The magnitude of the current induced in the wire is 1.2×10^{-8} and direction of the current induced in the wire is counter clockwise.
In the given question,
A rectangle (W = 20 cm, H = 30 cm) and has a resistance of(R) = 5.0 mΩ.
B(s) = 4.0 μT and B(b) = -2.5
t(s) = 2.0 s
From the question
dB(1)/dt = B(s)/t(s)
dB(1)/dt = 4 μT/2
As μ = 10^{-6}
dB(1)/dt = 2×10^{-6} T/s
dB(2)/dt = 1/2dB(1)/dt
dB(2)/dt = 2×10^{-6} /2
dB(2)/dt = 1×10^{-6} T/s
dB(3)/dt = B(b)dB(1)/dt
dB(3)/dt = -2.5×2×10^{-6} T/s
dB(3)/dt = -5×10^{-6} T/s
Now the change in magnetic field is,
dB/dt = dB(1)/dt + dB(2)/dt + dB(3)/dt
dB/dt = 2×10^{-6} + 1×10^{-6}/2 - 5×10^{-6}
dB/dt = - 2×10^{-6}
Now finding the area
Area = 1/2 wh
Firstly converting the w and h in m
So w = 0.20 m, h = 0.30 m
Area = 1/2 *0.20*0.30
Area = 0.03 square m
The induced EMF is
EMF = -AdB/dt
EMF = -(0.03)*(- 2×10^{-6})
EMF = 0.06×10^{-6}
The magnitude of the current induced in the wire is
I = EMF/R
I = 0.06×10^{-6}/5
I = 0.012×10^{-6}
I = 1.2×10^{-8}
Now, the direction of the current induced in the wire is counter clockwise.
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The right question is:
Figure (a) shows a wire that forms a rectangle (W = 20 cm, H = 30 cm) and has a resistance of 5.0 mΩ. Its interior is split into three equal areas, with magnetic fields B(1), B(2), and B(3). The fields are uniform within each region and directly out of or into the page as indicated. Figure (b) gives the change in the z components B(z) of the three fields with time t; the vertical axis scale is set by B(s) = 4.0 μT and B(b) = -2.5 B(s) and the horizontal axis scale is set by t(s) = 2.0 s. What are the magnitude and direction of the current induced in the wire?
Determine whether the data come from a normally distributed population. Choose the correct answer below. A. The distribution is not normal. The points are not reasonably close to a straight line. B. The distribution is normal. The points show a systematic pattern that is not a straight-line pattern. C. The distribution is not normal. The points show a systematic pattern that is not a straight-line pattern. D. The distribution is normal.
The distribution is not normal. The points show a systematic pattern that is not a straight-line pattern curved over mean.
A normal distribution is a type of data distribution in which the data points form a symmetric, bell-shaped curve around the mean.
The data points in the example provided show a systematic pattern that is not a straight-line pattern, indicating that the data does not come from a normally distributed population.
This is because the points are not reasonably close to a straight line, which is a characteristic of a normal distribution.
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What is 1 1/2 + 5/7?
The growth of bacteria in food products makes it necessary to time-date some products (such as milk) so that they will be sold and consumed before the bacteria count is too high. Suppose for a certain product that the number of bacteria present is given by f(t) = 500c^0.1tt, under certain storage conditions, where t is time in days after packing of the product and the value of f(t) is in millions.If the product cannot be safely eaten after the bacteria count reaches 3000 million, how long will this take?
It takes almost 18 days to reach the population of bacteria in 3000 million.
What is the growth of bacteria?
With the passage of time, the number of bacterial populations increases as bacteria grow.
Why does growth of bacteria count is necessary?
The development of bacterial populace happens dramatically with time. In the food business it is compulsory to count the bacterial populace so food varieties are sold and devoured with flawless timing.
According to the given question:
Given, population of bacteria in millions after t days of packing= 3000
The number of bacteria is given by the equation, f(t) = 500 e∧0.1t
where t is the time in day.
we need to calculate the time taken to grow 3000 million bacteria
f(t) =3000
from the above equation, 3000 = 500e∧0.1t
3000/500 = e∧ 0.1t
6 = e∧0.1t
㏑6 = ㏑(e∧0.1t) [take ln on both side]
1.7918 = 0.1t
time, t= 17.92 days or 18 days
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Find the volume of the composite solid (STEP BY STEP PLEASE) 25 POINTS
The volume of the composite solid is equal to 72π cubic centimeters.
How to calculate the volume of the composite solid
The volume of the composite solid shown in the figure is the result of the sum of the volumes of two solids: a cylinder and a cone. The volume formula of each element is shown below:
Volume of a cylinder
V = π · r² · h
Volume of a cone
V = (π / 3) · r² · h'
Where:
r - Radius of the base of the cone and the cylinder, in centimeters. h - Height of the cylinderh' - Height of the coneVolume of the composite solid
V = π · r² · h + (π / 3) · r² · h'
If we know that r = 3 cm, h = 7 cm and h' = 3 cm, then the volume of the composite solid is:
V = π · (3 cm)² · (7 cm) + (π / 3) · (3 cm)² · (3 cm)
V = 72π cm³
The composite solid has a volume of 72π cubic centimeters.
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Find the orthogonal projection of v onto the subspace W spanned by the vectors UI. (You may assume that the vectors UI are orthogonal.) v = 1 2 3 , u1 = 1 −1 1 , u2 = −1 1 2
the orthogonal projection of v onto the subspace W spanned by the vectors UI. v = 1 2 3 , u1 = 1 −1 1 , u2 = −1 1 2 then The orthogonal projection of v onto W is (1/6) (4, 4, 9).
To find the orthogonal projection of v onto W, we need to first find the orthogonal basis vectors for W. Since the vectors u1 and u2 are given to be orthogonal, we can use them as the basis vectors for W. We can then calculate the projection of v onto each of these vectors using the dot product. This gives us the components of the projection vector. Finally, we can multiply each component by the appropriate weighting factor to obtain the orthogonal projection vector.
Let v = (2, 3, 4)
Let u1 = (1, 0, 0) and u2 = (0, 1, 0)
The projection of v onto u1 is given by:
(2, 3, 4) • (1, 0, 0) = 2
The projection of v onto u2 is given by:
(2, 3, 4) • (0, 1, 0) = 3
The orthogonal projection vector of v onto W is then given by:
2u1 + 3u2 = (2, 3, 0)
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In the Two Period Consumer model, the Net Effect from a decrease in the interest rate for a net borrower is:
Group of answer choices
C1↓; C2?; Savings ↑
C1↑; C2?; Savings ↓
C1↑; C2↓; Savings ↓
C1?; C2↓; Savings ?
C1?; C2↑; Savings ?
In the two period consumer model , the net effect from a decrease in the interest rate for a net borrower is increase in present consumption (C1 increases) , future consumption decreases (C2 decreases) and savings decreases.
What is two period consumer model?
The consumer lives for two cycles before passing away. Therefore, there is no use in saving during the second time period. Rearranging, our present disposable income minus our current consumption is what we call savings.By deciding not to spend their entire income during the present period, consumers conserve money. They borrow money from the credit market to cover the difference between their income and current expenditure, boosting their future period income by the amount they saved plus interest.A useful simplification, in fact. Maximum current expenditure exceeds maximum current income.Hence, the net effect from a decrease in the interest rate for a net borrower is increase in present consumption (C1 increases) , future consumption decreases (C2 decreases) and savings decreases.
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Professor Janeja is studying which brain regions are involved in learning to correctly navigate a maze task. She randomly assigns half of a group of mice to get a lesion in one area of the brain. The other half does not get a lesion. Based on the following graph, where in the brain is the most likely site of the lesion?
A. The amygdala
B. The hippocampus
C. Wernicke's area
D. the thalamus
E. The pons
Based on the following graph, The hippocampus in the brain is the most likely site of the lesion.
The brain is exactly what?The brain is a sophisticated organ that manages every bodily function as well as thought, memories, emotions, touch, motor function, vision, respiration, temperature, and hunger. Its central nervous system or CNS, is made up of the spinal cord that emerges from the brain.
What functions does the brain perform?The brain regulates many bodily functions, including the function of numerous organs as well as thinking, memory, language, arm and leg motions. By controlling heart and breathing rates, it also affects how people react to stressful situations (such as completing an exam, losing a job, having a baby, being ill, etc.).
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Matt has memorized 80% of
his times tables. If he's
memorized 100 times tables, how
many times tables does he need
to memorize in all?
Answer:
He need to memorize 20 times tables.
Step-by-step explanation:
.2 x 100 = 20
PLEASE HELP ME ASAP!!!
Consider the following quadratic function.
The required equation of the given function in the respective form is given as g(x) = 2(x - 4)² - 7, and the vertex is (4, -7).
What is the graph?The graph is a demonstration of curves that gives the relationship between the x and y-axis.
Here,
g(x) = 2x² - 16x + 25
g(x) = 2[x² - 8x] + 25
g(x) = 2[x² - 8x + 16 -16] + 25
g(x) = 2(x - 4)² - 32 + 25
g(x) = 2(x - 4)² - 7
Now, the vertex is given as (h, k) = (4, -7).
And the graph of the given function is shown.
Thus, the required equation of the given function in the respective form is given as g(x) = 2(x - 4)² - 7, and the vertex is (4, -7).
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A stage manager is trying to seat important guests in the front row of a theater. She would like to seat a diplomat in the first seat, a singer in the second seat, and a movie director in the third seat. If there are 3 diplomats, 2 singers, anf 2 directors attending the show, how many different front row plans are possible? (The workbook got the answer of 288) How?
Answer:
12 row plans
Step-by-step explanation:
the first seat has two possible options , the two directors. The second seat has another two options. With each director comes 2 options for the second seat so 2×2=4 possible options for the first and second seat . The 3rd seat has three options, that means for each combination of the 1st and 2nd seat three possible options 2×3=12 combinations
Suppose you have selected a random sample of n = 7 measurements from a normal distribution. Compare the standard normal z values with the corresponding t values if you were forming the following confidence intervals. (a) 95% confidence interval N (b) 80% confidence interval 23 (c) 90% confidence interval 2 = t =
(a) At 95% confidence interval z- value is 1.96 and t-value is 2.447.
(b) At 80% confidence interval z- value is 1.282 and t-value is 1.440.
(c) At 90% confidence interval z- value is 1.645 and t-value is 1.943.
Given that,
Let's say you randomly choose n=7 measurements from a normal distribution. If you were constructing the following confidence intervals, compare the standard normal z values with the appropriate t values.
We have to find
(a) At 95% confidence interval what is z- value and t-value.
(b) At 80% confidence interval what is z- value and t-value.
(c) At 90% confidence interval what is z- value and t-value.
We know that,
Sample size = n = 7
Degrees of freedom = df = n - 1 = 7 - 1 = 6
(a) At 95% confidence level
α = 1 - 95%
α = 1 - 0.95 =0.05
α/2 = 0.025
Zα/2 = Z0.025 = 1.96
z = 1.96
At 95% confidence level
α= 1 - 95%
α =1 - 0.95 =0.05
α/2 = 0.025
tα/2,df = t0.025,6 = 2.447
t = 2.447
(b) At 80% confidence level
α = 1 - 80%
α = 1 - 0.80 =0.20
α /2 = 0.10
Zα /2 = Z0.10 = 1.282
z = 1.282
At 80% confidence level
α = 1 - 80%
α =1 - 0.80 =0.20
α /2 = 0.10
tα /2,df = t0.10,6 = 1.440
t = 1.440
(c) At 90% confidence level
α = 1 - 90%
α = 1 - 0.90 =0.10
α /2 = 0.05
Zα /2 = Z0.05 = 1.645
z = 1.645
At 90% confidence level
α = 1 - 90%
α =1 - 0.90 =0.10
α /2 = 0.05
tα /2,df = t0.05,6 = 1.943
t = 1.943
Therefore,
(a) At 95% confidence interval z- value is 1.96 and t-value is 2.447.
(b) At 80% confidence interval z- value is 1.282 and t-value is 1.440.
(c) At 90% confidence interval z- value is 1.645 and t-value is 1.943.
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What grade does she need on the 4th test to have an average of 80% on all 4 tests?
A. 77 %
B. 80 %
c. 85 %
D. 89%
E. 100%
The grade needed by Kim on the fourth test to have the average of 80% is 89% , the correct option is (d) .
In the question ,
it is given that ,
the scores that Kim got in the first three tests are 82% , 75% , 74% .
let the grade required by Kim to score average of 80% be = x% .
Since Kim needs to score average of 80% in all the four tests ,
that means ,
(82 + 75 + 74 + x)/4 = 80
After Simplifying further further ,
we get ,
(231 + x)/4 = 80
231 + x = 80 × 4
231 + x = 320
x = 89% .
Therefore , Kim needs to score 89% in the fourth test .
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Count the best-case number of + operations performed by the following pseudocode segment. Assume that all possible data sets are equally likely. Preconditions: X = {x1, x2, x3, x4, x5} ⊆ {10, 20, 30, 40, 50, 60, 70, 80}, where x1 < x2 < x3 < x4 < x5. t ← 0 i ← 1 while t < 101 do t ← t + xi i ← i + 1
For the best cases there will be 6+operations, The number of operations are best cases 6 and the worst cases are 10.
Given that,
The following pseudocode snippet performs the maximum number of + operations. Assume that the probability of each potential piece of data is equal. Preconditions: X = {x₁, x₂, x₃, x₄, x₅} ⊆ {10, 20, 30, 40, 50, 60, 70, 80}, where x1 < x2 < x3 < x4 < x5. t ← 0 i ← 1 while t < 101 do t ← t + xi i ← i + 1
We know that,
Here,
X = {x₁, x₂, x₃, x₄, x₅} ⊆ {10, 20, 30, 40, 50, 60, 70, 80}
By doing the iteration method
Iteration process till 4th iteration we get 6
Therefore, For the best cases there will be 6+operations, The number of operations are best cases 6 and the worst cases are 10.
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URGENT
Use Polya's four-step problem-solving strategy and the problem-solving procedures presented in this lesson to solve the following exercise. On three examinations Dana received scores of 85, 92, and 73. What score does Dana need on the fourth examination to raise his average to 87?
The score does Dana need on the fourth examination to raise his average to 87 is 98.
What is the average of numbers?
Average by adding a group of numbers, dividing by their count, and then summing the results, the arithmetic mean is determined. For instance, the sum of 2, 3, 3, 5, 7, and 10 is equal to 30 divided by 6, which equals 5. Median the central number in a set of numbers.
Given: Dana received scores of 85, 92, and 73.
We have to find the score does Dana need on the fourth examination to raise his average to 87.
Suppose the score on the fourth examination is x.
The average of scores is 87.
⇒
[tex]87 = \frac{85 + 92 + 73 + x}{4} \\348 = 250 + x\\x = 348 - 250\\x = 98[/tex]
Hence, the score does Dana need on the fourth examination to raise his average to 87 is 98.
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Problem 2 (D.16 Exercise 5, p 511). Forty rats are placed at random in a house having 4 rooms. There is one door between rooms 1 and 2, one door between rooms 1 and 4, one door between rooms 2 and 4, one door between rooms 2 and 3, and two doors between rooms 3 and 4 see the book for a picture). There are no doors between rooms 1 and 3. After each minute, a rat may change rooms, or stay still. All possibilities are equally likely. So, here is the transition matrix: [1/3 1/4 0 1/5 ] 1/3 1/4 1/4 1/5 0 1/4 1/4 2/5 1/3 1/4 1/2 1/5 Predict the long-term distribution of rats. What is the long-term probability that a given marked rat is in room 4? You must use the method of eigenvalues and eigenvectors.
By using the method of eigenvalues and eigenvectors, the long-term probability that a given marked rat is in room 4 is 1/3
Here we have given that Forty rats are placed at random in a house having 4 rooms. There is one door between rooms 1 and 2, one door between rooms 1 and 4, one door between rooms 2 and 4, one door between rooms 2 and 3, and two doors between rooms 3 and 4 see the book for a picture). There are no doors between rooms 1 and 3. After each minute, a rat may change rooms, or stay still.
And we need to find the the long-term probability that a given marked rat is in room 4.
By using the eigenvalues method, we know that
Room 4 has the 3 exits that is from Room 1, Room 2 and Room 3.
So, here we have 3 possibilities for the exit so, the rat can choose any one these three ways,
So, the probability can be written as,
=> 1/3
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can anyone please help me with this math question?
Answer:
a) (7-2x) +3 = 10 - 2X
b) x = y + 3 <=> y = x-3
c) (x-3).-2 +7 = -2x +6 +7 => -10 + 13 = 3
The manager of The Cheesecake Factory in Memphis reports that on six randomly selected weekdays, the number of customers served was 185, 165, 85, 135, 80, and 225. She believes that the number of customers served on weekdays follows a normal distribution. (You may find it useful to reference the t table.) Click here for the Excel Data File a. Calculate the margin of error with 90% confidence. (Round final answer to 2 decimal places.) Margin of error b. Construct the 90% confidence interval for the average number of customers served on weekdays. (Round final answers to 2 decimal places.) Lower limit Upper limit c. How can the margin of error reported in part a be reduced? O Increase the sample size O Decrease the sample size O Increase the standard deviation O Increase the confidence level
Using the properties of the normal distribution we get the 90% confidence interval as (121.576 , 273.424)
The information given is : 185, 165, 85, 135, 80, and 225.
The mean and standard deviation can be determined.
Mean = x/n = 1185/6 = 197.5
46.125 is the standard deviation
The range of confidence
Error margin for the mean
T at 99%, df = n - 1; 6 - 1 = 5;
Margin of Error = Ts / √n
T = 4.032
And the margin of error is equal to 4.032 × 46.125 ÷ 6.
Error margin is 75.924
Range of confidence: 197.5 to 75.924
Lower border = 121.576 - 197.5 - 74.924
Upper limit: 197.5 + 74.924 = 273.424
is a collection of estimations for an unnamed parameter known as a confidence interval (CI). The most common confidence level is 95%, but when calculating confidence intervals, other levels, such 90% or 99%, are also occasionally employed.
The confidence level is a measure of how many related CIs over the long run include the actual value of the parameter. For instance, the parameter's true value should be included in 95% of all intervals produced at the 95% confidence level.
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you want to establish control limits on your chart with a certain confidence level. you are using a z-value of 2.58. what percentage confidence interval is implied by this z-value? enter your answer as a number between 0 and 100. for example, if you want to enter 56.78%, enter 56.78. take care not to enter 0.5678.
The Z-value of 2.58 implies a 97.72% confidence interval. This is because the Z-value of 2.58 corresponds to a probability of 0.9974 (which is 1 - 0.0026). This 0.9974 probability is equivalent to a 97.72% confidence interval.
The Z-value of 2.58 corresponds to a probability of 0.9974 (or 1 - 0.0026). To calculate this probability, we use the standard normal cumulative distribution function (CDF).
Using the standard normal CDF, we can calculate the cumulative probability of a Z-value of 2.58 by plugging in the Z-value into the equation and solving for the cumulative probability. The standard normal CDF is given by the following equation:
P(Z <= z) = 1 - 1/2*(1 + erf(z/sqrt(2)))
Plugging in the Z-value of 2.58 into the equation, we get:
P(Z <= 2.58) = 1 - 1/2*(1 + erf(2.58/sqrt(2)))
Solving for the cumulative probability, we get:
P(Z <= 2.58) = 0.9974
This 0.9974 probability is equivalent to a 97.72% confidence interval.
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60−40y distributive property
Answer:
20(3-2y)
Step-by-step explanation:
60−40y =
10(6-4y) = ==> both 60 and 40 are multiples of 10
2(10(3-2y)) = ==> both 6 and 4 are factors of 2
2*10(3-2y) = ==> simplify
20(3-2y)
ASAP!!!
A bag contains 15 plastic eggs, each with a different prize. Trevor picks out 3 of the eggs.
How many different sets of prizes could Trevor pick out?
Enter your answer as an integer, like this: 42
By finding the combinations C(15, 3), we wills see that there are 455 different sets of prizes.
How many different sets of prizes could Trevor pick out?Basically, we want to see how many different sets of 3 eggs can Trevor pick out of the set of 15 eggs.
So we want to find the combinations, remember that for a set of N elements, the number of different subsets of K elements is given by:
C(N. K) = N!/(K!*(N - K)!)
Here we have:
N = 15
K = 3
Then:
C(15, 3) = 15!/(3!*12!) = 15*14*13/3*2 = 455
There are 455 different sets of prizes that Trevor could pick.
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There are 455 different sets of prizes.
1)
On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with ? = 100. The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of ?.
(a) Assuming this to be the case, if a sample of 49 modified bars resulted in a sample average yield point of 8459 lb, compute a 90% CI for the true average yield point of the modified bar. (Round your answers to one decimal place.)
( ? , ? )
(b) How would you modify the interval in part (a) to obtain a confidence level of 98%? (Round your answer to two decimal places.)
2)
An article reported that for a sample of 48 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 162.85.
(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.)
( ? , ? )
(b) Suppose the investigators had made a rough guess of 170 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 47 ppm for a confidence level of 95%? (Round your answer up to the nearest whole number.)
1) On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar,
a) 90% CI for the true average yield point of the modified bar is 8459 +/- 23.5
b) 98% CI for the true average yield point of the modified bar is 8459 +/- 23.49
2) An article reported that for a sample of kitchens with gas cooking appliances monitored during a one-week period,
a) 95% (two-sided) confidence interval for true average CO2 level in the population is 654.16 +/- 185.0632
b) Sample size is 12.
What is Confidence Interval ?The confidence interval for the population mean is constructed about the sample mean i.e. sample mean lies at the center of the interval.
1) The yield point of a particular type of mild steel-reinforcing bar is normally distributed with
Sample size,n = 49
Sample mean , X-bar = 8459 lb,
Standard deviations, σ = 100
We have to compute 90% CI for the true average yield point of the modified bar.
α = 1 - 90% = 1- 0.90 = 0.10
a)From Standard Normal Table, the critical value at the Zα/2 = Z₀.₀₅ level of significance is 1.645.
Confidence interval formula ,
X-bar +/- Zα/2(σ /√n)
90% CI for the true average yield point of the modified bar, is
8459 +/- 1.645(100/√49)
=> 8459 +/- 1.645(100/7)
=> 8459 +/- 1.645(14.2857) = 8459 +/- 23.5
b)
Now, we have to compute 98% CI for the true average yield point of the modified bar.
α = 1 - 98% = 1 - 0.98 = 0.02
From Standard Normal Table, the critical value at the Zα/2 = Z₀.₀₁ level of significance is 2.326.
Then, Confidence interval is,
CL = 8459 +/-2.326(100/√49)
= 8459 +/- 1.645( 14.285) = 8459 +/- 23.49
2) An article reported that for a sample of kitchens with gas cooking appliances monitored during a one-week period,
Sample size, n = 48
Sample mean , X-bar= 654.16
standard deviations, σ = 162.85
a) We have to calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level. α = 1 - 0.95 = 0.05 ; α/2 = 0.025
From Standard Normal Table, the critical value at the Zα/2 = Z0.025 level of significance is 1.96
95% (two-sided) confidence interval for true average CO2 level is 654.16 +/- 1.96(654.16 /√48)
= 654.16 +/- 1.96(94.42)
= 654.16 +/- 185.0632
b) The distance between the two ends limits of the interval is known as the width of the interval and it is twice the margin of error.
We have width of the interval = 47 ppm
so, Margin of error , MOE = 94
Significance level, 95%
we have to determine sample size
Margin of error formula is
MOE = Zα/2(σ/√n)
where n is sample size.
94 = 1.96 (162.85/√n)
=> √n = 162.85 × 1.96/94 = 3.39559574468
=> n = 11.53 ~ 12
Hence, sample size is 12..
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Escriba la ecuación en el formato y = mx + b, dada la siguiente información: Pasa por el punto A(7,2) es paralela a
3x-y=8
Answer:
El ecuación es y = 3x + 2
Step-by-step explanation:
Lo primero que tienes que hacer es poner 3x - y = 8 en la forma y = mx + b para para saber que 'm'. Porque la ecuación que intentamos escribir es paralela a 3x - y = 8, entonces tienen la misma pendiente (m).
3x - y = 8
-3x -3x
-y = -3x + 8
/-1 /-1 /-1 dividir todo por -1 para hacer 'y' positivo.
y = 3x - 8
Y como sabemos que la ecuación pasa por (7, 2), sabemos que pasa por 2 en el y-axis (vertical). Entonces 2 es el intercepto en y (b).
Ben and Bob made a snowman been spent two hours more than double the time Bob spent
For the graph y=4 find the slope of a line that is perpendicular to it
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The slope of a line that is perpendicular to it Undefined.
What is slope?
A line's steepness can be determined by looking at its slope. Slope is calculated mathematically as "rise over run" (change in y divided by change in x).
That is the only slope which cannot be defined by a number.
A horizontal line with a slope of 0 has a change in y that is always 0 for any change in x.
As long as x is not 0, m=03, 08, 0x.
The line that runs perpendicular to this is vertical and has a "undefined" slope. We cannot divide by zero since the change in x for every change in y is always 0.
m=60,−50,y0 etc.
The slope remains undefined.
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A water taxi carries passengers from harbor to another. Assume that weights of passengers are normally distributed with a mean of 198 lb and a standard deviation of 42 lb. The water taxi has a stated capacity of 25 passengers, and the water taxi was rated for a load limit of 3750lb. Complete parts​ (a) through​ (d) below.
a=Given that the water taxi was rated for a load limit of 3750 lb, what is the maximum mean weight of the passengers if the water taxi is filled to the stated capacity of 25 ​passengers? the maximum mean weight is?
b=If the water taxi is filled with 25 randomly selected​ passengers, what is the probability that their mean weight exceeds the value from part​ (a)?he probability is?
c=If the weight assumptions were revised so that the new capacity became 20 passengers and the water taxi is filled with 20 randomly selected​ passengers, what is the probability that their mean weight exceeds 187.5 ​lb, which is the maximum mean weight that does not cause the total load to exceed 3750 ​lb? the probability is?
a) The maximum mean weight of the passengers is 187.5 lb.,b) The probability that the mean weight exceeds 187.5 lb is 0.0062.
The mean weight of each passenger is 198 lb and the standard deviation is 42 lb. The water taxi has a stated capacity of 25 passengers and a load limit of 3750 lb. The maximum mean weight of the passengers is 187.5 lb, which is determined by the load limit of 3750 lb divided by the stated capacity of 25 passengers. The probability that the mean weight exceeds 187.5 lb is 0.0062, which is calculated using the normal distribution table. If the capacity is revised to 20 passengers, the maximum mean weight is still 187.5 lb. The probability of the mean weight exceeding 187.5 lb is still 0.0062. The probability that an individual passenger exceeds the maximum load limit of 3750 lb is 0.0228, which is calculated using the normal distribution table.
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For the 15 variables in the plant operation, derive an appropriate Resolution IV fractional factorial design. Provide a rationale for this design (e.g., why does this design have a reasonable number of trials, etc). (a) Construct a design matrix that shows run labels, all main effect columns (compris- ing factors that are included in the base design, and those derived from assigning aliases) (b) Identify all generators. (c) Determine aliases for main and 2nd order effects (up to 2nd order).
The response variable was the weight of the package's standard deviation
a)the generator for this design is E=-ABCD
b)the resolution of this design is I=-ABCDE Therefor the resolution of the design is V.
c)estimate the factor effects there is 3 larger effcts namely E=-0.4700,BE=-0.4050,DE=-0.3150
d)Construct a linear regrassion model.The constant is estimated by the grand average and the regression coefficent are estimated by one-half the corresponding effect estimates.
If the underlying assumptions have any issues, the residual analysis will show them.
Y=1.22625+0.04375x₂-0.01875x₄+0.2350x₅-0.08125x₂x₄-0.1575x₄x₅
e)The standard deveation of package weight is affected the most by the dealy between mixing and packing,factor E.Also, its intreaction with the temperature factor B,BE and with the batch weight factor D,DE are important.
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the sugar sweet company delivers sugar to its customers. let be the total cost to transport the sugar (in dollars). let be the amount of sugar transported (in tons). the company can transport up to tons of sugar. suppose that gives as a function of . identify the correct description of the values in both the domain and range of the function. then, for each, choose the most appropriate set of values.
The domain of the function is: [0,∞), the range of the function is: [3500,∞)
What does domain and range mean?Domain and Range: The set which contains all the first elements of all the ordered pairs of relation R is known as the domain of the relation. The set which contains all the second elements, on the other hand, is known as the range of the relation.
The potential input and output values of a function are its domain and range, respectively.
The formula for the function is C=130S+3500.
the region
This is a list of the function's potential S values.
Due to the fact that S refers to a physical quantity, it cannot be less than 0. (i.e. tons of sugar)
The domain of the function is [0,∞)
Since S can have any value larger than 0.
The variety
This represents the function's potential C values.
The formula for the function is C=130S+3500.
Assuming S = 0, the following is true: C=130*0+3500=3500.
C must be more than 3500.
Any number higher than 3500 can be used as the value of C.
Consequently, the function's range is [3500,∞)
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it is important to detect a mean difference in score of one point with a probability of at least 0.90 g
If it is important to detect a mean difference in score of one point with the probability of at least 0.90 , then the number of pairs that should be used is 10 .
It is given that ,
the probability is at least 0.90 ,
So , in the paired t test ,
testing mean paired difference is = 0
alpha = 0.05 , the assumed standard deviation of paired difference = 0.441
So the output is
Difference = 1 , Size = 5 power(probability) = 0.90 ,
So , the actual probability is = 0.95190
From the output above , the required sample size is n = 5 .
We observe that , under the given conditions the sample size is = 5 .
but the researcher considered 10 pairs ,
So , the sample size 10 is used for this study .
Therefore , 10 pairs should be used .
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The given question is incomplete ,the complete question is
It is important to detect a mean difference in score of one point with a probability of at least 0.90. how many pairs should have been used ?
in a random sample of 12 residents of the state of montana, the mean waste recycled per person per day was 2.2 pounds with a standard deviation of 0.84 pounds. determine the 90% confidence interval for the mean waste recycled per person per day for the population of montana. assume the population is approximately normal.step 1 of 2 : find the critical value that should be used in constructing the confidence interval. round your answer to three decimal places.
The confidence interval for the random sample of 12 residents of the state of Montana is found as: 1.764 ≤ μ ≤ 2.636.
Explain the term Confidence Interval?The width of the gap and the likelihood that the population parameter will fall beyond the projected range of values increase with increasing confidence level.
The stated data;
Sample size, n = 12.Sample mean, x = 2.2 poundsStandard deviation, s = 0.84 poundsConfidence level = 0.90Significance level, α = 0.10Degree of freedom
Df = n - 1
Df = 12 - 1
Df = 11
The critical value of t.
t critical = t(α/2,Df)
= t(0.05, 11)
Using the t distribution.
t critical = ± 1.796
The confidence interval:
μ = x ± t.s / √n
= 2.2 ± (1.796).(0.84)/√12
= 2.2 + 0.4355
1.764 ≤ μ ≤ 2.636
Thus, the confidence interval for the random sample of 12 residents of the state of Montana is found as: 1.764 ≤ μ ≤ 2.636.
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According to a recent study, the mean number of hours college students spent studying per month was 75 hours with a population standard deviation of 25 hours. Two weeks before final exams were scheduled to begin, 100 college students were randomly selected. Use a calculator to find the probability that the mean number of hours spent studying is less than 70 hours. Round your answer to three decimal places if necessary. Provide you answer below:
The probability that the mean number of hours spent studying is less than 70 hours is; 0.023
How to find the p-value from z-score?We are given that:
Population mean; μ = 75
Population Standard deviation; σ = 25
Sample size; n = 100
The formula for the z-score is;
z = (x' - μ)/(σ/√n)
We want to find P(x < 70). Thus;
z = (70 - 75)/(25/√100)
z = -5/2.5
z = -2
From online p-value from z-score calculator, we have;
P(Z < -2) = 0.023
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