Answer:
(1) True / False: (a) To reduce the margin of error in a stratified sample, we must design strata so that the units within each stratum are as similar as possible.
R = True, since to have a more exact margin you have to eliminate the errors, and as we do it by running error by error, in order to obtain a point on the graph where the products are within the positive margin and you can obtain a graph less error than the beginning.
(b) To reduce the margin of error in a group sample, we must design groups so that the units within each group are as similar as possible.
R = if this type of procedure is reliable because it helps you to correct the errors of the procedure, but to be able to do it you have to have very well accounted for, each piece of information, whether negative or positive, of the study being carried out and from there to create data that help reduce the margin of error.
2 x (c^2 -5) for c=4
Answer:
22
Step-by-step explanation:
PEMDAS
4^2 = 16
16 - 5 = 11
11 x 2 = 22
Answer:
22
Step-by-step explanation:
2 (c^2 -5)
Let c=4
2 (4^2 -5)
PEMDAS
Parentheses first,
(4^2 -5)
Exponents
16-5 =11
Replace into expression
2(11)
22
A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. Suppose that this chart is being used in phase I and the averages from a set of m samples or subgroups from this process are plotted on this chart. What is the probability that at least one of the averages will plot outside the control limits when m
Answer:
Check the explanation
Step-by-step explanation:
Ans=
A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134
M = 10: 1 – 0.9973^10 = 0.0267
M = 20: 1 – 0.9973^20 = 0.0526
M = 30: 1 – 0.9973^30 = 0.0779
M = 50: 1 – 0.9973^50 = 0.126
18)
Ans=
Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values
A random sample of ten 2011 sports cars is taken and their city mileage is recorded. The results are as follows: 20 21 25 21 21 23 31 32 28 26 Assuming the population distribution is normal, compute E, the margin of error for the t interval, for a 90% confidence interval for m, the population mean of the city mpg for 2011 sports cars.
Answer:
[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]
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[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n 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 mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=24.8[/tex]
The sample deviation calculated [tex]s=4.367[/tex]
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=10-1=9[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that [tex]t_{\alpha/2}=1.833[/tex]
And the margin of error would be given by:
[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]
cuanto es dos cuartos mas un octavo
Answer:
0.625
Step-by-step explanation:
PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!! 28 POINTS!
Cassie received a 5%-off coupon and a $5.00-off coupon from a department store. She visits the department store during a tax-free sale and plans to spend no more than $33.95. She also plans to use both of the coupons she received on her purchase. If this situation is modeled by the inequality below, what must be the original purchase total, x, before the discounts are applied?
0.95x - $5.00 < $33.95
A.
x < $38.95
B.
x < $30.74
C.
x < $30.47
D.
x < $41.00
Answer:
D.
Step-by-step explanation:
Rearrange
[tex]0.95x - 5.00 < 33.95[/tex]
[tex]0.95x<38.95[/tex]
[tex]x<\frac{38.95}{0.95}[/tex]
[tex]x<41[/tex]
A box contains 6 black socks and 4 red socks. What is the probability of picking two black socks from the bag without replacement?
Answer:
43/91 or 47%
Step-by-step explanation:
If 22x = 23, what is the value of x?
Answer:
1.045
Step-by-step explanation:
Answer:
3/2
Step-by-step explanation:
Jenny buys a television for the sale
price of $72.59. The television
normally costs $145.18. What percent is
being saved?
State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a 95% confidence level to a 90% confidence level. b. Changing from a sample size of 25 to a sample size of 250. c. Changing from a standard deviation of 20 pounds to a standard deviation of 30 pounds.
Answer:
A) Confidence Interval will become narrower. B) Confidence Interval will become narrower. C) Confidence Interval will become broader.
Step-by-step explanation:
Confidence Interval is the probable range around sample statistic, in which the population parameter is expected to lie.
Confidence Level shows the average percentage level of confidence interval, expected to contain population parameter. Lower confidence level implies narrower Confidence Interval
Bigger sample size reduces margin error (sample statistic, population parameter difference). Parameter-statistic proximity implies: narrower confidence interval around statistic, expected to contain parameter.
Standard Deviation is a measure of dispersion, spread. So, higher standard deviation implies more spread & broader confidence interval.
A company purchases a small metal bracket in containers of 5,000 each. Ten containers have arrived at the unloading facility, and 250 brackets are selected at random from each container. The fraction nonconforming in each sample are 0, 0, 0, 0.004, 0.008, 0.020, 0.004, 0, 0, and 0.008. Do the data from this shipment indicate statistical control
Answer:
Do the data from this shipment indicate statistical control: No
Step-by-step explanation:
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
= (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10
= 0.044/10
= 0.0044
Calculating the lower control limit (LCL) using the formula;
LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n
= 0.0044 - 3*√(0.0044(1-0.0044))
= 0.0044- (3*0.0042)
= 0.0044 - 0.01256
= -0.00816 ∠ 0
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
= 0.0044 + 3*√(0.0044(1-0.0044))
= 0.0044+ (3*0.0042)
= 0.0044 + 0.01256
=0.01696∠ 0
Do the data from this shipment indicate statistical control: No
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment do not indicate statistical control.
The data from this shipment does not indicate statistical control.
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
[tex]\frac{(0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)}{10}\\=\frac{0.044}{10}\\=0.0044[/tex]
Calculating the lower control limit (LCL) using the formula;
LCL
= (x-bar) - 3*√(x-bar(1-x-bar))/n
[tex]= 0.0044 - 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044- (3*0.0042)\\= 0.0044 - 0.01256\\= -0.00816[/tex]
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
[tex]= 0.0044 + 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044+ (3*0.0042)\\= 0.0044 + 0.01256\\=0.01696[/tex]
Do the data from this shipment indicate statistical control:
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment does not indicate statistical control.
Learn more about mean: https://brainly.com/question/12892403
When someone is on trial for suspicion of committing a crime, the hypotheses are: H0 : innocent HA : guilty Which of the following is correct? Question 40 options: Type I error is convicting an innocent person. Type I error is acquitting a guilty person. Type II error is acquitting an innocent person. Type II error is convicting an innocent person.
Answer:
(A) Type I error is convicting an innocent person.
Step-by-step explanation:
A type I error is the rejection of a true null hypothesis.A type II error is the non-rejection of a false negative/null hypothesis.When someone is on trial for suspicion of committing a crime, the hypotheses are:
[tex]H_0[/tex] : innocent
[tex]H_A[/tex] : guilty
From the given options, A Type I error is convicting an innocent person. If the null hypothesis holds (i.e. a person is innocent) nut we still go ahead to convict the person, we have rejected a true null hypothesis.
A six-sided die is rolled, and the number N on the uppermost face is recorded. Then a fair coin is tossed N times, and the total number Z of heads to appear is observed. Determine the mean and variance of Z by viewing Z as a random sum of N Bernoulli random variables. Determine the probability mass function of Z, and use it to find the mean and variance of Z.
Answer:
1. Mean is 1.75
2. The variance is 1.6042
3.
The distribution function is:
Z Z/K
0 21/128
1 5/16
2 33/128
3 1/6
4 29/384
5 1/48
6 1/384
Step-by-step explanation:
The mean of Z is given as:
E(Z) =Σ6, k=0 Kp (Z = k)
Σ6,k=0 K 1/6 Σ6, n=k (n k) (1/2)^n
=( 0(21/128) + 1(5/16) + 2( 33/128) + 3 (1/6) + 4 (29/384) + 5 (1/48) + 6 (1/384))
=7/4
=1.75
Thus, the mean Z is 1.75
The variance of Z is given as:
Var (Z) = E (Z^2) - (E (Z)) ^2
Therefore,
E(Z^2) = Σ 6, k=0 K^2P ( Z=K)
= ( 0(21/128 + 1(5/16) + 4(33/128) + 9(1/6) + 16(29/384) + 25(1/48) + 36(1/384))
=14/3
Var (Z) = 14/7 - (7/4)^2
= 14/7 - 49/16
=77/48
=1.6042
Thus, the variance is 1.6042
The probability of mass function is given as:
P(Z=k) = 1/6 Σ 6, n=k (n k) (1/2)^n
The distribution function is
Z Z/K
0 21/128
1 5/16
2 33/128
3 1/6
4 29/384
5 1/48
6 1/384
Yoselin has 7 boxes of coins. Each box has 28 coins. How many coins does 1 point
Yoselin have in all? Choose best equation. *
28-7=21
28x7=196
28+7=35
28/7=4
Answer:
28x7=196 (Pls give Brainliest)
The following is the (edited) output for the test: A Two-sample T-Test and CI for the data. Sample 1(M) has N=112, Mean=7.38, StDev=6.95, SE Mean=0.66. Sample 2 (F) has N=101, Mean=7.15, StDev=6.31, SE Mean = 0.63. The difference is mu (1) - mu (2) and its estimate is 0.230000. The 95% lower bound for difference is -1.271079. The T-Test of difference: T-Value = 0.25, P- Value =0.400, DF=210. From the output we learn that: (i) The data provide sufficient evidence to reject H0 and to conclude that the mean depression score for male teens is larger than that of female teens. (ii) The data provide sufficient evidence to conclude that male and female teens do not differ in mean depression score. (iii) The data do not provide sufficient evidence to conclude that the mean depression score of male teens is larger than that of female teens. (iv) The data do not provide sufficient evidence to reject H0, so we accept it, and conclude that male and female teens do not differ in mean depression score.
Answer:
Step-by-step explanation:
Hello!
You have the output:
Two-Sample T-Test and Cl
Sample N Mean StDev SE Mean
1(M) 112 7.38 6.95 0.66
2(F) 101 7.15 6.31 0.63
Difference = mu (1) - mu (2)
Estimate for difference: 0.230000
95% lower bound for difference: -1.271079
T-Test of difference: T-value = 0.25 P-Value = 0.400 DF= 210
This output summarizes the information of the two samples and indicates the order the populations where studied.
It also informs you of the value of the statistic under the null hypothesis and the p-value.
Unfortunately, there is no information on the type of hypotheses that were tested, i.e. if they where two-tailed or one-tailed, in the latter case, there is no information if it was left-tailed or right-tailed). Likewise is not specified if the test was done for a specific value of the parameter. (for example μ₁ - μ₂ = 0 or μ₁ - μ₂ = θ₀)
For these reasons, the data provided by the output isn't enough to conclude any hypothesis.
From all the provided answers the one more likely to be correct is:
(iii) The data do not provide sufficient evidence to conclude that the mean depression score of male teens is larger than that of female teens.
I hope this helps!
Answer:
iv
Step-by-step explanation:
Since the p-value (0.4) is greater than the significance level (0.05), we can conclude that the result is not significant. This means that there is no enough statistical evidence to reject the null hypothesis H0. Therefore, we must accept it and conclude that the mean depression score for male and female teens do not differ.
Find the inner product for (-4,9,8) . (3,2,-2) and state whether the vectors are perpendicular.
a
-10; no
c.
10; no
b.
-10; yes
d.
10; yes
Answer:
-10; no
Step-by-step explanation:
-4*3 + 9*2 + 8*-2 = -10-10 does not equal 0 so it is not perpendicular
Final answer:
The inner product of the vectors (-4,9,8) and (3,2,-2) is -10. Since the inner product is not zero, the vectors are not perpendicular. Therefore, the correct answer is: -10; no.
Explanation:
The inner product (also known as the dot product) of two vectors (-4,9,8) and (3,2,-2) is calculated by multiplying the corresponding components of the two vectors and summing the result:
Inner product = (-4)×3 + 9×2 + 8×(-2)
= -12 + 18 - 16
= -10
To determine if the vectors are perpendicular, we check if their inner product is zero. Since the inner product in this case is -10, not zero, the vectors are not perpendicular.
It is estimated that the city of Houston, Texas, will double in size every 35 years. If the population is currently 2.4 million people, how many
people will there be in 20 years?
OA.
4.8 million
OB. 8.36 million
OC. 3.77 million
D.
3.57 million
Final answer:
To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Using the formula New Population = Current Population x 2^(Number of Doubling Cycles), the new population will be approximately 4.8 million. Option A .
Explanation:
To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Since the city doubles in size every 35 years, in 20 years it will go through 20/35 of a doubling cycle.
To calculate how many people there will be, we use the formula:
New Population = Current Population x 2^(Number of Doubling Cycles)
Plugging in the values, we have:
New Population = 2.4 million x 2^(20/35)
Using a calculator, we find that the new population will be approximately 4.8 million. Therefore, the answer is 4.8 million.
Solve the right triangle. Round to two decimal places
A=20 , b=6.00
Answer:14
Step-by-step explanation:
PLEASE HELP
Find the volume and surface area of the 3-dimensional figure below.
Volume=
Surface Area=
Answer:
volume: 18 ft³surface area: 42 ft²Step-by-step explanation:
The volume of a cuboid is the product of its dimensions:
V = LWH = (3 ft)(3 ft)(2 ft)
V = 18 ft³
The area is the sum of the areas of the faces. Since opposite faces have the same area, we can figure the area from ...
A = 2(LW +H(L+W)) = 2((3 ft)(3 ft) +(2 ft)(3 ft +3 ft)) = 2(9 ft² +12 ft²)
A = 42 ft²
this uses pythagorean theorem
Step-by-step explanation:
x²=a²+b²
x=√6²+12²
x=√180
x=3√2v
y²=16²+12²
y=√400
y=20
Answer: the answer for rafter 1 is 13.4 and the answer for rafter 2 is 20
Step-by-step explanation: I just know
How many faces does the shape have
Answer:
5 faces
Step-by-step explanation:
4 triangular, 1 square
A circle has a radius of 6. An arc in this circle has a central angle of 48 degrees. What is the length of the arc
Write an exact, simplified answer in terms of pi.
Answer:
[tex] \huge \pi \: units [/tex]
Step-by-step explanation:
[tex]l = \frac{48 \degree}{360 \degree} \times 2\pi \: r \\ \\ = \frac{2}{24} \times 2 \times \pi \times 6 \\ \\ = \frac{24}{24} \times \pi \\ \\ = \pi \: units \\ [/tex]
What is the y intercept of this line and how did you find it?
(1,4) (6,-1)
Answer:
The slope is -1
Step-by-step explanation:
Let's find the slope between your two points.
(1,4);(6,−1)
(x1,y1)=(1,4)
(x2,y2)=(6,−1)
Use the slope formula:
m= y2−y1/x2−x1 = −1−4/6−1
= −5/5
= −1
Hope this is a good explanation :)
Find the area of the figure.
5.5 cm
20 cm
The area of the figure is
Answer:
25.5 cm²
Step-by-step explanation:
5.5 cm × 20 cm = 25.5 cm²
Answer:i think the answer is 110 but it would be more helpful if i knew what kind of shape it is
Step-by-step explanation:
to find the area of a square or rectangle(assuming this is a square or rectangle) you multiply the base by the height
During a 12-hour period, the temperature in a city dropped from a high of 66°F to a low of −29°F. What was the range of the temperatures during this period?
Answer:
95
Step-by-step explanation:
66-66=0
0-29=-29
66+29=95
The range of the temperatures during this period is 95.
Given the following data:
Time period = 12 hoursHighest temperature = 66°FLowest temperature = −29°FTo determine the range of the temperatures during this period:
Range is simply calculated by taking the difference between the highest number and the lowest number in a sample.
Mathematically, range is given by the formula;
[tex]Range = highest \;number -lowest \;number[/tex]
Substituting the given parameters into the formula, we have;
[tex]Range = 66-(-29)\\\\Range =66+29[/tex]
Range = 95
Therefore, the range of the temperatures during this period is 95.
Read more: https://brainly.com/question/16205323
5 people want to evenly share a 1/3 pound bag of peanuts. How many pounds should each person get?
Answer:
1 / 15
Step-by-step explanation:
1/3 / 5 =
1/ 15
Final answer:
Each person will get 1/15 of a pound of peanuts when a 1/3 pound bag is shared evenly among 5 people.
Explanation:
To figure out how much each of the 5 people should get from a 1/3 pound bag of peanuts, we need to divide the total weight of the peanuts by the number of people. This gives us:
1/3 pound ÷ 5 = 1/15 pound per person.
This means each person will get 1/15 of a pound of peanuts. In other calculations such as the candy survey, determining percent uncertainty, or unit conversions as in Michaela's party scenario, a similar process of division or unit conversion is applied to find the answer.
the experimental probability that an SUV will pass by andis store is 0.4. If 500 cars pass by andis store, how many can she expect to be SUVs?
Answer:
50
Step-by-step explanation:
Answer:
the answer is 200
Step-by-step explanation:
Researchers are investigating the effectiveness of leg-strength training on cycling performance. A sample of 7 men will be selected to participate in a training program that lasts for one month. Peak power during cycling will be recorded for each man both before training and after training. The mean difference in times will be used to construct a 95 percent confidence interval for the mean difference in the population.
Answer:
A. The interval will be narrower if 15 men are used in the sample.
Step-by-step explanation:
Hello!
When all other things remain the same, which of the following statements about the width of the interval is correct?
A. The interval will be narrower if 15 men are used in the sample.
B. The interval will be wider if 15 men are used in the sample.
C. The interval will be narrower if 5 men are used in the sample.
D. The interval will be narrower if the level is increased to 99% confidence.
E. The interval will be wider if the level is decreased to 90% confidence.
Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.
The general structure for the CI is "point estimate"±" margin fo error"
Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)
The margin of error of the interval is:
d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
1) The sample size changes, all other terms of the interval stay the same.
As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.
↓d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↑n)
↑d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↓n)
Correct option: A. The interval will be narrower if 15 men are used in the sample.
2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:
↓d= ↓[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
↑d= ↑[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
I hope it helps!
Answer:
The interval will be narrower if 15 men are used in the sample.
Step-by-step explanation:
A statistics lecturer poses the following question to her students as homework: 'Suppose I collected a sample and calculated the sample proportion. If I construct a 90% confidence interval for the population proportion and a 95% confidence interval for the population proportion, which of these intervals will be wider?' Three students provide their answers as follows: Tim: 'The 90% confidence interval will be wider.' Trevor: 'The 95% confidence interval will be wider.' Tracy: 'There is not enough information to tell. Either interval could be wider.'
Answer: The 95% confidence interval will be wider.
Step-by-step explanation:
Confidence interval for population proportion is written as
Sample proportion ± margin of error
margin of error = z score × √pq/n
The z score is determined by the confidence level. The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%
This means that with all other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.
The higher the margin of error, the wider the confidence interval. Therefore,
The 95% confidence interval will be wider.
The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'
Given that,
If construct a 90% confidence interval for the population proportion,
And a 95% confidence interval for the population proportion,
Tim: 'The 90% confidence interval will be wider.'
Trevor: 'The 95% confidence interval will be wider.
Tracy: 'There is not enough information to tell.
Either interval could be wider.'
We have to determine,
Which of these intervals will be wider.
According to the question,
Three students provide their answers as follows:
Tim: 'The 90% confidence interval will be wider.
'Trevor: 'The 95% confidence interval will be wider.
'Tracy: 'There is not enough information to tell. Either interval could be wider.'
Therefore, Confidence interval for population proportion is written as
Sample proportion ± margin of error
Margin of error [tex]= z score \times \frac{ \sqrt{pq}}{n}[/tex]
The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%.
Other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.
The higher the margin of error, the wider the confidence interval.
Therefore, The 95% confidence interval will be wider.
Hence, The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'
For more information about Probability click the link given below.
https://brainly.com/question/15688515
(1 point) Write the given second order equation as its equivalent system of first order equations. u′′+6.5u′−1.5u=8sin(3t),u(1)=−3,u′(1)=−1.5 Use v to represent the "velocity function", i.e. v=u′(t). Use u and v for the two functions, rather than u(t) and v(t). (The latter confuses webwork. Functions like sin(t) are OK.)
Answer:
The equivalent system of first-order differential equations is:
[tex]\[ \begin{cases}u' = v \\v' = 8\sin(3t) - 6.5v + 1.5u\end{cases} \][/tex]
with the initial conditions:
[tex]\[ \begin{cases}u(1) = -3 \\v(1) = -1.5\end{cases} \][/tex]
To convert the given second-order differential equation into an equivalent system of first-order equations, we follow these steps:
1. Identifying the original second-order differential equation:
u'' + 6.5u' - 1.5u = 8sin(3t)
2. Then we introduce a new variable v to represent the first derivative of u:
v = u'
3. Now ,we write the second-order equation as a system of first-order equations:
The original equation is:
u'' + 6.5u' - 1.5u = 8sin(3t)
Using the new variable v = u' , the second derivative u'' can be written as v'. Therefore, we have:
v' + 6.5v - 1.5u = 8sin(3t)
4. Formulating the system of first-order equations:
The first equation comes directly from the definition of v:
u' = v
The second equation comes from the rewritten second-order equation:
v' = 8sin(3t) - 6.5v + 1.5u
5. Let us include the initial conditions:
The initial conditions provided are:
u(1) = -3
u'(1) = -1.5
Since v = u', this translates to:
v(1) = -1.5
6. Writing the system with initial conditions:
The system of first-order equations is:
[tex]\[ \begin{cases} u' = v \\ v' = 8\sin(3t) - 6.5v + 1.5u \end{cases} \][/tex]
With the initial conditions:
[tex]\[ \begin{cases} u(1) = -3 \\ v(1) = -1.5 \end{cases} \][/tex]
Show that y=sin(t) is a solution to (dydt)2=1−y2. Enter your answers below in terms of the independent variable t in the order in which the terms were given. Be sure you can justify your answer.
Answer:
y = sin(t) is a solution to the differential equation
(dy/dt)² = 1 - y²
Step-by-step explanation:
Given (dy/dt)² = 1 - y²
Suppose y = sin(t) is a solution, then it satisfies the differential equation.
That is
[d(sin(t))]² = 1 - y²
Let y = sin(t)
dy/dt = d(sin(t)) = cos(t)
(dy/dt)² = cos²t
But cos²t + sin²t = 1
=> 1 - sin²t = cos²t
So
(dy/dt)² = 1 - sin²t
Since sin²t = (sint)² = y²,
we have
(dy/dt)² = 1 - y²
as required.
The differential equation becomes [tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]
Given the function;
[tex]y = sint[/tex]Take the differential of the function
[tex]\frac{dy}{dt} = cost[/tex]Square both sides of the equation to have:
[tex](\frac{dy}{dx} )^2 = (cost)^2[/tex]
Recall from trigonometry identity that [tex]sin^2t + cos^2t = 1[/tex]
Hence, [tex]cos^2t = 1- sin^2t[/tex]
Replace into the differential expression to have:
[tex](\frac{dy}{dx} )^2 = 1-sin^2t[/tex]
Recall that y = sin(t). On replacing, the differential equation becomes:
[tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]
learn more on differential equation here: https://brainly.com/question/18760518