is at its highest point, we sub that time into the position function to see where it is at that time:
s(3.3) = [tex]-3(3.3)^2+20(3.3)+2[/tex] and
s(3.3) = 35.3 meters.
Now onto the more difficult way...completing the square. Begin by setting the position function equal to 0 and then move over the constant to get:
[tex]-3t^2+20t=-2[/tex] Since the leading coefficient is not a 1 (it's a 3), we have to factor out the 3, leaving us with:
[tex]-3(t^2-\frac{20}{3}t)=-2[/tex] Now the rule is to take half the linear term, square it, and add it to both sides. Our linear term is [tex]\frac{20}{3}[/tex] and half of that is [tex]\frac{20}{6}[/tex]. Squaring that:
[tex](\frac{20}{6})^2=\frac{400}{36}=\frac{100}{9}[/tex]. We will add that in to both sides. On the left it's easy, but on the right we have to take into account that we still have that -3 sitting out front, refusing to be ignored. So we have to multiply it in when we add it to the right. Doing that gives us:
[tex]-3(t^2-\frac{20}{3}t+\frac{100}{9})=-2-\frac{100}{3}[/tex] We will clean this up a bit now. The reason we do this is because on the left we have created a perfect square binomial which will give us the time we are looking for to answer this question. Simplifying the right and at the same time writing the perfect square binomial gives us:
[tex]-3(t-\frac{20}{6})^2=-\frac{106}{3}[/tex] Now the last step is to move the constant back over and set the quadratic back equal to y:
[tex]y=-3(t-\frac{20}{6})^2+\frac{106}{3}[/tex]. The vertex of this quadratic is
[tex](\frac{20}{6},\frac{106}{3})[/tex] where
[tex]\frac{20}{6}=3.3[/tex] as the time it takes for the ball to reach its max height of
[tex]\frac{106}{3}=35.3[/tex] meters.
I'd say if you plan on taking calculus cuz you're not there yet, you'll see that many of these types of problems become much simpler when you know it!
What is the mean of this data? 7,5,5,3,2,2
Answer:
4
Step-by-step explanation:
The mean is the average of a data set. It can be found by adding up all of the values in a data set and then dividing it by the number of values in the data set.
The values in this data set;
[tex]7,5,5,3,2,2[/tex]
The number of values in this data set,
[tex]6[/tex]
Find the mean;
[tex]\frac{sum\ of\ vlaues}{number\ of\ values}[/tex]
[tex]=\frac{7+5+5+3+2+2}{6}\\\\=\frac{24}{6}\\\\=4[/tex]
5. Determine the total area of the figure below.
4 ft!
12 ft
10 ft
3 ft
9 ft
14. The Elizabeth Tower is 320 feet tall. At what time or times during your ride on the London Eye are you at the same height as the top of the tower? Show your work. (4 points: 2 points for finding the correct time(s), 2 points for work shown)
Answer:
Ok so on a clock there is 12 numbers where 12 is on top so at 12 am and 12 pm noon and midnight you will be at the top of the clock
Hope This Helps!!!
During the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.
To determine the time(s) during the ride on the London Eye when you are at the same height as the top of the Elizabeth Tower (commonly known as Big Ben), we need to consider the height of the London Eye and its rotational motion.
Given that the Elizabeth Tower is 320 feet tall, we need to find the position of the London Eye when its height aligns with the top of the tower.
The London Eye has a height of 443 feet, and it completes one full rotation in approximately 30 minutes (or 1800 seconds). This means that it moves at a constant angular velocity of 360 degrees per 1800 seconds.
To find the time(s) when the heights align, we can set up a proportion:
(Height of the Elizabeth Tower) / (Height of the London Eye) = (Angle covered by the London Eye) / 360 degrees
Substituting the given values:
320 / 443 = (Time to align) / 1800
Simplifying the equation:
(Time to align) = (320 / 443) * 1800
Calculating the value:
(Time to align) ≈ 1303.16 seconds
Converting the time to minutes and seconds:
(Time to align) ≈ 21 minutes and 43.16 seconds
Therefore, during the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.
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