Answer:
a) There is a 63.02% probability that no one has done an one time fling.
b) There is a 36.98% probability that at least one person has done a one time fling
c) There is a 99.15% pprobability that no more than two people have done a one time fling.
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they have done an one time fling, or they have not. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 9, p = 0.05[/tex]
A. no one has done a one time fling
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.05)^{0}.(0.95)^{9} = 0.6302[/tex]
There is a 63.02% probability that no one has done an one time fling.
B. at least one person has done a one time fling
Either no one has done a one time fling, or at least one person has. The sum of the probabilities of these events is decimal 1.
So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - `(X = 0) = 1 - 0.6302 = 0.3698[/tex]
There is a 36.98% probability that at least one person has done a one time fling
c. no more than two people have done a one time fling
This is
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 1) = C_{9,1}.(0.05)^{1}.(0.95)^{8} = 0.2985[/tex]
[tex]P(X = 2) = C_{9,2}.(0.05)^{2}.(0.95)^{7} = 0.0628[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.6302 + 0.2985 + 0.0628 = 0.9915[/tex]
There is a 99.15% pprobability that no more than two people have done a one time fling.
In a group of 9 adults, the probability that no one has done a 'one-time fling' is 0.630, the probability that at least one person has done it is 0.370, and the probability that no more than two people have done it is approximately 0.853.
Explanation:The question here involves probability given certain conditions. The probability of an event happening can be found by dividing the number of times the event occurs by the total number of occurrences. Here we have the probability of the 'one-time fling' happening as 5% or 0.05, the complement of which (it not happening) is 1 - 0.05 = 0.95.
A. The probability of none out of nine friends having done a one-fling time is (0.95)^9 = 0.630. So, there’s a 0.630 chance that none of them have done a one-time fling.
B. The probability of at least one person having done a one-time fling would be the complement of no one having done it, which equals 1 - (the probability of no one having done it), 1 - 0.630 = 0.370.
C. The probability of no more than two people having done a one-time fling is found by adding the probabilities of exactly zero, one, and two persons having done it. Using the binomial probability formula, this would amount to 0.630 (for 0 person) + 9(0.05)*(0.95)^8 (for one person) + 36*(0.05)^2*(0.95)^7 (for two people) = 0.853, approximately.
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A boiler has five identical relief valves. The probability that any particular valve will open on demand is 0.93. Assume independent operation of the valves. Calculate P(at least one valve opens). (Round your answer to eight decimal places.)
Answer:
There is a 99.99998% probability that at least one valve opens.
Step-by-step explanation:
For each valve there are only two possible outcomes. Either it opens on demand, or it does not. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 5, p = 0.93[/tex]
Calculate P(at least one valve opens).
This is [tex]P(X \geq 1)[/tex]
Either no valves open, or at least one does. The sum of the probabilities of these events is decimal 1. So:
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.93)^{0}.(0.07)^{5} = 0.0000016807[/tex]
Finally
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000016807 = 0.9999983193[/tex]
There is a 99.99998% probability that at least one valve opens.
Answer:
0.930000
Step-by-step explanation:
An automobile manufacturer claims that its van has a 39.0 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating. After testing 120 vans, they found a mean MPG of 38.8. Assume the standard deviation is known to be 2.2. A level of significance of 0.02 will be used. Find the value of the test statistic. Round your answer to 2 decimal places. Enter the value of the test statistic.
Final answer:
To evaluate the manufacturer's claim, a one-sample t-test can be conducted. The test statistic can be calculated using the given information.
Explanation:
To conduct a hypothesis test to evaluate the manufacturer's claim, we can use a one-sample t-test. The null hypothesis (H0) is that the mean MPG is equal to 39.0, and the alternative hypothesis (Ha) is that the mean MPG is not equal to 39.0. We can calculate the test statistic using the formula:
test statistic = (sample mean - population mean) / (standard deviation / sqrt(sample size))
Using the given values:
sample mean = 38.8
population mean = 39.0
standard deviation = 2.2
sample size = 120
Substituting these values, the test statistic is:
test statistic = (38.8 - 39.0) / (2.2 / sqrt(120))
Sarah can bicycle a loop around the north part of Lake Washington in 2 hours and 50 minutes. If she could increase her average speed by 1 km/hr, it would reduce her time around the loop by 7 minutes. How many kilometers long is the loop? (Round your answer to two decimal places.)
Answer:
67.84 km
Step-by-step explanation:
let her initial speed be 'u'
Time with speed u = 2 hours 50 minutes = 2.83 hours
Distance = speed × Time
therefore,
Distance = u × 2.83 ............(1)
when speed increase by 1 km/h
speed, v = u + 1
Time taken = 2 hours 50 minutes - 7 minutes
= 2 hours 43 minutes
= 2.7167 hours
Therefor,
Distance = ( u + 1 ) × 2.7167 .............(2)
since, the distance in both the cases will be same
therefore,
from 1 and 2
u × 2.83 = ( u + 1 ) × 2.7167
or
u × 1.0417 = u + 1
or
0.0417u = 1
or
u = 23.97 km/hr
Therefore,
Distance of loop = u × 2.83
= 23.97 × 2.83
= 67.84 km
The question involves calculating the distance of a loop that a bicyclist rides in two different conditions. Using speed, time and distance relation, we first set up two equations based on the given scenarios and solve them to find the distance. The distance turns out to be approximately 25.41 km.
Explanation:This problem can be solved using the relationship between speed, distance and time. Let's denote the distance of the loop around Lake Washington as D km and her original average speed as S km/hr. Initially, she covers the distance in 2 hours and 50 minutes (which is equal to 2.83 hours when converted into hours).
So, we have the first equation as: D = S x 2.83
Then, she increases her speed by 1 km/hr, and it reduces the total time by 7 minutes (which comes out to be 0.117 hours), resulting in a new time of 2.83 - 0.117 = 2.713 hours to cover the same distance. With her increased speed (S + 1 km/hr), this gives us the second equation as: D = (S + 1) x 2.713
Setting the equations equal to each other, we solve for S and subsequently find D. After solving, we get D approximately equal to 25.41 km.
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A poll asked people in a certain country, "Thinking about your finances today, would you say that it is now a good time or a bad time to take a lengthy vacation?
The choices were: "Good time" or "Bad time."
a. What kind of variable is the response?
Answer:
Qualitative variable
Step-by-step explanation:
The response variable is qualitative because it can be categorize as "Good time" or "Bad time". The time is also rated as "Good" or "Bad" for taking a lengthy vacation and so it involves ordinal scale of measurement.
We can say that our response variable is qualitative and ordinal. In short response variable is of qualitative type of variable and the scale of measurement it involves is ordinal scale.
In a sample of three people, the first person has a score 5 points above the mean, and the second person has a score 3 points above the mean.
How does the third persons score compare to the mean?
Answer:
8 points below the mean.
Step-by-step explanation:
Let 'X' be the score of the third person, and let 'M' be the mean score of the sample.
If the other two people scored 5 and 3 above the mean, in order to maintain consistency, the following expression must be true:
[tex]M+5+(M+3)+X =3M\\X=3M-2M-8\\X=M-8[/tex]
Therefore, the third person has a score 8 points below the mean.
Given that the other two people's scores are collectively 8 points above the mean, the third person's score must be 8 points below the mean to balance out the total deviation.
Explanation:This question is about determining the score of the third person in a sample relative to the mean of the sample. We know that the total deviation from the mean must equal zero because the mean is the average of all the scores. Given the first person has a score 5 points above the mean, and the second person has a score 3 points above the mean, the total deviation above the mean is (5+3)=8 points. For the score deviations to balance out to zero, the third person's score must be 8 points below the mean.
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A specific brand of gourmet chocolate candy contains 7.00 gg of dietary fat in each 22.7-gg piece. How many kilograms of dietary fat are in a box containing 4.00 lblb of candy? Express your answer numerically in kilograms.
Answer:
635599482kg
Step-by-step explanation:
A gourmet chocolate candy has 7.00 gg of dietary fat in each 22.7-gg piece
for the 22.7 piece , the gg is 7 x 22.7 = 158.9gg in total
1gg = 2204622.6218488 lbs
158.9gg in total = 158.9 x 2204622.6218488 = 350314534.612
for the box of 4.00 lblb of candy
it is 350314534.612 x 4 = 1401258138.45lb
since our answer is needed in kilograms we convert 1401258138.45lb to kilograms
1lb = 0.453592kg
1401258138.45lb = 1401258138.45lb x 0.453592kg = 635599481.513 ≈ 635599482kg
(b) For one each of your concrete and your hypothetical populations, give an example of a probability question and an example of an inferential statistics question.
Hi, you haven't provided the concrete and hypothetical populations, I'll be using people with red hair in Texas as our population, but you can use a similar approach for the population you decide to use.
Answer:
Probability question: For a known group of people in Texas, what is the probability that we select two people with red hair? Inferential statistics question: From a large population, selected randomly, we found that five of them have red hair, what is the total proportion of people with red hair in Texas?Explanation:
As you can see from the questions, the main difference between statics and probability has to do with knowledge, what do we know about our population? For the first question (probability) we talked about a known group of people, and base on our knowledge we ask about the probability of selecting people with red hair. For the second question (statistics) we are asked to make an inference about all the population based just on a sample (we don't have all the knowledge), also we say that the last question is Inferential because the data come from a large population selected randomly.
Final answer:
An example of a probability question could involve finding the likelihood of a statistics student using a product weekly based on a survey sample. Meanwhile, an inferential statistics question might use a t-test to infer differences in protein value between two lots of hay.
Explanation:
In the context of probability and inferential statistics, an example of a probability question for a concrete population might be: "If a random sample of 25 statistics students was surveyed, and six reported using a certain product within the past week, what is the probability that any given statistics student uses the product weekly?"
An example of an inferential statistics question using hypothetical population data could be: "Using a t-test to compare the mean protein value of two lots of alfalfa hay, can we infer that there is a significant difference between their protein contents?" The t-test determines if the observed difference in means is likely due to chance or if it is statistically significant.
These examples illustrate how probability is used to model uncertainty and random events, while inferential statistics are employed to make educated guesses about a population from sample data, such as calculating a confidence interval or testing a hypothesis.
A professor gives a statistics exam. The exam has 100 possible points. The scores for the students in the second classroom are as follows: 88 84 84 52 100 84 92 84 88 Calculate the sample size, n, and the sample mean, M.
Answer: Sample size n is 9.
The sample mean M is 84.
Step-by-step explanation:
Given : A professor gives a statistics exam.
The exam has 100 possible points.
The scores for the students in the second classroom are as follows:
88 84 84 52 100 84 92 84 88
Since , the number of scores of different students= 9
∴ Sample size is n= 9.
Now , Sample mean = [tex]\dfrac{\text{Sum of all data values}}{\text{Sample size}}[/tex]
[tex]=\dfrac{88+ 84 +84 +52 +100+ 84+ 92+ 84+ 88 }{9}\\\\=\dfrac{756}{9} =84[/tex]
Hence, the sample mean M is 84.
We have that the sample size, n, and the sample mean, M. is mathematically given as
n=9M_n=88From the question we are told
A professor gives a statistics exam.
The exam has 100 possible points. The scores for the students in the second classroom are as follows: 88 84 84 52 100 84 92 84 88 Calculate the sample size, n, and the sample mean, M.MeanGenerally the equation for the Mean is mathematically given as
[tex]M=\sum/n\\\\\Therefore\\\\M=\frac{88 +84 +84+ 52+ 100+ 84+ 92+ 84+ 88 }{9}\\\\M=84[/tex]
Therefore
New mean
M_n=84+4
M_n=88
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43. A one sample test for a proportion is being performed. If the critical value is 2.33 and the test statistics is 1.37, a) The null hypothesis should not be rejected. b) The alternative hypothesis cannot be rejected. c) The null hypothesis should be rejected. d) The sample size should be decreased.
Answer:
Option A) The null hypothesis should not be rejected.
Step-by-step explanation:
We are given the following in the question:
A one sample test for a proportion is being performed.
Critical Value = 2.33
Test statistic = 1.37
Since the critical value is positive and the calculated test statistic is positive, thus it is a right tailed test for a proportion.
The null and alternate hypothesis can be written as:
[tex]H_{0}: p \leq p_0\\H_A: p > p_0[/tex]
Since the calculated test statistic is less than the critical value, we fail to reject the null hypothesis.
We accept the null hypothesis.
Thus, the correct answer is
Option A) The null hypothesis should not be rejected.
First-order differential equations need not have unique solutions satisfying a given initial condition.
a)Prove that there are infinitely many different solutions of the differential equations x- = x^1/3 satisfying x(0) = 0.
b)Discuss the corresponding situation that occurs for x- = x/t, x(0) = x_0.
c)Discuss the situation that occurs for x- = x/t^2, x(0) = 0.
Answer:
a) No Unique solution
b) No Unique solution
c) No unique solution
Step-by-step explanation:
The step by step explanation is as shown in the attachment below
Let Fequalsleft angle 2 xy plus z squared comma x squared plus yz comma 2 xz plus StartFraction y squared Over 2 EndFraction right angle and let C be the circle r(t)equalsleft angle 6 sine t comma 4 sine t comma 3 cosine t right angle, for 0less than or equalstless than or equals2pi. Evaluate ModifyingBelow Contour integral With Upper C Bold Upper F times d Bold r using any method.
Answer:
[tex]\oint_C F \,dr = 0[/tex]
Step-by-step explanation:
In this problem, we have a vector field
[tex]F = \left \langle 2x-z^2,x^2+yz,-2xz +\frac{y^2}{2} \right \rangle[/tex].
We need to find the line integral
[tex]\oint_C F \,dr[/tex]
where [tex]C[/tex] is a circle
[tex]r(t) = \left \langle 2 \cos t, 4\sin t, 5 \cos t \right \rangle, \quad 0 \leq t \leq 2 \pi[/tex].
As we can see, the vector filed [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] and its component functions have continuous partial derivatives.
First, we need to find the curl of the vector filed [tex]F[/tex].
[tex]\text{curl} F = \begin{vmatrix}i & j & k\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\2xy-z^2 & x^2 + yz & \frac{y^2}{2} - 2xz\\\end{vmatrix}[/tex]
Therefore,
[tex]\text{curl}F = i \left( \frac{\partial}{\partial y} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( x^2+yz\right)\right) - j \left( \frac{\partial}{\partial x} \left( \frac{y^2}{2} - 2xz\right) - \frac{\partial}{\partial z} \left( 2xy-z^2\right)\right)\\ \phantom{12356} +k \left( \frac{\partial}{\partial x} \left( x^2+yz\right) - \frac{\partial}{\partial y} \left( 2xy-z^2\right)\right)[/tex]
Now, we can easily calculate the needed partial derivatives and we obtain
[tex]\text{curl} F = i(y-y) - j(-2z + 2z) + k(2x-2x) = 0i + 0j+0k = \langle 0,0,0 \rangle[/tex]
So, the vector field [tex]F[/tex] is defined [tex]\forall (x,y,z) \in \mathbb{R}^3[/tex] , its component functions have continuous partial derivatives and [tex]\text{curl} F = 0[/tex] .Therefore, by a well-known theorem, [tex]F[/tex] is a conservative field.
Since [tex]C[/tex] is a closed path, we obtain that
[tex]\oint_C F \,dr = 0[/tex]
A recent article reported that a job awaits only one in three new college graduates. (1 in 3 means the proportion is .333) A survey of 200 recent graduates revealed that 80 graduates had jobs. At the .02 significance level, we will conduct a hypothesis test to determine if we can conclude if a larger proportion of graduates have jobs than previously reported.
a.What will be the value of our critical value?
b.What will be the value of our test statistic?
c.can we conclude that a larger proportion of graduates have jobs than reported in the article?
Answer:
a) [tex]2.05[/tex]
b) [tex]z = 2.01[/tex]
c) No, we cannot conclude that a larger proportion of graduates have jobs than reported in the article.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 200
p = 0.333
Alpha, α = 0.02
Number of graduates had jobs , x = 80
First, we design the null and the alternate hypothesis
[tex]H_{0}: p = 0.333\\H_A: p > 0.333[/tex]
This is a one-tailed(right) test.
b) Formula:
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{80}{200} = 0.4[/tex]
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
Putting the values, we get,
a) [tex]z = \displaystyle\frac{0.4-0.333}{\sqrt{\frac{0.333(1-0.333)}{200}}} = 2.01[/tex]
Now, [tex]z_{critical} \text{ at 0.02 level of significance } = 2.05[/tex]
c) Since, the calculated z statistics less than the critical value, we fail to reject the null hypothesis and accept it.
Thus, same proportion of graduates have jobs as compared to previously reported.
Thus, we conclude that there is not enough evidence to support the claim that a larger proportion of graduates have jobs than previously reported.
A fish tank contains eight algae-eaters, five catfish, and six tilapia fish. A storekeeper randomly selects two fish (one after another without replacement) from the tank for a customer. Find the probability that both fish selected are catfish.
Answer:
P = 10/171
P = 0.058
The probability that both fish selected are Catfish is 0.0058
Step-by-step explanation:
Given:
The tank contains;
Algae eaters = 8
Catfish = 5
Tilapia fish = 6
Total number of fish = 8+5+6= 19
To determine the probability that both fish selected are Catfish P;
P = 5/19 × 4/18
( The number of the total number of fishes, and Catfish reduce by one after each selection because it is selected without replacement)
P = 10/171
P = 0.058
The equation r(t)=cos(5t)i + sin(5t)j, 0t≥0 describes the motion of a particle moving along the unit circle. Answer the following questions about the behavior of the particle. a. Does the particle have constant speed? If so, what is its constant speed? b. Is the particle's acceleration vector always orthogonal to its velocity vector? c. Does the particle move clockwise or counterclockwise around the circle? d. Does the particle begin at the point left parenthesis 1 comma 0 right parenthesis(1,0)?
Answer:
a) 5 units/s
b) yes
c) counter-clockwise
d) yes
Step-by-step explanation:
part a
r(t) = cos (5t) i + sin (5t)j
v(t) = dr(t) / dt = -5sin(5t) i + 5cos(5t)j
[tex]mag( v(t)) = \sqrt{(-5sin(5t))^2 + (5cos(5t))^2} \\mag( v(t)) = \sqrt{25sin^2(5t) + 25cos^2(5t)} \\ \\mag( v(t)) = \sqrt{25*(sin(5t)^2 + cos(5t)^2)} \\\\mag( v(t)) = \sqrt{25} \\\\mag( v(t)) = 5 units/s[/tex]
Hence, the particle has a constant speed of 5 units/s
part b
a(t) = dv(t) / dt = -25cos(5t) i - 25sin(5t)j
To check orthogonality of two vectors their dot product must be zero
a(t) . v(t) = (-25cos(5t) i - 25sin(5t)j) . (-5sin(5t) i + 5cos(5t)j)
= 125cos(5t)*sin(5t) -125cos(5t)*sin(5t)
= 0
Yes, the particles velocity vector is always orthogonal to acceleration vector.
part c
Use any two values of t and compute results of r(t)
t = 0 , r(0) = 1 i
t = pi/2, r(0) = j
Hence we can see that the particle moves counter-clockwise
part d
Find the value r(t) at t=0
r(0) = cos (0) i + sin (0) j
r(0) = 1 i + 0 j
Yes, the particle starts at point ( 1, 0)
A baseball leaves the hand of a pitcher 6 vertical feet above home plate and 60 feet from home plate. Assume the x-axis is a straight on the flat ground from mound to the plate, the y-axis lies on the groundfrom the 3rd to 1st base and the z axis is vertical height.
(a) In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of 130i +0j-3k ft/s (roughly 90mi/hr). How far above the ground is the ball when it cross the home plate.
(b) How long does it take for the pitch to arrive at home plate?
(c) what vertical velocity component should the pitcher use so that the pitch crosses home plate exactly 3 feet above the ground?
(d) A simple model to describe the curve of the baseball assumes that the spin of the ball produces a constant acceleration (in the y direction) of c ft/s^(2). Assume a pitcher throws a curve ball with c=8ft/s^(2) (1/4 the acceleration of gravity). how far does the ball move inthe y-directionby the time it reaches home plate, assuming an initial velocity of 130i +0j -3k ft/s ?
(e) Using the above information, does the ball curve more in the first half of its trip to second or in the second half?
(f) How does this effect better?
(g) Suppose the pitcher releases the ball from an initial position of <0,-3,6> with an initial velocity of 130i +0j-3k. What value of the spin parameter c is needed to put the ball over home plater passing through the point <60,0,3>?
In the absence of all forces except gravity, the ball is approximately 0.28 feet below the ground when it crosses the home plate.
Explanation:(a) In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of 130i +0j-3k ft/s (roughly 90mi/hr). How far above the ground is the ball when it cross the home plate.
To determine the vertical distance the ball is above the ground when it crosses home plate, we need to find the time it takes to reach the home plate using the vertical component of the initial velocity. The initial vertical velocity is -3 ft/s, and the vertical acceleration due to gravity is -32 ft/s2. Using the equation y = y0 + v0yt + 0.5at2, where y is the vertical distance above the ground, y0 is the initial vertical position, v0y is the initial vertical velocity, t is the time, and a is the acceleration due to gravity, we can solve for t. Plugging in the given values, we get:
y = 0 + (-3)t + 0.5(-32)t2
Simplifying the equation, we get 32t2 - 3t = 0. Factoring out t, we get t(32t - 3) = 0. So, either t = 0 (which is the initial time when the ball is released) or 32t - 3 = 0. Solving for t, we find that t = 3/32 seconds. Since the ball is in the air for this amount of time, we can substitute this value back into the equation to find the vertical distance above the ground when the ball crosses the home plate:
y = 0 + (-3)(3/32) + 0.5(-32)(3/32)2
Simplifying, we get y = -9/32 feet, which is approximately -0.28 feet above the ground. Therefore, the ball is 0.28 feet below the ground when it crosses the home plate.
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P (A or B ). P(A or B) indicates _______. Choose the correct answer below. A. the probability that A and B both occur in the same trial. B. the probability that in a single trial, event A occurs, event B occurs, or they both occur. C. the probability that event A occurs in one trial followed by event B in another trial. D. the probability that event A or event B does not occur in a single trial.
Answer:
Option B is right
Step-by-step explanation:
P(A or B)
In set theory if A and B are two sets we have either A occurs or B occurs or both as
P(A or B) = P(AUB)
This implies that A occurs or B occurs
Probability is calculated as P(A)+P(B)-P(AB)
P(AB) here represents both occurring and this is subtracted as this was added two times
Option A is wrong because this is AB
Option C is wrong because it is not necessary A occurs first
D is wrong because this is A'UB' where A' represents the complement of A
So correct answer out of four options is
B. the probability that in a single trial, event A occurs, event B occurs, or they both occur.
The notation P(A or B) in probability specifies the probability that in a single trial, either event A happens, event B happens, or they both happen.
Explanation:The notation P(A or B) in probability theory denotes the probability that event A occurs, event B occurs, or both events occur in a single trial. If we look at the four options provided, option B fits this definition. Hence, P(A or B) indicates the probability that in a single trial, event A occurs, event B occurs, or they both occur. It does not imply that both events A and B have to occur concurrently in the same trial (option A), nor that event A has to occur in one trial followed by event B in another trial (option C), and certainly not the probability that event A or event B does not occur in a single trial (option D).
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What function do you know from calculus is such that its first derivative is itself? Its first derivative is a constant multiple of itself? Write each answer in the form of a first-order differential equation with a solution.
Answer:
First Part:
exponential function [tex]e^t[/tex] is the one whose first order derivative is the function itself.
Second Part:
[tex]y=ce^{At}\\y'=Ay[/tex]
Third Part:
[tex]y=ce^t\\y'=ce^t\\y'=y[/tex]
Step-by-step explanation:
First Part:
In calculus exponential function [tex]e^t[/tex] is the one whose first order derivative is the function itself.
Where:
t is independent variable.
Derivative is represented as:
[tex]y=ce^t\\y'=\frac{d(ce^t)}{dt} \\y'=ce^t\\y'=y[/tex]
Where:
c is any number.
Second Part:
Consider the constant A.
The function will become:
[tex]y=ce^{At}\\y'=\frac{d(ce^{At})}{dt} \\y'=cAe^{At}\\y'=Ay[/tex]
Third Part:
Derivative is represented as:
[tex]y=ce^t\\y'=\frac{d(ce^t)}{dt} \\y'=ce^t\\y'=y[/tex]
Where:
c is any number.
Final answer:
The differential equation is [tex]\( \frac{df}{dx} = Cf(x) \),[/tex] with the solution[tex]\( f(x) = Ce^x \), where \( C \)[/tex] is a constant.
Explanation:
The function that satisfies both conditions is [tex]\( f(x) = Ce^x \), where \( C \)[/tex] a constant. Its first derivative is[tex]\( f'(x) = Ce^x \),[/tex]which is a constant multiple of itself. This can be expressed as a first-order differential equation:
[tex]\[ \frac{df}{dx} = Cf(x) \][/tex]
with the solution:
[tex]\( f(x) = Ce^x \), where \( C \)[/tex]
This differential equation represents exponential growth or decay, depending on the sign of [tex]\( C \). For \( C > 0 \)[/tex], the function exponentially grows as [tex]\( x \) increases, while for \( C < 0 \)[/tex]it exponentially decays. The solution captures phenomena like population growth, radioactive decay, or compound interest, making it a fundamental concept in calculus and applied mathematics.
A biased coin gives a head probability P the expected number of times does one need to make to get the pattern of HTH is 36.25 what is the value of P rounded to the nearest 0.01 .
Answer: P = 0.2
Step-by-step explanation:
let us define the expectations as;
Eнтн = no of flips to obtain HTH
Eн,тн = no of flips to obtain HTH where H is flipped
Eнт,н = no of flips to obtain HTH where HT is flipped
so let P and q represent the success and failure of probabilities
this gives;
Eнтн = 2 + P²Eн,тн + PqEнт,н + PqEн,тн + q²Eнтн
Eн,тн = 1 + PEн,тн + qEнт,н
Eн,тн = (1 + qEнт,н) / q
Eнт,н = 1 + p*0 + qEнтн = 1 + qEнтн
from this expression we have that;
Eнтн = (2 +(P² +Pq) (1/q + 1) +Pq) / (1-qP² + 2Pq² + q²)
E(x) = 1/P + 1/P²q
= 36.25
therefore the probability is P = 0.2
P = 0.2
Find the particular solution that satisfies the differential equation and the initial condition.
1. f '(x) = 8x, f(0) = 7
2. f '(s) = 14s − 12s3, f(3) = 1
Answer:
1) [tex] y =4x^2 +7[/tex]
2) [tex] y =7s^2 -3s^4 +181[/tex]
Step-by-step explanation:
Assuming that our function is [tex] y = f(x)[/tex] for the first case and [tex] y=f(s)[/tex] for the second case.
Part 1
We can rewrite the expression like this:
[tex] \frac{dy}{dx} =8x[/tex]
And we can reorder the terms like this:
[tex] dy = 8 x dx[/tex]
Now if we apply integral in both sides we got:
[tex] \int dy = 8 \int x dx[/tex]
And after do the integrals we got:
[tex] y = 4x^2 +c[/tex]
Now we can use the initial condition [tex] y(0) =7[/tex]
[tex] 7 = 4(0)^2 +c, c=7[/tex]
And the final solution would be:
[tex] y =4x^2 +7[/tex]
Part 2
We can rewrite the expression like this:
[tex] \frac{dy}{ds} =14s -12s^3[/tex]
And we can reorder the terms like this:
[tex] dy = 14s -12s^3 dx[/tex]
Now if we apply integral in both sides we got:
[tex] \int dy = \int 14s -12s^3 ds[/tex]
And after do the integrals we got:
[tex] y = 7s^2 -3s^4 +c[/tex]
Now we can use the initial condition [tex] y(3) =1[/tex]
[tex] 1 = 7(3)^2 -3(3)^4 +c, c=1-63+243=181[/tex]
And the final solution would be:
[tex] y =7s^2 -3s^4 +181[/tex]
Choose the accurate description of why it is not correct to make a pie chart for the social usage hours or the entertainment usage hours.
Answer:
The question is incomplete or missing some part, but I have answered it appropriately.
Step-by-step explanation:
The question is incomplete, but I will give a detailed description of where pie chart are appropriate to use. A pie chart as we known is a divided circle containing breakdown d and this are developed into a table from calculations and then a chart is developed. The circle in a pie chart are divided into sectors where the angle of each sector is directly proportional to the frequency of the item or data that is being represented.
The process is done by finding the sum of all the frequencies, the sum is then used to divide 360 degree which is the total degree in a complete circle. thereafter each of the frequencies is used to multiply this result to arrive at the sectoral angle for each item. for example, the breakdown of weekkly expenditures of a house wife. the expensiture are marked with the amount spent on each. this information can be represented or illustrated on pie chart.
It should be noted that a pie chart is only suitable for a single set of data and not more than one date as is the case of histogram where more than one grouped or un-grouped data maybe represented or ilustrated on a histogram. if this is the case, pie chart are best employed when making a description between a single part of a data and not the whole set.
Of all the numbers whose difference is 36, find the two that have the minimum product
Answer:
54,18
Step-by-step explanation:
Let one of the numbers be x. The other number be represented as 36-x
(x-(36-x ) = 36
open brackets
x - 36 + x = 36
x + x = 36+ 36 = 72
2x = 72
x = 72/2 = 36
x = 36
The product can then be represented as y = x(36-x) or y=36x-x²
The maximum or minimum is always on the axis of symmetry which has the formula x=-b/2a.
In our case, the axis of symmetry is -36/-2, so x=18.
If one number is 18 and the 2 numbers differentiate by 36, the other number is 18 + 36 = 54
So the 2 numbers are 18 and 54 and the minimum product is 972
The two numbers are 18 and 54 and the minimum product is 972
What is the fundamental principle of multiplication?If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.
Let suppose one of the numbers be x and the other number be represented as 36-x
Then the equation would be;
(x-(36-x ) = 36
Now open brackets;
x - 36 + x = 36
x + x = 36+ 36 = 72
2x = 72
x = 72/2 = 36
x = 36
Therefore, The product can be represented as y = x(36-x) or y=36x-x²
Here maximum or minimum is always on the axis of symmetry which is the formula x=-b/2a.
In our case; the axis of symmetry is -36/-2,
Then x=18.
If one number is 18 and the 2 numbers differentiate by 36 then the other number must be 18 + 36 = 54
Hence the two numbers are 18 and 54 and the minimum product is 972
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According to Newton's law of cooling, if an object at temperature T is immersed in a medium having the constant temperature M, then the rate of change of T is proportional to the difference of temperature M-T. This gives the differential equation dT/dt=k(M-T)Solve the differential equation for T.
Answer:
[tex] T(t) = M -C_1 e^{-kt}[/tex]
And as we can see that represent the solution for the differential equation on this case.
Step-by-step explanation:
For this case we have the following differential equation:
[tex] \frac{dT}{dt}= k(M-T)[/tex]
We can rewrite this expression like this:
[tex] \frac{dT}{M-T} = k dt[/tex]
We can us the following susbtitution for the left part [tex] u = M-T[/tex] then [tex] du= -dt[/tex] and if we replace this we got:
[tex] \frac{-du}{u} = kdt[/tex]
We can multiply both sides by -1 and we got;
[tex] \frac{du}{u} =-k dt[/tex]
Now we can integrate both sides and we got:
[tex] ln |u| = -kt + C[/tex]
Where C is a contant. Now we can exponetiate both sides and we got:
[tex] u(t) =e^{-kt} *e^C = C_1 e^{-kt}[/tex]
Where [tex] C_1 = e^C[/tex] is a constant. And now we can replace u and we got this:
[tex] M-T = C_1 e^{-kt}[/tex]
And if we solve for T we got:
[tex] T(t) = M -C_1 e^{-kt}[/tex]
And as we can see that represent the solution for the differential equation on this case.
The solution to the differential equation representing Newton's law of cooling is calculated by rearranging the equation and then integrating both sides. The general solution for the temperature of an object over time is given by T = M - Ce^-kt where M is the ambient temperature, T is the temperature of the object, C is a constant, and k is the constant of proportionality.
Explanation:To solve the given differential equation which represents Newton's law of cooling, we shall proceed with integrating both sides. This is in the form of a separable first order differential equation and can be written in the form as follows:
dT / (M - T) = k dt
By re-arranging and integrating both sides, we can find the solution. When we integrate both sides, we get:
- ln |M - T| = kt + C
Where C is the constant of integration. Simplifying further:
T = M - Ce-kt
This is the general solution to the temperature T of an object over time t according to Newton's law of cooling.
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A van starts off 191 miles directly north from the city of Morristown. It travels due east at a speed of 25 miles per hour. After travelling 119 miles, how fast is the distance between the van and Morristown changing?
Answer:
Distance between the van and Morristown is changing at the rate of 13.22 miles per hour.
Step-by-step explanation:
From the figure attached,
Van starts from C (City of Morristown), reaches the point A 191 miles due North and then it travels with a speed of 25 miles per hour due East from A towards B.
We have to calculate the rate of change of distance BC, when the van reaches point B which is 119 miles away from A.
By Pythagoras theorem in the triangle ABC,
[tex]BC^{2}=AB^{2}+AC^{2}[/tex]
Distance AC is constant equal to 191 mi.
By differentiating the equation with respect to time 't'
[tex]2BC.\frac{d(BC)}{dt}=0+2AB.\frac{d(AB)}{dt}[/tex]
[tex]BC.\frac{d(BC)}{dt}=AB.\frac{d(AB)}{dt}[/tex]
Since BC² = (119)²+ (191)²
BC = √50642 = 225.04 miles
From the differential equation,
[tex](225.03).\frac{d(BC)}{dt}=119\times 25[/tex] [Since [tex]\frac{d(AB)}{dt}=25[/tex] miles per hour]
[tex]\frac{d(BC)}{dt}=13.22[/tex] miles per hour
Therefore, distance between the van and Morristown is changing at the rate of 13.22 miles per hour.
Following are the published weights (in pounds) of all of the team members of Football Team A from a previous year.178; 203; 212; 212; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174;185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270;280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230;250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265Organize the data from smallest to largest value.Part (a)Find the median.Part (b)Find the first quartile. (Round your answer to one decimal place.)Part (c)Find the third quartile. (Round your answer to one decimal place.)Part (d)The middle 50% of the weights are from to .Part (e)If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why? (choose one)-The above data would be a sample of weights because they represent a subset of the population of all football players.-The above data would be a sample of weights because they represent all of the players from one year. -The above data would be a population of weights because they represent all of the football players.-The above data would be a population of weights because they represent all of the players on a team.Part (f)If our population were Football Team A, would the above data be a sample of weights or the population of weights? Why? (choose one)-The data would be a sample of weights because they represent all of the professional football players.-The data would be a population of weights because they represent all of the players on Football Team A. - The data would be a sample of weights because they represent all of the players on Football Team A.-The data would be a population of weights because they represent all of the professional football players.Part (g)Assume the population was Football Team A. Find the following. (Round your answers to two decimal places.)(i) the population mean, ? ______(ii) the population standard deviation, ? _____(iii) the weight that is 3 standard deviations below the mean ____(iv) When Player A played football, he weighed 229 pounds. How many standard deviations above or below the mean was he? _____standard deviations _______ (above or below) the meanPart (h)That same year, the average weight for Football Team B was 240.08 pounds with a standard deviation of 44.38 pounds. Player B weighed in at 209 pounds. Suppose Player A from Football Team A weighed 229 pounds. With respect to his team, who was lighter, Player B or Player A? How did you determine your answer? (choose one)-Player A, because he is more standard deviations away from his team's mean weight.-Player B, because he is more standard deviations away from his team's mean weight. -Player A, because Football Team A has a higher mean weight.-Player B, because Football Team B has a higher mean weight.Expert Answer
Answer:
a. 241; b. 206.0; c. 272.0; d. 174 to 232; other answers (see below).
Step-by-step explanation:
First, at all, we need to organize the data from the smallest to the largest value:
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
This step is important to find the median, the first quartile, and the third quartile. There are numerous methods to find the median and the other quartiles, but in this case, we are going to use a method described by Tukey, and it does not need calculators or software to estimate it.
Part a: MedianIn this case, we have 53 values (an odd number of values), the median is the value that has the same number of values below and above it, so what is the value that has 26 values below and above it? Well, in the organized data above this value is the 27th value, because it has 26 values above and below it:
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
The median is 241.
Part b: First QuartileFor the first quartile, we need to calculate the median for the lower half of the values from the median previously obtained. Since we have an odd number of values (53), we have to include the median in this calculation. We have 27 values (including the median), so the "median" for these values is the value with 13 values below and above it.
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241
Since we are asking to round the answer to one decimal place, the first quartile is 206.0
Part c: Third QuartileWe use the same procedure used to find the first quartile, but in this case, using the upper half of the values.
241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
So, the third quartile is 272.0
Part d: The middle 50%The second quartile is the median and "50% of the data lies below this point" (Quartile (2020), in Wikipedia). Having this information into account, 50% of the weights lies below the median 241.
Thus, the middle 50% of the weights are from 174 to 232.
174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241
Part e: Sample or Population 1If the population were all professional football players, the right option is:
The above data would be a sample of weights because they represent a subset of the population of all football players.
It represents a sample. Supposing Football Team A are all professional, they can be considered a sample from all professional football players.
Part f: Sample or Population 2If the population were Football Team A, the right option is:
The data would be a population of weights because they represent all of the players on Football Team A.
Part g: population mean and morePart iThe mean for the population of weights of Football Team A is the sum of all weights (12529 pounds) divided by the number of cases (53).
[tex] \\ \mu = \frac{12529}{53} = 236.39622641509433 [/tex]
[tex] \\ \mu = \frac{12529}{53} = 236.40 [/tex]
Part iiThe standard deviation for the population is:
[tex] \\ \sigma = \sqrt(\frac{(x_{1}-\mu)^2 + (x_{2}-\mu)^2 + ... + (x_{n}-\mu)^2}{n}) [/tex]
In words, we need to take either value, subtract it from the population mean, square the resulting value, sum all the values for the 53 cases (in this case, the value is 74332.67924), divide the value by 53 (1402.50338) and take the square root of it (37.45001).
Then, the population standard deviation is 37.45.
Part iiiThe weight that is 3 standard deviations below the mean can be obtained using the following formula:
[tex] \\ z = \frac{x - \mu}{\sigma} [/tex]
[tex] \\ \mu= 236.40\;and\;\sigma=37.45[/tex]
Then, in the case of three standard deviations below the mean, z = -3.
[tex] \\ -3 = \frac{x - 236.40}{37.45} [/tex]
[tex] \\ x = -3*37.45 + 236.40 [/tex]
[tex] \\ x = 124.05 [/tex]
Part ivFor the player that weights 209 pounds:
[tex] \\ z = \frac{x - \mu}{\sigma} [/tex]
[tex] \\ z = \frac{229 - 236.40}{37.45} [/tex]
[tex] \\ z = -0.19 [/tex]
Part h: Comparing Weights of Team A and Team BFor Team B, we have a mean and a standard deviation of:
[tex] \\ \mu=240.08\;and\;\sigma=44.38[/tex]
And Player B weighed 229 pounds.
[tex] \\ z = \frac{x - \mu}{\sigma} [/tex]
[tex] \\ z = \frac{229 - 240.08}{44.38} [/tex]
[tex] \\ z = -0.70 [/tex]
This value says that Player B is lighter with respect to his team than Player A, because his weight is 0.70 below the mean of Football Team B, whereas Player A has a weight that is closer to the mean of his team. So, the answer is:
Player B, because he is more standard deviations away from his team's mean weight.
"If a ball is thrown into the air with a velocity of 44 ft/s, its height in feet t seconds later is given by y = 44t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following. (i) 0.5 seconds ft/s (ii) 0.1 seconds ft/s (iii) 0.05 seconds ft/s (iv) 0.01 seconds ft/s (b) Estimate the instantaneous velocity when t = 2. ft/s"(c) Find the average velocity for the time period beginning whent = 2 and lasting 0.05 second. ft/s(d) Find the average velocity for the time period beginning whent = 2 and lasting 0.01 second. ft/s(e) Estimate the instantaneous velocity when t = 2.
Answer:
i) -28 ft/s
ii) -21.6 ft/s
iii) -20.8 ft/s
iv) -20.16 ft/s
b) -20 ft/s
Step-by-step explanation:
if v represents vertical velocity then if the height y is
y = 44*t − 16*t²
and the instant velocity v is the derivative with respect to the time
v= dy/dt = 44*(1) - 16*(2*t) = 44- 32*t
while the average velocity is va= (y-y₀)/(t-t₀)
where t₀ = 2 and y₀=y(t₀) = 44*2 − 16*2² = 24
then
va= (y-y₀)/(t-t₀) = (44*t − 16*t² - 24)/(t-2)
for
i) t= 0.5s + 2 s= 2.5 s
va = (44* 2.5 − 16* 2.5 ² - 24)/( 2.5 -2) = -28 ft/s
ii) t= 0.1s + 2 s= 2.1 s
va = (44* 2.1 − 16* 2.1 ² - 24)/( 2.1 -2) = -21.6 ft/s
iii) t= 0.05 s + 2 s= 2.05 s
va = (44* 2.05 − 16* 2.05 ² - 24)/( 2.05 -2) = -20.8 ft/s
iv) t= 0.01s + 2 s= 2.01 s
va = (44* 2.01 − 16* 2.01 ² - 24)/( 2.01 -2) = -20.16 ft/s
b) the instantaneous velocity when t=2
v (t=2) = 44- 32*(2) = -20 ft/s
Answer:
i) The first -28 ft/s
ii) Then second -21.6 ft/s
iii) Thirds one is -20.8 ft/s
iv) Then -20.16 ft/s
b) -20 ft/s
Step-by-step explanation:When if v represents vertical velocity then if the height y isthen y = 44*t − 16*t²After that and the instant velocity v is the derivative with respect to the timeNow v= dy/dt = 44*(1) - 16*(2*t) = 44- 32*tAlso that while the average velocity is va= (y-y₀)/(t-t₀)After that where t₀ = 2 and y₀=y(t₀) = 44*2 − 16*2² = 24then va= (y-y₀)/(t-t₀) = (44*t − 16*t² - 24)/(t-2)for (i) t= 0.5s + 2 s= 2.5 sNow va = (44* 2.5 − 16* 2.5 ² - 24)/( 2.5 -2) = -28 ft/sNext (ii) t= 0.1s + 2 s= 2.1 sAfter that va = (44* 2.1 − 16* 2.1 ² - 24)/( 2.1 -2) = -21.6 ft/sNow (iii) t= 0.05 s + 2 s= 2.05 sThen va = (44* 2.05 − 16* 2.05 ² - 24)/( 2.05 -2) = -20.8 ft/siv) t= 0.01s + 2 s= 2.01 sAfter that va = (44* 2.01 − 16* 2.01 ² - 24)/( 2.01 -2) = -20.16 ft/sWhen b) the instantaneous velocity then t=2Now v (t=2) = 44- 32*(2) = -20 ft/sLearn more about:
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We have a bag filled with 201 marbles, of which 100 of them are blue and 101 of them are red. Every turn, we remove 2 marbles from the bag. If the two marbles are of the same color, we remove the two marbles but add a blue marble into the bag. If the two marbles are of different colors, we remove the two marbles and add a red marble into the bag. What is the color of the last marble in the bag?
Answer:
Red Marble is the color in the bag
Step-by-step explanation:
Total number of marbles ; 201
number of red marbles ; 101
number of blue marbles ; 101
Every turn, 2 marbles of the same color are removed, and add a blue marbleif two marbles are different color are removed , and add a red marbleFrom the analysis of the question; there are three possibilities;first possibility ; From what is said as the condition, it can be inferred that the number of blue marbles is reduced by a factor of 1 , while the number of red marbles remains constant. this shows that the two marbles of the same color that was removed is most likely BLUE.
Second possibility ; Two red marbles are added (of the same color) and add a blue marbles, implies the the red marbles reduce by a factor of 2 while the blue marbles increase by a factor of 1. this is also possible.
Third possibility ; Two marbles of different color implies 1R and 1B , and then a red marble is added. this shows that the number of red marble still remains constant ( 1R removed, 1R added), while the number of blue reduce by 1 ( remove 1B, and not replaced back).
From the various possibilities, it can be inferred that the number of red marbles is either increasing by a factor of 2 or remains constant. this further shows that in all three , 1R marble will always remain in the bag no matter the possibility as such in all three scenerios, it further shows that if there is only a marble remaining, it would have been the RED MARBLE WHICH IS AUTOMATICALLY THE LAST MARBLE from the analysis done above.
In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
y''-y=0, y1(t)=e^t , y2(t)=cosht
Answer:
For First Solution: [tex]y_1(t)=e^t[/tex]
[tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.
For 2nd Solution:[tex]y_2(t)=cosht[/tex]
[tex]y_2(t)=cosht[/tex] is the solution of equation y''-y=0.
Step-by-step explanation:
For First Solution: [tex]y_1(t)=e^t[/tex]
In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.
[tex]y_1(t)=e^t[/tex]
First order derivative:
[tex]y'_1(t)=e^t[/tex]
2nd order Derivative:
[tex]y''_1(t)=e^t[/tex]
Put Them in equation y''-y=0
e^t-e^t=0
0=0
Hence [tex]y_1(t)=e^t[/tex] is the solution of equation y''-y=0.
For 2nd Solution:
[tex]y_2(t)=cosht[/tex]
In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.
[tex]y_2(t)=cosht[/tex]
First order derivative:
[tex]y'_2(t)=sinht[/tex]
2nd order Derivative:
[tex]y''_2(t)=cosht[/tex]
Put Them in equation y''-y=0
cosht-cosht=0
0=0
Hence [tex]y_2(t)=cosht[/tex] is the solution of equation y''-y=0.
The functions [tex]y(t)=e^t[/tex] and [tex]y(t)=cosht[/tex] are solutions of given differential equation.
We have to prove that given function is a solution of the differential equation.
Given function, [tex]y(t)=e^t[/tex]
Given differential equation are, [tex]y''-y=0[/tex]
So that, [tex]y'(t)=e^t , y''(t)=e^t[/tex]
Substitute values in given differential equation.
[tex]e^t - e^t=0[/tex]
Thus, function [tex]y(t)=e^t[/tex] is solution of given differential equation.
Another function is, [tex]y(t)=cosht[/tex]
So that, [tex]y'(t)=sinht , y''(t)=cosht[/tex]
Substitute values in given differential equation.
[tex]cosht-cosht=0[/tex]
Thus, function [tex]y(t)=cosht[/tex] is solution of given differential equation.
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Show that the given set of functions is orthogonal with respect to the given weight on the prescribed interval. Find the norm of each function in the set.
Answer/Explanation
The complete question is:
Show that the set function {1, cos x, cos 2x, . . .} is orthogonal with respect to given weight on the prescribed interval [- π, π]
Step-by-step explanation:
If we make the identification For ∅° (x) = 1 and ∅n(x) = cos nx, we must show that ∫ lim(π) lim(-π) .∅°(x)dx = 0 , n ≠0, and ∫ lim(π) lim(-π) .∅°(x)dx = 0, m≠n.
Therefore, in the first case, we have
(∅(x), ∅(n)) ∫ lim(π) lim(-π) .∅°(x)dx = ∫ lim(π) lim(-π) cosn(x)dx
This will therefore be equal to :
1/n sin nx lim(π) lim(-π) = 1/n [sin nπ - sin(-nπ)] = 0 , n ≠0 (In the first case)
and in the second case, we have,,
(∅(m) , ∅(n)) = ∫ lim(π) lim(-π) .∅°(x)dx
This will therefore be equal to:
∫ lim(π) lim(-π) cos mx cos nx dx
Therefore, 1/2 ∫ lim(π) lim(-π)( cos (m+n)x + cos( m-n)x dx (Where this equation represents the trigonometric function)
1/2 [ sin (m+n)x / m+n) ]+ [ sin (m-n)x / m-n) ] lim(π) lim(-π) = 0, m ≠ n
Now, to go ahead to find the norms in the given set intervals, we have,
for ∅°(x) = 1 we have:
//∅°(x)//² = ∫lim(π) lim(-π) dx = 2π
So therefore, //∅°(x)//² = √2π
For ∅°∨n(x) = cos nx , n > 0.
It then follows that,
//∅°(x)//² = ∫lim(π) lim(-π) cos²nxdx = 1/2 ∫lim(π) lim(-π) [1 + cos2nx]dx = π
Thus, for n > 0 , //∅°(x)// = √π
It is therefore ggod to note that,
Any orthogonal set of non zero functions {∅∨n(x)}, n = 0, 1, 2, . . . can be normalized—that is, made into an orthonormal set by dividing each function by its norm. It follows from the above equations that has been set.
Therefore,
{ 1/√2π , cosx/√π , cos2x/√π...} is orthonormal on the interval {-π, π}.
Determine the decision criterion for rejecting the null hypothesis in the given hypothesis test. Describe the values of the test statistic that would result in rejection of the null hypothesis. A cereal company claims that the mean weight of the cereal in its packets is at least 14 oz. You wish to test this claim at the 0.02 level of significance. The mean weight for the random sample of 45 cerial packets is 13.8 oz with a standard deviation of 0.3 oz. What criterion would be used for rejecting the null hypothesis, that oz ? Reject H0 if test statistic < 2.05. Reject H0 if test statistic > -2.05. Reject H0 if test statistic < -2.05. Reject H0 if test statistic < -2.33.
Answer:
We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution
Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is [tex]z_{crit}=-2.05[/tex]
We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"Reject H0 if test statistic < -2.05
And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=13.8[/tex] represent the sample mea n
[tex]s=0.3[/tex] represent the sample standard deviation
[tex]n=45[/tex] sample size
[tex]\mu_o =68[/tex] represent the value that we want to test
[tex]\alpha=0.02[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is at least 14 oz, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 14[/tex]
Alternative hypothesis:[tex]\mu < 14[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{13.8-14}{\frac{0.3}{\sqrt{45}}}=-4.47[/tex]
Critical region
We can assume that the sample size is large enough to use the z distribution as an approximation of the t distribution
Now we need to find on the z distribution a value that accumulates 0.02 of the area on the left and this value is [tex]z_{crit}=-2.05[/tex]
We can use the following excel code to verify it: "=NORM.INV(0.02,0,1)"
So then the correct answer for this case would be:
Reject H0 if test statistic < -2.05
And for our case our calculated value <-2.05 so we have enough evidence to reject the null hypothesis at 2% of significance.
The decision criterion to reject the null hypothesis that the mean weight is at least 14 oz at the 0.02 level of significance is to reject H0 if the test statistic is less than -2.33, as derived from the standard normal z-table.
Explanation:To determine the criterion for rejecting the null hypothesis in a hypothesis test like the one presented - where a cereal company claims that the mean weight of the cereal in its packets is at least 14 oz - we use the level of significance (alpha). Since the level of significance in this problem is 0.02, we refer to the standard normal z-table to find the critical z-value that corresponds to this significance level for a one-tailed test, which in this case is a left-tailed test because we are checking if the mean weight is less than the claim (14 oz).
Using the z-table, we find that the critical z-value for an alpha of 0.02 in a one-tailed test is approximately -2.33. If our calculated test statistic is less than this critical value, we would reject the null hypothesis in favor of the alternative. Therefore, the decision criterion is: Reject H0 if test statistic < -2.33.
Since the given options are 'Reject H0 if test statistic < 2.05', 'Reject H0 if test statistic > -2.05', 'Reject H0 if test statistic < -2.05', and 'Reject H0 if test statistic < -2.33', the correct criterion for this test, considering the level of significance is 0.02, is 'Reject H0 if test statistic < -2.33'.
Evaluate the given expression and express the result using the usual format for writing numbers (instead of scientific notation). 29 Upper C 329C3 29 Upper C 329C3equals=nothing
Answer:
29C3 = 3654
Step-by-step explanation:
What is applied here is combination that n - combination of r which is a subscript and the n - is a superscript.
nCr = nPr/r!where nPr = n!/(n-r)!nCr = n!/(n-r)! r!And we are told to find 29C3, apply the formula and your results will be out.Attached is the detailed explanation.