Answer:
The correct answer is A. Being American but working in Argentina, John is an expatriate manager.
Step-by-step explanation:
An expatriate is a person who, temporarily or permanently, resides in a country different from the country in which he was born. The term is commonly used in the case where companies send their professionals or workers to their delegations abroad.
The probability of buying a movie ticket with a popcorn coupon is 0.629 and without a popcorn coupon is 0.371. If you buy 29 movie tickets, we want to know the probability that more than 16 of the tickets have popcorn coupons. Consider tickets with popcorn coupons as successes in the binomial distribution. Give the numerical value of the parameter p in this binomial distribution scenario.
Answer:
The probability[tex]0.75095[/tex] and the parameter [tex]p=0.629[/tex]
Step-by-step explanation:
The formula for probability in a binomial distribution is:
[tex]b(x;n,p)=\frac{n!}{x!(n-x)!}\ast p^{x}\ast(1-p)^{n-x}[/tex]
where p is the probability of success (ticket with popcorn coupon), n is the number of trials (tickets bought) and x the number of successes desired. In this case p=0.629 (probability of buying a movie ticket with coupon), n=29, and x=17,18,19, ...29.
[tex]b(17;29,0.629)=\frac{29!}{17!(29-17)!}\ast0.629^{17}\ast(1-0.629)^{29-17}=0.133\,25\\ b(18;29,0.629)=\frac{29!}{18!(29-18)!}\ast0.629^{18}\ast(1-0.629)^{29-18}=0.150\,61[/tex]
[tex]b(19;29,0.629)=\frac{29!}{19!(29-19)!}\ast0.629^{19}\ast(1-0.629)^{29-19}=0.147\,84\\ b(20;29,0.629)=\frac{29!}{20!(29-20)!}\ast0.629^{20}\ast(1-0.629)^{29-20}=0.125\,32 \\ b(21;29,0.629)=\frac{29!}{21!(29-21)!}\ast0.629^{21}\ast(1-0.629)^{29-21}=0.091\,06 \\ b(22;29,0.629)=\frac{29!}{22!(29-22)!}\ast0.629^{22}\ast(1-0.629)^{29-22}=0.056\,14[/tex]
[tex]b(23;29,0.629)=\frac{29!}{23!(29-23)!}\ast0.629^{23}\ast(1-0.629)^{29-23}=2.896\,8\times10^{-2} \\ b(24;29,0.629)=\frac{29!}{24!(29-24)!}\ast0.629^{24}\ast(1-0.629)^{29-24}=1.227\,8\times10^{-2}\\ b(25;29,0.629)=\frac{29!}{25!(29-25)!}\ast0.629^{25}\ast(1-0.629)^{29-25}=4.163\,4\times10^{-3} \\ b(26;29,0.629)=\frac{29!}{26!(29-26)!}\ast0.629^{26}\ast(1-0.629)^{29-26}=1.085\,9\times10^{-3} \\ b(27;29,0.629)=\frac{29!}{27!(29-27)!}\ast0.629^{27}\ast(1-0.629)^{29-27}=2.045\,7\times10^{-4}[/tex]
[tex]b(28;29,0.629)=\frac{29!}{28!(29-28)!}\ast0.629^{28}\ast(1-0.629)^{29-28}=2.477\,4\times10^{-5} \\ b(29;29,0.629)=\frac{29!}{29!(29-29)!}\ast0.629^{29}\ast(1-0.629)^{29-29}=1.448\,3\times10^{-6}[/tex]
The probability of more than 16 is equal to the sum of the probability of x=17, 17,18,19, ...29.
[tex]b(x>16;29,0.629)=0.13325+0.15061+0.14784+0.12532+0.09106+0.05614+2.8968\times10^{-2}+1.2278\times10^{-2}+4.1634\times10^{-3}+1.0859\times10^{-3}+2.0457\times10^{-4}+2.4774\times10^{-5}+1.4483\times10^{-6}=0.75095[/tex]
The numerical value of the parameter p in this binocular distribution scenario, where getting a popcorn coupon is defined as a success, is 0.629.
Explanation:In this binomial distribution scenario, we are considering buying a movie ticket with a popcorn coupon as a success. The probability of success (p) is given as 0.629. Therefore, in this context, the numerical value of the parameter p for the binomial distribution is 0.629.
This means that each time a ticket is bought, there is a 0.629 chance (or 62.9%) of getting a popcorn coupon with the ticket. This probability stays constant with each new ticket purchase. In other words, the purchase of one ticket does not influence the likelihood of the outcome of the next ticket. This makes the scenario suitable to be modeled using a binomial distribution.
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If a ball is drawn from a bag containing 13 red balls numbered 1-13 and 5 white balls numbered 14-18. What is the probability that a. the ball is not even numbered? b. the ball red and even numbered? c. the ball red or even numbered? d. the ball is neither red or even numbered?
Answer:
a. 50%
b. 33%
c. 17% (I'm assuming the exercise is wrong and it has to say "white" instead of "red", because if not is the same as b.)
d. 67%
Step-by-step explanation:
a. We have a total of 18 balls, 13 are red and 5 are white. They are numbered from 1 to 18. In this case, we don't care about the color of the ball, we just need it to be not even. We have to count how many not even numbers are between 1 and 18, that is 9. So, the chances of getting a ball not even numbered are 9 in 18, that's
[tex]\frac{9}{18}*100=50\%[/tex]
b. Now we do care about the color of the ball. The red balls are numbered from 1 to 13, so we have 6 balls even numbered. That makes the chances 6 in 18 (we still have 18 in total), that's
[tex]\frac{6}{18}*100=33\%[/tex]
c. (I'm assuming the exercise is wrong and it has to say "white" instead of "red", because if not is the same as b.)
The white balls are numbered from 14 to 18, so we have 3 balls even numbered. That makes the chances 3 in 18,
[tex]\frac{3}{18}*100=17\%[/tex]
d. Let's notice that "the ball is neither red or even numbered" is the complement (exactly the opposite) of "the ball is red and even numbered", that means
100% = Probability (ball red and even numbered) + Probability (ball neither red or even numbered)
So, Probability (ball neither red or even numbered) = 100% - Probability (ball red and even numbered) = 100% - 33% = 67%
An automobile emissions testing center has six inspectors and tests 50 vehicles per hour. Each inspector can inspect 12 vehicles per hour. How many inspectors would the center require to have a target utilization of 90 percent?
Answer:
The center require to have a 4.6 inspectors, but this on you decide if you have 4 or 5 inspectors, it may not have 4.6 people working.
So then if you decide to have 5 people their utilization percentage is 83.3%, but if you decide to have 4 people their utilization rate will be 104.17% at the risk of defaulting on demand.
Step-by-step explanation:
1. Define your variables
Demand rate= 50 vehicles per hour
Service rate = 12 vehicles per hour per inspector
Inspectors= a
2. Use the formule of the center´s utilization
U= [tex]\frac{Demand}{(Service) X a}[/tex]
0,9= [tex]\frac{50}{(12) X a }[/tex]
0,9a=[tex]\frac{50}{12}[/tex]
a=[tex]\frac{50}{12X0,9}[/tex]
a= 4.6
3. According to the center´s utilization formula the center require 4,6 inspectors, but approaches 5 because people cannot be divided. With this numbers of inspectors the utilization is:
U= [tex]\frac{Demand}{(Service) X (No. inspectors)}[/tex]
U= [tex]\frac{50}{(12) X (5)}[/tex]
U= 83,3%
4. Other option that you will be used is that the center require 4 inspectors. With this numbers of inspectors the utilization is:
U= [tex]\frac{Demand}{(Service) X (No. inspectors)}[/tex]
U= [tex]\frac{50}{(12) X (4)}[/tex]
U= 104,17%
The center would require 6 inspectors to have a target utilization of 90 percent.
To determine the number of inspectors required to have a target utilization of 90 percent, we can use the formula:
Number of inspectors = Total vehicles per hour / Vehicles inspected per inspector per hour / Target utilization
Given that the center tests 50 vehicles per hour, and each inspector can inspect 12 vehicles per hour, the formula becomes:
Number of inspectors = 50 / 12 / 0.9 = 5.56
Since we cannot have a fraction of an inspector, we round up to the nearest whole number. Therefore, the center would require 6 inspectors to have a target utilization of 90 percent.
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A study was conducted to measure the effectiveness of a diet program that claims to help manage weight. Subjects were randomly selected to participate. Before beginning the program, each participant was given a score based on his or her fitness level. After six months of following the diet, each participant received another score. The study wanted to test whether there was a difference between before and after scores. What is the correct alternative hypothesis for this analysis?
Answer: u (sub d) is inequal to zero
Step-by-step explanation: Because this is a paired t-test, our alternative hypothesis would be u(sub d) is inequal to zero.
Final answer:
The alternative hypothesis for a study on the effectiveness of a diet program would express the expectation of a statistically significant change in fitness level scores, likely a decrease if lower scores indicate better fitness, after participating in the program compared to before.
Explanation:
The correct alternative hypothesis for a study that aims to measure the effectiveness of a diet program in managing weight would look at whether there is a statistically significant difference in the fitness level scores of the participants before and after following the program. As the question suggests evaluating the effectiveness of a diet program, we are particularly interested in seeing an improvement, which would mean expecting a lower score after the program if the score represents a measure where lower is better.
Thus, the alternative hypothesis (H1) should reflect the expectation of improvement. If the fitness score is such that a lower score indicates better fitness, the alternative hypothesis would be:
H1: The mean fitness level score after the program is lower than the mean fitness level score before the program.
This implies that the diet program is effective in improving fitness levels. On the contrary, if a higher fitness score indicates better fitness, the alternative hypothesis would be framed to reflect an expected increase in the score after the program.
The mean per capita income is 15,451 dollars per annum with a variance of 298,116. What is the probability that the sample mean would differ from the true mean by less than 22 dollars if a sample of 350 persons is randomly selected? Round your answer to four decimal places.
Answer: 0.5467
Step-by-step explanation:
Let X be the random variable that represents the income (in dollars) of a randomly selected person.
Given : [tex]\mu=15451[/tex]
[tex]\sigma^2=298116\\\\\Rightarrow\ \sigma=\sqrt{298116}=546[/tex]
Sample size : n=350
z-score : [tex]\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
To find the probability that the sample mean would differ from the true mean by less than 22 dollars, the interval will be
[tex]\mu-22,\ \mu+22\\\\=15,451 -22,\ 15,451 +22\\\\=15429,\ 15473[/tex]
For x=15429
[tex]z=\dfrac{15429-15451}{\dfrac{546}{\sqrt{350}}}\approx-0.75[/tex]
For x=15473
[tex]z=\dfrac{15473-15451}{\dfrac{546}{\sqrt{350}}}\approx0.75[/tex]
The required probability :-
[tex]P(15429<X<15473)=P(-0.75<z<0.75)\\\\=1-2(P(z<0.75))=1-2(0.2266274)=0.5467452\approx0.5467[/tex]
Hence, the required probability is 0.5467.
4> Solve by using Laplace transform: y'+5y'+4y=0; y(0)=3 y'(o)=o
Answer:
[tex]y=3e^{-4t}[/tex]
Step-by-step explanation:
[tex]y''+5y'+4y=0[/tex]
Applying the Laplace transform:
[tex]\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0[/tex]
With the formulas:
[tex]\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)[/tex]
[tex]\mathcal{L}[y']=s\mathcal{L}[y]-y(0)[/tex]
[tex]\mathcal{L}[x]=L[/tex]
[tex]s^2L-3s+5sL-3+4L=0[/tex]
Solving for [tex]L[/tex]
[tex]L(s^2+5s+4)=3s+3[/tex]
[tex]L=\frac{3s+3}{s^2+5s+4}[/tex]
[tex]L=\frac{3(s+1)}{(s+1)(s+4)}[/tex]
[tex]L=\frac3{s+4}[/tex]
Apply the inverse Laplace transform with this formula:
[tex]\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}[/tex]
[tex]y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}[/tex]
Which equation correctly describes the relationship between the measures of the angles and arcs formed by the intersecting secants?
m∠1=1/2(mAB−mEF)
m∠1=1/2(mAB+mEF)
m∠1=1/2mAB
m∠1=mAB+mEF
Answer:
m∠1=1/2(mAB+mEF)
Step-by-step explanation:
The measure of the angle is half the sum of the intercepted arcs.
Answer:
[tex]m\angle 1 = \frac{1}{2}(m\angle AB+m\angle EF)[/tex]
Step-by-step explanation:
When two chords intersect each other inside a circle, the measure of the angle formed is one half the sum of the measure of the intercepted arcs.
Here, the chords FA and BE intersected each other inside the circle,
Also, angle 1 is the angle formed by the intersection,
Thus, from the above statement,
[tex]m\angle 1 = \frac{1}{2}(m\angle AB+m\angle EF)[/tex]
Second option is correct.
For a certain candy, 20% of the pieces are yellow, 15% are red, 20% are blue, 20% are green, and the rest are brown. a) If you pick a piece at random, what is the probability that it is brown? it is yellow or blue? it is not green? it is striped? b) Assume you have an infinite supply of these candy pieces from which to draw. If you pick three pieces in a row, what is the probability that they are all brown? the third one is the first one that is red? none are yellow? at least one is green?
Answer:
Step-by-step explanation:
Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).
Brown: [tex]\frac{25}{100}[/tex]Yellow or Blue: [tex]\frac{20}{100} +\frac{20}{100} = \frac{40}{100}[/tex] ....add the numeratorsNot Green: [tex]\frac{80}{100}[/tex].... since the sum of all the rest is 80%Stiped: [tex]\frac{25}{100}[/tex] .... there are 0 striped candies.Assuming the ratios/percentages of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.
3 Browns: [tex]\frac{25*25*25}{100*100*100} = \frac{15,625}{1,000,000} = \frac{1.5625}{100}[/tex]the 1st and 3rd are red while the middle is any. We multiply 15% * (total of all minus red which is 85%) * 15% like so.[tex]\frac{15*85*15}{100*100*100} = \frac{19,125}{1,000,000} = \frac{1.9125}{100}[/tex]
None are Yellow: multiply the percent of all minus yellow three times.[tex]\frac{80*80*80}{100*100*100} = \frac{512,000}{1,000,000} = \frac{51.2}{100}[/tex]
At least 1 green: multiply the percent of green by 100% twice, since the other two can by any[tex]\frac{20*100*100}{100*100*100} = \frac{200,000}{1,000,000} = \frac{20}{100}[/tex]
A true false test with 10 questions is given. Compute the probability of scoring exactly 80% by guessing
Answer: 0.04395
Explanation:
Given: 10 true-false questions.
So, we will have 50% chances (probability = 0.5) of being correct.
Prob( Exactly 80% score) = Prob (exactly 8 answers correct)
As we observe, if X= number of correct answers, then X~ Binomial (n=10, p=0.5)
So, Prob( Exactly 80% score) = Prob (exactly 8 answers correct)
=[tex]\binom{10}{8}\times(1/2)^{8}}\times(1/2)^{2}[/tex]
= 0.0439453125
= 0.04395
The time needed to complete a final examination in a particular college course is normally distributed with a mean of 79 minutes and a standard deviation of 8 minutes. Answer the following questions.
What is the probability of completing the exam in one hour or less (to 4 decimals)?
What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes (to 4 decimals)?
Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time (to the nearest whole number)?
Answer: a) 0.0088
b) 0.2997
c) 5
Step-by-step explanation:
Given : Mean : [tex]\mu = 79[/tex] minutes
Standard deviation : [tex]\sigma = 8[/tex] minutes
The formula for z-score :
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
a) For x = 60 minutes
[tex]z=\dfrac{60-79}{8}=-2.375[/tex]
The p-value =[tex]P(z\leq-2.375)=0.0087745\approx0.0088[/tex]
b) For x = 75 minutes
[tex]z=\dfrac{75-79}{8}=-0.5[/tex]
The p-value =[tex]P(60<x<75)=P(-2.375<z<-0.5)[/tex]
[tex]=P(-0.5)-P(-2.375)=0.3085-0.0088=0.2997[/tex]
c) For x = 90 minutes
[tex]z=\dfrac{90-79}{8}=1.375[/tex]
The p-value =[tex]P(z>1.375)=1-P(z<1.375)[/tex]
[tex]=1-0.9154342=0.0845658[/tex]
If the number of students in the class = 60 .
Then , the number of students will be unable to complete the exam in the allotted time =[tex]0.0845658\times60=5.073948\approx5[/tex]
The probability of completing the exam in one hour or less is 0.0087. The probability that the exam is completed in more than 60 minutes but less than 75 minutes is 0.2998. We expect about 5 students to not finish the exam in the given 90 minutes.
Explanation:In statistics, when a data set is normally distributed, we use a z-score to describe the position of a raw score in terms of its distance from the mean, when measured in standard deviation units. The formula to calculate a z-score is Z = (X - μ) / σ, where X is the raw score, μ is the mean, and σ is the standard deviation.
To answer the questions:
Probability of completing the exam in one hour or less: Here, we need to calculate the z-score for 60 minutes (which is one hour) using the given mean (79 minutes) and standard deviation (8 minutes). Using the Z score formula, Z = (60-79)/8 = -2.375. You would then look up this z-score in a Z-table (also known as standard normal table) to find the probability, which is around 0.0087 to four decimal places. So the probability of completing the exam in one hour or less is 0.0087.Probability that a student will complete the exam in more than 60 minutes but less than 75 minutes: We need to calculate the z-scores for 60 minutes and 75 minutes. We know the z-score for 60 minutes from before is -2.375. The z-score for 75 minutes is (75-79)/8 = -0.5. The probabilities in the Z-table for these z-scores are about 0.0087 and 0.3085 respectively. We need to subtract the two probabilities to get the answer: 0.3085 - 0.0087 = 0.2998. So the probability that the exam is completed in more than 60 minutes but less than 75 minutes is 0.2998.Expected number of students unable to complete the exam in the 90 minutes examination period: Here we need to find the probability that a student will take more than 90 minutes to finish the exam. The z-score for 90 minutes is (90-79)/8 = 1.375. The probability associated with this z-score in the Z-table is about 0.9157. This essentially means the probability of completing the exam in 90 minutes or less is 0.9157. So, the probability of not completing in time is 1 - 0.9157 = 0.0843. If there are 60 students in the class, we expect about 60*0.0843 = 5.058, which rounds to about 5 students, not to finish the exam in the given 90 minutes time.Learn more about Normal Distribution here:https://brainly.com/question/34741155
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A sample of ????=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of ????=$4 . We obtain a 95% confidence interval as (14.43,17.57) . Which action will not reduce the margin of error?
Answer:
"Decreasing the sample size."
Step-by-step explanation:
The margin of error is :
[tex]ME=z\times \frac{\sigma }{\sqrt{n}}[/tex]
Therefore, as n decreases, margin of error increases.
In the question we are asked " Which action will not reduce the margin of error."
Therefore, the correct option for which action will not reduce the margin of error is "Decreasing the sample size."
Find a simplified weighted voting system which is equivalent to
[8: 9, 3, 2, 1] and
[20: 8, 6, 3, 2, 1].
Answer: The explanation is as follows:
Step-by-step explanation:
(a) [8: 9, 3, 2, 1]
q = 8
Here, coalition is as follows:
[P1, P2, P3, P4] = [9, 3, 2, 1]
for the above coalition, the combined weight is
[P1, P2, P3, P4] = 9+3+2+1 = 15 ⇒ combined weight
For simplified weighted voting system;
q = combined weight ⇒ both the terms have to be equal for a simplified weighted voting system.
But, here 8 ≠ 15
∴ It is not a simplified weighted voting system.
(b) [20: 8, 6, 3, 2, 1]
q = 20
Here, coalition is as follows:
[P1, P2, P3, P4, P5] = [8, 6, 3, 2, 1]
for the above coalition, the combined weight is
[P1, P2, P3, P4, P5] = 8+6+3+2+1 = 20 ⇒ combined weight
For simplified weighted voting system;
q = combined weight
Since, 20 = 20
∴ It is a simplified weighted voting system.
Math help ASAP!! Also both drop down boxes are the same.
Answer:
Domain: amount of fuel in the airplane's tank (in gallons)
The set of all real numbers from 0 to 200
Range: weight of airplane (In pounds)
The set of all real numbers from 3000 to 4400
Step-by-step explanation:
We have the following function
[tex]W=7F+3000[/tex]
Where W represents the weight of the plane in pounds and F represents the amount of fuel in gallons.
The domain of a function is the set of values "F" that can be entered in a function W(F) to obtain an output value of W.
In this case the range of the function W(F) is the whole set of values [tex]W_1, W_2, W_3, ..., W_n[/tex] that are obtained for [tex]F_1, F_2, F_3, ..., F_n[/tex]
Note that, in this case, equation W(F) is used to obtain the weight of the airplane from the amount of fuel F.
Then the domain of the function is the amount of fuel in the airplane tank (in gallons). Since the tank can only hold up to 200 gallons, and there are no negative volume units, then the domain is all real numbers between 0 and 200.
The range of the function is the weight of the plane (in pounds). Note that the minimum weight of the airplane with 0 gallons of fuel is 3000 pounds and the maximum weight with the full tank is 4400 pounds.
Then the range is all real numbers between 3000 and 4400
FH←→ is tangent to circle E at point F.
What is the measure of ∠EFH?
80º
90º
160º
180º
Check the picture below.
Answer: SECOND OPTION.
Step-by-step explanation:
It is important to remember that a tangent to a circle is a line that touches it at one point. This point is called "Point of tangency". By definition, the angle between the tangent and the radius is 90 degrees.
In this case you can observe that EF is the radius of this circle, therefore, the angle between EF and FH measures 90 degrees.
Based on this, you can say the following:
[tex]\angle EFH=90\°[/tex]
This matches with the second option.
M1Q6.) How many degrees should be used to represent convertables in the Pie Graph?
Answer:
80 degrees
Step-by-step explanation:
The entire circle represents 72 and the "slice of pie" is represented by a portion:
16/72 * 100
= 0.22 * 100
= 22.222%
.
22.222% of 360 degrees
= .22222 * 360
= 80 degrees
Each portion can be represented by 16/72
16/72 = 0.22
0.22... * 100% = 22.22...%
0.2222 * 360 = 80
Therefore, the answer is 80 degrees.
Best of Luck!
I need the answer to this math question.
1) Divide 251 days 21 hours by 13.
Then round to the nearest hundredth as necessary.
Answer: 465
Step-by-step explanation:
251 (days) x 24 (hours) = 6,024 hours
6,024+21 hours= 6,045
6045/13=465
It is estimated that one third of the general population has blood type A A sample of six people is selected at random. What is the probability that exactly three of them have blood type A?
Answer: 0.2195
Step-by-step explanation:
Binomial distribution formula :-
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of x successes in the n independent trials of the experiment and p is the probability of success.
Given : The probability of that the general population has blood type A = [tex]\dfrac{1}{3}[/tex]
Sample size : n=6
Now, the probability that exactly three of them have blood type A is given by :-
[tex]P(3)=^6C_3(\dfrac{1}{3})^3(1-\dfrac{1}{3})^{6-3}\\\\=\dfrac{6!}{3!3!}(\dfrac{1}{3})^3(\dfrac{2}{3})^{3}\\\\=0.219478737997\approx0.2195[/tex]
Therefore, the probability that exactly three of them have blood type A = 0.2195
A survey conducted by the U.S. department of Labor found the 48 out of 500 heads of households were unemployed. Compute a 99% confidence interval for the proportion of unemployed heads of households in the population. Round to three decimal places.
Answer:
(0.062, 0.130)
Step-by-step explanation:
Sample size = n = 500
Number of heads that were unemployed = x = 48
Proportion of heads that were unemployed = p = [tex]\frac{x}{n}=\frac{48}{500}=0.096[/tex]
Proportion of heads that were not unemployed = q = 1 - p = 1 - 0.096 = 0.904
Confidence Level = 99%
z-value for 99% confidence level = z = 2.58
The confidence interval about a population proportion is calculated as:
[tex](p-z\sqrt{\frac{pq}{n}} , p+z\sqrt{\frac{pq}{n}})[/tex]
Using the values, we get:
[tex](0.096-2.58\sqrt{\frac{0.096 \times 0.904}{500}},0.096+2.58\sqrt{\frac{0.096 \times 0.904}{500}})\\\\ = (0.062,0.130)[/tex]
Thus, 99% confidence interval for the proportion of unemployed heads of households in the population is (0.062, 0.130)
The 99% confidence interval for the proportion of unemployed heads of households in the population is approximately 0.096 ± 0.029.
To compute the 99% confidence interval for the proportion of unemployed heads of households, we can use the formula:
Confidence interval = sample proportion ± margin of error
1. Find the sample proportion:
Divide the number of unemployed heads of households (48) by the total number of heads of households surveyed (500).
Sample proportion = 48 / 500 = 0.096
2. Calculate the margin of error:
The margin of error depends on the level of confidence and the sample size. For a 99% confidence level, we need to find the critical value, which corresponds to 99% confidence and 500 as the sample size.
The critical value for a 99% confidence level and 500 as the sample size is approximately 2.576.
Margin of error = critical value * sqrt((sample proportion * (1 - sample proportion)) / sample size)
Margin of error = 2.576 * sqrt((0.096 * (1 - 0.096)) / 500) ≈ 0.029
3. Calculate the confidence interval:
Confidence interval = sample proportion ± margin of error
Confidence interval = 0.096 ± 0.029
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A given binomial experiment has n=100 trials and p=1/3. Is it more likely to get x=20 successes or x=45 successes. Why?
Answer:
The P(x=45) is more that the P(x=20). Therefore x=45 successes is more likely to get.
Step-by-step explanation:
Given information: n=100 and p=1/3.
According to the binomial distribution, the probability of getting r success in n trials is
[tex]P(x=r)=^nC_rp^rq^{n-r}[/tex]
where, n is total trials, p is probability of success and q is probability of failure.
Total trials, n = 100
Probability of success, p = [tex]\frac{1}{3}[/tex]
Probability of failure, q = [tex]1-\frac{1}{3}=\frac{2}{3}[/tex]
The probability of 20 successes is
[tex]P(x=20)=^{100}C_{20}\times (\frac{1}{3})^{20}\times (\frac{2}{3})^{100-20}[/tex]
[tex]P(x=20)=\frac{100!}{20!(100-20)!}\times (\frac{1}{3})^{20}\times (\frac{2}{3})^{80}\approx 0.001257[/tex]
The probability of 45 successes is
[tex]P(x=45)=^{100}C_{45}\times (\frac{1}{3})^{45}\times (\frac{2}{3})^{100-45}[/tex]
[tex]P(x=45)=\frac{100!}{45!(100-45)!}\times (\frac{1}{3})^{45}\times (\frac{2}{3})^{55}\approx 0.004296[/tex]
The P(x=45) is more that the P(x=20). Therefore x=45 successes is more likely to get.
I need the answer to these math questions.
1) Multiply 8 minutes 31 seconds by 17.
Answer:
2 h 16 min 23 sec
Step-by-step explanation:
Hello
the time is expressed in the sexadecimal system, which uses the number 60 as an arithmetic base,hence
1 min=60 sec
1 hora =60 min
Now, we have
[tex](8 min + 31 sec)*17=136 min +527 sec\\\\\\we\ need\ to\ convert\ this\ in\ our\ base\ 60\, using\ a\ rule\ o\ three\\\\ 60 min=1\ hour\\136 min=x ?\\\\x=\frac{ 136 h}{60}\\ x=2.26 hours\\\\[/tex]
we take the whole part as an hour, and the decimal part is multiplied by 60 to get minutes
Step 1
[tex]8min*17=136 min =2.26 h\\\\2.26h = 2\ h + 0.26h(\frac{60 min}{1 h}) \\2.26h =2h+15.6 min\\\\[/tex]
we repeat the procedure to leave the minutes as a whole part
[tex]2.26\ h =2\ h+15\ min + 0.6\ min*(\frac{60 \sec}{1 m} )\\2.26\ h =2\ h\ 15\ min\ 36\ sec[/tex]
Step 2
[tex]\\527\s*(\frac{1 min}{ 60\ sec})=8.78\ min\\ \\8.78\ min= 8\min\0.78\ min\\8.78\ min=8\ min\ 0.78\min(\frac{60\ seg}{min})\\8.78\min=8\ min \ 47\ sec\\\\now, add\\\\8 min *17 =2\ h\ 15\ min\ 36\ sec\\31 sec *17 =8\ min \ 47\ sec\\(8\ min\ 31\ sec)*17=2\ h\ 15\ min\ 83\ sec\\83 s(\frac{1 min}{60 sec})=1.38 min\\1.38\ min\ =1\ min\ 0.38\ min*(\frac{60 sec}{1\ min})\\1.38\ min=1\min\ 23 s.\\( 8min 31 sec)*17=2 h 16 min 23 sec[/tex]
Have a great day
Solve the system for the exact special solution y = y(x): (keep the fraction and the square root without decimals.) 1. ydx + x[ In(x) - In(y) - 1]dy = 0 and y(1) = e for In(e) = 1.
Assume a solution of the form [tex]\Psi(x,y)=C[/tex]. Differentiating both sides gives
[tex]\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0[/tex]
with [tex]\Psi_x=y[/tex] and [tex]\Psi_y=x(\ln x-\ln y-1)[/tex].
Divide both sides by [tex]x[/tex] and we have
[tex]\dfrac yx\,\mathrm dx+(\ln x-\ln y-1)\,\mathrm dy=0[/tex]
Notice that
[tex]\left(\dfrac yx\right)_y=\dfrac1x[/tex]
[tex]\left(\ln x-\ln y-1\right)_x=\dfrac1x[/tex]
so the ODE is exact. Now we can look for a solution [tex]\Psi[/tex] with
[tex]\Psi_x=\dfrac yx[/tex]
[tex]\Psi_y=\ln x-\ln y-1[/tex]
Integrating the first PDE with respect to [tex]x[/tex] gives
[tex]\Psi(x,y)=y\ln x+f(y)[/tex]
and differentiating this with respect to [tex]y[/tex] gives
[tex]\Psi_y=\ln x+f'(y)=\ln x-\ln y-1\implies f'(y)=-\ln y-1\implies f(y)=-y\ln y+C[/tex]
So this ODE has general solution
[tex]y\ln x-y\ln y=C[/tex]
Given that [tex]y(1)=e[/tex], we have
[tex]e\ln1-e\ln e=C\implies C=-e[/tex]
so the particular solution is
[tex]y(\ln x-\ln y)=-e[/tex]
[tex]y\ln\dfrac xy=-e[/tex]
[tex]\boxed{y\ln\dfrac yx=e}[/tex]
Use the graph of a function f to determine x or y to the nearest integer, as indicated. Some problems may have more than one answer.
5=f(x)
To find the x for a given function f(x) =5, use the graph of the function and search for places where the y-coordinate is 5. The x-coordinates of these points are the solutions.
Explanation:The problem is asking for the value of x when f(x) = 5. To solve this problem, you would examine the graph of the function and look for the point(s) where the y-coordinate (the function value) is 5; the corresponding x-coordinate(s) would be your answer. For example, if seen directly above the number five on the y-axis, the line crosses at x=3, then x=3 is your solution. If it crosses again at x=-2, then x=-2 is another solution. Always remember that some problems may indeed have more than one answer, especially with functions that are not linear.
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Find the inverse of h(x) = [tex]\frac{2x+6}{5}[/tex]
show work please!
Answer:
The inverse of h(x) is [tex]\frac{5x-6}{2}[/tex]
Step-by-step explanation:
* Lets explain how to make the inverse of a function
- To find the inverse of a function we switch x and y and then solve
for new y
- You can make it with these steps
# write g(x) = y
# switch x and y
# solve for y
# write y as [tex]g^{-1}(x)[/tex]
* Lets solve the problem
∵ [tex]h(x)=\frac{2x+6}{5}[/tex]
# Step 1
∴ [tex]y=\frac{2x+6}{5}[/tex]
# Step 2
∴ [tex]x=\frac{2y+6}{5}[/tex]
# Step 3
∵ [tex]x=\frac{2y+6}{5}[/tex]
- Multiply each side by 5
∴ 5x = 2y + 6
- Subtract 6 from both sides
∴ 5x - 6 = 2y
- Divide both sides by 2
∴ [tex]y=\frac{5x-6}{2}[/tex]
# Step 4
∴ [tex]h^{-1}(x)=\frac{5x-6}{2}[/tex]
Find the point on the plane 4x+3y+z=10 that is nearest to (2,0,1). What are the values of x, y, and z for the point? x= 28 / 13 y = 3 / 26 z= 27 / 26 (Type integers or simplified fractions.)
To find the point on the plane that is nearest to (2,0,1), we minimize the squared distance between the two points using partial derivatives and set them equal to 0. The values of x, y, and z for the point are x = 28/13, y = 3/26, and z = 27/26.
To find the point on the plane that is nearest to (2,0,1), we need to find the coordinates that satisfy the equation 4x+3y+z=10 and minimize the distance between the point and (2,0,1).
This can be done by minimizing the squared distance between the two points. Using the formula for distance, we get the squared distance as:
d^2 = (x-2)^2 + y^2 + (z-1)^2
To minimize the squared distance, we can find the partial derivatives with respect to x, y, and z and set them equal to 0.
Solving these equations, we find that x = 28/13, y = 3/26, and z = 27/26.
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The probability that a randomly selected teenager watched a rented video at least once during a week was 0.75. What is the probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week? (Round your answer to four decimal places.)
Answer:
0.7564
Step-by-step explanation:
Let X be the event of watching a rented video at least once during a week,
Given,
The probability of watching a rented video at least once during a week was, p = 0.75,
So, the probability of not watching a rented video at least once during a week was, q = 1 - p = 0.25,
Binomial distributive formula,
[tex]P(x)=^nC_x p^x q^{n-x}[/tex]
Where,
[tex]^nC_x=\frac{n!}{x!(n-x)!}[/tex]
Hence, the probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week,
P(X ≥ 5) = P(X=5) + P(X=6 )+ P(X=7)
[tex]=^7C_5 0.75^5 0.25^{7-5}+^7C_6 0.75^6 0.25^{7-6}+^7C_7 0.75^7 0.25^{7-7}[/tex]
[tex]=21 (0.75)^5 (0.25)^2 + 7 (0.75)^6 0.25 + 0.75^7[/tex]
[tex]=0.756408691406[/tex]
[tex]\approx 0.7564[/tex]
The probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week is 0.3015.
Explanation:The probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week can be calculated using the binomial probability distribution formula:
P(X ≥ k) = 1 - P(X < k)
where X is the number of teenagers who watched a rented movie at least once, k is the minimum number of teenagers (5 in this case), and P(X < k) is the probability that less than k teenagers watched a rented movie at least once.
In this case, the probability that a randomly selected teenager watched a rented video at least once during a week is 0.75. Therefore, the probability that a randomly selected teenager did not watch a rented video at least once is 1 - 0.75 = 0.25.
Using the binomial probability distribution formula, we can calculate the probability that less than 5 teenagers watched a rented movie at least once:
P(X < 5) = C(7, 0) * (0.25)^0 * (0.75)^7 + C(7, 1) * (0.25)^1 * (0.75)^6 + C(7, 2) * (0.25)^2 * (0.75)^5 + C(7, 3) * (0.25)^3 * (0.75)^4 + C(7, 4) * (0.25)^4 * (0.75)^3
where C(n, r) is the number of combinations of n items taken r at a time:
C(n, r) = n! / (r! * (n-r)!)
Substituting the values and evaluating the expression, we get:
P(X < 5) = 0.698486328125
Therefore, the probability that at least 5 teenagers in a group of 7 watched a rented movie at least once last week is:
P(X ≥ 5) = 1 - P(X < 5) = 1 - 0.698486328125 = 0.301513671875
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a. Given a = 13 and b = 195, find b div a and b mod a. b. Given a = 24 and b = 377, find b div a and b mod a.
Answer:
A. 15 and 0. B. 15 and 17
Step-by-step explanation:
A. a= 13 and b = 195.
b div a = 195 div 13. The result is the integer part of the division, so
195 div 13 = 15.
b mod a = 195 mod 13. The result is the residue of the division. In this case the division is exact, so
195 mod 13 = 0.
B. a = 24 and b=377.
b div a = 377 div 24. The result is the integer part of the division, so
377 div 24 = 15.
b mod a = 377 mod 24. The result is the residue of the division. In this case the division is not exact, so
377 = 24*15+17, then
377 mod 24 = 17.
F Find an equation for a circle san istying the gve a) Center (-1,4), passes through (3,7) Center (-1,4). passes through (3,7
Answer:
[tex](x+1)^{2} +(y-4)^{2} =25[/tex]
Step-by-step explanation:
In order to find the equation of the circle, first we need to know the circle's general equation, which is:
[tex](x-h)^{2} +(y-k)^{2} =r^{2}[/tex] where:
(h,k) is the center of the circle and r the radius of the circle.
Because the problem has given the center (-1,4) then h=-1 and k=4.
We need to find now the radius:
Using the distance equation: [tex]distance=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}[/tex] and because we have the center coordinates and an extra point (3,7) we can find the radius as:
[tex]distance=\sqrt{(3-(-1))^{2}+(7-4)^{2}}[/tex]
[tex]distance=\sqrt{4^{2}+3^{2}}[/tex]
[tex]distance=\sqrt{16+9}[/tex]
[tex]distance=\sqrt{25}[/tex]
[tex]distance=5[/tex] which means r=5
In conclusion, the equation for the given circle is [tex](x+1)^{2} +(y-4)^{2} =5^{2}[/tex] which also, can be written as [tex](x+1)^{2} +(y-4)^{2} =25[/tex]
Find and classify any equilibrium solutions and then sketch typical solution curves to the differential equation: dx /dt = x 2 − 5x + 4.
Answer:
[tex]x=1+\dfrac{3}{1-Ke^{3t}}[/tex]
Step-by-step explanation:
Given that
[tex]\dfrac{dx}{dt}=x^2-5x+4[/tex]
This is a differential equation.
Now by separating variables
[tex]\dfrac{dx}{x^2-5x+4}=dt[/tex]
[tex]\dfrac{dx}{(x-1)(x-4)}=dt[/tex]
[tex]\dfrac{1}{3}\left(\dfrac{1}{x-4}-\dfrac{1}{x-1}\right)dx=dt[/tex]
Now by integrating both side
[tex]\int\dfrac{1}{3}\left(\dfrac{1}{x-4}-\dfrac{1}{x-1}\right)dx=\int dt[/tex]
[tex]\dfrac{1}{3}\left(\ln(x-4)-\ln(x-1)\right )=t+C[/tex]
Where C is the constant
[tex]\dfrac{1}{3}\ln\dfrac{x-4}{x-1}=t+C[/tex]
[tex]\dfrac{x-4}{x-1}=Ke^{3t}[/tex] K is the constant.
[tex]x=1+\dfrac{3}{1-Ke^{3t}}[/tex]
So the solution of above differential equation is
[tex]x=1+\dfrac{3}{1-Ke^{3t}}[/tex]
PLEASE HELP ME GET THESE FINISHED
Answer:
g(-0.5) = -1
g(0.2) = 0
g(0.5) = 1
Step-by-step explanation:
We are given the value of g(x) for which the x is defined.
Solving
g(-0.5) = -1
As given g(x) = -1 if -1.5 ≤ x ≤ 0.5
g(0.2) = 0
As given g(x) = 0 if -0.5 < x < 0.5
g(0.5) = 1
As given g(x) = 1 if 0.5 ≤ x < 1.5
The probability that Mary will win a game is 0.02, so the probability that she will not win is 0.98. If Mary wins, she will be given $160; if she loses, she must pay $16. If X = amount of money Mary wins (or loses), what is the expected value of X? (Round your answer to the nearest cent.)
Given:
Probability of winning, P(X) = 0.02
Probability of losing, P([tex]\bar{X}[/tex]) = 0.98
Wining amount = $160
Losing amount = $16
Step-by-step explanation:
Let the expected amount of money win be 'X'
Expected value of X, E(X) = Probability of winning, P(X).Probability of winning, P(X) - Probability of losing, P([tex]\bar{X}[/tex]).Losing amount
Now,
E(X) = ([tex]0.02\times 160 - 0.98\times 16[/tex])
E(X) = -12.48
Expected value of X = -12.48
Expected loss value = $12.48 loss