Answer:
The probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with
Mean [tex]\mu=26[/tex] mpg and standard deviation [tex]\sigma=12[/tex] mpg.
Number of sample n=36
To find : What is the probability the average mpg is less than 28 mpg?
Solution :
Applying z-score formula,
[tex]z=\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The portability is given by, [tex]P(X<28)[/tex]
[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{28-26}{\frac{12}{\sqrt{36}}})[/tex]
[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{2}{\frac{12}{6}})[/tex]
[tex]=P(z<\dfrac{2}{2})[/tex]
[tex]=P(z<1)[/tex]
Using z-table,
[tex]=0.8413[/tex]
Therefore, the probability the average mpg is less than 28 mpg is 0.8413.
Answer:
he probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with
Mean mpg and standard deviation mpg.
Number of sample n=36
To find : What is the probability the average mpg is less than 28 mpg?
Solution :
Applying z-score formula,
The portability is given by,
Using z-table,
Therefore, the probability the average mpg is less than 28 mpg is 0.8413.
Step-by-step explanation:
Lee is running around the perimeter of a circular track at a rate of 10 ft/sec. The track has a radius of 100 yards. After 10 seconds, Lee turns and runs along a radial line to the center of the circle. Once he reaches the center, he turns and runs along a radial line to his starting point on the perimeter. Assume Lee does not slow down when he makes these two turns.
(a) Sketch a picture of the situation.
(b) How far has Lee traveled once he returns to his starting position? circuit? enclosed by Lee's path.
(c) How much time will elapse during Lee's
(d) Find the area of the pie shaped sector.
The picture is given below. The perimeter, total time, and area of the circuit will be 700 feet, 70 seconds, and 100 square meters, respectively.
What is the arc length of the sector?Let r is the radius of the sector and θ be the angle subtended by the sector at the center.
Then the arc length of the sector of the circle will be
Arc = (θ/2π) 2πr
Then the area of the sector of the circle will be
Arc = (θ/2π) 2πr
We know that 100 yards = 300 feet.
The perimeter covered in 10 seconds will be given as,
10 x 10 = (θ/2π) 2π(300)
θ = 19.1°
The perimeter of the circuit will be given as,
P = 100 + 300 + 300
P = 700 feet
The total time is given as,
Time = 700 / 10
Time = 70 seconds
The area of the shape will be given as,
A = (19.1/360) 2π(300)
A = 100 square feet
The picture is given below. The perimeter, total time, and area of the circuit will be 700 feet, 70 seconds, and 100 square meters, respectively.
More about the arc length of the sector link is given below.
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Suppose you had to use exactly 200 m of fencing to make either one square enclosure or two separate square enclosures of any size you wished. What plan would give you the least area? What plan would give you the greatest area?
Answer:
Greatest area: One square enclosure with side 50 m
Least area: Two square enclosures with side 25 m each
Step-by-step explanation:
We know we have 200 m of fencing to make the required enclosures. Since the fence will run surrounding any enclosure, it is called the perimeter.
The perimeter for any square area of side z is computed as
P=4z
And its area is
[tex]A=z^2[/tex]
Now, let's analyze both options (shown in the figure below)
Option 1: One square enclosure
Knowing P=200 m, we can determine the length of the side
z=200 m /4 = 50 m
The area is easily computed
[tex]A=50^2=2500 m^2[/tex]
Option 2: Two separate square enclosures of any size
Let's say the side of one of them is x and the side of the other one is y
Assuming both enclosures have no sides in common, the total perimeter is
P=4x+4y
We have 200 m to make the job, so
4x+4y=200
Or equivalently
x+y=50 => y=50-x
The total area of both enclosures is
[tex]A=x^2+y^2[/tex]
Replacing the expression of y
[tex]A=x^2+(50-x)^2[/tex]
To know what the best value is for x to maximize or minimize the area, we use derivatives with respect to x
[tex]A'=2x+2(50-x)(-1)[/tex]
[tex]A'=2x-100+2x=4x-100[/tex]
We equate A'=0 to find the critical point
4x-100=0
x=25 m
Since y=50-x
y=25 m
And the total area is
[tex]A=25^2+25^2=1250\ m^2[/tex]
Note: if we set any other combination for x and y, say x=20 m and y=30m we would get greater areas
[tex]A=20^2+30^2=1300\ m^2[/tex]
The first option gives us the greatest area of 2500 m^2 and the second option has the least area of 1250 m^2
To get the greatest area with 200 meters of fencing, construct one single square enclosure, which will provide an area of 2500 m^2. For the least area, construct two separate square enclosures, which will give a total area of 312.5 m^2.
Explanation:The question is asking how to use 200 meters of fencing to create square enclosures for the least and greatest area. For a given perimeter, a square has the largest area of any rectangle, so if we want to maximize the area enclosed, we would create one large square.
To find the size of the square, we would divide the total length of the fence by the number of sides of our square. So, 200 m / 4 sides = 50 m per side. Hence, the area of this square would be 50 m * 50 m = 2500 m^2.
For the least area, we would make 2 separate squares, each with half the fencing. So, each square would have 50 m of fence, or 12.5 m per side. The area for each square would then be 12.5 m * 12.5 m = 156.25 m^2, for a total area of 312.5 m^2 when you add both squares together.
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Some experts believe that 24% of all freshwater fish in a country have such high levels of mercury that they are dangerous to eat. Suppose a fish market has 350 fish tested, and 80 of them have dangerous levels of mercury. Test the hypothesis that this sample is not from a population with 24% dangerous fish, assuming that this is a random sample. Use a significance level of 0.05. State the null and alternative hypothesis, determine the z-test statistic, find the p-value, does the conclusion support or reject the null.
Final answer:
To assess if the sample of fish with high mercury levels supports the hypothesis of a 24% proportion in the population, we set up null and alternative hypotheses, calculate the z-test statistic, compare it to a critical value, compute the p-value, and then make a conclusion based on whether the p-value is less than the chosen significance level of 0.05.
Explanation:
To test the hypothesis about the percentage of freshwater fish with high levels of mercury, we must first establish our null hypothesis (H0) and alternative hypothesis (Ha). The null hypothesis is that the true proportion of fish with dangerous mercury levels is 24% (H0: p = 0.24), while the alternative hypothesis is that the proportion is not 24% (Ha: p ≠ 0.24).
We then calculate the z-test statistic using the sample proportion (p' = 80/350 = 0.2286) and the assumed population proportion under H0 (p = 0.24), along with the standard deviation of the sampling distribution (σp = sqrt[p(1-p)/n]). This result is compared to the critical z-score of 1.96, which corresponds to our significance level (α) of 0.05 in a two-tailed test.
The p-value is calculated to determine the probability of observing our sample statistic, or one more extreme if the null hypothesis is true. If this p-value is less than 0.05, we reject H0 and conclude that the sample provides enough evidence against the null hypothesis.
The graph of f(x) = x2 is translated to form
g(x) = (x – 2)2 – 3.
On a coordinate plane, a parabola, labeled f of x, opens up. It goes through (negative 2, 4), has a vertex at (0, 0), and goes through (2, 4).
Which graph represents g(x)?
On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).
On a coordinate plane, a parabola opens up. It goes through (negative 3, 4), has a vertex at (negative 2, 3), and goes through (negative 1, 4).
On a coordinate plane, a parabola opens up. It goes through (1, 4), has a vertex at (2, 3), and goes through (3, 4).
On a coordinate plane, a parabola opens up. It goes through (negative 4, 1), has a vertex at (negative 2, negative 3), and goes through (0, 1).
The correct answer is: A,
On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).
Step-by-step explanation:
Figure represents graph of [tex]f(x) = x^{2}[/tex] and [tex]g(x) = (x-2)^{2} -3[/tex]
Here, [tex]f(x) = x^{2}[/tex] is " Translated " or " Transformation " to [tex]g(x) = (x-2)^{2} -3[/tex]
In process of transformation,
You should remember that shape of curve remain same and only changes we get in vertex shift
Now, Vertex can be shift in two direction, we are going to discuss both the cases
(A). Shifting of Vertex in X-Axis:
A new function g(x) = f(x - c) represents to X-axis shift and In graph of f(x), Curve is shifted c units along right side of the X-axis
(B). Shifting of Vertex in Y-axis:
A new function g(x) = f(x) + b represents to Y-axis shift and In graph of f(x), Curve is shifted b units along the upward direction of Y-axis
Looking at the figure, You can see that vertex of f(x) is shifted 2 Units in X-axis and Negative 3 units in Y-axis and result into g(x)
Now, [tex]g(x) = (x-2)^{2} -3[/tex] = f(x-2) + (-3)
Therefore, new vertex we get is (2,-3)
Also, [tex]g(x) = (x-2)^{2} -3[/tex]
[tex]g(0) = (0-2)^{2} -3[/tex]
[tex]g(0) = (4} -3[/tex]
[tex]g(0) = 1 [/tex]
So. g(x) passes through (0,1)
The correct answer is: A On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).
Answer:
Graph A is the correct answer
Step-by-step explanation:
I juust took the test and got it right :D hope this helps
A group of climbers begin climbing at an elevation of 5000 feet and ascend at a steady rate of 1500 vertical feet per hour. This situation can be modeled by a linear function. Identify the independent and dependent variables.
Final answer:
The independent variable is time and the dependent variable is elevation in the context of a linear function modeling climbers ascending a mountain at a rate of 1500 feet per hour.
Explanation:
The situation described by the student can be modeled by a linear function in which the independent variable is time and the dependent variable is elevation.
As the group of climbers begins at an elevation of 5000 feet and climbs at a steady rate of 1500 feet per hour, the relationship between the time spent climbing and the elevation gained is direct.
The function that models this situation is f(t) = 1500t + 5000, where 't' represents the time in hours, and 'f(t)' represents the elevation in feet.
Find a recurrence relation for the number of sequences of 1s, 3s, and 5s whose terms sum to n. (b) Repeat part (a) with the added condition that no 5 can be followed by a 1. (c) Repeat part (a) with the condition of no subsequence of 135
Answer:
The solution has been given in the following attachment .
Step-by-step explanation:
High Tech, Inc randomly tests its employees about company policies. Last year in the 400 random tests conducted, 14 employees failed the test.A. Develop a 99% confidence interval for the proportion of applicants that fail the test. (round answers to 3 decimal places) Confidence interval for the proportion mean is between _____ and ____B. Would it be reasonable to conclude 5% of the employees cannot pass the employee test? Yes or No
Answer:
Confidence interval for the proportion mean is between 0.0113 and 0.0587. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test.
Step-by-step explanation:
We have a large sample size of n = 400 random tests conducted. Let p be the true proportion of employees who failed the test. A point estimate of p is [tex]\hat{p} = 14/400 = 0.035[/tex], we can estimate the standard deviation of [tex]\hat{p}[/tex] as [tex]\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.035(1-0.035)/400}=0.0092[/tex]. A [tex]100(1-\alpha)%[/tex] confidence interval is given by [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n[/tex], then, a 99% confidence interval is [tex]0.035\pm z_{0.005}0.0092[/tex], i.e., [tex]0.035\pm (2.5758)(0.0092)[/tex], i.e., (0.0113, 0.0587). [tex]z_{0.005} = 2.5758[/tex] is the value that satisfies that there is an area of 0.005 above this and under the standard normal curve. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test, because this inverval contain 0.05.
The 99% confidence interval for the proportion of employees that fail the test is between 0.016 and 0.054. Since 5% is within this range, it is reasonable to conclude that 5% of the employees cannot pass the test.
Explanation:To compute a 99% confidence interval for the proportion of employees that fail the test, we first need to calculate the sample proportion (p). Here, 14 employees failed the tests out of 400, so p = 14/400 = 0.035. The 99% confidence interval requires Z-score of 2.576 (as 99% of the data lies within 2.576 standard deviations of the mean in a normal distribution).
The estimation error (E) can be calculated using the formula E = Z * √( (p*(1-p)) / n), where n is the total number of tests. Substituting the values, E = 2.576 * √(0.035 * (1 - 0.035) / 400) = 0.019
So, the 99% confidence interval is (p - E, p + E) = (0.035 - 0.019, 0.035 + 0.019) = (0.016, 0.054). Thus, we are 99% confident that the true proportion of employees that fail the test is between 0.016 and 0.054.
Given that 5% (or 0.05) is within the 99% confidence interval we calculated, it would be reasonable to conclude that 5% of the employees cannot pass the employee test.
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A study was interested in determining if eating milk chocolate lowered someone's cholesterol levels. Ten people's cholesterol was measured. Then, each of these individuals were told to eat 100g of milk chocolate every day and to eat as they normally did. After two weeks, their cholesterol levels were measured again. Is there evidence to support that their cholesterol levels went down? How should we write the alternative hypothesis? (mud = the population mean difference= before - after)
A. Ha: mud = 0B. Ha: mud > 0C. Ha: mud < 0D. Ha: mud does not equal 0
The alternative hypothesis whereby mud = the population mean difference is expressed as; Ha: mud > 0
What is the Alternative hypothesis?
An alternative hypothesis is defined as one in which the observers or researchers anticipate a difference (or an effect) between two or more variables.
Now, in this case, the null hypothesis is that eating milk chocolate lowered someone's cholesterol levels. This means the alternate hypothesis is that eating milk chocolate increases someone's cholesterol levels.
Thus, alternative hypothesis in this case is expressed as;
Ha: mud > 0
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Use the model for projectile motion, assuming no air resistance.
A baseball, hit above the ground, leaves the bat at an angle of 45 degrees and is caught by an outfielder 3 feet above the ground and 300 feet from home plate.
What is the initial speed of the ball, and how high does it rise?
Answer:
s=29.93m/s
h=22.88m
Step-by-step explanation:
we must find the initial speed, we will determine its position (x-y).
x component [tex]s=0+v_{0}cos\alpha.t+0=v_{0}cos\alpha.t[/tex]
y component [tex]h=0+v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}\\[/tex] since the ball is caught at the same height then h=0
[tex]h=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0\\v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0;v_{0}sin\alpha.t=\frac{1}{2}gt^{2}\\t=\frac{2v_{0}sin\alpha}{g}\\[/tex]
where t= flight time;[tex]s=v_{0}cos\alpha.t[/tex], replacing t:
[tex]v_{0}=\sqrt{\frac{sg}{sin2\alpha}}[/tex]
[tex]s=v_{0}cos\alpha(\frac{2sin\alpha }{g})=\frac{v_{0} ^{2}2sin\alpha.cos\alpha}{g}=\frac{v_{0} ^{2}sin(2\alpha)}{g}]
: the values must be taken to the same units
[tex]300ft*0.3048m/ft=91.44m[/tex]
[tex]v_{0}=\sqrt{\frac{91.44m*9.8\frac{m}{s^{2}}}{sin2(45)}}=\sqrt{895.112(\frac{m}{s} )^{2} }=29.93\frac{m}{s}[/tex]
To calculate the height you should know that this is achieved when its component at y = 0
[tex]v_{y}=v_{0}sin\alpha-gt=0;gt=v_{0}sin\alpha\\\\ t=\frac{v_{0}sin\alpha }{g}\\h=v_{0}sin\alpha .t-\frac{1}{2}gt^{2}[/tex]
replacing t;[tex]h=v_{0}sin\alpha(\frac{v_{0}sin\alpha}{g})-\frac{1}{2}g(\frac{v_{0sin\alpha}}{g}) ^{2}\\[/tex]
finally
[tex]h=\frac{(v_{0}sin\alpha)^{2}}{2g}=\frac{(29.95*sin45)^{2}}{2*9.8}=22.88m[/tex]
You are given the polar curve r = cos(θ) + sin(θ)
a) List all of the points (r,θ) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of and limit yourself to (r > or equal to 0) and θ: [0, 2pi] If two or more points share the same value of r, list those starting with the smallest value of θ.
Point 1 (r,θ): ?
Point 2 (r,θ): ?
Point 3 (r,θ): ?
b) repeat this with a tangent vertical line
Answer:
a) 1.30656∠(3π/8), 0.541196∠(15π/8)
b) 1.30656∠(π/8), 0.541196∠(5π/8)
Step-by-step explanation:
The critical points can be found in polar coordinates by considering ...
[tex]\displaystyle{dy\over dx}={dy/d\theta\over dx/d\theta}={r\cos\theta + r'\sin\theta\over -r\sin\theta + r'\cos\theta} \quad\text{where $r'=dr/d\theta$}[/tex]
We can simplify the effort a little bit by rewriting r as …
[tex]r=\sqrt{2}\sin{(\theta+\pi /4)}[/tex]
Then, filling in function and derivative values, we have …
[tex]\dfrac{dy}{dx}=\dfrac{\sqrt{2}(\sin{(\theta+\pi /4)}\cos{(\theta)}+\cos{(\theta+\pi /4)}\sin{(\theta)})}{\sqrt{2}(-\sin{(\theta+\pi /4)}\sin{(\theta)}+\cos{(\theta+\pi /4)}\cos{(\theta)})}\\\\=\dfrac{\sin{(2\theta+\pi /4)}}{\cos{(2\theta+\pi /4)}}\\\\\dfrac{dy}{dx}=\tan{(2\theta +\pi /4)}[/tex]
__
(a) For horizontal tangents, dy/dx = 0, so we have …
[tex]\tan{(2\theta+\pi /4)}=0\\\\2\theta+\dfrac{\pi}{4}=k\pi \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}-\dfrac{\pi}{8}[/tex]
We can use reference angles for the “r” expressions and write the two horizontal tangent point (r, θ) values of interest as …
[tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{3\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{15\pi}{8})[/tex]
__
(b) For vertical tangents, dy/dx = undefined, so we have …
[tex]2\theta+\dfrac{\pi}{4}=k\pi +\dfrac{\pi}{2} \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}+\dfrac{\pi}{8}[/tex]
Again using reference angles for “r”, the two vertical tangent point values of interest are …
[tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{5\pi}{8})[/tex]
__
The attached graph shows the angle values in degrees and the radius values as numbers. The points of tangency are mirror images of each other across the line y=x.
To find points with horizontal tangents for the polar curve r = cos(θ) + sin(θ), one must set the derivative with respect to θ to zero; for vertical tangents, find where the derivative is undefined. The angles found are then substituted back into the equation to determine r values, and points are listed within the specified domain of θ and r.
The question deals with finding points where the tangent lines to the polar curve r = cos(θ) + sin(θ) are horizontal or vertical. To find where the tangent lines are horizontal, we need to look for points where the derivative of the function with respect to θ is zero, since a horizontal tangent line would have a slope of zero. Conversely, for vertical tangent lines, we want to find where the derivative is undefined or infinite.
A general method for finding this for a polar curve involves taking the derivative of r with respect to θ and setting it equal to zero for horizontal tangents, and finding where it is undefined for vertical tangents:
Horizontal tangent: [tex]\frac{dr}{dθ}[/tex]= 0
Vertical tangent: [tex]\frac{dr}{dθ}[/tex] is undefined
Once the relevant angles are determined, we substitute them back into the original equation to find the corresponding r value, ensuring that we only consider points within the given domain of θ from 0 to 2π and r ≥ 0.
For the given polar curve r = cos(θ) + sin(θ), we take the derivative of r with respect to θ, which gives us -sin(θ) + cos(θ). Setting this equal to zero yields angles where horizontal tangents occur. Points where this derivative is undefined will give us the angles for vertical tangents. The corresponding points (r, θ) are then listed according to the instructions.
Suppose that we wanted to estimate the true average number of eggs a queen bee lays with 95 percent confidence. The margin of error we are willing to accept is 0.5. Suppose we also know that s is about 10. At minimum, what sample size should we use ?
Answer:
At least 1537 samples needed to estimate the true average number of eggs a queen bee lays with 95 percent confidence
Step-by-step explanation:
Minimum sample size required can be found using the formula
N≥[tex](\frac{z*s}{ME} )^2[/tex] where
N is the sample sizez is the corresponding z-score of 95% confidence level (1.96)s is the standard deviation (10)ME is the margin of error (0.5)then N≥[tex](\frac{1.96*10}{0.5} )^2[/tex] =1536.64
Then, at least 1537 samples needed to estimate the true average number of eggs a queen bee lays with 95 percent confidence
The minimum sample size needed is 1537 for a margin of error of 0.5.
What is margin of error?Margin of error is used to determine by what value there is deviation from the real value. Margin of error (E) is given by:
[tex]E=Z_\frac{\alpha}{2} *\frac{standard\ deviation}{\sqrt{sample\ size} }[/tex]
The 95% confidence level have a z score of 1.96. Hence for E = 0.5, standard deviation = 10. hence:
[tex]0.5=1.96*\frac{10}{\sqrt{sample\ size}}\\ \\sample\ size=1537[/tex]
The minimum sample size needed is 1537 for a margin of error of 0.5.
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The top three countries in oil consumption in a certain year are as follows: the United States, China, and Japan. In millions of barrels per day, the three top countries consume 32.2 million barrels of the world’s oil. The United States consumes 12 million more barrels a day than China. China consumes 3.2 million barrels a day than Japan. How many barrels of the world oil consumption did the United States, Japan, and China consume
Answer: Consumption of,
Japan = 4.6 million barrel,
China = 7.8 million barrel,
The US = 19.8 million barrel.
Step-by-step explanation:
Let x be the quantity of oil per day consumed by Japan,
∵ China consumes 3.2 million barrels a day than Japan.
So, the consumption of China = ( x + 3.2) million Barrel,
Also, The United States consumes 12 million more barrels a day than China.
So, the consumption of the US = (x + 3.2 + 12) = (x + 15.2) million barrel,
Thus, the total consumption of these three countries
= x + x + 3.2 + x + 15.2
= (3x + 18.4) million barrel,
According to the question,
3x + 18.4 = 32.2
3x = 32.2 - 18.4
3x = 13.8
⇒ x = 4.6
Hence, Japan, China and the US consume 4.6 million barrel, 7.8 million barrel and 19.8 million barrel respectively.
A garden hose having an internal diameter of 0.740 in. ( 1.8796 cm ) is connected to a lawn sprinkler that consists merely of an enclosure with 36 holes, each 0.055 in. ( 0.1397 cm ) in diameter. If the water in the hose has a speed of 4.00 ft/s ( 121.920 cm/s ), at what speed does it leave the sprinkler holes?
Answer:
It will leave the sprinkler at speed of [tex]v_2=613.87m/sec[/tex]
Step-by-step explanation:
We have given internal diameter of the garden hose [tex]d_1=0.740in=1.8796cm[/tex]
So radius [tex]r_1=\frac{d_1}{2}=\frac{1.8796}{2}=0.9398cm[/tex]
So area [tex]A_1=\pi r_1^2=3.14\times 0.9398^2=2.7733cm^2[/tex]
Water in the hose has a speed of 4 ft/sec
So [tex]v_1=4ft/sec=121.92cm/sec(As\ 1ft/sec\ =30.48cm/sec)[/tex]
Number of holes n = 36
Diameter of each hole [tex]d_2=0.1397cm[/tex]
So radius [tex]r_2=0.0698cm[/tex]
So area [tex]A_2=\pi r^2=3.14\times 0.0698^2=0.0153cm^2[/tex]
From continuity equation
[tex]A_1v_1=nA_2v_2[/tex]
[tex]2.7733\times 121.92=36\times 0.0153\times v_2[/tex]
[tex]v_2=613.87m/sec[/tex]
Please help asap 20 points plus brainlest
5\6 of a pound + 3\4 of a pound = ??
Answer: 19/12 pounds or 1.583 pounds
Step-by-step explanation:
Let a pound be represented by x. So x = 1 pound.
5/6 of a pound is the same as multiplying 5/6 × x. It becomes 5/6 × x = 5x/6
3/4 of a pound is the same as multiplying 3/4 × x. It becomes 3/4 × x = 3x/4
We want to look for 5/6 of a pound + 3/4 of a pound.
It becomes
5x/6 + 3x/4
Taking lowest common multiple(LCM) of 12, it becomes
= (2×5x + 3×3x) / 12
= (10x +9x)/12
= 19x /12
Substituting x = 1 pound,
It becomes 19/12 pounds
Expressing in decimal, 19/12 = 15.83
We can also solve from the beginning in terms of decimals
5/6 pounds = 0.833 pounds
3/4pounds = 0.75 pounds
5/6 pounds + 3/4 pounds = 0.833 + 0.75 = 1.583 pounds
Shelby bought a 2-ounce tube of blue paint. She used 2/3 ounce to paint the water, 3/5 ounce to paint the sky, and some to paint a flag. After that, she has 2/15 ounce left. How much paint did Shelby use to paint her flag?
Answer:
The amount of ounce used for paint is [tex]\frac{9}{15} ounce[/tex]
Step-by-step explanation:
used amount for water is [tex]\frac{2}{3}[/tex] ounce.
used amount for sky is [tex]\frac{3}{5}[/tex] ounce.
thus, the total amount used is [tex]\frac{2}{3}[/tex] + [tex]\frac{3}{5}[/tex]
= [tex]\frac{19}{15}[/tex]
Thus the total unused amount is 2 - [tex]\frac{19}{15}[/tex] = [tex]\frac{11}{15}[/tex] .
But the remaining amount is [tex]\frac{2}{15}[/tex].
Thus used amount for flag is [tex]\frac{11}{15}[/tex] - [tex]\frac{2}{15}[/tex]
= [tex]\frac{9}{15}[/tex]
graph f(x), -f(x) and y= 4x ^ 2 cos(x)
Answer:
See attached image for the graph of the function
Step-by-step explanation:
Notice that this is the product of a power function ([tex]4x^2[/tex]) times the trigonometric and periodic function cos(x). So the zeros (crossings of the x axis will be driven by the values at which they independently give zero. That is the roots of the power function (only x=0) and the many roots of the cos function: [tex]x= \frac{\pi}{2} , \frac{3\pi}{2} ,...[/tex], and their nagetiva values.
Notice that the blue curve in the graph represents the original function f(x), with its appropriate zeros (crossings of the x-axis), while the orange trace is that of "-f(x)". Of course for both the zeroes will be the same, while the rest of the curves will be the reflection over the x-axis since one is the negative of the other.
The value given below is discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. Probability of fewer than 6 passengers who do not show up for a flight Choose the correct answer below. A. The area between 5.5 and 6.5 B. The area to the left of 6.5 C. The area to the right of 6.5 D. The area to the right of 5.5 E. The area to the left of 5.5
Answer:
E. The area to the left of 5.5 and D. The area to the right of 5.5
Step-by-step explanation:
Please see attachment .
Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91.The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The day" subscript refers to the statistics day students. The "night subscript refers to the statistics night students. Assume that the standard deviations are equal. A concluding statement is:
a. There is sufficient evidence to conclude that statistics night students' mean on Exam 2 is better than the statistics day students' mean on Exam 2.
b. There is insufficient evidence to conclude that the statistics day students' mean on Exam 2 is better than the statistics night students' mean on Exam 2.
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
d. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
Answer:
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
Step-by-step explanation:
Given that a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2.
Group Group One Group Two
Mean 75.8600 75.4100
SD 16.9100 19.7300
SEM 2.8583 3.2436
N 35 37
*SEM is std error/sqrt n
Mean difference = 0.4500
[tex]H_0: \bar x = \bar y\\H_a: \bar x \neq \bar y[/tex]
(two tailed test)
Std error for difference = 4.342
Test statistic t = [tex]\frac{0.45}{4.342} \\=0.1036[/tex]
df =70
p value = 0.9178
Since p >0.05 we accept H0
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
Final answer:
The concluding statement is that there is insufficient evidence to conclude a significant difference in the means of statistics day students and statistics night students on Exam 2.
Explanation:
The hypothesis test being conducted in this scenario is a two-sample t-test. The null hypothesis, denoted as H0, states that there is no significant difference between the mean scores of statistics day students and statistics night students on Exam 2. The alternative hypothesis, denoted as Ha, states that there is a significant difference between the means of the two groups. To determine whether there is sufficient evidence to support the alternative hypothesis, we can compare the t-statistic with the critical t-value from a t-distribution table.
In this case, since the standard deviations are assumed to be equal, we can calculate the pooled standard deviation and use it to calculate the t-statistic. With the given sample means, standard deviations, and sample sizes, the calculated t-statistic is -0.246. The critical t-value for a two-tailed test with a significance level of 0.05 and 70 degrees of freedom is approximately 1.994. Since the calculated t-statistic (-0.246) falls within the range between -1.994 and 1.994, we fail to reject the null hypothesis.
Therefore, the concluding statement is:
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2.
An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273–279) "Arthroscopic meniscal repair with an absorbable screw: results and surgical technique" showed that only 25 out of 37 tears (67.6%) located between 3 and 6 mm from the meniscus rim were healed.
(a) Calculate a 99% two-sided confidence interval on the proportion of such tears that will heal. Round the answers to 3 decimal places.
(b) Calculate a 99% lower confidence bound on the proportion of such tears that will heal. Round the answer to 3 decimal places.
Answer:
63
Step-by-step explanation:
The lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)
How to find the confidence interval for population proportion from large sample?Suppose we're given that:
Favorable Cases X (in count, in sample)Sample Size N Level of significance = [tex]\alpha[/tex]Then, the sample proportion of favorable cases is:
[tex]\hat{p} = \dfrac{X}{N}[/tex]
The critical value at the level of significance [tex]\alpha[/tex] is [tex]Z_{1- \alpha/2}[/tex]
The corresponding confidence interval is:
[tex]CI = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right)[/tex]
We need to construct the 99% confidence interval for the population proportion.
We have been provided with the following information about the number of favorable cases(the number of tears located between 3 and 6 mm from the meniscus rim were healed):
Favorable Cases X =25Sample Size N =37The sample proportion is computed as follows, based on the sample size N = 37 and the number of favorable cases X = 25
[tex]\hat p = \displaystyle \frac{X}{N} = \displaystyle \frac{25}{37} = 0.676[/tex]
The critical value for [tex]\alpha = 0.01[/tex] is [tex]z_c = z_{1-\alpha/2} = 2.576[/tex]
The corresponding confidence interval is computed as shown below:
[tex]\begin{array}{ccl} CI & = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right) \\\\ \\\\ & = & \displaystyle \left( 0.676 - 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}}, 0.676 + 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}} \right) \\\\ \\\\ & = & (0.477, 0.874) \end{array}[/tex]
Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.477<p<0.874, which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.477, 0.874).
Thus, the lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)
Learn more about confidence interval for population proportion here:
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Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) for 0 ≤ t ≤ 2π), if the density (in grams/cm) of the wire at any point is equal to the square of the distance from the origin to the point
Answer:
Mass
[tex]\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)[/tex]
Center of mass
Coordinate x
[tex]\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]
Coordinate y
[tex]\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]
Coordinate z
[tex]\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]
Step-by-step explanation:
Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions
x(t) = t
y(t) = 4 cos(t)
z(t) = 4 sin(t)
for 0 ≤ t ≤ 2π
If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W
[tex]m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt[/tex]
The density D(x,y,z) is given by
[tex]D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16[/tex]
on the other hand
[tex]||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}[/tex]
and we have
[tex]m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)[/tex]
The center of mass is the point [tex](\bar x,\bar y,\bar z)[/tex]
where
[tex]\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)[/tex]
We have
[tex]\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)[/tex]
so
[tex]\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]
[tex]\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi[/tex]
[tex]\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]
[tex]\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi[/tex]
[tex]\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]
Final answer:
To find the mass and center of mass of the wire loop in the shape of a helix, we need to integrate the density function along the length of the wire. The formula for the mass of a wire is m = ∫(ρ dl), where ρ is the density function and dl is an infinitesimally small segment of the wire. To find the center of mass, we can use the formulas Cx = (1/m) ∫(x ρ dl), Cy = (1/m) ∫(y ρ dl), and Cz = (1/m) ∫(z ρ dl), where x, y, and z are the parametric equations and ρ is the density function.
Explanation:
To find the mass and center of mass of the wire loop in the shape of a helix, we need to integrate the density function along the length of the wire. Since the density is equal to the square of the distance from the origin to the point, we can use the formula for the mass of a wire:
m = ∫(ρ dl)
Where ρ is the density function and dl is an infinitesimally small segment of the wire. In this case, we have:
ρ = r² = (x² + y² + z²)
Using the given parametric equations for x, y, and z, we can substitute them into the formula and integrate along the range of t:
m = ∫(t² + (4cos(t))² + (4sin(t))²) dt
Next, to find the center of mass, we can use the formula:
Cx = (1/m) ∫(x ρ dl)
Cy = (1/m) ∫(y ρ dl)
Cz = (1/m) ∫(z ρ dl)
We can substitute the parametric equations for x, y, and z, and the density function into these formulas and integrate along the range of t to find the center of mass.
In a little over a month, from June 5, 1879, to July 2, 1879, Albert Michelson measured the velocity of light in air 100 times (Stigler, Annals of Statistics, 1977). Today we know that the true value is 299,734.5 km/sec. Michelson’s data have a mean of 299,852.4 km/sec with a standard deviation of 79.01. (a) Find a two-sided 95% confidence interval for the true mean (the true value of the speed of light). (b) What does the confidence interval say about the accuracy of Michelson’s measurements?
Answer:
(a) According to the Albert Michelson's measurements, two-sided 95% confidence interval for velocity of light in the air would be 299,852.4±15.5
(b) Since true velocity of light in the air is not included in the 95% confidence interval, we can say that Michelson’s measurements were not accurate.
Step-by-step explanation:
Confidence Interval can be calculated using M±ME where
M is the sample mean velocity of light in air (299,852.4)ME is the margin of error from the meanAnd margin of error (ME) from the mean can be calculated using the formula
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the corresponding statistic in 95% confidence level (1.96)s is the sample standard deviation (79.01)N is the sample size (100)Then ME=[tex]\frac{1.96*79.01}{\sqrt{100} }[/tex] ≈ 15.5
According to the Albert Michelson's measurements, 95% confidence interval for velocity of light in the air would be 299,852.4±15.5
The diameter of bushings turned out by a manufacturing process is a normally distributed random variable with a mean of 4.035 mm and a standard deviation of 0.005 mm. A sample of 25 bushings is taken once an hour. (a) Within what interval should 95 percent of the bushing diameters fall? (Round your answers to 4 decimal places.) The 95% confidence interval is from to (b) Within what interval should 95 percent of the sample means fall? (Round your answers to 5 decimal places.) The 95% confidence interval is from to (c-1) What conclusion would you reach if you saw a sample mean of 4.020? The sample came from a population that a population mean equal to 4.035. (c-2) What conclusion would you reach if you saw a sample mean of 4.055? The sample came from a population that a population mean equal to 4.035.
A) the z score for 95% is 1.96
Multiply by the deviation:
1.96 x 0.005 = 0.0098
Now add and then subtract that from the mean:
4.035 - 0.0098 = 4.0252
4.035 + 0.0098 = 4.0448
The interval is (4.0252, 4.0448)
B) 4.035 +/- 1.96 x sqrt( 0.005/sqrt(25))
= 4.035 +/-0.00196
Answer: ( 4.033, 4.037)
C) the conclusion is that both 4.020 and 4.055 are out of the range.
4.020 is below the lowest range and 5.055 is higher than highest range.
The 95% confidence interval for the sample means is; CI = (4.033, 4.037)
What is the Confidence Interval?A) We are given;
Mean; x' = 4.035 mm
standard deviation; σ = 0.005 mm
sample size; n = 25
Formula for confidence interval is;
CI = x' ± z(σ/√n)
The z score for 95% confidence level is 1.96. Thus;
CI = 4.035 ± 1.96(0.005)
CI = 4.035 ± 0.0098
CI = (4.025, 4.045)
B) From the formula for confidence interval earlier stated, we have;
CI = 4.035 ± 1.96(0.005/√(25))
CI = 4.035 ± 0.00196
CI = (4.033, 4.037)
C)i) The conclusion is that both 4.020 is out of the range of the confidence interval.
ii) The conclusion is that 4.055 is higher than the confidence interval
Read more about Confidence Interval at; https://brainly.com/question/17097944
Why i got low score?
Answer:
There is exactly one real root.
Step-by-step explanation:
There are two parts to arrive at the solution.
(i) The polynomial has at least one real root.
(ii) The polynomial has exactly one real root.
We prove (i) using Intermediary Value Theorem.
f(x) = x³ + x - 1 = 0 is a polynomial. So, it is continuous.
At x = 1, f(1) = 1
At x = o, f(0) = -1
Since, there is a change of sign it should have crossed through zero.
Now, to prove there is exactly one real root we use Rolle's theorem.
Let us assume there are two real roots to the polynomial, say 'a' and 'b'.
Then f(a) = 0 and f(b) = 0.
⇒ f(a) = f(b)
To use Rolle's theorem we need the function to be continuous, differentiable and for any two points a,b f(a) = f(b) there should exist a 'c' such that f'(c) = 0.
Now, f'(x) = 3x² + 1
Note that f'(x) is always greater than equal to 1.
It can never be zero for any c. This contradicts Rolle's Theorem. o, our assumption that two real roots exist must be wrong.
Hence, we conclude that there is exactly one real root to the polynomial.
Based on information from a large insurance company, 68% of all damage liability claims are made by single people under the age of 25. A random sample of 53 claims showed that 41 were made by single people under the age of 25. Does this indicate that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company? State the null and alternate hypothesis then give the test statistic and your conclusion.
a)z = 2.326; reject Hoat the 5% significance level
b)z = 1.826; reject Hoat the 5% significance level
c)z = -2.326; reject Hoat the 5% significance level
d)z = 2.326; fail to reject Hoat the 5% significance level
e)z = -1.826; fail to reject Hoat the 5% significance level
Answer:
z=1.461, fail to reject the null hypothesis since [tex]p_v>\alpha[/tex] at 5% of singificance.
Step-by-step explanation:
1) Data given and notation
n=53 represent the random sample taken
X=41 represent the adults with damage liability claims by single people under the age of 25.
[tex]\hat p=\frac{41}{53}=0.774[/tex] estimated proportion of adults with damage liability claims by single people under the age of 25.
[tex]p_o=0.68[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company (68%).:
Null hypothesis:[tex]p\leq 0.68[/tex]
Alternative hypothesis:[tex]p > 0.68[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.774 -0.68}{\sqrt{\frac{0.68(1-0.68)}{53}}}=1.461[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a unilateral right tailed test the p value would be:
[tex]p_v =P(z>1.461)=0.072[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with damage liability claims by single people under the age of 25 is not significantly higher from 0.68 or 68% .
Which sounds more reasonable for the weight of 250 grams ? A - a toy truck B- a pickup truck or C- a monster truck
Answer:
Toy truck. 1 lb is equal to 453 grams so a toy truck would be less than a lb
Answer:
Step-by-step explanation:
was
(1 point) The number of households with cable TV service in a certain community, N, begins at only 55 households and has seen a fivefold increase every 17 years. Give the constants a, b, and T so that N is represented by a function of the form N=abt/T, where t is the time in years since N was first measured.
Answer:
a=55
b=5
T=17
Step-by-step explanation:
The general form of the equation is:
[tex]N(t)=ab^{\frac{t}{T}}[/tex]
For t = 0:
[tex]N(0)=ab^{\frac{0}{T}}\\N(0) = a = 55\\a=55[/tex]
Since there has been a fivefold increase after 17 years, at t = 17, N(17) = 55*5
[tex]N(17)=55b^{\frac{17}{T}}\\55*5 = 55b^{\frac{17}{T}}\\b^{\frac{17}{T}} = 5[/tex]
If at every 17*n years there in an increase of 5^n, one can deduct that the values for T and b are respectively 17 and 5:
[tex]b^{\frac{t}{T}}= 5^{\frac{17n}{17}}[/tex]
Therefore, the function that represents N(t) is:
[tex]N(t)=55*5^{\frac{t}{17}}[/tex]
The constants for the function representing the number of households with cable TV service are: a = 55 (initial number of households), b = 5 (fivefold increase), and T = 17 (the number of years for each fivefold increase). The function is N = 55 x 5^(t/17).
The student is asking to define the constants a, b, and T for the function representing the number of households N with cable TV service. The function is of the form N = abt/T, where t is the time in years since the households were first measured, N initially is 55, and the number of households increases fivefold every 17 years.
To find constants a, b, and T, first note that a is the initial value of households with cable TV service, so a = 55. Since the number increases fivefold every 17 years, b is the factor of increase which is 5. The time period for this increase, T, is every 17 years, so T = 17.
Therefore, the function representing the number of households with cable TV service is N = 55 x 5t/17.
Curls and divergences * Calculate the curl and the divergence of each of the following vec- tor fields. If the curl turns out to be zero, try to discover a scalar function φ of which the vector field is the gradient.
(a) F=(x+y,−x+y,−2z);
(b) G=(2y,2x+3z,3y);
(c) H=(x2 −z2,2,2xz).
Answer:
a) [tex]\nabla F=0, \nabla\times F=(0,0,-2)[/tex]
b) [tex]\nabla F=0, \nabla\times F=(0,0,0)[/tex], [tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].
c) [tex]\nabla F=4x, \nabla\times F=(0,-4z,0)[/tex]
Step-by-step explanation:
Remember, if F= <f,g,h> is a vector field and [tex]\nabla[/tex] is the operator [tex]\nabla=<\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}>[/tex]
the divergence of F is [tex]\nabla F=<\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}>\cdot F= \frac{\partial f}{\partial x}+\frac{\partial g}{\partial y} +\frac{\partial h}{\partial z}[/tex]the curl of F is [tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\f&g&h\end{array}\right] )[/tex]a) [tex]F=(x+y,-x+y,-2z)[/tex]
The divergence of F is
[tex]\nabla F=\frac{\partial}{\partial x}(x+y)+\frac{\partial}{\partial y}(-x+y)+\frac{\partial}{\partial z} (-2z)=1+1-2=0[/tex]
The curl of F is
[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x+y&-x+y&-2z\end{array}\right])\\=i(\frac{\partial}{\partial y}(-2z)-\frac{\partial}{\partial z}(-x+y))-j(\frac{\partial}{\partial x}(-2z)-\frac{\partial}{\partial z}(x+y))+k(\frac{\partial}{\partial x}(-x+y)-\frac{\partial}{\partial y}(x+y))=0i-0j-2k=(0,0,-2)[/tex]
b) [tex]F=(2y,2x+3z,3y)[/tex]
The divergence of F is
[tex]\nabla F=\frac{\partial}{\partial x}(2y)+\frac{\partial}{\partial y}(2x+3z)+\frac{\partial}{\partial z} (3y)=0+0+0=0[/tex]
The curl of F is
[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\2y&2x+3z&3y\end{array}\right])\\=i(\frac{\partial}{\partial y}(3y)-\frac{\partial}{\partial z}(2x+3z))-j(\frac{\partial}{\partial x}(3y)-\frac{\partial}{\partial z}(2y))+k(\frac{\partial}{\partial x}(2x+-3z)-\frac{\partial}{\partial y}(2y))=0i-0j+0k=(0,0,0)[/tex]
Since the curl of F is 0 the we will try find f such that the gradient of f be F.
Since [tex]f_x=2y[/tex], [tex]f=2xy+g(y,z)[/tex].
Since [tex]f_y=2x+3z[/tex], [tex]2x+3z=f_y=2x+g_y(y,z)\\3z=g_y[/tex].
Since [tex]f_z=3y[/tex] and [tex]f_z=3y+h'(z)[/tex], then [tex]h'(z)=0[/tex]. Thins means that [tex]h(z)=C, C\in\mathbb{R}[/tex]
Therefore,
[tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].
c) [tex]H=(x^2-z^2,2,2xz)[/tex]
The divergence of F is
[tex]\nabla F=\frac{\partial}{\partial x}(x^2-z^2)+\frac{\partial}{\partial y}(2)+\frac{\partial}{\partial z} (2xz)=2x+0+2x=4x[/tex]
The curl of F is
[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2-z^2&2&2xz\end{array}\right])\\=i(\frac{\partial}{\partial y}(2xz)-\frac{\partial}{\partial z}(2))-j(\frac{\partial}{\partial x}(2xz)-\frac{\partial}{\partial z}(x^2-z^2))+k(\frac{\partial}{\partial x}(2)-\frac{\partial}{\partial y}(x^2-z^2))=0i-(2z-(-2z))j-0k=(0,-4z,0)[/tex]
Find the Laplace transform of the given function; a and b are real constants. f(t) = eat sinh bt Your answer should be an expression in terms of a, b and s. L{f(t)}(s) = F(s) =
The Laplace transform of the function f(t) = eat sinh bt is F(s) = (e^a)/((s - a)^2 + b^2) after manipulating and simplifying using the standard Laplace transform rules.
Explanation:The Laplace transform of the function f(t) = eat sinh bt is given by the formula:
F(s) = L{f(t)}(s) = Int0->infinity[ e^(at - s)*sinh(bt) dt]
By splitting the hyperbolic sine function, sinh(bt), into exponentials, we get:
F(s) = 0.5*Int0->infinity [ (e^(-s + a + b)*t - e^(-s + a - b)*t) dt]
As a result, after applying Laplace transform rules and simplifying, we get:
F(s) = (e^a)/((s - a)^2 + b^2).
Learn more about Laplace Transform here:https://brainly.com/question/31481915
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A pizza parlor offers a choice of 12 different toppings. How many 4-topping pizzas are possible? (no double orders of toppings are allowed)
In this situation, does the order of toppings matter? Yes/No?
There are ____ possible 4-topping pizzas.
9514 1404 393
Answer:
No; 495
Step-by-step explanation:
For many folks, the order of application of toppings will not matter. For those folks, the number of possible 4-topping pizzas is
C(12,4) = 495 = 12!/(8!×4!)
There are 495 possible 4-topping pizzas.
A large manufacturing firm tests job applicants who recently graduated from college. The test scores are normally distributed with a mean of 500 and a standard deviation of 50. Management is considering placing a new hire in an upper level management position if the person scores in the upper 6 percent of the distribution. What is the lowest score a college graduate must earn to qualify for a responsible position?
Answer:
The lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 50
We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.06.
P(X > x) = 6% = 0.06
[tex]P( X > x) = P( z > \displaystyle\frac{x - 500}{50})=0.06[/tex]
[tex]= 1 - P( z \leq \displaystyle\frac{x - 500}{50})=0.06 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 500}{50})=1-0.06=0.94 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z < 1.555) = 0.94[/tex]
[tex]\displaystyle\frac{x - 500}{50} = 1.555\\x = 577.75[/tex]
Hence, the lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.