Stokes' theorem says the integral of the curl of [tex]\vec F[/tex] over a surface [tex]S[/tex] with boundary [tex]C[/tex] is equal to the integral of [tex]\vec F[/tex] along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that
[tex]\displaystyle\iint_S\nabla\times\vec F\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r[/tex]
We have
[tex]\vec F(x,y,z)=x^2\,\vec\imath+4x\,\vec\jmath+z^2\,\vec k[/tex]
[tex]\implies\nabla\times\vec F(x,y,z)=4\,\vec k[/tex]
Parameterize the ellipse [tex]S[/tex] by
[tex]\vec s(u,v)=\dfrac{u\cos v}{\sqrt5}\,\vec\imath+\dfrac{u\sqrt5\sin v}4\,\vec\jmath[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial\vec u}\times\dfrac{\partial\vec s}{\partial\vec v}=\dfrac u4\,\vec k[/tex]
Then the flux of the curl is
[tex]\displaystyle\iint_S4\,\vec k\cdot\dfrac u4\,\vec k\,\mathrm dA=\int_0^{2\pi}\int_0^1u\,\mathrm du\,\mathrm dv=\boxed{\pi}[/tex]
An eight-sided die, which may or may not be a fair die, has four colors on it; you have been tossing the die for an hour and have recorded the color rolled for each toss. What is the probability you will roll an orange on your next toss of the die? Express your answer as a simplified fraction or a decimal rounded to four decimal places.
Answer:
Probability of an Orange on the next toss
= (23/60) = 0.3833
Step-by-step explanation:
Orange: 46
Brown: 23
Green: 32
Yellow: 19
Probability of an Orange on the next toss
= n(orange colours obtained in the tosses) ÷ n(number of tosses)
n(orange colours in the tosses) = 46
Total number of tosses = 46 + 23 + 32 + 19
= 120.
Probability of an Orange on the next toss
= (46/120) = (23/60) = 0.3833
Hope this Helps!!!!
My computer is not letting me take a good quality of picture, but explanation and answer ?
Answer:
There will be 6 buses needed; the number of buses has to be a whole number.
Step-by-step explanation:
So we know that there is 232 people, and each bus holds 44 people. To find how many buses we need, you divide 232/44. Then, you get 5 with 12 extra. So you add one more bus, but that bus isn't full; there will only be 12 people inside it.
The reason why the number of buses has to be a whole number is because a bus has to be a whole bus, even though the bus isn't full. The maximum capacity of the bus is 44, but you can have 12 people inside it.
Solve the following:
1)If 32x=2, find x
2) simplify 9-1\2
Answer:
the answers are
1) 1/16
2) -18
Answer:
[tex]1)32x = 2 \\ \frac{32x}{32} = \frac{2}{32} \\ x = \frac{1}{16} [/tex]
[tex]2)9 - \frac{1}{2} \\ \frac{9}{1} \frac{ \times }{ \times } \frac{2}{2} - \frac{1}{2} \\ \frac{18}{2} - \frac{1}{2} \\ = \frac{17}{2} \\ = 8 \frac{1}{2} [/tex]
The price-earnings (PE) ratios of a sample of stocks have a mean value of 10.5 and a standard deviation of 3. If the PE ratios have a normal distribution, use the Empirical Rule (also called the 68-95-99.7 Rule) to estimate the percentage of PE ratios that fall between:
Answer:
a) 68% falls within 7.5 and 13.5
b) 95% falls within 4.5 and 16.5
c) 99.7% falls within 1.5 and 19.5
Step-by-step explanation:
Given that:
mean (μ) = 10.5, Standard deviation (σ) = 3.
The Empirical Rule (also called the 68-95-99.7 Rule) for a normal distribution states that all data falls within three standard deviations. That is 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
Therefore using the empirical formula:
68% falls within the first standard deviation (µ ± σ) = (10.5 ± 3) = (7.5, 13.5)
95% within the first two standard deviations (µ ± 2σ) = (10.5 ± 2(3)) = (10.5 ± 6) = (4.5, 16.5)
99.7% within the first three standard deviations (µ ± 3σ) = (10.5 ± 3(3)) = (10.5 ± 9) = (1.5, 19.5)
The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.
Answer:
(a) [tex]X\sim N(0.10,\ 0.02^{2})[/tex].
(b) The probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.
(c) The 60th percentile is 11%.
Step-by-step explanation:
The complete question is:
The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.
A. X ~ N( , )
B. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11%.
C.Find the 60th percentile for this distribution.
Solution:
The random variable X is defined as the THC content for a bag of marijuana that is sold on the street.
(a)
The random variable X is Normally distributed with mean, μ = 0.10 and standard deviation, σ = 0.02.
Thus, [tex]X\sim N(0.10,\ 0.02^{2})[/tex].
(b)
Compute the value of P (X > 0.11) as follows:
[tex]P(X>0.11)=P(\frac{X-\mu}{\sigma}>\frac{0.11-0.10}{0.02})\\=P(Z>0.50)\\=1-P(Z<0.50)\\=1-0.69146\\=0.30854\\\approx 0.31[/tex]
*Use a z-table.
Thus, the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.
(c)
The pth percentile is a data value such that at least p% of the data-set is less than or equal to this data value and at least (100 - p)% of the data-set are more than or equal to this data value.
Let x represent the 60th percentile value.
The, P (X < x) = 0.60.
⇒ P (Z < z) = 0.60
The value of z for this probability is,
z = 0.26.
Compute the value of x as follows:
[tex]z=\farc{x-\mu}{\sigma}\\0.26=\farc{x-0.10}{0.02}\\x=0.10+(0.26\times 0.02)\\x=0.1052\\x\approx 0.11[/tex]
Thus, the 60th percentile is 11%.
A bag of marbles contains 3 yellow, 6 blue, 1 green, 8 red, 2 orange. Find the probability: P(yellow, then orange WITHOUT replacement)
Answer:
P(yellow, then orange WITHOUT replacement) = 0.0158
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes
This problem asks:
The probability of first getting a yellow marble, then an orange marble, without replacemtn.
Yellow marble first:
3 yellow marbles out of 3+6+1+8+2 = 20 marbles.
So P(a) = 3/20.
Then the orange marble:
Now the yellow marble is retired.
2 orange marbles out of 19.
So P(b) = 2/19
Probability:
P(yellow, then orange WITHOUT replacement) = P(a)P(b) = (3/20)*(2/19) = 0.0158
Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8) (a) the tangent plane (b) the normal line (x(t), y(t), z(t)) =
Let [tex]f(x,y,z)=x+y+z-8e^{xyz}[/tex]. The tangent plane to the surface at (0, 0, 8) is
[tex]\nabla f(0,0,8)\cdot(x,y,z-8)=0[/tex]
The gradient is
[tex]\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)[/tex]
so the tangent plane's equation is
[tex](1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8[/tex]
The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by [tex]t[/tex], then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation
[tex](1,1,1)t+(0,0,8)=(t,t,t+8)[/tex]
or [tex]x(t)=t[/tex], [tex]y(t)=t[/tex], and [tex]z(t)=t+8[/tex].
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
Final answer:
This detailed answer explains how to find the equation of the tangent plane and the normal line to a given surface at a specified point. The parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]
Explanation:
To find the equations of the tangent plane and the normal line to the given surface x + y + z = 8exyz at the specified point 0, 0, 8) we first need to find the partial derivatives of x + y + z - 8exyz with respect to x, y, and z.
Given the surface equation:
[tex]\[ x + y + z = 8exyz \][/tex]
Let's denote the function as [tex]\(F(x, y, z) = x + y + z - 8exyz\).[/tex]
(a) To find the equation of the tangent plane, we use the gradient vector of F at the point 0, 0, 8 as the normal vector to the tangent plane.
The gradient vector of F is given by:
[tex]\[ \nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \][/tex]
Let's calculate the partial derivatives:
[tex]\[ \frac{\partial F}{\partial x} = 1 - 8eyz \][/tex]
[tex]\[ \frac{\partial F}{\partial y} = 1 - 8exz \][/tex]
[tex]\[ \frac{\partial F}{\partial z} = 1 - 8exy \][/tex]
Now, evaluate these partial derivatives at the point (0, 0, 8):
[tex]\[ \frac{\partial F}{\partial x} = 1 - 8e(0)(0) = 1 \][/tex]
[tex]\[ \frac{\partial F}{\partial y} = 1 - 8e(0)(0) = 1 \][/tex]
[tex]\[ \frac{\partial F}{\partial z} = 1 - 8e(0)(0) = 1 \][/tex]
So, the gradient vector of [tex]\(F\) at \((0, 0, 8)\) is \( \nabla F = (1, 1, 1) \).[/tex]
Therefore, the equation of the tangent plane is:
[tex]\[ x + y + z = 8 \][/tex]
(b) To find the equation of the normal line, we use the same gradient vector [tex]\( \nabla F = (1, 1, 1) \)[/tex] as the direction vector for the line.
Given the point \((0, 0, 8)\) on the surface, the parametric equations of the normal line are:
[tex]\[ x(t) = 0 + t(1) = t \][/tex]
[tex]\[ y(t) = 0 + t(1) = t \][/tex]
[tex]\[ z(t) = 8 + t(1) = 8 + t \][/tex]
So, the parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]
The same researcher wants to see if recently widowed spouses leave the house more after they perform some activity (e.g., Bingo, Support Group, etc.). She wants to compare participants before and after a group activity. Which type of t-test should she use?
Answer:
paired t-test
Step-by-step explanation:
Paired t-test is used to measure before and after effect. For example, a researcher wants to check the effect of new teaching method and he wants to compares marks of students before new teaching method and marks of students before new teaching method.
In the given example there are two groups participants before group activity and participants after group activity. A researcher wants to investigate that whether recently widowed spouses leave house after performing group activity or not. The observations in these two groups are paired naturally and we take the differences of these two groups are considered as random sample. This type of t-test is known as paired t-test.
Factorize 12x² + 15xy
Answer:
[tex]12 {x}^{2} + 15xy \\ = 3x(4x +5 y)[/tex]
hope this helps you...
You spin the spinner shown below once. Each sector shown has an equal area.
What is p (not squirrel)
Answer: 0.8
Step-by-step explanation:
There are 5 possible equally sized sections meaning there is a 0.2 or 20% of landing on each section. We can subtract the probability of getting a squirrel from the whole thing (1 as probability always equals 1) to determine the probability of not squirrel.
P(squirrel) = 0.2
1 - P(squirrel) = P(not a squirrel)
1 - 0.2 = 0.8
Answer: the anserw is 0.8
Step-by-step explanation:
The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,119. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.
Answer:
The smallest sample size needed is 49.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.
This is n for which [tex]M = 500, \sigma = 2119[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]500 = 1.645*\frac{2119}{\sqrt{n}}[/tex]
[tex]500\sqrt{n} = 1.645*2119[/tex]
[tex]\sqrt{n} = \frac{1.645*2119}{500}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645*2119}{500})^{2}[/tex]
[tex]n = 48.6[/tex]
Rounding up
The smallest sample size needed is 49.
Answer:
[tex]n=(\frac{1.640(2119)}{500})^2 =48.31 \approx 49[/tex]
So the answer for this case would be n=49 rounded up to the nearest integer
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=37500[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=2119[/tex] represent the sample standard deviation
n represent the sample size
Solution to the problem
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =500 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.640[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.640(2119)}{500})^2 =48.31 \approx 49[/tex]
So the answer for this case would be n=49 rounded up to the nearest integer
"The Crunchy Potato Chip Company packages potato chips in a process designed for 10.0 ounces of chips with an upper specification limit of 10.5 ounces and a lower specification limit of 9.5 ounces. The packaging process results in bags with an average net weight of 9.8 ounces and a standard deviation of 0.12 ounces. The company wants to determine if the process is capable of meeting design specifications."
Answer:
The process does not meets the specifications.
Step-by-step explanation:
It is provided that the Crunchy Potato Chip Company packages potato chips in a process designed for 10.0 ounces of chips.
The value of upper specification limit is, USL = 10.5 ounces.
The value of lower specification limit is, LSL = 9.5 ounces.
The mean weight of the chips bags is, [tex]\bar X=9.8\ ounces[/tex].
And the standard deviation weight of the chips bags is, [tex]\sigma=0.12\ ounces[/tex].
Compute the value of process capability as follows:
[tex]C_{p_{k}}=min \{C_{p_{L}},\ C_{p_{U}}\}[/tex]
Compute the value of [tex]C_{p_{L}}[/tex] as follows:
[tex]C_{p_{L}}=\frac{\bar X-LSL}{3 \sigma}=\frac{9.8-9.5}{3\times 0.12}=0.833[/tex]
Compute the value of [tex]C_{p_{K}}[/tex] as follows:
[tex]C_{p_{K}}=\frac{USL-\bar X}{3 \sigma}=\frac{10.5-9.8}{3\times 0.12}=1.944[/tex]
The value of process capability is:
[tex]C_{p_{k}}=min \{C_{p_{L}},\ C_{p_{U}}\}[/tex]
[tex]=min\{0.833,\ 1.944\}\\=0.833[/tex]
The value of process capability lies in the interval 0 to 1. So the process is not capable of meeting design specifications.
Thus, the process does not meets the specifications.
Final answer:
The process capability of The Crunchy Potato Chip Company's packaging process has been questioned. By calculating the process capability index (Cpk), we find that the process capability is below the desirable threshold, suggesting that the process might not consistently meet the specified design requirements.
Explanation:
The student's query pertains to the process capability of The Crunchy Potato Chip Company's packaging process. This concept is part of quality control in process management and is measured by the process capability index (Cpk). In this case, the company has set specifications for a bag to contain between 9.5 and 10.5 ounces of chips and aims to have an average of 10.0 ounces. However, with a mean of 9.8 ounces and a standard deviation of 0.12 ounces, we need to calculate whether the process falls within the acceptable limits and how often it meets design specifications. To do this, the Cpk value is computed, which considers the mean, standard deviation, and the proximity of the mean to the closest specification limit.
Calculating the Process Capability Index (Cpk)
The formula for Cpk is:
Cpk = minimum of [(USL - mean) / (3 * sigma), (mean - LSL) / (3 * sigma)]
Where:
USL = Upper Specification Limit
LSL = Lower Specification Limit
mean = average net weight of the bags
sigma = standard deviation
Here, USL is 10.5 ounces, LSL is 9.5 ounces, mean is 9.8 ounces, and sigma is 0.12 ounces. Substituting these values:
Cpk = minimum of [(10.5 - 9.8) / (3 * 0.12), (9.8 - 9.5) / (3 * 0.12)]
Cpk = minimum of [1.9444, 0.8333]
The smaller of the two results is 0.8333, indicating that this is the Cpk value. A Cpk of 1.33 or higher is generally considered good, while a Cpk less than 1 indicates that the process may not be capable of meeting design specifications consistently. In this case, with a Cpk of 0.8333, The Crunchy Potato Chip Company's process may need improvement.
You are a personnel director and are interested in predicting the Number of Shares of Company Stock (Y) using the Number of Years Employed with the Company (X). You randomly selected 8 employees and found that the average number of shares is 525 and the average number of years employed is 22.5. If the slope (b1) is 20.0, what is the least squares estimate of the intercept (b0)
Answer:
intercept=b0=75
Step-by-step explanation:
The least squares estimate of the intercept b0 can be computed as
b0=ybar-b1*xbar.
ybar=average number of shares of company stock=525.
xbar= average number of years employed=22.5.
slope=b1=20.
Thus,
intercept=b0=ybar-b1*xbar
intercept=b0=525-20*22.5
intercept=b0=525-450
intercept=b0=75.
Thus, the estimate of intercept b0=75.
7 divided by 4 long division
We survey a random sample of American River College students and ask if they drink coffee on a regular basis. The 90% confidence interval for the proportion of all American River College students who drink coffee on a regular basis is (0.262, 0.438). What will be true about the 95% confidence interval for these data? Group of answer choices The 95% confidence interval is narrower than the 90% confidence interval. The 95% confidence interval is wider than the 90% confidence interval. The two intervals will have the same width. It is impossible to say which interval will be wider.
Answer:
Correct option:
The 95% confidence interval is wider than the 90% confidence interval.
Step-by-step explanation:
The (1 - α)% confidence interval for the population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The width of this interval is:
[tex]W=UL-LL\\=2\times z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The width of the interval is directly proportional to the critical value.
The critical value of a distribution is based on the confidence level.
Higher the confidence level, higher will be critical value.
The z-critical value for 95% and 90% confidence levels are:
[tex]90\%: z_{\alpha /2}=z_{0.05}=1.645\\95\%: z_{\alpha /2}=z_{0.025}=1.96[/tex]
*Use a z-table.
The critical value of z for 95% confidence level is higher than that of 90% confidence level.
So the width of the 95% confidence interval will be more than the 90% confidence interval.
Thus, the correct option is:
"The 95% confidence interval is wider than the 90% confidence interval."
Imagine you conducted a study to look at the association between whether an expectant mother eats breakfast (or not) and the gender of her baby. Cramér's V = .22. How would you interpret this value? (Hint: Cramér's V can range between 0 and 1.) There was a small to medium association between baby gender and whether the mother ate breakfast every day. 22% of the variation in frequency counts of baby gender (boy or girl) can be explained by whether or not the mother ate breakfast every day. 2.2% of the variation in frequency counts of baby gender (boy or girl) can be explained by whether or not the mother ate breakfast every day. There is a medium to large association between the gender of the baby and whether or not the mother ate breakfast every day.
Answer:
: The percentage of newborn infants who are exclusively breastfed at the time of hospital discharge.
Data Source: Rhode Island Department of Health, Center for Health and Data Analysis, Newborn Developmental Risk Screening Program Database and Maternal and Child Health Database.
Footnotes: The year indicated is the mid-year of a five-year period of data.
Step-by-step explanation:
Cramér's V value of .22 in the study indicates a small to medium association between whether the mother ate breakfast and the gender of her baby.
Explanation:In the study, Cramér's V measuress the strength of association between two nominal variables, in this case, the mother's breakfast-eating habits and the gender of her baby. The value of Cramer’s V ranges from 0 (indicating no association) to 1 (indicating a perfect association)). Given the value of Cramér's V at .22 for this study, there is a small to medium association between whether a mother ate breakfast and the gender of her baby. It does not mean that 22% of the variation in baby gender can be explained by the mother's breakfast-eating habits. Rather, it quantifies the extent of the association between these two factors.
Learn more about Cramér's V here:https://brainly.com/question/32264601
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The two dot plots represent a sample of the number of
people in households in two towns. Which statements
are true about the data sets?
Check all that apply.
Both have the same number of data points.
Both means are between 3 and 4.
O Both have the same median.
Both have the same range.
Westwood has less variability than Middleton.
The answer is:
Both have the same number of data points
Both means are between 3 and 4
Westwood has less variability than Middleton
Answer:
The right anwers are : A B and E
Step-by-step explanation:
right on edg!
Convert 100 USD to rubles. Use exchange rate, where 1 USD equals 32.5 rubles.
Answer:
3250 rubles.
Step-by-step explanation:
1 USD = 32.5 rubles so...
1 * 100 USD = 32.5 * 100 rubles.
The answer is 3250 rubles.
Answer:
3250 rubles
Step-by-step explanation:
1 USD = 32.5 rubles
100 USD = 100 × 32.5
= 3250 rubles
Find the derivative.
StartFraction d Over dt EndFraction Integral from 0 to t Superscript 8 StartRoot u cubed EndRoot du
a. by evaluating the integral and differentiating the result.
b. by differentiating the integral directly.
Answer:
Using either method, we obtain: [tex]t^\frac{3}{8}[/tex]
Step-by-step explanation:
a) By evaluating the integral:
[tex]\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du[/tex]
The integral itself can be evaluated by writing the root and exponent of the variable u as: [tex]\sqrt[8]{u^3} =u^{\frac{3}{8}[/tex]
Then, an antiderivative of this is: [tex]\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}[/tex]
which evaluated between the limits of integration gives:
[tex]\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}[/tex]
and now the derivative of this expression with respect to "t" is:
[tex]\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}[/tex]
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
[tex]g(x)=\int\limits^x_a {f(t)} \, dt[/tex]
is continuous on [a,b], differentiable on (a,b) and [tex]g'(x)=f(x)[/tex]
Since this this function [tex]u^{\frac{3}{8}[/tex] is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
[tex]\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}[/tex]
Write the equation of the line that goes through the point (-4,7) and has a slope of -5.
Answer:
y= -5x-13
Step-by-step explanation:
start with the linear equation
y= mx+b
indentify what you already have.
x= -4 and y= 7 and m(slope)= -5
We are solving for b, so plug in what you have.
7= -5(-4)+b
Simplify.
7= 20+b
Subtract 20 on both sides.
-13= b
Rewrite your equation with the x and y values.
y= -5x-13
The trucking company has a policy that any truck with a mileage more than 2.5 standard deviations above the mean should be removed from the road and inspected. What is the probability that a randomly selected truck from the fleet will have to be inspected? ROUND YOUR ANSWER TO 4 DECIMAL PLACES
Answer:
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Step-by-step explanation:
Z-score
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
What is the probability that a randomly selected truck from the fleet will have to be inspected?
Values of Z above 2.5. So this probability is 1 subtracted by the pvalue of Z = 2.5.
Z = 2.5 has a pvalue of 0.9938
1 - 0.9938 = 0.0062
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Answer:
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the mileage of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]
Decompose 0.42 into tenths and hundredths
Answer:
0.40 + 0.02
Step-by-step explanation:
4 tenths plus 2 hundredths
The diameters of a batch of ball bearings are known to follow a normal distribution with a mean 4.0 in and a standard deviation of 0.15 in. If a ball bearing is chosen randomly, find the probability of realizing the following event: (a) a diameter between 3.8 in and 4.3 in, (b) a diameter smaller than 3 9 in, (c) a diameter larger than 4.2 in
Answer:
a) 88.54% probability of a diameter between 3.8 in and 4.3 in
b) 25.14% probability of a diameter smaller than 3.9in
c) 90.82% probability of a diameter larger than 4.2 in
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 4, \sigma = 0.15[/tex]
(a) a diameter between 3.8 in and 4.3 in,
This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.
X = 4.3
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.3 - 4}{0.15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
X = 3.8
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.8 - 4}{0.15}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918
0.9772 - 0.0918 = 0.8854
88.54% probability of a diameter between 3.8 in and 4.3 in
(b) a diameter smaller than 3 9 in,
This is the pvalue of Z when X = 3.9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.9 - 4}{0.15}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a pvalue of 0.2514
25.14% probability of a diameter smaller than 3.9in
(c) a diameter larger than 4.2 in
This is 1 subtracted by the pvalue of Z when X = 4.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.2 - 4}{0.15}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
90.82% probability of a diameter larger than 4.2 in
Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000 If we record the selling price of 40 randomly selected US homes what will be the shape of the distribution of sample means what will be the mean of this distribution what will be the standard deviation of this distribution
Answer:
The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Population:
Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000
Sample of 40
Shape approximately normal
Mean 350000
Standard deviation [tex]s = \frac{160000}{\sqrt{40}} = 25298[/tex]
The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.
Answer:
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The mean would be:
[tex] \mu_{\bar X} =350000[/tex]
And the standard deviation would be:
[tex]\sigma_{\bar X} =\frac{160000}{\sqrt{40}}= 25298.221[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the selling price of a population, and for this case we know the following info
Where [tex]\mu=350000[/tex] and [tex]\sigma=160000[/tex]
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The mean would be:
[tex] \mu_{\bar X} =350000[/tex]
And the standard deviation would be:
[tex]\sigma_{\bar X} =\frac{160000}{\sqrt{40}}= 25298.221[/tex]
please help me ASAP!!
Answer:
x = 16
Step-by-step explanation:
The sum of measures of these three angles of any triangle is equal to 180 °
54 + 90 + (2x + 4) = 180
144 + 2x + 4 = 180
148 + 2x = 180
2x = 180 - 148
2x = 32
[tex]x= \frac{32}{2} \\\\x=16[/tex]
Focus on triangle KLM. Since JKLM is a rectangle, angle KLM is right. Since the sum of the interior angles of every triangle is 180, we deduce that
[tex]\hat{K}+\hat{L}+\hat{M}=180[/tex]
Now plug the values for each angle and you have
[tex]54+90+(2x+4)=180[/tex]
Move 54 and 90 to the right side to get
[tex]2x+4=36 \iff 2x=32 \iff x=16[/tex]
Shawndra said that it is not possible to draw a trapezoid that is a rectangle, Explain your answer using the properties of quadrilaterals. Help FAST please. TELL ME WHAT TO DO AND SAY 30 points
Answer:
She is wrong
Rectangle can be a trapezoid
Step-by-step explanation:
A trapezoid is a quadrilateral which has atleast one pair of parallel sides.
A rectangle has two pairs of parallel sides, so it is a trapezoid too.
All rectangles are trapezoid.
But all trapezoids are not rectangles
Cuál es el valor de la expresión -2+3(8+-13)?
Answer:
-17
Step-by-step explanation:
-2+3(8+-13)
-2+3(8-13)
-2+3(-5)
-2-15
-17
[tex]\text{Solve:}\\\\-2+3(8+-13)\\\\\text{Solve the variables inside of the parenthesis}\\\\-2+3(-5)\\\\-2-15\\\\\boxed{-17}[/tex]
I think c is the right answer am I correct ?
Answer: yes
-5 is not greater than 4
-5
Step-by-step explanation:
A sample of 16 elements is selected from a population with a reasonably symmetrical distribution. The sample mean is 100 and the sample standard deviation is 40. We can say that we are 90% confident that is between what two numbers? 12. What is the upper number? Maintain three decimal points in your calculations and give at least two decimal points in your answer.
Answer:
We can say that we are 90% confident that the population mean is between 29.876 and 170.124.
The upper number is 170.124.
Step-by-step explanation:
We have the sample's standard deviation, so we use the students' t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 16 - 1 = 15
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95([tex]t_{95}[/tex]). So we have T = 1.7531
The margin of error is:
M = T*s = 40*1.7531 =70.124
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 100 - 70.124 = 29.876
The upper end of the interval is the sample mean added to M. So it is 100 + 70.124 = 170.124
We can say that we are 90% confident that the population mean is between 29.876 and 170.124.
The upper number is 170.124.
Answer:
[tex]100-1.753\frac{40}{\sqrt{16}}=82.470[/tex]
[tex]100+ 1.753\frac{40}{\sqrt{16}}=117.530[/tex]
So on this case the 90% confidence interval would be given by (82.470;117.530)
And the upper value would be 117.530 for this case
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=100[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=40 represent the sample standard deviation
n=16 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=16-1=15[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,15)".And we see that [tex]t_{\alpha/2}=1.753[/tex]
Now we have everything in order to replace into formula (1):
[tex]100-1.753\frac{40}{\sqrt{16}}=82.470[/tex]
[tex]100+ 1.753\frac{40}{\sqrt{16}}=117.530[/tex]
So on this case the 90% confidence interval would be given by (82.470;117.530)
What is/are the factor(s) in the "Sub-U-Like" study?
A. 30-second, 60-second and 90-second commercials
B. The number of students
C. Craving for Sub-U-Like
D. The length of the television program
E. Length and frequency of the commercial.
F. One, three, or five commercials during the 50-minute television program
Answer:
E. Length and frequency of the commercial.
Remaining Details of the Question:
A study investigated the effect of the length and the repetition of TV advertisements on students' desire to eat at a Sub-U-Like sandwich franchise. Sixty students watched a 50- minute television program that showed at least one commercial for Sub-U-Like during advertisement breaks. Some students saw a 30-second commerical, some saw a 60-second commercial, others a 90-second commerical. The same commerical was shown one, three, or five times during the program. After the viewing, each student was asked to rate their craving for a Sub-U-Like sandwich from the set ("Don't want to eat", "Neutral", "Want to eat"}
Explanation:
During the course of an experimental study, it is important to identify the variables of interest. These variables are referred to as the factors. Factors are controlled independent variables in an experimental study that can be manipulated by the experimenter.
In this case, it was indicated that the study investigated the effect of the length and the repetition of TV advertisements on students' desire to eat at a Sub-U-Like sandwich franchise. Therefore, the factors in the "Sub-U-Like" study are length and frequency of the commercial.
Final answer:
The factors in the "Sub-U-Like" study include the length and frequency of the commercial, the number of students, and the number of commercials during the TV program. Factors in the "Sub-U-Like" study, while not explicitly defined in the examples provided, can generally include variables such as length and frequency of commercials and the number of commercials shown, all of which can influence viewers' behaviors and perceptions in psychological or marketing research.
Explanation:
Factors in the "Sub-U-Like" study:
Factor A: Length and frequency of the commercial
Factor B: The number of students
Factor F: One, three, or five commercials during the 50-minute television program