A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 800 ft/s when it has risen 3000 ft. (Round your answers to three decimal places.)
(a) How fast is the distance from the television camera to the rocket changing at that moment?
ft/s
(b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?
rad/s
Answers
============================================
Explanation for part (a)
t = time in secondsx = horizontal distance from the camera to the launch pady = vertical distance from the launch pad to the rocket's locationz = distance from camera to the rocket at time tAll distances mentioned are in feet.
We'll have a right triangle which allows us to apply the pythagorean theorem. Refer to the diagram below.
a^2+b^2 = c^2
x^2+y^2 = z^2
Apply the derivative to both sides with respect to t. We'll use implicit differentiation and the chain rule.
[tex]x^2+y^2 = z^2\\\\\frac{d}{dt}[x^2+y^2] = \frac{d}{dt}[z^2]\\\\\frac{d}{dt}[x^2]+\frac{d}{dt}[y^2] = \frac{d}{dt}[z^2]\\\\2x*\frac{dx}{dt}+2y*\frac{dy}{dt}=2z*\frac{dz}{dt}\\\\x*\frac{dx}{dt}+y*\frac{dy}{dt}=z*\frac{dz}{dt}\\\\[/tex]
Now we'll plug in (x,y,z) = (4000,3000,5000). The x and y values are given. The z value is found by use of the pythagorean theorem. Ie, you solve 4000^2+3000^2 = z^2 to get z = 5000. Or you could note that this is a scaled copy of the 3-4-5 right triangle.
We know that dx/dt = 0 because the horizontal distance, the x distance, is not changing. The rocket is only changing in the y direction. Or you could say that the horizontal speed is zero.
The vertical speed is dy/dt = 800 ft/s and it's when y = 3000. It's likely that dy/dt isn't the same value through the rocket's journey; however, all we care about is the instant when y = 3000.
Let's plug all that in and isolate dz/dt
[tex]4000*0+3000*800=5000*\frac{dz}{dt}\\\\2,400,000=5000*\frac{dz}{dt}\\\\\frac{dz}{dt} = \frac{2,400,000}{5000}\\\\\frac{dz}{dt} = 480\\\\[/tex]
At the exact instant that the rocket is 3000 ft in the air, the distance between the camera and the rocket is changing by an instantaneous speed of 480 ft/s.
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Explanation for part (b)
Again, refer to the diagram below.
We have theta (symbol [tex]\theta[/tex]) as the angle of elevation. As the rocket's height increases, so does the angle theta.
We can tie together the opposite side y with the adjacent side x with the tangent function of this angle.
[tex]\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(\theta) = \frac{y}{x}[/tex]
Like before, we'll apply implicit differentiation. This time we'll use the quotient rule as well.
[tex]\tan(\theta) = \frac{y}{x}\\\\\frac{d}{dt}[\tan(\theta)] = \frac{d}{dt}\left[\frac{y}{x}\right]\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{\frac{dy}{dt}*x - y*\frac{dx}{dt}}{x^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{800*4000 - 3000*0}{4000^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{3,200,000}{16,000,000}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{32}{160}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}\\\\[/tex]
Let's take a brief detour. We'll return to this later. Recall earlier that [tex]\tan(\theta) = \frac{y}{x}\\\\[/tex]
If we plug in y = 3000 and x = 4000, then we end up with [tex]\tan(\theta) = \frac{3}{4}\\\\[/tex] which becomes [tex]\tan^2(\theta) = \frac{9}{16}[/tex]
Apply this trig identity
[tex]\sec^2(\theta) = 1 + \tan^2(\theta)[/tex]
and you should end up with [tex]\sec^2(\theta) = 1+\frac{9}{16} = \frac{25}{16}[/tex]
So we can now return to the equation we want to solve
[tex]\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{25}{16}*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{d\theta}{dt} = \frac{1}{5}*\frac{16}{25}\\\\\frac{d\theta}{dt} = \frac{16}{125}\\\\\frac{d\theta}{dt} = 0.128\\\\[/tex]
This means that at the instant the rocket is 3000 ft in the air, the angle of elevation theta is increasing by 0.128 radians per second.
This is approximately 7.334 degrees per second.
The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s
Let x represent the distance from the camera to the rocket and let h represent the height of the rocket.
a)
[tex]x^2=h^2+4000^2\\\\2x\frac{dx}{dt}=2h\frac{dh}{dt} \\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\At \ h=3000\ ft, \frac{dh}{dt}=800\ ft/s;\\x^2=3000^{2} +4000^2\\x=5000\\\\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\5000\frac{dx}{dt}=3000*800\\\\\frac{dx}{dt}=480\ ft/s[/tex]
b)
[tex]tan(\theta)=\frac{h}{4000} \\\\h=4000tan(\theta)\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt} \\\\\\At\ h=3000\ ft;\\\\tan\theta = \frac{3000}{4000}=\frac{3}{4} \\\\sec^2(\theta)=1+tan^2(\theta)=1+(\frac{3}{4})^2=\frac{25}{16} \\\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt}\\\\800=4000*\frac{25}{16}* \frac{d\theta}{dt}\\\\\frac{d\theta}{dt}=0.128\ rad/s[/tex]
Hence, The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s
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According to government data, the probability than an adult never had the flu is 19%. You randomly select 70 adults and ask if he or she ever had the flu. Decide whether you can use the normal distribution to approximate the binomial distribution, If so, find the mean and standard deviation, If not, explain why. Round to the nearest hundredth when necessary.
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Answer:
Since [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal distribution can be used to approximate the binomial distribution.
The mean is 13.3 and the standard deviation is 3.28.
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
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The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
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The probability than an adult never had the flu is 19%.
This means that [tex]p = 0.19[/tex]
You randomly select 70 adults and ask if he or she ever had the flu.
This means that [tex]n = 70[/tex]
Decide whether you can use the normal distribution to approximate the binomial distribution
[tex]np = 70*0.19 = 13.3 \geq 10[/tex]
[tex]n(1-p) = 70*0.81 = 56.7 \geq 10[/tex]
Since [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal distribution can be used to approximate the binomial distribution.
Mean:
[tex]\mu = E(X) = np = 70*0.19 = 13.3[/tex]
Standard deviation:
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{70*0.19*0.81} = 3.28[/tex]
The mean is 13.3 and the standard deviation is 3.28.
Help me this question
Answers
Answer:
(a) 218.6 N
(b) 97.14 N
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When the system is in equilibrium, the net torque on the system is zero.
AC = 1.5 m, CD = 2.3 m, DB = 5 - 1.5 - 2.3 = 1.2 m
Let the centre of gravity of plank is at G.
AG = 2.5 m, CG = 2.5 - 1.5 = 1 m, GB = 2.5 m
(a) Let the reaction at C is R and at D is R'.
R + R' = 29 x 9.8 = 284.2 N ... (1)
Take the torque about C.
29 x 9.8 x CG = R' x GD
29 x 9.8 x 1 = R' x 1.3
R' = 218.6 N
(b) Take the torque about D.
6 x 9.8 x AD = R x CD
6 x 9.8 x (1.5 + 2.3) = R x 2.3
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In an experiment, you choose to have two randomly assigned groups. In one, you take measurements both pretest and posttest; with the second, a posttest-only measure. This describes which task of conducting an experiment
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Answer:
The answer is "Specific treatment levels".
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Complete Question
Four spinners are spun. Spinner 1 has outcomes {1,2,3,4,5,6,7,8} Spinner 2 has outcomes {1,2,3,4,5,6} Spinner 3 has outcomes {1,2,3,4,5,6} Spinner 4 has outcomes {1,2,3,4,5} The outcomes for each spinner are equally likely. S is the sum of the numbers that come up on the spinners. What is the expected value of S?
Answer:
[tex]E(s)=14.5[/tex]
Step-by-step explanation:
From the question we are told that:
Spinner 1 ={1,2,3,4,5,6,7,8}
Spinner 2= {1,2,3,4,5,6}
Spinner 3 = {1,2,3,4,5,6}
Spinner 4 {1,2,3,4,5}
Generally the equation for expected outcome is mathematically given by
[tex]E(s)=\sum P(x).x[/tex]
Where
[tex]x=\frac{n(n+1)}{2}[/tex]
For Spinner 1
[tex]E(s_1)=\sum \frac{1}{8}*\frac{8(8+1)}{2}[/tex]
[tex]E(s_1)=4.5[/tex]
For Spinner 2
[tex]E(s_2)=\sum \frac{1}{6}*\frac{6(6+1)}{2}[/tex]
[tex]E(s_2)=3.5[/tex]
For Spinner 3
[tex]E(s_2)=E(s_3)[/tex]
For Spinner 3
[tex]E(s_4)=\sum \frac{1}{6}*\frac{6(6+1)}{2}[/tex]
[tex]E(s_4)=3[/tex]
Therefore The Expected Value
[tex]E(s)=\sum E(s 1..4)[/tex]
[tex]E(s)=4.5+2(3.5)+3[/tex]
[tex]E(s)=14.5[/tex]
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Answer:
B
Step-by-step explanation:
5³.5^×
simply means
5³×5^×
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multiplication is addition
5 is common
so 5(³+×)
hence 5^3+×
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[tex]3-5\left(a-4\right)[/tex]
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[tex]=3-5a+20[/tex]
[tex]=-5a+23[/tex]
OAmalOHopeO
[tex]\huge\text{Hey there!}[/tex]
[tex]\large\text{3 - 5(a - 4)}\\\\\huge\text{\underline{\underline{DISTRUBUTE -5 within the parentheses}}}\\\\\large\text{3 - 5(a) - 5(-4)}\\\large\text{= 3 - 5a + 20}\\\\\huge\text{\underline{\underline{COMBINE the LIKE TERMS}}}\\\large\text{-5a + (3 + 20)}\\\large\text{= \bf -5a + 23}\\\\\\\boxed{\boxed{\huge\text{Therefore, your answer is: \bf -5a + 23}}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!} \\\\\\\frak{Amphitrite1040:)}[/tex]
PLEASE HELPP ASAP!!
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Hey there!
Given; The Line BC contains points B (4, -5) and C (3, 2).
And the Line DE contains points D (2,0) and E (9, 1)
Note: Use double point formula for finding the equation and then find slopes of both then put the condition for perpendicular lines and parallel lines.
From line BC;
The points are B (4, -5) and C (3, 2).
Using double point formula;
[tex](y - y1) = \frac{y2 - y1}{x2 - x1}(x - x1) [/tex]
Keep all the value;
[tex](y + 5) = \frac{2 + 5}{3 - 4} (x - 4)[/tex]
Simplify it;
[tex]y + 5 = - 7x + 28[/tex]
Therefore, the equation is y = -7x+23........(I)And slope(m1) is -7 {comparing the equation (I) with y=Mx+c}
Again;
The points D (2,0) and E (9, 1)
Using double point formula;
[tex](y - y1) = \frac{y2 - y1}{x2 - x1} (x - x1)[/tex]
Keep all values;
[tex](y - 0) = \frac{9 - 2}{1 - 2} (x - 2)[/tex]
[tex]y = - 7x + 14[/tex]
Therefore, the equation is y = -7x+14......(ii)And the slope (m2) is -7. {comparing the equation (ii) with y= mx+c}
Check:
For parallel lines:
m1= m2
-7 = -7 (true)
Therefore, the lines are parallel.
Hope it helps!
A manufacturer knows that their items have a normally distributed length, with a mean of 18.2 inches, and standard deviation of 3.9 inches. If 2 items are chosen at random, what is the probability that their mean length is less than 21.9 inches
Answers
Answer:
0.9099 = 90.99% probability that their mean length is less than 21.9 inches.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 18.2 inches, and standard deviation of 3.9 inches.
This means that [tex]\mu = 18.2, \sigma = 3.9[/tex]
2 itens:
This means that [tex]n = 2, s = \frac{3.9}{\sqrt{2}}[/tex]
What is the probability that their mean length is less than 21.9 inches?
This is the p-value of Z when X = 21.9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21.9 - 18.2}{\frac{3.9}{\sqrt{2}}}[/tex]
[tex]Z = 1.34[/tex]
[tex]Z = 1.34[/tex] has a p-value of 0.9099.
0.9099 = 90.99% probability that their mean length is less than 21.9 inches.
