Answer:
-1024
Step-by-step explanation:
Moivre's theorem allows to easily obtain trigonometric formulas that express the sine and cosine of a multiple angle as a function of the sine and cosine of a simple angle.
De Moivre's theorem can be applied to any complex number [tex]z[/tex]
Where:
[tex]z\in Z[/tex]
Let:
[tex](1+i)=z\\n=20[/tex]
According to Demoivres Theorem, If:
[tex]z=||z||(cos(\theta)+isin(\theta))[/tex]
Then:
[tex]z^n=||z||^n(cos(n\theta)+isin(n\theta))[/tex]
For a complex number [tex]z[/tex]:
[tex]z=a+bi[/tex]
Its magnitude and angle are given by:
[tex]||z||=\sqrt{a^2+b^2} \\\\\theta=arctan(\frac{b}{a} )[/tex]
So:
[tex]||z||=\sqrt{1^2+1^2} =\sqrt{2}[/tex]
[tex]\theta=arctan(\frac{1}{1} )=45^{\circ}[/tex]
Therefore, using De Moivre's theorem:
[tex]z^n=(\sqrt{2} )^{20}(cos(20*45)+isin(20*45))\\\\z^n=(\sqrt{2} )^{20}(cos(900)+isin(900))\\\\z^n=1024(-1+i(0))\\\\z^n=1024(-1)\\\\z^n=(1+i)^{20}=-1024[/tex]
C on e2020
i did it *dab*
Consider the following sets of sample data:
A:
$36,900
, $19,400, $22,200, $21,900, $35,300, $20,500, $35,400, $24,000, $37,700, $35,300, $38,300, $29,600, $26,000, $38,400
B:
2.1
, 5.0, 3.5, 3.7, 2.5, 2.1, 3.7, 4.6, 2.7, 4.1, 1.7
For each of the above sets of sample data, calculate the coefficient of variation, CV. Round to one decimal place.
Answer:
The coefficient of variation for A is 24.6%.
The coefficient of variation for B is 33.7%.
Step-by-step explanation:
The coefficient of variation (CV) is well defined as the ratio of the standard deviation to the mean. It exhibits the degree of variation in association to the mean of the population.
The formula to compute the coefficient of variation is,
[tex]CV=\frac{SD}{Mean}\times 100\%[/tex]
Consider the data set A.
Compute the mean of the data set A as follows:
[tex]Mean_{A}=\frac{1}{n}\sum X[/tex]
[tex]=\frac{1}{14}\times [36900+19400+...+26000+38400]\\=30064.2857[/tex]
Compute the standard deviation of the data set A as follows:
[tex]SD_{A}= \sqrt{ \frac{ \sum{\left(x_i - Mean_{A}\right)^2 }}{n-1} }[/tex]
[tex]= \sqrt{ \frac{ 712852142.8571 }{ 14 - 1} } \\\approx 7405.051[/tex]
Compute the coefficient of variation for A as follows:
[tex]CV=\frac{SD_{A}}{Mean_{A}}\times 100\%[/tex]
[tex]=\frac{7405.051}{30064.2857}\times 100\%\\=24.6\%[/tex]
The coefficient of variation for A is 24.6%.
Consider the data set B.
Compute the mean of the data set B as follows:
[tex]Mean_{B}=\frac{1}{n}\sum X[/tex]
[tex]=\frac{1}{11}\times [2.1+5.0+...+4.1+1.7]\\=3.2455[/tex]
Compute the standard deviation of the data set B as follows:
[tex]SD_{B}= \sqrt{ \frac{ \sum{\left(x_i - Mean_{B}\right)^2 }}{n-1} }[/tex]
[tex]= \sqrt{ \frac{ 11.9873 }{ 11 - 1} } \\\approx 1.0949[/tex]
Compute the coefficient of variation for B as follows:
[tex]CV=\frac{SD_{B}}{Mean_{B}}\times 100\%[/tex]
[tex]=\frac{1.0949}{3.2455}\times 100\%\\=33.7\%[/tex]
The coefficient of variation for B is 33.7%.
10/9 = 3/q
solve for q
Answer:
Q= 27/10
Step-by-step explanation:
Answer: [tex]q=\frac{27}{10}[/tex]
Step-by-step explanation:
[tex]\frac{10}{9}=\frac{3}{q}[/tex]
Multiply by q.
[tex]q*\frac{10}{9}=\frac{3}{q}*q[/tex]
[tex]\frac{10}{9}q=3[/tex]
Now multiply by the reciprocal of [tex]\frac{10}{9}[/tex] which is the inverted fraction. [tex]\frac{9}{10}[/tex]
[tex]\frac{9}{10}*\frac{10}{9}q=3*\frac{9}{10}[/tex]
[tex]q=\frac{27}{10}[/tex]
Email marketing, the online counterpart of direct mail marketing, can be a highly personalized, very targeted means of communication. With its very low cost and wide reach, you feel that email marketing must play a major role in Caffè Gustoso's online advertising campaign.
The owners want to send as many emails as possible to potential customers. However before you send out the first email, you have to choose an email permission strategy.
Select an option from the choices below:
a. Opt-in: Request the consumer's permission before sending any Caffé Gustoso email correspondence.
b. Unsolicited: Send unsolicited Caffe Gustoso email to a large number of potential customer.
c. Opt-out: Allow consumers to quest to be removed from have been contacted
Answer:
c. Opt-out: Allow consumers to quest to be removed from have been contacted.
Step-by-step explanation:
The Caffe Gutoso is considering a marketing strategy to promote its business. The owner want to send emails to its potential customers. Before sending the email the sender must insert email permission strategy as send an email from Caffe Gutoso and place and Opt-out option for those customers who are not willing to receive any email from the Caffe.
The mean potassium content of a popular sports drink is listed as 138 mg in a 32-oz bottle. Analysis of 40 bottles indicates a sample mean of 136.9 mg. (a) State the hypotheses for a two-tailed test of the claimed potassium content.
Answer:
The value for p <0.1 i.e. 0.02088 that means ,we had success in test on potassium content
Step-by-step explanation:
Given:
Means: 138 mg
sample size =40
New mean=136.9 mg
To find : Hypothesis on test.
Solution:
We know that ,
Z=(X-mean)/standard deviation)/{(sqrt(sample size)}
Consider the standard deviation of 3.00 mg
Z=(136.9-138)/(3/sqrt(40))
=-1.1/0.4743
Z=-2.3192
two tailed test
=2{(1-p(<z))}
P value for Z=2.3192 is 0.98956
=2{1-0.98956}
=0.02088
The value for p <0.1
we had success in test on potassium content.
The hypothesis test for the sports drink's potassium content involves a null hypothesis (H0) that the mean is equal to the claimed 138 mg, and an alternative hypothesis (Ha) that it is not equal.
Explanation:The question involves conducting a hypothesis test for a sports drink's potassium content.
Establishing hypotheses for a two-tailed test involves setting a null hypothesis (H0) that the mean potassium content is equal to the claimed amount, and an alternative hypothesis (Ha) that the mean potassium content is not equal to the claimed amount.
Specifically:
H0: μ = 138 mg (The mean potassium content is 138 mg as listed.)Ha: μ ≠ 138 mg (The mean potassium content is not 138 mg as listed.)This is a basic procedure in statistical testing where the null hypothesis often represents a statement of 'no effect' or 'no difference', and the alternative hypothesis contradicts this assumption.
evaluate the expression when y=-6
y squared+7y+4
Answer:
In the equation of a straight line (when the equation is written as "y = mx + b"), the slope is the number "m" that is multiplied on the x, and "b" is the y-intercept (that is, the point where the line crosses the vertical y-axis). This useful form of the line equation is sensibly named the "slope-intercept form".
Step-by-step explanation:
This is just something to keep in mind
Answer:
-2
Step-by-step explanation:
Well, lt me hope the expression is complete amd correct.
Let's go on with the information you submitted.
y² + 7y + 4
When y= -6.
Let's substitute the value, y= -6 in the expression. We'll have to be careful with the signs.
y² + 7y + 4
-6² + 7(-6) + 4
36 - 42 + 4
40 - 42 =
-2
Find the common difference for the sequence shown.
56, 49, 42, 35, ...
-7
-12
7
Answer:
-7
Step-by-step explanation:
Given : sequence 56, 49, 42, 35, ...
To find : Find the common difference.
Solution : We have given that 56, 49, 42, 35, ...
By the common difference of an arithmetic sequence is:
d = a_{2} -a_{1}a2−a1
Where, a_{1}a1 is the first term and a_{1}a1 is the second term.
Then common difference = 49 -56
= -7 .
To get an estimate of consumer spending over a holiday season in 2009, 436 randomly sampled American adults were surveyed. Their spending for the six-day period after Thanksgiving, spanning the Black Friday weekend and Cyber Monday, averaged $84.71. A 95% confidence interval based on this sample is ($80.31, $89.11). Which of the following statements are true? Select all that apply. A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate. In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger. This confidence interval is not valid since the distribution of spending in the sample data is right skewed. The margin of error is $4.4. We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11. This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436. 95% of random samples have a sample mean between $80.31 and $89.11. We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.
Answer:
-A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate.
-This confidence interval is not valid since the distribution of spending in the sample data is right skewed.
-The margin of error is $4.4.
-This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.
-We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.
Step-by-step explanation:
A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate.
TRUE. The 90% confidence is less strict in its probability of having the mean within the interval, so it is narrower than the 95% CI. It relies more in the information given by the sample.
In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger.
FALSE. The margin of error is z*σ/(n^0.5). So to reduce it by two thirds, the sample size n needs to be 3^2=9 times larger.
This confidence interval is not valid since the distribution of spending in the sample data is right skewed.
FALSE. There is no information about the skewness in the sample.
The margin of error is $4.4.
TRUE. The margin of error is (89.11-80.31)/2=$4.4.
We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11.
FALSE. The CI is related to the populations mean. We are 95% confident that the average spending of the population is between $80.31 and $89.11.
This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.
TRUE. This happens accordingly to the Central Limit Theorem.
95% of random samples have a sample mean between $80.31 and $89.11.
FALSE. The confidence interval refers to the population mean.
We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.
TRUE. This is the conclusion that is looked for when constructing a confidence interval.
mean of 153 176 249 150 620 242
Answer: 265
Step-by-step explanation:
Add up all the numbers and divide this value by the total number of terms,in this case 6 terms.
(153+ 176+ 249+ 150+ 620+ 242)/6 =265
If 3ab = c, then solve for a
To solve for an in the equation 3ab = c, divide both sides by 3b, yielding a = c / (3b), with the condition that b is not zero.
Explanation:To solve the equation 3ab = c for a, we need to isolate an on one side of the equation. Starting with the original equation:
3ab = c
We divide both sides of the equation by 3b to get:
a = c / (3b)
This is the solution for an in terms of b and c. It's important to note that b cannot be zero, as division by zero is undefined.
What is pi? Explain it in your own words.
Answer:
pi is the ratio if the circumference over the diameter
Step-by-step explanation:
What is the solution to the equation One-fourth x minus one-eighth = StartFraction 7 Over 8 EndFraction + one-half x?
Answer:
The solution to the equation is
x = -4
Step-by-step explanation:
We want to find the solution to the equation
(1/4)x - 1/8 = 7/8 + (1/2)x
First, add -(1/2)x + 1/8 to both sides of the equation.
(1/4)x - 1/8 - (1/2)x + 1/8 = 7/8 + (1/2)x - (1/2)x + 1/8
[1/4 - 1/2]x = 7/8 + 1/8
x(1 - 2)/4 = (7 + 1)/8
-(1/4)x = 8/8
-(1/4)x = 1
Multiply both sides by -4
x = -4
Answer:
x=-4
Step-by-step explanation:
Daniel believes that people perform better in the barbell curl, on average, if they are encouraged by a coach. He recruited 29 subjects to participate in an experiment and randomly assigned them into two groups. Daniel gave one group verbal encouragement during the exercise and was quiet the exercise for the other group. He recorded the total number of barbell curls each subject was able to complete before setting the bar down.
(a) Explain the purpose of random assignment in this experiment.
(b) Compare these distributions,
(c) State the hypotheses Daniel should use to test his belief about receiving encouragement during exercise. Make sure to define any parameters you use.
(d) Identify the significance test Daniel should use to analyze the results of his experiment and show that the conditions for this test are met.
(e) The P-value for Daniel's test is 0.107. What conclusion should Daniel make at the a=0.05 significance level?
Answer:
a) to remove bias in the study
b) The number of barbell curls performed by group who was encourged was higher than the group who was not encouraged
c) H0: mean of barbell curls for group encouraged= mean of barbell curls for group not encouraged
Ha: mean of barbell curls for group encouraged≠ mean of barbell curls for group not encouraged
d) two sample t-test.
Requirement has been met as there are two separate groups whose means have to be compared. Median of box plot is equal to mean. Standard Deviation is equal to range divided by four. Sample size will have to be assumed
e) Encouragement affects the number of barbell curl performed by an individua
Step-by-step explanation:
a) random assignemnt removes bias in any study.
b) box plot is attached
c) null hypthesis and alternate hypothesis would compare means of the two distributions
d) As we are comparing two groups, we will use two sample t-test.
The median of box plot is equal to mean for both groups.
The rnge/4 is equal to standard deviation.
mean of encouraged group= 30
mean of not encoouraged group=19
SD of encouraged group= (75-5)/4= 15
SD of not encouraged group= (60-0)/4= 15
Assume number of inidivudals in encouraged group =15
assume number of individuals in not encouraged group = 14
e) test statistic= (mean1-mean2)/√(SD1²/(N1-1)+ SD2²/(N2-1))
= (30-19)/(√(15²/14+ 15²/13)
= 1.9734
Since test statistic is greater than p-value, null hypothesis is rejected.
Encouragement affects the number of barbell curl performed by an individual
Find the value of y+5 given that -3y-3=6
Simplify
Suppose that a study of elementary school students reports that the mean age at which children begin reading is 5.8 years with a standard deviation of 0.7 years. (2 pts) Step 1. If a sampling distribution is created using samples of the ages at which 61 children begin reading, what would be the mean of the sampling distribution of sample means
Answer:
5.8 years
Step-by-step explanation:
The sampling distribution of the sample means has a mean that is equal to mean of the population from which the sample has been drawn.
study of elementary school students reports that the mean age at which children begin reading is 5.8 years with a standard deviation of 0.7 years.
This means the population mean is
[tex] \mu = 5.8 \: years[/tex]
and the population standard deviation is
[tex] \sigma = 0.7 \: years[/tex]
If a sampling distribution is created using samples of the ages at which 61 children begin reading, the mean of the sampling distribution of the sample mean will be
[tex]5.8 \: years[/tex]
The mean of the sampling distribution of sample means can be calculated by dividing the mean of the population by the square root of the sample size.
Explanation:To find the mean of the sampling distribution of sample means, we need to divide the mean of the population by the square root of the sample size. In this case, the mean of the population is 5.8 years and the sample size is 61 children. So, the mean of the sampling distribution of sample means can be calculated as:
Mean of sampling distribution of sample means = 5.8 years / sqrt(61) ≈ 0.745 years
Learn more about Mean of sampling distribution of sample means here:https://brainly.com/question/29451043
#SPJ3
Someone please help
????????
Answer:
use the distance formula,
by using it, prove the adjacent sides of the quadrilateral DEFG
hence, DEFG would be a rhombus
In a study of 346 comma 145 cell phone users, it was found that 26 developed cancer of the brain or nervous system. Assuming that cell phones have no effect, there is a 0.000113 probability of a person developing cancer of the brain or nervous system. We therefore expect about 40 cases of such cancer in a group of 346 comma 145 people. Estimate the probability of 26 or fewer cases of such cancer in a group of 346 comma 145 people. What do these results suggest about media reports that cell phones cause cancer of the brain or nervous system?
Answer:
a) P ( X ≤ 26 ) = 0.0134
b) The significance of the cell phone causing brain cancer is 1.34 or 0.0134
Step-by-step explanation:
Solution:-
- From the entire population of cell phone users os size N = 346,145 media reported n = 26 people who developed cancer of brain or nervous system.
- The probability of a person developing cancer of the brain or nervous system, assuming that cell phones have no effect, is p = 0.000113.
- Denote a random variable X: The number people who developed cancer of brain or nervous system are normally distributed.
- The expected or mean number of people who developed cancer of brain or nervous system are u = 40.
- The standard deviation ( s ) for the distribution would be:
[tex]s = \sqrt{u*( 1 - p ) }\\\\s = \sqrt{40*( 1 - 0.000113 ) }\\\\s = 6.3242\\[/tex]
- The random variate X follows normal distribution with parameters:
X ~ Norm ( 40 , 6.3242^2 )
- The probability of X ≤ 26 cases from the total population of N = 346,145 can be determined by evaluating the Z-score standard value of the test statistics:
[tex]P ( X \leq 26 ) = P ( Z \leq [ \frac{X - u }{s} ) \\\\P ( Z \leq [ \frac{ 26 - 40 }{6.3242} ) = P ( Z \leq -2.21371 )[/tex]
- Using standard normal table compute the probability on the left side of Z-score value -2.21371. Hence,
P ( X ≤ 26 ) = 0.0134
- The probability of the less than 26 number of cases who developed cancer of the brain or nervous system is 0.0134 as per media reports.
- So 1.34% of population is a case brain cancer due to cell phones according to media reports.
Which is significant less than the expected number of cases ( 40 ) that occur regardless of cell phone effect.
Answer: The significance of the cell phone causing brain cancer is 1.34 or 0.0134.
Solve (x - 5)^2 = 3
Answer:
c
Step-by-step explanation:
√(x-5)²=±√3
x-5=±√3
x=5±√3
To solve the quadratic equation (x - 5)² = 3, take the square root of both sides to give x - 5 = ± √3, then add 5 to each side to give x = 5 ± √3. The two possible solutions are x = 5 + √3 and x = 5 - √3.
Explanation:The question is asking to solve the equation (x - 5)² = 3. This is a quadratic equation in the form (x-a)² = b. To solve it, you should start by taking the square root on both sides.
Step 1: √[(x - 5)²] = √[3]
This leads to: x - 5 = ± √3
Step 2: Followed by adding 5 to both sides
x = 5 ± √3
So there are two possible solutions: x = 5 + √3 and x = 5 - √3.
Learn more about quadratic equation here:
https://brainly.com/question/26090334
#SPJ3
What is the answer?
A. Infinite Number of Solutions
B. (3, 3)
C. (3, -3)
D. (-3, 3)
Answer:
It's either B or A
Step-by-step explanation:
If I had to choose one I would pick B because I did the math; and logically, that answer makes the most sense.
Simplify 18 - 2[X + (x - 5)].
8- 4x
28 - 4x
28 - 2x
Answer:
28 - 4x
Step-by-step explanation:
[tex]18 -2 [x + (x - 5)] \\ = 18 - 2 [x + x - 5] \\ = 18 - 2 [2x - 5] \\ = 18 -4x + 10 \\ = 28 - 4x \\ [/tex]
HELP ASAP PLEASE!!!
Given the figure below, which of the following points name a line segment, a line, or a ray?
a. point B and point C
b. point A and point C
c. point D and point A
d. point E and point A
U may pick more than one.
Find the sum of 526+213 by breaking apart the addenda into hundreds
Mr. Gordon has 13 girls and 14 boys in his fourth period algebra class. One person is chosen at random.
What is the probability that the person chosen is a boy?
A. 1/14
B. 1/27
C. 13/14
D. 13/27
E. 14/27
Answer:
E
Step-by-step explanation:
It's E because you have to add 13 and 14 and as you can see there's 14 boys so it's E.
The probability that the person chosen is a boy is 14/27.
Number of girls = 13
Number of boys = 14
Total students = 27
What is probability?Probability is a measure of the likelihood of occurrence of an event.
P(event) = Favorable outcomes / total outcomes
Favourable outcomes = 14 because there are 14 boys
Total outcomes =27
So, the probability that the person chosen is a boy will be given by:
P(boy)=14/27
So, the probability that the person chosen is a boy is 14/27.
Hence, the probability that the person chosen is a boy is 14/27.
To get more about probability visit:
https://brainly.com/question/24756209
Shiffon Electronics manufactures music player. Its costing system uses two cost categories, direct materials and conversion costs. Each product must pass through the Assembly Department, the Programming department, and the Testing Department. Direct materials are added at the beginning of the production process. Conversion costs are allocated evenly throughout production. Shiffon Electronics uses weighted-average costing. The following information is available for the month of March 2017 for the Assembly department. Work in process, beginning inventory 340 units Conversion costs (25% complete) Units started during March 860 units Work in process, ending inventory: 140 units Conversion costs (60% complete) The cost details for the month of March are as follows: Work in process, beginning inventory: Direct materials $349,000 Conversion costs $360,000 Direct materials costs added during March $703,500 Conversion costs added during March $1,130,000 What amount of conversion costs is assigned to the ending Work-in-Process account for March
Answer:
Possible options:
A.$182,343
B.$122,792
C.$109,406
D.$270,629
Answer C
Step-by-step explanation:
Under this method the percentage completion done in the Beginning Inventory is ignored while calculating the equivalent completed units during the current period. Beginning Inventory Units are treated as fresh units introduced for production.
Cost Accounted for:
- The cost of opening work-in-progress and cost of the current period are aggregated and the aggregate cost is divided by output in terms of completed equivalent units.
- The units of Beginning Inventory of WIP and their cost are taken in full under this method.
The typical lifespan for various mammal species in captivity (L) in years has been related to averageadult size (M) in kilograms according to the regression equation seen below. A typical adult Meerkat weighs about 0.9 kilograms. What is the predicted lifespan of a Meerkat incaptivity, according to this equation? * 2 points 11.6 years 2.4 years 14.1 years 2.6 years 28.8 years
Complete Question:
The typical lifespan for various mammal species in captivity (L) in years has been related to average adult size (M) in kilograms according to the regression equation seen below.
[tex]ln L = 2.468 + 0.2 (lnM)[/tex]
A typical adult Meerkat weighs about 0.9 kilograms. What is the predicted lifespan of a Meerkat incaptivity, according to this equation? * 2 points 11.6 years 2.4 years 14.1 years 2.6 years 28.8 years
Answer:
Option A) L = 11.6 years
Step-by-step explanation:
From the given equation:
[tex]ln L = 2.468 + 0.2 (lnM)[/tex]..........(1)
Average adult size, M = 0.9 kg
Putting the value of M into the regression equation in (1)
[tex]ln L = 2.468 + 0.2 (ln0.9)\\ln L = 2.468 + (-0.02107)\\ln L = 2.447\\L = e^{2.447} \\L = 11.55 years[/tex]
The question is an illustration of regressions.
The lifespan of a typical adult Meerkat that weighs about 0.9 kilograms is (a) 11.6 years
The regression equation is given as:
[tex]\ln(L) = 2.468 + 0.2\ln(M)[/tex]
From the question, we have:
M = 0.9
Substitute 0.9 for M in [tex]\ln(L) = 2.468 + 0.2\ln(M)[/tex]
[tex]\ln(L) = 2.468 + 0.2\ln(0.9)[/tex]
Take natural logarithm of 0.9
[tex]\ln(L) = 2.468 + 0.2\times -0.1054[/tex]
[tex]\ln(L) = 2.468 -0.02108[/tex]
[tex]\ln(L) = 2.44692[/tex]
Take exponents of both sides
[tex]L = e^{2.44692}[/tex]
[tex]L = 11.553[/tex]
Approximate
[tex]L = 11.6[/tex]
Hence, the lifespan is (a) 11.6 years
Read more about regression equations at:
https://brainly.com/question/7656407
Moose Drool Makes Grass More Appetizing Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study27 has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is 0.183. The standard error for this estimate is 0.016.
Give notation for the quantity being estimated, and define any parameters used.
Give notation for the quantity that gives the best estimate, and give its value.
Give a 95% confidence interval for the quantity being estimated. Interpret the interval in context.
3.68 Bisphenol A in Your Soup Cans Bisphenol A (BPA) is in the lining of most canned goods, and recent studies have shown a positive association between BPA exposure and behavior and health problems. How much does canned soup consumption increase urinary BPA concentration? That was the question addressed in a recent study34 in which consumption of canned soup over five days was associated with a more than 1000% increase in urinary BPA. In the study, 75 participants ate either canned soup or fresh soup for lunch for five days. On the fifth day, urinary BPA levels were measured. After a two-day break, the participants switched groups and repeated the process. The difference in BPA levels between the two treatments was measured for each participant. The study reports that a 95% confidence interval for the difference in means (canned minus fresh) is 19.6 to 25.5 μg/L.
Is this a randomized comparative experiment or a matched pairs experiment? Why might this type of experiment have been used?
What parameter are we estimating?
Interpret the confidence interval in terms of BPA concentrations.
If the study had included 500 participants instead of 75, would you expect the confidence interval to be wider or narrower?
3.70 Effect of Overeating for One Month: Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows.35 Researchers recruited 18 healthy and normal-weight people with an average age of 26. For a four-week period, participants increased calorie intake by 70% (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight.
What is the relevant parameter?
How could we find the actual exact value of the parameter?
Give a 95% confidence interval for the parameter and interpret it.
Give the margin of error and interpret it.
In the first study, moose drool reduces the toxin levels on grass. In the second study, canned soup consumption increases urinary BPA concentrations. In the third study, overeating for four weeks leads to long-term weight gain.
Explanation:Question 1:
The notation for the quantity being estimated is the mean level of the toxin ergovaline on the grass (in ppm). Scientists estimate the mean level after treatment with moose drool to be 0.183 ppm.
The quantity that gives the best estimate is the mean level of the toxin ergovaline on the grass after treatment with moose drool (0.183 ppm).
A 95% confidence interval for the quantity being estimated is (0.183 - 1.96 * 0.016, 0.183 + 1.96 * 0.016), which is approximately (0.152, 0.214) ppm. This means there is a 95% chance that the true mean level of the toxin ergovaline on the grass after treatment with moose drool is between 0.152 and 0.214 ppm.
Question 2:
This is a matched pairs experiment because each participant is subjected to both treatments (canned soup and fresh soup). It was used to eliminate individual differences and control for variables that might affect the urinary BPA concentrations.
The parameter being estimated is the difference in means (canned minus fresh) of urinary BPA concentrations.
The 95% confidence interval for the difference in means is (19.6, 25.5) μg/L. This means that we are 95% confident that the true difference in mean urinary BPA concentrations between canned soup and fresh soup is between 19.6 and 25.5 μg/L.
If the study had included 500 participants instead of 75, we would expect the confidence interval to be narrower because a larger sample size reduces the standard error, leading to a more precise estimate.
Question 3:
The relevant parameter is the mean weight gain for participants two and a half years after the experiment.
The actual exact value of the parameter cannot be found unless we measure the weight gain for all individuals in the population.
A 95% confidence interval for the parameter is (6.8 - 1.96 * 1.2, 6.8 + 1.96 * 1.2), which is approximately (4.44, 9.16) lbs. This means there is a 95% chance that the true mean weight gain for participants two and a half years after the experiment is between 4.44 and 9.16 lbs.
The margin of error is 1.96 * 1.2 lbs, which is approximately 2.35 lbs. This means that the estimated mean weight gain may vary by up to 2.35 lbs from the true mean weight gain.
Learn more about Biology here:https://brainly.com/question/12400350
#SPJ12
What is the Simplify expression of (15+39)÷6
Answer:
9
Step-by-step explanation:
[tex] \frac{(15 + 39)}{6} \\ \frac{54}{6 } \\ ans = 9 \\ [/tex]
Which expressions represent 1 less than the product of 3 and 4?
Answer:3+4-1
Step-by-step explanation:
The compressive strength of concrete is normally distributed with mu = 2500 psi and sigma = 50 psi. A random sample of n = 8 specimens is collected. What is the standard error of the sample mean?
Round your final answer to three decimal places (e.g. 12.345).
The standard error of the sample mean is __ psi.
Answer:
The standard error of the sample mean is _17.677_ psi.
Step-by-step explanation:
Explanation:-
A random sample of n = 8 specimens is collected.
Given sample size is n = 8
Given mean of the population 'μ' = 2500 psi
standard deviation 'σ' = 50 psi
Let x⁻ is the mean of the observed sample
Standard error of the sample mean = [tex]\frac{S.D}{\sqrt{n} }[/tex] ...(i)
Given Population of standard error (S.D) 'σ' = 50 psi
Now substitute all values in (i)
[tex]S.E = \frac{50}{\sqrt{8} } =17.677[/tex]
Conclusion:-
The standard error of the sample mean is _17.677psi.
The standard error of the sample mean for a sample of 8 concrete specimens is approximately 17.68 psi
To calculate the standard error of the sample mean, we use the formula:
[tex]\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} \][/tex]
Where:
- [tex]\sigma[/tex] is the standard deviation of the population.
- n is the sample size.
Given:
- [tex]\sigma[/tex] = 50 psi
- n = 8
Substituting these values into the formula:
[tex]\[ \text{Standard Error} = \frac{50}{\sqrt{8}} \][/tex]
[tex]\[ \text{Standard Error} = \frac{50}{2.828} \][/tex]
[tex]\[ \text{Standard Error} \approx 17.68 \][/tex]
So, the standard error of the sample mean is approximately 17.68 psi.
basketball players score 19 times in one game he scored a total of 33.2 of each point to point shot in one of each free-throw how many to point shots did he make how many free throws
The answer is 17 free throws because I said so and that’s the correct answer
Answer:
my answer was gunna be his but that person answered first so ill just sound like i copied that person
Step-by-step explanation:
As a result of complaints by the staff about noise, the coffee and recreation area for student interns at OHaganBooks will now be in a 512 square foot rectangular area in the headquarter's basement against the southern wall. (The specified area was arrived at in complex negotiations between the student intern representative and management.) The construction of the partition will cost $16 per foot for the north wall and $4 per foot for the east and west walls. What are the dimensions of the cheapest recreation area that can be made?
Answer:
The dimensions are 32 ft by 16 ft
Step-by-step explanation:
Area of the coffee and recreation room=512 square foot
LB=512
L=512/B
Perimeter of the Room = Perimeter of north wall+Perimeter of east wall+ Perimeter of west wall =L+2B (West and East are opposite)
Cost of the Perimeter=16L+4(2B)
[tex]C(B)=16(\frac{512}{B})+8B\\C(B)=\frac{8192+8B^2}{B}[/tex]
To minimize cost, first, we take the derivative of C(B)
Using quotient rule
[tex]C^{'}(B)=\frac{8B^2-8192}{B^2}[/tex]
Setting the derivative equal to zero
[tex]8B^2-8192=0\\8B^2=8192\\B^2=1024\\B=32 ft[/tex]
[tex]L=\frac{512}{B}=\frac{512}{32}=16ft[/tex]
The dimension of the cheapest recreation area will be 32 ft by 16 ft