Answer:
P(A&B) = 0.4
Explanation:
Because it is a random process and there are no special constraints the probability for everybody is the same, the probability of choosing a particular site is 1/7, the person originally seated in chair number seven has 5/7 chance of not seating in chair number six and seven, the same goes for the person originally seated in chair number six; Because we want the probability of the two events happening, we want the probability of the intersection of the two events, and because the selection of a chair change the probability for the others (Dependents events) the probability P(A&B) = P(A) * P(B/A) where P(A) is 5/7 and the probability of choosing the right chair after the event A is 4/7, therefore, P(A&B) = 4/7*5/7 = 0.4.
If the events were independent the probability would be 0.51.
********Please show work*********
Kenen loves trains, especially those that run on narrow-gauge tracks. (The gauge
of a track measures how far apart the rails are.) He has decided to build a model
train of the Rio Grande, a popular narrow-gauge train.
Use the following information to help him know how big his model should be:
* The real track has a gauge of 3 feet (36 inches).
* His model railroad track has a gauge of 3/4 inches.
*The Rio Grande train he wants to model has driving wheels that measure 44 inches high.
Your Task: With your team, discuss what you know about the model train Kenen will build.
1) What scale factor should he use?
2) What will be the height of the driving wheels of his model?
Since the real track has a gauge of 3 feet but the model railroad track has a gauge of 3/4 inches, the scale factor must be 1/4, because you must follow the rule
"true measurement * scale factor = model measurement"
In fact, we have
[tex]3\cdot\dfrac{1}{4}=\dfrac{3}{4}[/tex]
This implies that the original driving wheels height, 44 inches, must be scaled down to
[tex]44\cdot\dfrac{1}{4}=11[/tex]
inches.
The deck for a card game is made up of 108 cards. Twenty-five each are red, yellow, blue, and green, and eight are wild cards. Each player is randomly dealt a seven-card hand.
(a) What is the probability that a hand will contain exactly two wild cards?
(b) What is the probability that a hand will contain two wild cards, two red cards, and three blue cards?
(a) The probability that a hand will contain exactly two wild cards is 0.076.
(b) The probability that a hand will contain two wild cards, two red cards, and three blue cards is 0.0007.
Let's solve these problems using the concept of combinations in probability.
Remember, [tex]C(n, k)[/tex] denotes the number of ways to choose k items from a set of n items, and is calculated as
[tex]C(n,k)=\frac{n!}{k!(n-k)!}[/tex]
where "!" denotes factorial. For example, [tex]5! = 5 \times 4 \times 3 \times 2 \times 1[/tex]
(a) The probability that a hand will contain exactly two wild cards:
The total number of ways to choose 7 cards out of 108 is [tex]C(108, 7)[/tex].
The number of ways to choose 2 wild cards out of 8 is [tex]C(8, 2)[/tex].
The number of ways to choose the remaining 5 cards out of the 100 non-wild cards is [tex]C(100, 5)[/tex].
So, the probability is [tex]\frac{C(8, 2) \times C(100, 5)}{C(108, 7)} \approx 0.076[/tex]
(b) The probability that a hand will contain two wild cards, two red cards, and three blue cards:
The number of ways to choose 2 wild cards out of 8 is [tex]C(8, 2)[/tex].
The number of ways to choose 2 red cards out of 25 is [tex]C(25, 2)[/tex].
The number of ways to choose 3 blue cards out of 25 is [tex]C(25, 3)[/tex].
So, the probability is [tex]\frac{C(8, 2) \times C(25, 2) \times C(25, 3)}{C(108, 7)} \approx 0.0007[/tex]
Find M. Write your answer in simplest radical
Answer:
(√6/√2)ft
Step-by-step explanation:
cos 45 = m / √6 ft
m = cos 45 x √6 ft
m = (1 / √2) x √6 ft = (√6/√2)ft
Find the probability for the experiment of drawing two marbles (without replacement) at random from a bag containing one green, two yellow, and three red marbles.
1. Both marbles are red.
2. Both marbles are yellow.
3. Neither marble is yellow.
4. The marbles are of different colors.
Final answer:
To find the probability of drawing two marbles without replacement, calculate the probability of each event separately and multiply them together.
Explanation:
To find the probability of drawing two marbles without replacement, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
1. Both marbles are red:
First, we calculate the probability of drawing a red marble on the first draw (3 red marbles out of 6 total marbles). Then, we calculate the probability of drawing a red marble on the second draw, given that the first marble drawn was red (2 red marbles out of 5 total marbles remaining). The probability is (3/6) * (2/5) = 1/5.
2. Both marbles are yellow:
First, we calculate the probability of drawing a yellow marble on the first draw (2 yellow marbles out of 6 total marbles). Then, we calculate the probability of drawing a yellow marble on the second draw, given that the first marble drawn was yellow (1 yellow marble out of 5 total marbles remaining). The probability is (2/6) * (1/5) = 1/15.
Solve the system using the substitution or elimination method. How many solutions are there to this system?
Answer:
[tex] -3*(3y+2) + 9y = -6[/tex]
[tex] -9y -6 + 9y = -6[/tex]
[tex]-6=-6[/tex]
So then as we can see we can have infinite solutions.
[tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]
Step-by-step explanation:
Assuming the following system of equations:
[tex] 2x-6y =4[/tex] (1)
[tex] -3x+9y =-6[/tex] (2)
For this case we can use the substitution method in order to find the possible solutions for the system.
If we solve for x from equation (1) we got:
[tex] 2x = 6y +4[/tex]
[tex] x = 3y +2 [/tex] (3)
Now we can replace equation (3) into equation (2) and we got:
[tex] -3*(3y+2) + 9y = -6[/tex]
[tex] -9y -6 + 9y = -6[/tex]
[tex]-6=-6[/tex]
So then as we can see we can have infinite solutions.
And the possible solutions are for a fixed value of x, we can solve y from equation (3) and we got:
[tex] y = \frac{x-2}{3}[/tex]
So the solution would be: [tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]
The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions (BLS website, January 2014). Suppose a sample of 400 U.S. workers is collected in 2014 to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membership increased in 2014. H0: p Ha: p b. If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test (to 4 decimals)? c. At α = .05, what is your conclusion?
Answer:
a) Null hypothesis: [tex] p \leq 0.113[/tex]
Alternative hypothesis: [tex] p >0.113[/tex]
b) [tex]p_v =P(z>1.07)=0.1423[/tex]
c) So the p value obtained was a high low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of workers belonged to unions is not significantly higher than 0.113.
Step-by-step explanation:
Part a
For this case we want to check the following system of hypothesis:
Null hypothesis: [tex] p \leq 0.113[/tex]
Alternative hypothesis: [tex] p >0.113[/tex]
Part b
Data given and notation
n=400 represent the random sample taken
X=52 represent the workers belonged to unions
[tex]\hat p=\frac{52}{400}=0.13[/tex] estimated proportion of workers belonged to unions
[tex]p_o=0.113[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.13 -0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}}=1.07[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>1.07)=0.1423[/tex]
Part c
So the p value obtained was a high low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of workers belonged to unions is not significantly higher than 0.113.
Suppose you received a score of 95 out of 100 on exam 1 . The mean was 79 and the standard deviation was 8 . If your score on exam 2 is 90 out of 100 , and the mean was 60 with a standard deviation of 15 , then you did:
better on exam 1 .
worse on exam 1 .
the same on both exams.
worse on exam 2
Answer:
You did the same on both exams.
Step-by-step explanation:
To compare both the scores, we need to compute the z scores of both the exams and then compare the values. The formula for z-score is:
Z = (X - μ)/σ
Where X = score obtained
μ = mean score
σ = standard deviation
For Exam 1:
Z = (95 - 79)/8
= 16/8
Z = 2
For Exam 2:
Z = (90 - 60)/15
= 30/15
Z = 2
The z-scores for both the tests are same hence the third option is correct i.e. you did the same on both exams.
Your performance on exam 1 and exam 2 can be compared using Z-scores, which measure how many standard deviations a score is from the mean. You scored 2 standard deviations above the mean on both exams, so you did the same on both exams.
Explanation:In this question, your performance on exams is being compared relative to the mean of the class scores and their standard deviation. This is a concept in statistics known as Z-scores. The Z-score tells us how many standard deviations an observation (your score) is from the mean. The formula for Z-score is (observation - mean) / standard deviation.
For exam 1 your Z - score is (95-79) / 8 which equals 2. This means you scored 2 standard deviations above the mean on exam 1. For exam 2, your Z-score is (90-60) / 15 which equals 2. Again, this means you scored 2 standard deviations above the mean on exam 2. Because your Z-score for both exams is the same, you did the same on both exams.
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Harry notes that the state sales tax went from 2% to 2.5%, which he says is not too bad because it's just a one-half percent increase. But Linda says that it really is bad because it's a 25% increase. Who's right, and why?
We are required to calculate the percentage increase in tax and determine who is right.
The percentage increase in tax is 25% and Linda is very correct
percentage increase = difference in tax /
percentage increase = difference in tax / old tax × 100
old tax = 2%
New tax = 2.5%
Difference = New tax - old tax
= 2.5% - 2%
= 0.5%
percentage increase = difference in tax /
percentage increase = difference in tax / old tax × 100
= 0.5% / 2% × 100
= 0.25 × 100
= 25%
Therefore, the percentage increase in tax is 25% and Linda is very correct
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The admission price was $1.00 in 1909. How much would the Speedway have had to charge in 1999 to match the purchasing power of $1 in 1909? In other words, how much was that in 1999? (Don't use a $ sign, use 2 decimal places.)
Answer: 13.04
Here are some consumer price indexes from the past 100+ years:
Year CPI
1909 9.1
1919 17.3
1929 17.1
1939 13.9
1949 23.8
1959 29.1
1969 36.7
1979 72.6
1989 118.3
1999 166.6
2009 214.5
2015 238.5
The admission price was $1.00 in 1909. How much would the Speedway have had to charge in 1989 to match the purchasing power of $1 in 1909? In other words, how much was that in 1989?
The time taken for a computer to boot up, X, follows a normal distribution with mean 30 seconds and standard deviation 5 seconds. What is the probability that a computer will take more than 42 seconds to boot up?
Answer:
0.008 is the probability that a computer will take more than 42 seconds to boot up.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 30 seconds
Standard Deviation, σ = 5 second
We are given that the distribution of time taken for a computer to boot up is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(computer will take more than 42 seconds to boot up)
P(x > 42)
[tex]P( x > 42) = P( z > \displaystyle\frac{42 - 30}{5}) = P(z > 2.4)[/tex]
[tex]= 1 - P(z \leq 2.4)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 42) = 1 - 0.992 = 0.008[/tex]
0.008 is the probability that a computer will take more than 42 seconds to boot up.
The article "Chances are you know someone with a tattoo, and he's not a sailor" included results from a survey of adults aged 18 to 50. The accompanying data are consistent with the summary values given in the article. Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities.
(A) P(tattoo)
(B) P(tattoo | age 18-29)
(C) P(tattoo | age 30-50)
(D) P(age 18-29 | tattoo)
At Least One Tattoo No Tattoo
Age 18-29 126 324
Age 30-50 54 396
Answer:
a) 0.2
b) 0.28
c) 0.12
d) 0.7
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
We have the following table:
At Least One Tattoo No Tattoo
Age 18-29 126 324
Age 30-50 54 396
So in total, there are
126 + 324 + 54 + 396 = 900 people
(A) P(tattoo)
This is the probability that a randomly selected person has a tattoo.
Desired outcomes:
126 + 54 = 180
180 people have at least one tattoo
Total outcomes:
There are 900 people.
P(tattoo) = 180/900 = 0.2
(B) P(tattoo | age 18-29)
This the probability that a person aged 18-29 has a tattoo
Desired outcomes:
126 people aged 18-29 have tattoos
Total outcomes:
126 + 324 = 450 people aged 18-29
P(tattoo | age 18-29) = 126/450 = 0.28
(C) P(tattoo | age 30-50)
This the probability that a person aged 30-50 has a tattoo
Desired outcomes:
54 people aged 18-29 have tattoos
Total outcomes:
54 + 396 = 450 people aged 30-50
P(tattoo | age 18-29) = 54/450 = 0.12
(D) P(age 18-29 | tattoo)
The probability that a tattoed person is 18-29.
Desired outcomes:
126 tattoed people are 18-29
Total outcomes
126 + 54 = 180 tattoed people
P(age 18-29 | tattoo) = 126/180 = 0.7
In a shipment of 58 vials, only 16 do not have hairline cracks. If you randomly select 2 vials from the shipment, what is the probability that none of the 2 vials have hairline cracks?
Answer:
0.0726 or 7.26%
Step-by-step explanation:
When choosing the first vial, there is a 16 in 58 chance that the vial does not have a hairline crack. When choosing the second vial, since on good vial was already picked, there is a 15 in 57 chance that the vial does not have a hairline crack. The probability that none of the 2 vials have hairline cracks is:
[tex]P = \frac{16}{58}*\frac{15}{57}\\P=0.0726[/tex]
There is a 0.0726 or 7.26% chance that none of the 2 vials have a hairline crack.
Keeping water supplies clean requires regular measurement of levels of pollutants. The measurements are indirect—a typical analysis involves forming a dye by a chemical reaction with the dissolved pollutant, then passing light through the solution and measuring its "absorbence." To calibrate such measurements, the laboratory measures known standard solutions and uses regression to relate absorbence and pollutant concentration. This is usually done every day. Here is one series of data on the absorbence for different levels of nitrates. Nitrates are measured in milligrams per liter of water.
Nitrates 50 50 100 200 400 800 1200 1600 2000 2000
Absorbence 7.0 7.6 12.7 24.0 47.0 93.0 138.0 183.0 231.0 226.0
The calibration process sets nitrate level and measures absorbence. The linear relationship that results is used to estimate the nitrate level in water from a measurement of absorbence.
a. What is the equation of the line used to estimate nitrate level?
b. What does the slope of this line say about the relationship between nitrate level and absorbence?
c. What is the estimated nitrate level in a water specimen with absorbence 40?
Answer:
a) Equation is
[tex]y = 0.1135x+1.590[/tex]
b) Slope = 0.1135 represents the change in y for a unit change in x
i.e. When nitrate content is increasedby 1, absorbence is increased by 0.1135
Step-by-step explanation:
Nitrates Absorbence
x y
50 7
50 7.6
100 12.7
200 24
400 47
800 93
1200 138
1600 183
2000 231
2000 226
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.999911043
R Square 0.999822094
Adjusted R Square 0.999799856
Standard Error 1.2890282
Observations 10
Coefficients
Intercept 1.589782721
x 0.113500259
we get regression line as
y = 0.1135x+1.590
a) Equation is
[tex]y = 0.1135x+1.590[/tex]
b) Slope = 0.1135 represents the change in y for a unit change in x
i.e. When nitrate content is increasedby 1, absorbence is increased by 0.1135
Suppose that you play the game with three different friends separately with the following results: Friend A chose scissors 100 times out of 400 games, Friend B chose scissors 20 times out of 120 games, and Friend C chose scissors 65 times out of 300 games. Suppose that for each friend you want to test whether the long-run proportion that the friend will pick scissors is less than 1/3.
1) Select the appropriate standardized statistics for each friend from the null distribution produced by applet.
-3.47 (100 out of 400; 25%), -4.17 (20 out of 120; 16.7%), -3.80 (65 out of 300; 21.7%)
-3.80 (100 out of 400; 25%), -3.47 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)
-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)
-4.17 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -3.47 (65 out of 300; 21.7%)
Answer:
Friend A
[tex]\hat p_A= \frac{100}{400}=0.25[/tex]
[tex]z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47[/tex]
Friend B
[tex]\hat p_B= \frac{20}{120}=0.167[/tex]
[tex]z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80[/tex]
Friend C
[tex]\hat p_C= \frac{65}{300}=0.217[/tex]
[tex]z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17[/tex]
So then the best solution for this case would be:
-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)
Step-by-step explanation:
Data given and notation
n represent the random sample taken
X represent the number of scissors selected for each friend
[tex]\hat p=\frac{X}{n}[/tex] estimated proportion of scissors selected for each friend
[tex]p_o=\frac{1}{3}=0.333[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion that the friend will pick scissors is less than 1/3 or 0.333, the system of hypothesis would be:
Null hypothesis:[tex]p\geq 0.333[/tex]
Alternative hypothesis:[tex]p < 0.333[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
Friend A
[tex]\hat p_A= \frac{100}{400}=0.25[/tex]
[tex]z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47[/tex]
Friend B
[tex]\hat p_B= \frac{20}{120}=0.167[/tex]
[tex]z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80[/tex]
Friend C
[tex]\hat p_C= \frac{65}{300}=0.217[/tex]
[tex]z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17[/tex]
So then the best solution for this case would be:
-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)
A 22 KHz baseband channel is used by a digital transmission system. Suppose ideal pulses are sent at the Nyquist rate, and the pulses can take 1024 levels. There is no noise in the system. What is the bit rate of this system
Answer:
Bit rate = 440 kBits/sec
Step-by-step explanation:
Band width = W= 22 kHz
Number of levels = L = 1024 levels
Bit per sample:
[tex]m=log_2 L\\\\m =log_2(1024)\\\\m=10 bits/sample[/tex]
Ideal pulses are sent at the Nyquist rate then bit rate = 2 x W x m
[tex]bit\,\,rate= 2\times 22\times 10^3\times 10\\\\bit\,\,rate= 440\times 10^3 bits\,sec^{-1}[/tex]
bit rate = 440 kBits/sec
Two solids are described in the list below.
One solid is a sphere and has a radius of 6 inches.
The other solid is a cylinder with a radius of 6 inches and a height of 6 inches.
what is the difference betwen the volumes in cubic inches of the solids in terms of pi
A.72pi
B.144pi
C.216pi
D.288pi
The difference between the volumes in cubic inches is option A) 72pi
Step-by-step explanation:
Volume of the sphere = 4/3 πr³radius r = 6 inchesVolume = 4/3 π(6)³
⇒ 4/3(216)π
⇒ 4[tex]\times[/tex]72π
⇒ 288π cubic inches
Volume of a cylinder = π r²hradius r = 6 inchesheight h = 6 inchesVolume = π(6)²(6)
⇒ 6³π
⇒ 216π cubic inches
Difference between the volumes = 288π - 216π = 72π
The difference in volume between the two solids is 226.08in^3
Data;
radius of sphere = 6inradius of cylinder = 6inheight of cylinder = 6inVolume of SphereThe volume of a sphere is given as
[tex]v = \frac{4}{3} \pi r^3\\[/tex]
Let's substitute the values and find the volume
[tex]v = \frac{4}{3}*3.14*6^3\\v = 904.32in^3[/tex]
Volume of CylinderThe formula of volume of a cylinder is given as
[tex]v = \pi r^2 h\\[/tex]
Let's substitute the values into the equation and solve
[tex]v = 3.14 * 6^2 * 6\\v = 678.24in^3[/tex]
The difference in volume between the two solids is
[tex]volume of sphere - volume of cylinder = 904.32 - 678.24 = 226.08in^3[/tex]
The difference in volume between the two solids is 226.08in^3
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Consider the following vector-valued function:~h(t) =〈2 sin(3t),3 cos(3t),√5 sin(3t)〉0≤t≤2π3This defines a smooth parameterized curve.(a) Find the unit tangent vector~T(t) for 0≤t≤2π3.(b) Find all of the values oftin the interval 0≤t≤2π3where~h(t) and~T(t) areorthogonal.(c) Show that the curve~h(t) lies on a sphere. What is the radius of the sphere?
Answer:
a) h'(t)= (6cos3t,-9sin3t,3[tex]\sqrt[]{5}[/tex]cos3t)
b) t=0.93994736+πn/3
c) Magnitude of h(t) is 3 which is a constant, so h(t) lies on a sphere
Step-by-step explanation:
Be my Valentine: The following frequency distribution presents the amounts, in dollars, spent for Valentine's Day gifts in a survey of 120 U.S. adults in a recent year. Approximate the mean amount spent on Valentine's Day gifts to two decimal places.Amount Frequency0-19.99 16 20.00-39.99 1340.00-59.99 21 60.00-79.99 19 80.00-99.99 12 100.00-119.99 10 120.00-139.99 7140.00-159.99 8160.00-179.99 7180,00-199.99 1200.00-219.99 3220.00-239.99 2240.00-259.99 1
Answer:
the mean is 82.75
Step-by-step explanation:
Amount Frequency Mid Point fx
0-19.99 16 9.995 159.92
20.00-39.99 13 29.995 389.935
40.00-59.99 21 49.995 1049.895
60.00-79.99 19 69.995 1329.905
80.00-99.99 12 89.995 1079.94
100.00-119.99 10 109.995 1099.95
120.00-139.99 7 129.995 909.965
140.00-159.99 8 149.995 1199.96
160.00-179.99 7 169.995 1189.965
180,00-199.99 1 189.995 189.995
200.00-219.99 3 209.995 629. 985
220.00-239.99 2 229.995 459.99
240.00-259.99 1 249.995 240.995
∑ 120 9930.4
Mean = ∑fx/∑x
Mean = 9930.4/120=82.7533 = 82.75
Answer: The mean amount spent on Valentine's day is;
$82.83
Step-by-step explanation: To find the mean amount we first arrange the numbers in a frequency table, then solve.
STEP1:
AMOUNT FREQUENCY
0-19.99 16
20.00-39.99 13
40.00-59.99 21
60.00-79.99 19
80.00-99.99 12
100.00-119.99 10
120.00-139.99 7
140.00-159.99 8
160.00-179.99 7
180,00-199.99 1
200.00-219.99 3
220.00-239.99 2
240.00-259.99 1
STEP 2: Find the center of each amount, to do this we have to find the average value of the amounts.
For the first amount is;
(0+19.99)/2 = 9.995
For the second amount is;
(20+39.99)/2 =29.995
Solving this for all the amount. Therefore the table comes
AMOUNT FREQUENCY
9.995 16
29.995 13
49.995 21
69.995 19
89.995 12
109.995 10
129.995 7
149.995 8
169.995 7
189.995 1
209.995 3
229.995 2
249.995 1
STEP 3: multiple each amount in step 2 with the frequency.
For the first amount;
9.995×16 = 159.92
For the second amount;
29.995×13= 389.935
For the third amount;
49.995×21= 1049.895
For the fourth amount;
69.99×19= 1329.81
For the fifth amount;
89.995×12=1079.94
For the six amount;
109.995×10= 1099.95
For the sixth amount;
129.995×7= 909.965
For the seventh amount;
149.995×8= 1199.96
For the eight amount;
169.995×7= 1189.965
For the ninth amount;
189.995×1= 189.995
For the tenth amount;
209.995×3= 629.985
For the eleventh amount;
229.995×2=459.99
For the twelveth amount;
249.995×1= 249.995
STEP 4: Sum up all the answers from the multiplication in step 3
Therefore;
159.92+389.935+1049.895+1329.81+1079.94+1099.95+909.965+1199.96+1189.965+189.995+629.985+459.99+629.985+459.99+249.995 = 9939.995
STEP 5: divide the sum of the value seen in step 4 with the total number of frequency to get the mean value.
The total number of frequency is 120
Therefore;
9939.305÷120=82.827541
Take the value to two decimal place, it becomes;
$82.83 this is the mean value of money spent on Valentine's day.
If a confidence interval is given from 45.82 up to 55.90 and the mean is known to be 50.86, what is the margin of error?
Answer:
[tex] ME = \frac{10.08}{2}= 5.04[/tex]
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=50.86[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex] \bar X \pm ME[/tex] (1)
Or equivalently:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (2)
Where the margin of error is given by:
[tex] ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
For this case we have the confidence interval limits given (45.82, 55.90)
We can find the width of the interval like this:
[tex] Width =55.90-45.82= 10.08[/tex]
And now the margin of error would be the half of the width since we assume that the confidence interval is symmetrical.
[tex] ME = \frac{10.08}{2}= 5.04[/tex]
A sample of 100 cars driving on a freeway during a morning commute was drawn, and the number of occupants in each car was recorded. The results were as follows: NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
Occupants 1 2 3 4 5
Number of Cars 71 16 8 3 2
a. For what proportion of cars was the number of occupants more than one standard deviation greater than the mean? (Round the final answer to two decimal places.)
Answer: 29%
Step-by-step explanation:
I considered the question to be: for what proportion of cars was the number of occupants more than 1 standard deviation and greater than the mean.
The first step is to calculate the weighted mean of the number of occupants in the cars. The next step is to determine the standard deviation using the formula √[ Σ ( xi - μ )² / N ]. Subsequently, identify the number of occupants that are more than one standard deviation greater than the mean.
Explanation:In this question, we're asked to calculate the proportion of cars that had more than one standard deviation above the mean number of occupants. First, we need to calculate the weighted mean (average) of the number of occupants in the cars. Based on the data, the mean can be calculated as:
Mean = (1*71 + 2*16 + 3*8 + 4*3 + 5*2) / 100 = 1.66
Next, we find the standard deviation. The standard deviation tells us how spread out the numbers are from the mean. Calculating standard deviation is a bit involved. The formula is √[ Σ ( xi - μ )² / N ]. Once you have these, look at the number of occupants that are more than one standard deviation greater than this mean.
Note: This is a statistical analysis question that requires knowledge of the concepts of mean and standard deviation.
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An elementary school is offering 2 language classes: one in Spanish (S) and one in French (F). Given that P(S) = 50%, P(F) = 40%, P(S ∪ F) = 70%, find the probability that a randomly selected student (a) is taking Spanish given that he or she is taking French; (b) is not taking French given that he or she is not taking Spanish. 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears.
Answer:
(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French = 0.5 .
(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish = 0.6 .
Step-by-step explanation:
We are given that an elementary school is offering 2 language classes ;
Spanish Language is denoted by S and French language is denoted by F.
Also we are given, P(S) = 0.5 {Probability of students taking Spanish language}
P(F) = 0.4 {Probability of students taking French language}
[tex]P(S\bigcup F)[/tex] = 0.7 {Probability of students taking Spanish or French Language}
We know that, [tex]P(A\bigcup B)[/tex] = [tex]P(A) + P(B) -[/tex] [tex]P(A\bigcap B)[/tex]
So, [tex]P(S\bigcap F)[/tex] = [tex]P(S) + P(F) - P(S\bigcup F)[/tex] = 0.5 + 0.4 - 0.7 = 0.2
[tex]P(S\bigcap F)[/tex] means Probability of students taking both Spanish and French Language.
Also, P(S)' = 1 - P(S) = 1 - 0.5 = 0.5
P(F)' = 1 - P(F) = 1 - 0.4 = 0.6
[tex]P(S'\bigcap F')[/tex] = 1 - [tex]P(S\bigcup F)[/tex] = 1 - 0.7 = 0.3
(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French is given by P(S/F);
P(S/F) = [tex]\frac{P(S\bigcap F)}{P(F)}[/tex] = [tex]\frac{0.2}{0.4}[/tex] = 0.5
(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish is given by P(F'/S');
P(F'/S') = [tex]\frac{P(S'\bigcap F')}{P(S')}[/tex] = [tex]\frac{1- P(S\bigcup F)}{1-P(S)}[/tex] = [tex]\frac{0.3}{0.5}[/tex] = 0.6 .
Note: 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears ; This question is incomplete please provide with complete detail.
Explain the meaning of each of the following. (a) lim x → −3 f(x) = [infinity] The values of f(x) can be made arbitrarily close to 0 by taking x sufficiently close to (but not equal to) −3. The values of f(x) can be made arbitrarily large by taking x sufficiently close to (but not equal to) −3.
Answer:
The right answer is option 3.
lim x → −3 f(x) = [infinity] means the values of f(x) can be made arbitrarily large by taking x sufficiently close to (but not equal to) −3.
Step-by-step explanation:
The limit of a function is a fundamental concept concerning the behavior of that function near a particular input.
A function f assigns an output f(x) to every input x. We say the function has a limit L at an input a: this means f(x) gets closer and closer to L as x moves closer and closer to a. More specifically, when f is applied to any input sufficiently close to a, the output value is forced arbitrarily close to L.
That is,
lim x → a f(x) = L
Hope this helps!
The limit lim x → −3 f(x) = [infinity] means that as x values get closer to -3 (without becoming -3), the value of the function f(x) goes towards infinity i.e., it grows without bound. This is akin to certain function behaviors near a value at which an asymptote is present. However, the second part about f(x) values getting close to 0 seems contradiction to the first statement.
Explanation:The statement lim x → −3 f(x) = [infinity] is related to a concept in Calculus known as a limit. When we say that the limit of f(x) as x approaches -3 is infinity, we mean that as we make x values closer and closer to -3 (without letting x actually be -3), the value of the function f(x) becomes larger and larger without bound, i.e., approaches infinity.
This is similar to some function behaviors near an asymptote. For example, the function y = 1/x has a vertical asymptote at x = 0, where y approaches infinity as x approaches zero from either direction. Here, as x gets arbitrarily close to 0, the value of y = 1/x gets arbitrarily large, or 'approaches infinity'.
On the other hand, when the question states, 'The values of f(x) can be made arbitrarily close to 0 by taking x sufficiently close to (but not equal to) -3', it signifies the tendency of the function values to get closer and closer to 0 as x gets closer to -3. This indicates a certain limit behavior, but it seems to be contradictory with the first part where the limit was stated to be infinity. It is important to scrutinize the function's properties and behavior around x = -3 carefully.
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If s is an increasing function, and t is a decreasing function, find Cs(X),t(Y ) in terms of CX,Y .
Answer:
C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).
Step-by-step explanation:
Let's introduce the cumulative distribution of (X,Y), X and Y :
F(X,Y)(x,y)=P(X≤x,Y≤y)
FX(x)=P(X≤x) FY(y)=P(Y≤y).Likewise for (s(X),t(Y)), s(X) and t(Y) :
F(s(X),t(Y))(u,v)=P(s(X)≤u
t(Y)≤v)Fs(X)(u)=P(s(X)≤u) Ft(Y)(v)=P(t(Y)≤v).Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :
F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))
The last step is obtained by applying the functions s and t since s preserves order and t reverses it.
This can be further transformed into
F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))
Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.
Now, to transform this into a statement about copulas, note that
C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))
Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,
we get
F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))
The left hand side is the copula C(X,Y), the right hand side still needs some work.
Note that
Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a
and likewise
Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b
Combining all results we obtain for the relationship between the copulas
C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).
Find the equation for the plane through the points Po(3,-2,5), Qo(-3,-1,-5), and Ro(0,-4,4) The equation of the plane is Type an equation.)
Answer:
- 21 x + 24 y + 15 z =120
Step-by-step explanation:
Given that
Po(3,-2,5), Qo (-3,-1,-5), and Ro (0,-4,4) ,These are the point in the space.
We know that equation of a plane is given as
[tex]\begin{vmatrix}x-x_1 & y-y_1 &z-z_1 \\ x_2-x_1 & y_2-y_1 &z_2-z_1 \\ x_3-x_1 &y_3-y_1 & z_3-z_1\end{vmatrix}=0\\[/tex]
[tex]\begin{vmatrix}x-0 & y+4 &z-4 \\ 3-0 & -2+4 &5-4 \\ -3-0 &-1+4 & -5-4\end{vmatrix}=0.[/tex]
[tex]\begin{vmatrix}x & y+4 &z-4 \\ 3 & 2 &1 \\ -3 &3 & -9\end{vmatrix}=0.[/tex]
Now by solving above determinate we get
x( -18 -3 ) -(y+4 ) ( -27 +3 ) + ( z- 4) (9+6) = 0
-21 x +24 y -24 x 4 + 15 z - 24 = 0
- 21 x + 24 y + 15 z -120 = 0
- 21 x + 24 y + 15 z =120
Therefore the equation of the plane will be
- 21 x + 24 y + 15 z =120
A new drug to treat psoriasis has been developed and is in clinical testing. Assume that those individuals given the drug are examined before receiving the treatment and then again after receiving the treatment to determine if there was a change in their symptom status. If the initial results showed that 2.0% of individuals entered the study in remission, 77.0% of individuals entered the study with mild symptoms, 16.0% of individuals entered the study with moderate symptoms, and 5.0% entered the study with severe symptoms calculate and interpret a chi-squared test to determine if the drug was effective treating psoriasis given the information below from the final examination.
Answer:
Step-by-step explanation:
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: The distribution of severity of psoriasis cases at the end and prior are same.
Alternative hypothesis: The distribution of severity of psoriasis cases at the end and prior are different.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = k - 1 = 4 - 1
D.F = 3
(Ei) = n * pi
Category observed Num expected num [(Or,c -Er,c)²/Er,c]
Remission 380 20 6480
Mild
symptoms 520 770 81.16883117
Moderate
symptoms 95 160 24.40625
Severe
symptom 5 50 40.5
Sum 1000 1000 6628.075081
Χ2 = Σ [ (Oi - Ei)2 / Ei ]
Χ2 = 6628.08
Χ2Critical = 7.81
where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and Χ2 is the chi-square test statistic.
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 6628.08.
We use the Chi-Square Distribution Calculator to find P(Χ2 > 19.58) =less than 0.000001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.
We reject H0, because 6628.08 is greater than 7.81. We have statistically significant evidence at alpha equals to 0.05 level to show that distribution of severity of psoriasis cases at the end of the clinical trial for the sample is different from the distribution of the severity of psoriasis cases prior to the administration of the drug suggesting the drug is effective.
The chi-square test is a statistical method that determines if there's a significant difference between observed and expected frequencies in different categories, such as symptom status in this clinical trial. Without post-treatment numbers, we can't run the exact test. However, if the test statistic exceeded the critical value, we could conclude that the drug significantly affected symptom statuses.
Explanation:This question pertains to the use of a chi-squared test, which is a statistical method used to determine if there's a significant difference between observed frequencies and expected frequencies in one or more categories. For this case, the categories are the symptom statuses (remission, mild, moderate, and severe).
To conduct a chi-square test, you first need to know the observed frequencies (the initial percentages given in the question) and the expected frequencies (the percentages after treatment). As the question doesn't provide the numbers after treatment, I can't perform the exact chi-square test.
If the post-treatment numbers were provided, you would compare them to the pre-treatment numbers using the chi-squared formula, which involves summing the squared difference between observed and expected frequencies, divided by expected frequency, for all categories. The result is a chi-square test statistic, which you would then compare to a critical value associated with a chosen significance level (commonly 0.05) to determine if the treatment has a statistically significant effect.
To interpret a chi-square test statistic, if the calculated test statistic is larger than the critical value, it suggests that the drug made a significant difference in the distribution of symptom statuses. If not, we can't conclude the drug was effective.
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A square matrix A is said to be idempotent iff A2 = A. (i) Show that if A is idempotent, then so is I − A. (ii) Show that if A is idempotent, then the matrix 2A − I is also invertible. Hint: Same as before, guess the inverse and check your answer with the definition of inverse.
Answer:
Step-by-step explanation:
Given that A is a square matrix and A is idempotent
[tex]A^2 = A[/tex]
Consider I-A
i) [tex](I-A)^2 = (I-A).(I-A)\\= I^2 -2A.I+A^2\\= I-2A+A\\=I-A[/tex]
It follows that I-A is also idempotent
ii) Consider the matrix 2A-I
[tex](2A-I).(2A-I)=\\4A^2-4AI+I^2\\= 4A-4A+I\\=I[/tex]
So it follows that 2A-I matrix is its own inverse.
A financial talk show host claims to have a 55.3 % success rate in his investment recommendations. You collect some data over the next few weeks, and find that out 10 recommendations, he was correct 3 times. If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successfu
Answer:
There is a 25.52% probability of observating 4 our fewer succesful recommendations.
Step-by-step explanation:
For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So 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 combinations 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]p = 0.553, n = 10[/tex]
If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:
This is
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003[/tex]
[tex]P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039[/tex]
[tex]P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219[/tex]
[tex]P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724[/tex]
[tex]P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567[/tex]
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552[/tex]
There is a 25.52% probability of observating 4 our fewer succesful recommendations.
A sample of 100 cars driving on a freeway during a morning commute was drawn, and the number of occupants in each car was recorded. The results were as follows: NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Occupants 1 2 3 4 5 Number of Cars 74 10 11 3 2 Find the sample standard deviation of the number of occupants. The sample standard deviation is 37.60 37.60 Incorrect . (Round the final answer to two decimal places.)
Answer:
[tex]E(X)=1*0.74 +2*0.1 +3*0.11+ 4*0.03 +5*0.02=1.49[/tex]
[tex]Var(X)=E(X^2)-[E(X)]^2 =3.11-(1.49)^2 =0.8899[/tex]
[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{0.8899}=0.943[/tex]
Step-by-step explanation:
For this case we have the following data given:
X 1 2 3 4 5
F 74 10 11 3 2
The total number of values are 100, so then we can find the empirical probability dividing the frequency by 100 and we got the followin distribution:
X 1 2 3 4 5
P(X) 0.74 0.10 0.11 0.03 0.02
Previous concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
In order to calculate the expected value we can use the following formula:
[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]
And if we use the values obtained we got:
[tex]E(X)=1*0.74 +2*0.1 +3*0.11+ 4*0.03 +5*0.02=1.49[/tex]
In order to find the standard deviation we need to find first the second moment, given by :
[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]
And using the formula we got:
[tex]E(X^2)=1^2 *0.74 +2^2 *0.1 +3^2 *0.11 +4^2 0.03 +5^2 *0.02=3.11[/tex]
Then we can find the variance with the following formula:
[tex]Var(X)=E(X^2)-[E(X)]^2 =3.11-(1.49)^2 =0.8899[/tex]
And then the standard deviation would be given by:
[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{0.8899}=0.943[/tex]
Guessing Answers Standard tests, such as the SAT, ACT, or Medical College Admission Test (MCAT), typically use multiple choice questions, each with five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to the first three questions. a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third is correct. That is, find P(WWC), where W denotes a wrong answer and C denotes a correct answer. b. Beginning with WWC, make a complete list of the different possible arrangements of two wrong answers and one correct answer, then find the probability for each entry in the list. c. Based on the preceding results, what is the probability of getting exactly one correct answer when the guesses are made?
Answer:
a) Probability of picking the first two answers wrong & the third answer correctly in that order, P(WWC) = 0.128
b) All possible outcomes = WWC, WCW, CWW
P(WWC) = 0.128; P(WCW) = 0.128; P(CWW) = 0.128
c) Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = 0.384
Step-by-step explanation:
P(Correct answer) = P(C) = number of correct options/total options available.
That is, P(C) = 1/5 = 0.2
P(Wrong answer) = P(W) = 1 - 0.2 = 0.8 or (number of wrong options)/(total options available) = 4/5 = 0.8
a) Using the multiple rule for independent events,
P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128
Probability of picking the first two answers wrong & the third answer correctly in that order = 0.128
b) All possible outcomes of picking two wrong answers and one right answer in whichever order for the first 3 questions are (WWC, WCW, CWW)
P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128
P(WCW) = P(W) × P(C) × P(W) = 0.8 × 0.2 × 0.8 = 0.128
P(CWW) = P(C) × P(W) × P(W) = 0.2 × 0.8 × 0.8 = 0.128
c) Using the addition rule for disjoint events,
Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = P(WWC) + P(WCW) + P(CWW) = 0.128 + 0.128 + 0.128 = 0.384
QED!
The probability of getting exactly one correct answer when guessing on three multiple-choice questions with five possible answers each is [tex]\(\frac{3}{25}\)[/tex].
a. To find the probability that the first two guesses are wrong and the third is correct, denoted as P(WWC), we use the multiplication rule for independent events. Since there are five possible answers for each question and only one correct answer, the probability of guessing wrong on one question is [tex]\(\frac{4}{5}\)[/tex] and the probability of guessing correctly is [tex]\(\frac{1}{5}\)[/tex]. Therefore, the probability of WWC is:
[tex]\[ P(WWC) = P(W) \times P(W) \times P(C) = \left(\frac{4}{5}\right) \times \left(\frac{4}{5}\right) \times \left(\frac{1}{5}\right) = \frac{16}{125} \][/tex]
b. The different possible arrangements of two wrong answers and one correct answer (WWC) are:
1. WWC
2. WWC
3. CWW
4. CW
5. WCW
6. WCW
For each of these arrangements, the probability is the same as calculated in part (a), which is \(\frac{16}{125}\).
c. Since there are six different ways to arrange two wrong answers and one correct answer, and each of these arrangements has the same probability, we multiply the probability of one such arrangement by the number of arrangements to find the total probability of getting exactly one correct
[tex]\[ P(\text{exactly one correct answer}) = 6 \times P(WWC) = 6 \times \frac{16}{125} = \frac{96}{125} \][/tex]
However, we must note that we have counted each arrangement twice because the order of the wrong answers (W) does not matter. Therefore, we need to divide the total by 2 to get the correct probability:
[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{2 \times 125} = \frac{48}{125} \][/tex]
Upon reviewing the calculations, it appears there was an error in the final step. We should not divide by 2 because the arrangements are distinct due to the position of the correct answer (C). The correct total probability is indeed [tex]\(\frac{96}{125}\)[/tex], but since we are looking for the probability of exactly one correct answer, we have to consider that there are three questions and the correct answer could be on any of them. Therefore, we have already accounted for all possible arrangements by multiplying by 6.
The correct probability of getting exactly one correct answer when guessing on three multiple-choice questions is:
[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{125} \][/tex]
However, this is not the final answer. We need to consider that there are three different ways to get exactly one correct answer out of three questions (CWW, WCW, WWC). Since these are mutually exclusive events, we add their probabilities:
[tex]\[ P(\text{exactly one correct answer}) = P(CWW) + P(WCW) + P(WWC) \][/tex]
[tex]\[ P(\text{exactly one correct answer}) = \frac{16}{125} + \frac{16}{125} + \frac{16}{125} \][/tex]
[tex]\[ P(\text{exactly one correct answer}) = 3 \times \frac{16}{125} \][/tex]
[tex]\[ P(\text{exactly one correct answer}) = \frac{48}{125} \][/tex]
Thus, the final correct probability is [tex]\(\frac{48}{125}\)[/tex], which simplifies to [tex]\(\frac{3}{25}\)[/tex].
Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceeds to fall straight down and ends up breaking the windshield of a car passing under the bridge. The car was 25 m away from the point of impact when the bolt began to fall down; unfortunately, the driver did not notice it and proceeded at constant speed of 21 m/s. How high is the bridge
Answer:
the bridge has a height y₀ = 6.94 m
Step-by-step explanation:
The position y of the loose bolt is given by (0,y) where
y = y₀ - 1/2*g*t²
where
y₀ = initial position of the bolt (height of the bridge) , g= gravity , t=time
and the position x of the car is given by (x,0) where
x= x₀ + v*t
where
x₀= initial position of the car
v= car's velocity
then in order for the bolt to hit the windshield they should be at x=0 and y=0 at the same time , then
0= x₀ + v*t
t= -x₀/v
replacing in the equation for y
0 = y₀ - 1/2*g*t²
0 = y₀ - 1/2*g*(-x₀/v)²
0 = y₀ - 1/2*g*x₀²/v²
y₀ = 1/2*g*x₀²/v²
replacing values
y₀ = 1/2*g*x₀²/v² = 1/2* 9.8m/s² * (-25 m)²/(21 m/s)² = 6.94 m
then the bridge has a height y₀ =6.94 m
We have assumed that
- The bolt has no horizontal velocity ( only vertical velocity) , starts from rest and neglected air friction
- Neglecting the height of the car , position of the windshield and size of the loose bolt