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.
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]
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]
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
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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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
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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4. A process has been developed that can transform ordinary iron into a kind of super iron called metallic glass. Metallic glass is three to four times stronger than the toughest steel alloys. To estimate the mean temperature, u, at which a particular type of metallic glass becomes brittle, 25 pieces of this metallic glass were randomly sampled from a recent production run. Each piece was subjected to higher and higher temperatures until it became brittle. The temperature at which brittleness first appeared was recorded for each piece in the sample. The following results were obtained: X 480°F and s 11°F Construct a 95% confidence interval to estimate . (20 points)
Answer:
The 95% confidence interval to estimate is 475.4592 to 484.5408
Step-by-step explanation:
Consider the provided information.
25 pieces of this metallic glass were randomly sampled from a recent production run.
That means the value of n is 25.
The degree of freedom is:
df = n-1
df = 25-1 = 24
X = 480°F and s = 11°F
We need to Construct a 95% confidence interval to estimate.
1-α=0.95
α=0.05
It is a two tail test with small sample size.
Determine the value of t by using Degrees of freedom and Significance level:
The required t value is 2.064
[tex]95\% CI=\bar x\pm t_c\times \frac{s}{\sqrt{n}}[/tex]
Substitute the respective values as shown:
[tex]95\% CI=480\pm 2.064\times \frac{11}{\sqrt{25}}[/tex]
[tex]95\% CI=480\pm 4.5408[/tex]
[tex]95\% CI=475.4592\ to\ 484.5408[/tex]
Hence, the 95% confidence interval to estimate is 475.4592 to 484.5408
Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval to estimate the temperature at which brittleness first appeared is given by (475.46 ºF, 484.54 ºF).
What is a t-distribution confidence interval?The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
[tex]\overline{x}[/tex] is the sample mean.t is the critical value.n is the sample size.s is the standard deviation for the sample.The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 25 - 1 = 24 df, is t = 2.0639.
The other parameters are as follows:
[tex]n = 25, \overline{x} = 480, s = 11[/tex]
Hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 480 - 2.0639\frac{11}{\sqrt{25}} = 475.46[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 480 + 2.0639\frac{11}{\sqrt{25}} = 484.54[/tex]
The 95% confidence interval is (475.46 ºF, 484.54 ºF).
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The weight (in pounds) for a population of school-aged children is normally distributed with a mean equal to 133 ± 25 pounds (μ ± σ). Suppose we select a sample of 100 children (n = 100) to test whether children in this population are gaining weight at a 0.05 level of significance. Part (a) What are the null and alternative hypotheses? H0: μ ≤ 133 H1: μ > 133 H0: μ = 133 H1: μ < 133 H0: μ = 133 H1: μ ≠ 133 H0: μ ≤ 133 H1: μ = 133
Answer:
Null hypothesis: [tex]\mu \leq 133[/tex]
Alternative hypothesis :[tex]\mu>133[/tex]
Step-by-step explanation:
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Data given and notation
[tex]\bar X[/tex] represent the mean breaking strength value for the sample
[tex]\sigma=25[/tex] represent the population standard deviation
[tex]n=100[/tex] sample size
[tex]\mu_o =133[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
Is a one right tailed test.
What are H0 and Ha for this study?
We want to test if the children in this population are gaining weight, so we want to test if the mean increase from the reference value or no.
Null hypothesis: [tex]\mu \leq 133[/tex]
Alternative hypothesis :[tex]\mu>133[/tex]
Final answer:
The correct null and alternative hypotheses for testing whether children are gaining weight compared to a known mean would be H0: μ = 133 and H1: μ > 133, respectively, representing a one-tailed test.
Explanation:
When a student is asking about the null and alternative hypotheses for a population's weight distribution, they are referring to the initial assumptions and contrasting propositions made in a statistical test.
Part (a) of the question suggests that we are trying to test if children in this population are gaining weight. Therefore, the correct set of hypotheses to test this claim would be:
H0: μ = 133 (the null hypothesis, stating that the mean weight is equal to 133 pounds)
H1: μ > 133 (the alternative hypothesis, stating that the mean weight is greater than 133 pounds).
This setup represents a one-tailed test because we are only interested in whether the weight has increased, not if it has changed in any direction.
A survey of 300 randomly selected high school students determined that 236 play organized sports.
(a) What is the probability that a randomly selected high school student plays organized sports?
(b) Interpret this probability.
Answer:
Step-by-step explanation:
(a) The probability that randomly selected high school student plays organized sport is 236/300 = 59/75 = 0.787 = 78.7%.
(b) The probability means that for every 75 random students selected, 59 of them plays organized sport.
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.
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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We don't know f (x) but we do know that x = 1 is a critical number of f' (x) and we know f" (x)=12x. Use this information to decide if f(x) has a relative minimum or maximum at x=1.
Answer:
relative minimum
Step-by-step explanation:
From f^''(x) = 12x we can take the integration to find out what f'(x) is:
[tex]f'(x) = 6x^2 + C[/tex]
Furthermore, we can substitute x = 1 for f''(x) to find out whether it's positive or negative
f''(1) = 12*1 = 12 > 0
So if x=1 is a critical point of f'(x) and f''(x=1) > 0 then that point is a relative minimum point
Using the second derivative test with the given information f"(x)=12x, it's concluded that f(x) has a relative minimum at x=1, as f"(1) is positive.
Explanation:To determine if f(x) has a relative minimum or maximum at x=1, we can use the given information that f"(x)=12x and that x=1 is a critical number of f'(x). Since the second derivative, f"(x), is 12x, at x = 1, it is positive (f"(1)=12(1)=12). According to the second derivative test, a positive second derivative at a critical point indicates that the function has a relative minimum at that point. Therefore, we can conclude that f(x) has a relative minimum at x=1.
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:
https://brainly.com/question/16427118
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.
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.
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.
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% .
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.
(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.
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]
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
A box contains 10 tags, numbered 1 through 10, with a different number on each tag. A second box contains 8 tags, numbered 20 through 27, with a different number on each tag. One tag is drawn at random from each box. What is the expected value of the sum of the numbers on the two selected tags?
(A) 13.5(B) 14.5(C) 15.0(D) 27.0(E) 29.0
Answer:
A
Step-by-step explanation:
The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a standard deviation of 1.91. Find the probability that a randomly selected individual will take less than 6 minutes to select a shoe purchase. Is this outcome unusual?
Answer: 0.0918, it is not unusual.
Step-by-step explanation:
Given : The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a standard deviation of 1.91.
i.e. [tex]\mu=8.54[/tex] minutes and [tex]\sigma= 1.91[/tex] minutes
Let x denotes the length of time a person takes to decide which shoes to purchase.
Formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Then, the probability that a randomly selected individual will take less than 6 minutes to select a shoe purchase will be :-
[tex]\text{P-value=}P(x<6)=P(\dfrac{x-\mu}{\sigma}<\dfrac{6-8.54}{1.91})\\\\\approx P(z<1.33)=1-P(z<1.33)\ \ \ [\becaus\ P(Z<-z)=1-P(Z<z)]\\\\=1-0.9082\ \ [\text{By using z-value}]=0.0918[/tex]
Thus , the required probability = 0.0918
Since, P-value (0.0918) >0.05 , it means this outcome is not unusual.
[Note : When a outcome is unusual then the probability of its happening is less than or equal to 0.05. ]
rewrite the statement in conditional form. lines with slopes 2/3 and -3/2 are perpendicular
Answer:
Step-by-step explanation:
The slope of a perpendicular line is the negative reciprocal of the slope of the original line. Let line x represent the original line and let line y represent the line that is perpendicular to line x
A conditional statement is an if - then statement. It is connected by a hypothesis statement and a conclusion statement. The hypothesis statement is "if the slope of line x is 2/3 and the slope of line y is -3/2 which is its negative reciprocal
The conclusion statement is " then, line x is perpendicular to line y
So the combined statement is
if the slope of line x is 2/3 and the slope of line y is -3/2 which is its negative reciprocal, then line x is perpendicular to line y
Determine the area under the standard normal curve that lies to the left of (a) Upper Z equals negative 1.45 commaZ=−1.45, (b) Upper Z equals 0.63Z=0.63, (c) Upper Z equals 1.48Z=1.48, and(d) Upper Z equals negative 1.37Z=−1.37. LOADING... Click the icon to view a table of areas under the normal curve. (a) The area to the left of Zequals=negative 1.45−1.45 is nothing. (Round to four decimal places as needed.)
Answer:
a) [tex]P(-1.45<Z<1.45)=0.853[/tex]
b) [tex]P(-1.63<Z<1.63)=0.8968[/tex]
c) [tex]P(-1.48<Z<1.48)=0.8612[/tex]
d) [tex]P(-1.37<Z<1.37)=0.8294[/tex]
Step-by-step explanation:
To find : Determine the area under the standard normal curve that lies ?
Solution :
a) In between Z=-1.45 and Z=1.45
i.e. [tex]P(-1.45<Z<1.45)[/tex]
Now, [tex]P(-1.45<Z<1.45)=P(Z<1.45)-P(Z<-1.45)[/tex]
Using Z-table,
[tex]P(-1.45<Z<1.45)=0.9265-0.0735[/tex]
[tex]P(-1.45<Z<1.45)=0.853[/tex]
b) In between Z=-1.63 and Z=1.63
i.e. [tex]P(-1.63<Z<1.63)[/tex]
Now, [tex]P(-1.63<Z<1.63)=P(Z<1.63)-P(Z<-1.63)[/tex]
Using Z-table,
[tex]P(-1.63<Z<1.63)=0.9484-0.0516[/tex]
[tex]P(-1.63<Z<1.63)=0.8968[/tex]
c) In between Z=-1.48 and Z=1.48
i.e. [tex]P(-1.48<Z<1.48)[/tex]
Now, [tex]P(-1.48<Z<1.48)=P(Z<1.48)-P(Z<-1.48)[/tex]
Using Z-table,
[tex]P(-1.48<Z<1.48)=0.9306-0.0694[/tex]
[tex]P(-1.48<Z<1.48)=0.8612[/tex]
d) In between Z=-1.37 and Z=1.37
i.e. [tex]P(-1.37<Z<1.37)[/tex]
Now, [tex]P(-1.37<Z<1.37)=P(Z<1.37)-P(Z<-1.37)[/tex]
Using Z-table,
[tex]P(-1.37<Z<1.37)=0.9147-0.0853[/tex]
[tex]P(-1.37<Z<1.37)=0.8294[/tex]
Finding the area under the standard normal curve to the left of specific Z-scores involves looking up these Z-values in a Z-table or inputting them into an appropriate calculator. The Z-scores in question range from -1.45 to 1.48 and the related cumulative probabilities represent the area under the curve.
Explanation:The question relates to finding the area under the standard normal curve, commonly referred to in statistics as Z-scores. These standardized scores indicate how many standard deviations away from the mean a particular point or score is located.
(a) Z equals negative 1.45, meaning it lies 1.45 standard deviations below the mean. When looking up this value in a standard Z-table (or using a calculator), you will find the associated cumulative probability (the area to the left under the curve).
The same process applies to (b) Z equals 0.63 (lies 0.63 standard deviations above the mean), (c) Z equals 1.48 (1.48 standard deviations above the mean) and (d) Z equals negative 1.37 (1.37 standard deviations below the mean). The exact cumulative probabilities vary with each Z-score and these represent the area under the curve to the left of each point.
Learn more about Z-Scores & Area under Normal Curve here:https://brainly.com/question/13734111
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$4000 is deposited in an account that pays an APR of 8.4% compounded annually. How long will it take for the balance to reach $120,000?
Answer: number of years that it will take for the balance to reach $120,000 is 42 years
Step-by-step explanation:
Initial amount deposited into the account is $4000. This means that the principal is $4000
P = 4000
It was compounded annually. This means that it was compounded once in a year. So
n = 1
The rate at which the principal was compounded is 8.4%. So
r = 8.4/100 = 0.084
Let the number of years that it will take for the balance to reach $120,000. It means that it was compounded for a total of t years.
Amount, A at the end of t years is $120,000
The formula for compound interest is
A = P(1+r/n)^nt
120000 = 4000(1 + 0.084/1)^1×t
120000/4000 = 1.084^t
30 = 1.084^t
t = 42 years
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.
sample of 60 account balances of a credit company showed an average balance of $1,165 and a standard deviation of $125. You want to determine if the mean of all account balances is significantly greater than $1,150. Assume the population of account balances is normally distributed. Compute the p-value for this test.
Answer:
z=0.930
[tex]p_v =P(z>0.930)=1-P(z<0.930)=1-0.824=0.176[/tex]
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=1165[/tex] represent the mean for the account balances of a credit company
[tex]s=125[/tex] represent the population standard deviation for the sample
[tex]n=60[/tex] sample size
[tex]\mu_o =1150[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean for account balances of a credit company is greater than 1150, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 1150[/tex]
Alternative hypothesis:[tex]\mu > 1150[/tex]
We don't know the population deviation, but the problem says the the distribution for the random variable is normal, so for this case we can use the z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{1165-1150}{\frac{125}{\sqrt{60}}}=0.930[/tex]
4) Calculate the P-value
Since is a one-side upper test the p value would be:
[tex]p_v =P(z>0.930)=1-P(z<0.930)=1-0.824=0.176[/tex]
In Excel we can use the following formula to find the p value "=1-NORM.DIST(0.93,0,1,TRUE)"
5) Conclusion
If we compare the p value with a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the balances of a credit company are significantly higher than $1150 at 0.05 of signficance.
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]
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
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