Final answer:
To find the probability of selecting one ace and one of either a ten, jack, queen, or king from a deck with replacement, calculate the probability for each draw and account for both possible orders of drawing these cards the final probability is 2 * ((4/52) * (16/52)) to account for both orders.
Explanation:
To calculate the probability that one card is an ace and the other is either a ten, jack, queen, or king when two cards are selected randomly from an ordinary deck, we approach this problem considering the combinations and probabilities of each event happening.
First, let's establish the total possibilities. Since we are replacing the card back into the deck after the first draw, the number of possibilities for each draw remains 52. So, the total number of ways to draw any two cards (with replacement) is 52 * 52.
The probability of drawing an ace (P(Ace)) on the first draw is 4/52 because there are 4 aces in a deck of 52 cards.
The probability of drawing a ten, jack, queen, or king (P(10/J/Q/K)) on the second draw is 16/52, as there are 4 of each of these cards in the deck, totaling 16 cards.
To find the probability of both events happening, you multiply the probabilities of each event: P(Ace) * P(10/J/Q/K). This gives you (4/52) * (16/52).
However, the order in which the ace and the 10/J/Q/K are drawn matters, as the ace could also be drawn second. This means we need to calculate the probability again with the ace being drawn second and the 10/J/Q/K card drawn first, which is the same calculation. Therefore, the final probability is 2 * ((4/52) * (16/52)) to account for both orders.
To calculate the probability of drawing an ace followed by a ten, jack, queen, or king, we find the probabilities of each event separately and multiply them since the events are independent, assuming the card is replaced after the draw.
Explanation:The probability of selecting one ace and another card which is a ten, jack, queen, or king involves understanding the composition of a standard deck and calculating the odds of each draw.
A standard deck has 52 cards with 4 aces and 16 face cards (including the tens). The chance of drawing an ace (event A) is therefore 4/52. For the second draw (event B), with the card being replaced after the first draw, the odds of drawing either a ten, jack, queen, or king from any suit remain identical, since we're back to having a full deck.
To find the overall probability, we calculate the probability of both events occurring, which involves multiplying the probabilities of each individual event, assuming they are independent. This is represented by: P(A and B) = P(A) * P(B).
For the specific example of drawing a four (event A) then a five (event B) when replacing the card back into the deck after each draw, the probability of event A is 1/13 since there are four fours in the deck and the probability of event B is also 1/13 for the same reason. Therefore, the combined probability is (1/13)*(1/13), which simplifies to 1/169.
The marginal cost of printing a poster when x posters have been printed is dc/dx = 1/2√x dollars. Find c(100) - c(1), the cost of printing posters 2 to 100
The cost of printing posters from 2 to 100, given the marginal cost function dc/dx = 1/2√x, is calculated by finding the anti-derivative (or integral) of the function from 1 to 100, resulting in a cost of $18.
Explanation:The given function dc/dx = 1/2√x represents the marginal cost, which is the derivative of the cost function c(x). To find the cost of printing from the 2nd to the 100th poster, we need to find the integral of the marginal cost from 1 to 100, not including the cost of the first poster. This is represented mathematically as ∫ from 1 to 100 of (1/2√x) dx. Here's how to solve this:
First, change 1/2√x to x(-1/2)/2.Now, find the anti-derivative of x^(-1/2)/2, which will be x(1/2), or √x.Using the Fundamental Theorem of Calculus, we first plug in the upper limit of 100, then subtract the result of plugging in the lower limit of 1.This results in 2*(√100 - √1) = 20 - 2 = $18. This is the cost for printing posters 2 to 100.
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simplify -1/64........
Answer:
- 2⁻⁶
Step-by-step explanation:
To simplify this we have to know the following rules.
(i) (xᵃ)ᵇ = xᵃᵇ
(ii) 1/xᵃ = x⁻ᵃ
Given: [tex]$ \frac{-1}{64} $[/tex]
= [tex]$ \frac{-1}{4^3} $[/tex]
= [tex]$ \frac{-1}{(2^2)^3} $[/tex]
= [tex]$ \frac{-1}{2^6} \hspace{10mm} $[/tex] [using (i)]
= [tex]$ 2^{-6} $[/tex] [using (ii)]
Hence, the simplified form would be: [tex]$ 2^{-6} $[/tex]
In which direction should the motorboat head in order to reach a point on the opposite bank directly east from the starting point? (The boat's speed relative to the water remains 4.30
Answer:
east
Step-by-step explanation:
Answer:
Step-by-step explanation:
Find the equation of the sphere passing through P(negative 4 comma 7 comma 6 )and Q(8 comma negative 3 comma 5 )with its center at the midpoint of PQ. ?
Answer: The equation of the sphere is:
(x-2)^2 + (y-2)^2 + (z-5/2)^2 = sqrt(245)/2
Step-by-step explanation:
The centre of the sphere is the midpoint of the diameter, which is
1/2
[(-4,7,6) + (8,-3,5)] =(2,2,5/2)
The length of the diameter is
=sqrt |(8,-3,5) - (-4,7,6)|^2
=sqrt (12^2 + (-10)^2 + (-1)^2)
=sqrt (144+100+1)
=sqrt(245)
so the radius of
the sphere is:
1/2(sqrt(245)) = sqrt(245)/2.
The equation of the sphere is:
(x-2)^2 + (y-2)^2 + (z-5/2)^2 = sqrt(245)/2
Identify the power of ten that defines each of these prefixes. Input your answers as 10* where x is the power of ten. kilo- nano- micro- centi- mega- milli-
Answer:
kilo is [tex]10^{3}[/tex]
nano is [tex]10^{-9}[/tex]
micro is [tex]10^{-6}[/tex]
centi is [tex]10^{-2}[/tex]
mega is [tex]10^{6}[/tex]
milli is [tex]10^{-3}[/tex]
Step-by-step explanation:
Examples of power of 10 are:
[tex]100 = 10^{2}[/tex]
[tex]10 = 10^{1}[/tex]
[tex]1 = 10^{0}[/tex]
[tex]0.1 = 10^{-1}[/tex]
[tex]0.01 = 10^{-2}[/tex]
kilo is 1000. So kilo is [tex]10^{3}[/tex]
Nano is 0.000000001. So nano is [tex]10^{-9}[/tex]
micro is 0.000001. So micro is [tex]10^{-6}[/tex]
centi is 0.01. So centi is [tex]10^{-2}[/tex]
Mega is 1000000. So mega is [tex]10^{6}[/tex]
milli is 0.001. So milli is [tex]10^{-3}[/tex]
The following are conversions.
[tex]\rm{kilo}=10^3\\\rm {nano}=10^{-9}\\\rm{micro}=10^{-6}\\\rm{centi}=10^{-2}\\\rm{Mega}=10^6\\\rm{milli}=10^{-3}[/tex]
On the basis of the power of 10 the following categories can be defined that are as follow:
[tex]\rm{kilo}=10^3\\\rm {nano}=10^{-9}\\\rm{micro}=10^{-6}\\\rm{centi}=10^{-2}\\\rm{Mega}=10^6\\\rm{milli}=10^{-3}[/tex]
The values of all are as follows.
kilo is 1000.
Nano is 0.000000001.
micro is 0.000001.
Centi is 0.01.
Mega is 1000000.
Milli is 0.001.
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In 2000, the average GPA for Rutgers College students was 3.15. A sample in 2002 showed that the average GPA for Rutgers College students was 3.46.
An alternative hypothesis for the GPA of Rutgers College students is:
A. The mean Rutgers College GPA is more than 3.15 and less than 3.46
B. The mean Rutgers College GPA is less than 3.46
C. The mean Rutgers College GPA has increased to more than 3.15
D. None of the above
Answer:
Null hypothesis: [tex]\mu_{2002}\leq 3.15[/tex]
Alternative hypothesis :[tex]\mu_{2002} > 3.15[/tex]
C. The mean Rutgers College GPA has increased to more than 3.15
Step-by-step explanation:
Data given and notation
[tex]\bar X_{2000}=3.15[/tex] represent the average score for Rutgers College in 2000
[tex]\bar X_{2002}=3.46[/tex] represent the average score for Rutgers College in 2002
[tex]s_{2000}[/tex] represent the sample standard deviation in 2000
[tex]s_{2002}[/tex] represent the sample standard deviation in 2002
[tex]n[/tex] sample size
State the null and alternative hypotheses. .
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu_{2002}\leq 3.15[/tex]
Alternative hypothesis :[tex]\mu_{2002} > 3.15[/tex]
For this case we need to take in count that the alternative hypothesis can't have and = sign, and based on the information a good hypothesis here would be if the average GPA increased between 2000 and 2002, so based on this the best option for this case would be:
C. The mean Rutgers College GPA has increased to more than 3.15
The best alternative hypothesis based on the data provided is that the mean GPA for Rutgers College students has increased to more than 3.15.
Explanation:The correct alternative hypothesis for the GPA of Rutgers College students based on the information given is: The mean Rutgers College GPA has increased to more than 3.15. This is because the average GPA in 2002 was reported to be 3.46, which is greater than the 2000's average of 3.15. If we are hypothesizing a change in the population mean, the new mean would be tested against the old mean. Since there was an increase observed, we hypothesize that the new mean GPA is greater than the one from 2000.
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Listed below are the amounts of weight change (in pounds) for 12 women during their first year of work after graduating from college. Positive values correspond to women who gained weight and negative values correspond to women who lost weight. Find the standard deviation for the given sample data. 1.0 -5 3.0 -8 14 -11 12 0 16 -2 12 7a. 1.0lbb. 8.1lbc. 2.0lbd. 3.0lb
Answer:
[tex] s= \sqrt{80.568}=8.97 [/tex]
[tex] Median =\frac{1+3}{2}=2[/tex]
Step-by-step explanation:
For this case we have the following data:
1.0 -5 3.0 -8 14 -11 12 0 16 -2 12 7
And ordering this data we have:
-11 -8 -5 -2 0 1 3 7 12 12 14 16
And we are interested in find the standard deviation for the sample data. In order to do this the first step is find the mean given by this formula:
[tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}=\frac{1-5+3-8+14-11+12+0+16-2+12+7}{12}=3.25[/tex]
Now with the sample mean we can calculate the sample variance with the following formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And if we replace we got:
[tex] s^2 =80.568[/tex]
And for the sample deviation we just need to take the square root of the sample variance and we got:
[tex] s= \sqrt{80.568}=8.97 [/tex]
The median on this case would be given by:
[tex] Median =\frac{1+3}{2}=2[/tex]
Using the positions 5 and 6 for the average since the sample size is an even number.
Final answer:
The standard deviation of the weight changes for the 12 women is approximately 8.1 pounds, after computing the variance and taking the square root.
Explanation:
To calculate the sample standard deviation of the given weight changes, we will follow these steps:
Calculate the mean (average) of the sample data.Subtract the mean from each data point and square the result.Sum all the squared values.Divide by the sample size minus one (n-1) to get the variance.Take the square root of the variance to find the standard deviation.Now let's apply these steps to the given data:
The data points are: 1.0, -5, 3.0, -8, 14, -11, 12, 0, 16, -2, 12, 7
The mean (step 1) is calculated as:
(1.0 - 5 + 3.0 - 8 + 14 - 11 + 12 + 0 + 16 - 2 + 12 + 7) / 12 = (49) / 12 ≈ 4.0833 pounds
Continuing with steps 2 to 5, the squared differences from the mean, their sum, and the variance are computed. Finally, the sample standard deviation is found to be approximately 8.1 pounds.
Therefore, the correct answer is option b, which is 8.1 pounds.
Obtain POS expressions for the following by using distributive laws and theorem of factoring, eliminate redundant terms.
(a) (2 points) (xy + x 0 z)
(b) (3 points) (ab0 f + a 0 cf)(a 0 bcf + c 0 f(d + e))
Answer:
a) (y+ Oz)
b) (ab + ac)(abc + cd + ce)
Step-by-step explanation:
a) xy + xOz (First find the common factor between xy and xOz, and it is x as it appears in both xy and xOz. Factoring out x will result in
x(y+ Oz)
There are no more common factors thus the answer is x(y + Oz)
Since the question was asking after removing redundant terms it will on be y + Oz
b) (ab0 f + a 0 cf)(a 0 bcf + c 0 f(d + e)) simplfying the expression
(ab0 f + a 0 cf)(aObcf + cOfd + cOfe) common factor Of is factored out as below
Of[ (ab + ac)(abc + cd + ce) ] therefore after removing redundant terms the answer is (ab + ac)(abc + cd + ce)
Find the sample standard deviation s for the following sample data. Round your answer to the nearest hundredth. 23 20 14 35 28
Answer:
The standard deviation for given sample is 7.97
Step-by-step explanation:
We are given the following data set:
23, 20, 14, 35, 28
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{120}{5} = 24[/tex]
Sum of squares of differences = 1 + 16 + 100 + 121 + 16 = 254
[tex]S.D = \sqrt{\dfrac{254}{4}} = 7.97[/tex]
The standard deviation for given sample is 7.97
To find the sample standard deviation for the data sets 23, 20, 14, 35, 28, we calculate the mean, subtract the mean from each data point and square the result, sum these squares, divide by one less than the sample size to find the variance, and finally take the square root to find the standard deviation which is approximately 7.97.
To calculate the sample standard deviation (s), follow these steps:
Find the mean (average) of the sample data.
Subtract the mean from each data point and square the result.
Sum all the squared values.
Divide this sum by the sample size minus one (n-1) to get the sample variance.
Take the square root of the sample variance to find the sample standard deviation.
Let's apply these steps to the given data: 23, 20, 14, 35, 28.
Mean = (23 + 20 + 14 + 35 + 28) / 5 = 120 / 5 = 24.
Subtract the mean and square: (23 - 24)² = 1, (20 - 24)² = 16, (14 - 24)² = 100, (35 - 24)² = 121, (28 - 24)² = 16.
Sum of squares = 1 + 16 + 100 + 121 + 16 = 254.
Variance = 254 / (5 - 1) = 254 / 4 = 63.5.
Standard Deviation = √63.5 ≈ 7.97 (rounded to the nearest hundredth).
The sample standard deviation s is approximately 7.97.
in a class there are
8 students who play football and cricket
4 students who do not play football or cricket
14 students who play football
20 students who play cricket
find the probability that a student chosen at random plays football or cricket or both
dont necessarily need an explanation but if you have a simple one i can understand then please do tell me x
Answer:
the probability that a student chosen at random plays football or cricket or both = [tex]\frac{1}{5} + \frac{2}{5} + \frac{4}{15} = \frac{13}{15}[/tex]
Step-by-step explanation:
i) 8 students play football and cricket
ii) 4 students do not play football or cricket
iii) total of 14 students play football.
iv) therefore the number of students who play only football is = 14 - 8 = 6
v) total of 20 students play cricket.
vi) therefore the number of students who play only cricket is = 20 - 8 = 12
vii) therefore the total number of students = 8 + 4 + 6 + 12 = 30
viii) the probability a student chosen at random plays football = [tex]\frac{6}{30} = \frac{1}{5}[/tex]
ix) the probability a student chosen at random plays cricket = [tex]\frac{12}{30} = \frac{2}{5}[/tex]
x) the probability a student chosen at random plays both football and cricket = [tex]\frac{8}{30} = \frac{4}{15}[/tex]
xi) therefore the probability that a student chosen at random plays football or cricket or both = [tex]\frac{1}{5} + \frac{2}{5} + \frac{4}{15} = \frac{13}{15}[/tex]
The probability that a student chosen at random plays football or cricket or both is [tex]\frac{13}{15}[/tex].
We have
Number of students play football and cricket = 8
Number of students do not play football or cricket = 4
Total Number of students play football = 14
Therefore, the number of students who play only football
= 14 - 8
= 6
Total Number of students play cricket = 20
Therefore, the number of students who play only cricket
= 20 - 8
= 12
So, the total number of students
= 8 + 4 + 6 + 12
= 30
Now, the probability that a student chosen at random plays football
[tex]=\frac{6}{30} \\=\frac{1}{5}[/tex]
The probability that a student chosen at random plays cricket
[tex]=\frac{12}{30} \\=\frac{2}{5}[/tex]
The probability a student chosen at random plays both football and cricket [tex]=\frac{8}{30} \\=\frac{4}{15}[/tex]
Therefore, the probability that a student chosen at random plays football or cricket or both
[tex]=\frac{1}{5} +\frac{2}{5}+\frac{4}{15}\\=\frac{3}{15} +\frac{6}{15}+\frac{4}{15}\\=\frac{13}{15}[/tex]
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A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that a) None of the LED light bulbs are defective? b) Exactly one of the LED light bulbs is defective? c) Two or fewer of the LED light bulbs are defective? d) Three or more of the LED light bulbs are not defective?
Answer:
a) There is a 59.87% probability that none of the LED light bulbs are defective.
b) There is a 31.51% probability that exactly one of the light bulbs is defective.
c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.
d) There is a 100% probability that three or more of the LED light bulbs are not defective.
Step-by-step explanation:
For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 10, p = 0.05[/tex]
a) None of the LED light bulbs are defective?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}*(0.05)^{0}*(0.95)^{10} = 0.5987[/tex]
There is a 59.87% probability that none of the LED light bulbs are defective.
b) Exactly one of the LED light bulbs is defective?
This is P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{10,1}*(0.05)^{1}*(0.95)^{9} = 0.3151[/tex]
There is a 31.51% probability that exactly one of the light bulbs is defective.
c) Two or fewer of the LED light bulbs are defective?
This is
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 2) = C_{10,2}*(0.05)^{2}*(0.95)^{8} = 0.0746[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884[/tex]
There is a 98.84% probability that two or fewer of the LED light bulbs are defective.
d) Three or more of the LED light bulbs are not defective?
Now we use p = 0.95.
Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.
So
[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]
[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 0) = C_{10,0}*(0.95)^{0}*(0.05)^{10}\cong 0[/tex]
[tex]P(X = 1) = C_{10,1}*(0.95)^{1}*(0.05)^{9} \cong 0[/tex]
[tex]P(X = 2) = C_{10,1}*(0.95)^{2}*(0.05)^{8} \cong 0[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0[/tex]
[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1[/tex]
There is a 100% probability that three or more of the LED light bulbs are not defective.
The question relates to binomial distribution in probability theory. The probabilities calculated include those of none, one, two or less, and three or more LED bulbs being defective out of a random sample of 10.
Explanation:
This question relates to the binomial probability distribution. A binomial distribution is applicable because there are exactly two outcomes in each trial (either the LED bulb is defective or it's not) and the probability of a success remains consistent.
a) In this scenario, 'none of the bulbs being defective' means 10 successes. The formula for probability in a binomial distribution is p(x) = C(n, x) * [p^x] * [(1-p)^(n-x)]. Plugging in the values, we find p(10) = C(10, 10) * [0.95^10] * [0.05^0] = 0.5987 or 59.87%.
b) 'Exactly one of the bulbs being defective' implies 9 successes and 1 failure. Following the same formula, we get p(9) = C(10, 9) * [0.95^9] * [0.05^1] = 0.3151 or 31.51%.
c) 'Two or less bulbs being defective' means 8, 9 or 10 successes. We add the probabilities calculated in (a) and (b) with that of 8 successes to get this probability. Therefore, p(8 or 9 or 10) = p(8) + p(9) + p(10) = 0.95.
d) 'Three or more bulbs are not defective' means anywhere from 3 to 10 successes. As the failure rate is low, it's easier to calculate the case for 0, 1 and 2 successes and subtract it from 1 to find this probability. This gives us p(>=3) = 1 - p(2) - p(1) - p(0) = 0.98.
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In what order should a boolean retrieval system merge the related postings for best query efficiency?
Answer:
For each n get its posting then AND them
Step-by-step explanation:
1. Lets presume we have multiple n terms
2. We get the posting from each n term
3. We use the function AND and apply it to each n term
4. We start with the tiniest set and we continue from there
5. We have a pair of n terms
6. We get the posting from 1st n
7. We get the posting from 2nd n
8. We apply n1 AND n2
Two observers collected frequency data for 10 two-minute intervals. They agreed on 8 of the intervals. What is the percentage of inter-rater reliability?
Answer:
The percentage of inter-rater reliability is 80%.
Step-by-step explanation:
The rate of inter-rater reliability is the division of the number of intervals in which they agreed number of total intervals.
In this problem, we have that:
There are 10 intervals.
The observers agreed on 8 of them.
So the rate of agreement is 8/10 = 0.8.
As a percentage, we multiply the rate by 100, so 0.8*100 = 80%.
The percentage of inter-rater reliability is 80%.
An aircraft seam requires 22 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability. (Round your answers to four decimal places.)
(a) If 19% of all seams need reworking, what is the probability that a rivet is defective?
(b) How small should the probability of a defective rivet be to ensure that only 9% of all seams need reworking?
Answer:
Part A:
[tex]p=0.0095[/tex]
Part B:
[tex]p=0.0043[/tex]
Step-by-step explanation:
Part A:
The number of rivets=22 rivets
Probability that no rivet is defective= (1-p)^22
The probability that at least one rivet is defective=1-(1-p)^22
For 19% of all seams need reworking, probability that a rivet is defective is given by:
[tex]1-(1-p)^{22}=0.19[/tex]
[tex](1-p)^{22}=0.81\\p=1-\sqrt[22]{0.81} \\p=0.0095[/tex]
Part B:
For 9% of all seams need reworking, probability of a defective rivet is:
[tex]1-(1-p)^{22}=0.09\\p=1-\sqrt[22]{0.91} \\p=0.0043[/tex]
To find the probability of a defective rivet in a seam and the smallest probability of a defective rivet to ensure a certain reworking percentage, we use the concept of independent events and probability calculations.
Explanation:(a) To find the probability that a rivet is defective:
Let p be the probability of a defective rivet.
Since 19% of seams need reworking, 19% of the seams have at least one defective rivet.
Therefore, 19% of all seams equals the probability that at least one rivet is defective:
P(at least one defective rivet) = 1 - P(no defective rivets) = 0.19
P(no defective rivets) = 1 - P(at least one defective rivet) = 1 - 0.19
P(no defective rivets) = 0.81
Since each rivet is defective independently of one another, the probability that a rivet is defective is:
p = 1 - P(no defective rivet)
p = 1 - 0.81
p = 0.19
Therefore, the probability that a rivet is defective is 0.19 or 19%.
(b) To find the smallest probability of a defective rivet:
Let p be the probability of a defective rivet that ensures only 9% of seams need reworking.
We need to find the value of p such that P(at least one defective rivet) = 0.09.
From part (a), we know that P(at least one defective rivet) = 1 - P(no defective rivets) = 0.19.
Therefore, we can set up the equation:
0.19 = 1 - (1 - p)22
Solving this equation will give us the smallest value of p that satisfies the condition.
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Manuel thinks the tens digit goes up by 1 in these numbers. Do you agree? Explain. 460,470,480, 490,500,510
Answer:
Manuel is correct, since adding 10 is the same as the tens digit going up by 1.
Step-by-step explanation:
It is important to know the concepts of units, tenths and cents.
For example
1 = 1 unit
10 = 1*10 + 0 = The tens digit is one the unit digit is 0
21 = 2*10 + 1 = The tens digit is two and the unit digit is 1.
120 = 1*100 + 2*10 + 0 = The cents digit is 1, the tens digit is two and the unit digit is 0.
So
Adding 1 is the same as the unit digit going up by 1.
Adding 10 is the same as the tens digit going up by 1.
Adding 100 is the same as the cents digit going up by 1.
In this problem, we have that:
460,470,480, 490,500,510
Each number is 10 added to the previous, that is, the tens digit going up by 1.
Manuel thinks the tens digit goes up by 1 in these numbers. Do you agree?
Manuel is correct, since adding 10 is the same as the tens digit going up by 1.
The statement was given by Manuel that the tens digit goes up by 1 in these numbers, is true.
What is the tens place in a number?The second place from the right of the number before the decimal is the tens place of a number.
If we look at the series of the number that is given to us then we will notice that in series 460,470,480, 490,500,510, the second place of the number is increasing, therefore, from 6 to 7 to 8 and further. In the last value, the tens value becomes one because after 490, when we add 10, the number will become 500.
Hence, the statement was given by Manuel that the tens digit goes up by 1 in these numbers, is true.
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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming these events are independent: a) What is the probability that both the student and the professor are late? b) What is the probability that the student is late given that the professor is on time? c) Now assume the events are not independent. The probability that the professor is late given that the student is late is 0.4 i. What is the probability that at least one of them is on time?ii. What is the probability that they are both on time?
Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
[tex]{A}'[/tex] & [tex]{B}'[/tex] is also independent
[tex]{A}'[/tex] = 1-0.8 = 0.2
[tex]{B}'[/tex] = 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
[tex]P(\frac{B'}{A})[/tex] = [tex]\frac{P(B'\cap A)}{P(A)}[/tex] = [tex]\frac{0.4\times 0.8}{0.8}[/tex] = 0.4
Part c)
Assume the events are not independent
Given Data
P[tex](\frac{{A}'}{{B}'})[/tex] = 0.4
=[tex]\frac{P({A}'\cap {B}')}{P({B}')}[/tex] = 0.4
[tex]P[/tex][tex]({A}'\cap {B}')[/tex] = 0.4 x P[tex]({B}')[/tex]
= 0.4 x 0.4 = 0.16
[tex]P({A}'\cap {B}')[/tex] = 0.16
i)
The probability that at least one of them is on time
[tex]P(A\cup B)[/tex] = 1- [tex]P({A}'\cap {B}')[/tex]
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P[tex](A\cap B)[/tex] = 1 - [tex]P({A}'\cup {B}')[/tex] = 1 - [tex][P({A}')+P({B}') - P({A}'\cap {B}')][/tex]
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56
The five-number summary of the ages of passengers on a cruise ship is listed below. Min 1 Q1 20Median 29 Q3 38Max 80 Consider the following two statements regarding outliers for this data and determine which, if any, are correct (i) There is at least one passenger whose age is a low outlier. (ii) There is at least one passenger whose age is a high outlier.a. Only statement (i) is correct. b. Only statement (ii) is correct. c. Both statements (i) and (ii) are correct. d. Neither statement (i) or (ii) is correct.
After calculating the interquartile range (IQR) and identifying the thresholds for outliers, it is clear that there are no low outliers but at least one high outlier, as the maximum age exceeds the high outlier threshold. Therefore, only statement (ii) is correct.
Explanation:The question asks to consider two statements regarding outliers in a five-number summary of the ages of passengers on a cruise ship, which includes a minimum age (Min), first quartile (Q1), median, third quartile (Q3), and maximum age (Max). The statements to consider are:
There is at least one passenger whose age is a low outlier.There is at least one passenger whose age is a high outlier.To determine whether these statements are correct, we use the interquartile range (IQR) method for identifying outliers. The IQR is calculated as Q3 - Q1. An age is considered a low outlier if it is less than Q1 - 1.5 * IQR, and it is a high outlier if it is greater than Q3 + 1.5 * IQR.
Using the given data:
Q1 = 20
Q3 = 38
IQR = Q3 - Q1 = 38 - 20 = 18
Therefore, the cutoff for low outliers is 20 - 1.5 * 18 = 20 - 27 = -7, and for high outliers, it is 38 + 1.5 * 18 = 38 + 27 = 65.
Since the minimum age is 1, which is well above -7, there are no low outliers. However, the maximum age is 80, which is above the high outlier threshold of 65. Hence, there is at least one high outlier.
The correct answer to the question given the provided data is:
Only statement (ii) is correct.
Find the expression for the electric field, E [infinity] , of the ring as the point P becomes very far from the ring ( x ≫ R ) in terms of the radius R , the distance x , the total charge on the ring q , and the constant k = 1 / ( 4 π ϵ 0 ).
Answer:
The expression of the field E as the point P becomes very far from the ring is:
[tex]\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x} \\\left \{ {{\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{1}{x^2}\vec{x} \mapsto x>0} \atop {\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{-1}{x^2}\vec{x} \mapsto x< 0 }} \right.[/tex]
Step-by-step explanation:
The Electric field expression is:
[tex]\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}[/tex]
To determine the asked expression we use limits. If we consider that x≫R, this is the same as considering the radius insignificant respect the x distance. Therefore we can considerate than from this distance X, the radius R tends to zero:
[tex]\displaystyle\lim_{R \to{}0}{\vec{E}(x)}=\lim_{R \to{}0}{\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}}\rightarrow\frac{q}{4\pi\epsilon_0} \frac{x}{(0^2+x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{x}{(x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{\cancel{x}}{|x|^{\cancel{3}}}\vec{x}=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x}[/tex]
The expression for the electric field of the ring as the point P becomes very far from the ring is [tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex].
How to estimate an electric field for a ring with an uniform chargeLet suppose that the ring has an uniform linear electric density ([tex]\lambda[/tex]). A formula for the electric field at point P ([tex]E[/tex]) in rectangular coordinates is shown below:
[tex]\vec E = (E_{x}, E_{y}, E_{z})[/tex] (1)
Where:
[tex]E_{x}[/tex] - Electric field in the x-direction.[tex]E_{y}[/tex] - Electric field in the y-direction.[tex]E_{z}[/tex] - Electric field in the z-direction.Each component of the electric field are defined by the following integral formulae:
[tex]E_{x} = \int\limits^{2\pi}_{0} {\sin \theta \cdot \cos \phi} \, dE[/tex] (2)
[tex]E_{y} = \int\limits^{2\pi}_{0} {\sin \theta\cdot \sin\phi} \, dE[/tex] (3)
[tex]E_{z} = \int\limits^{2\pi}_{0} {\cos \theta} \, dE[/tex] (4)
Where:
[tex]\theta[/tex] - Axial angle, in radians.[tex]\phi[/tex] - Radial angle, in radians.By Coulomb's law and trigonometric and geometric relationships, we expand and solve each integral as following:
[tex]E_{x} = \frac{R}{\sqrt{x^{2}+R^{2}}}\int\limits^{2\pi}_{0} {\cos \phi} \, dE = \frac{k\cdot \lambda\cdot R^{2}}{(x^{2}+R^{2})^{3/2}}\int\limits^{2\pi}_{0} {\cos \phi} \, d\phi = 0[/tex]
[tex]E_{y} = \frac{R}{\sqrt{x^{2}+R^{2}}}\int\limits^{2\pi}_{0} {\sin \phi} \, dE = \frac{k\cdot \lambda\cdot R^{2}}{(x^{2}+R^{2})^{3/2}}\int\limits^{2\pi}_{0} {\sin \phi} \, d\phi = 0[/tex]
[tex]E_{z} = \frac{k\cdot \lambda\cdot x \cdot R}{(x^{2}+R^{2})^{3/2}} \int\limits^{2\pi}_{0}\, d\phi = \frac{x\cdot k \cdot (2\pi\cdot \lambda\cdot R)}{(x^{2}+R^{2})^{3/2}} = \frac{x\cdot k\cdot Q}{(x^{2}+R^{2})^{3/2}}[/tex] (5)
Where [tex]k[/tex] is the electrostatic constant.
If [tex]x >> R[/tex], (5) is simplified into the following expression:
[tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex] (6)
Where [tex]Q[/tex] is the electric charge of the entire ring.
Please notice that (6) tends to be zero when [tex]x \to \infty[/tex]. The expression for the electric field of the ring as the point P becomes very far from the ring is [tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex]. [tex]\blacksquare[/tex]
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4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.
Answer:
[tex]\sum_{n=0}^9cos(\frac{\pi n}{2})=1[/tex]
[tex] \sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0[/tex]
[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}[/tex]
Step-by-step explanation:
[tex] \sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))[/tex]
[tex]=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})[/tex]
[tex]=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1[/tex]
2nd
[tex]\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}[/tex]
[tex]=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0[/tex]
3th
[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))[/tex]
[tex]=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}[/tex]
What we use?
We use that
[tex] e^{i\pi n}=cos(\pi n)+i sin(\pi n)[/tex]
and
[tex]\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}[/tex]
Geometric sum formulas are used to evaluate sums of a geometric series, with the result expressed in Cartesian form (a + bi) where a is the real part and bi is the imaginary part. The sum of a geometric series is calculated with the formula: Sum = a * (1 - r^n) / (1 - r), where a is the first term and r is the ratio. Please provide the specific sums for a detailed step-by-step calculation.
Explanation:The problem at hand revolves around the usage of geometric sum formulas to evaluate sums and to express the result in Cartesian form. The critical point to remember is that a geometric series is a series with a constant ratio between successive terms. The sum of the first 'n' terms of a geometric sequence can be calculated using the formula:
Sum = a * (1 - rⁿ) / (1 - r)
Assuming 'a' represents the first term in the series and 'r' is the ratio.To convert a complex number into Cartesian form, you simply map the real and imaginary parts of the number 'a + bi', where 'a' is the real part, and 'bi' is the imaginary part.Unfortunately, without the specifics of the sums you're looking to evaluate, it's impossible to give a concrete step-by-step calculation. However, understanding the formulas and how they're applied should provide you with a good start.
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Categorize these measurements associated with a robotics company according to level: nominal, ordinal, interval, or ratio. (a) Salesperson's performance: below average, average, above average. nominal ordinal interval ratio (b) Price of company's stock nominal ordinal interval ratio Incorrect: Your answer is incorrect. (c) Names of new products nominal ordinal interval ratio (d) Temperature (°F) in CEO's private office nominal ordinal interval ratio (e) Gross income for each of the past 5 years nominal ordinal interval ratio (f) Color of product packaging nominal ordinal interval
Answer:
a) ordinal
b) ratio
c) nominal
d) interval
e) ratio
f) nominal
Step-by-step explanation:
a) Salesperson performance is ordinal because it is ordered from below average to above average
b) Price of company stock is ratio because it has a defined zero point.
c) Names of new product is nominal because it can be classified into different categories yet cannot be ordered.
d) Temperature in CEO's office is interval because it don't have a defined zero point. In temperature zero temperature can exists.
e) Again gross income is ratio because it has a defined zero point. Zero gross income will means that there is no gross income.
f) Color of product packaging is again nominal because it can be classified into different categories yet cannot be ordered.
The measurements can be categorized as "ordinal, ratio, nominal, and interval."
Explanation:(a) Salesperson's performance: below average, average, above average. - Ordinal
(b) Price of company's stock - Ratio
(c) Names of new products - Nominal
(d) Temperature (°F) in CEO's private office - Interval
(e) Gross income for each of the past 5 years - Ratio
(f) Color of product packaging - Nominal
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The California State University (CSU) system consists of 23 campuses, from San Diego State in the south to Humboldt State near the Oregon border. A CSU administrator wishes to make an inference about the average distance between the hometowns of students and their campuses.
Describe and discuss several different sampling methods that might be employed.
Answer:
For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected [tex]p=\frac{1}{N}[/tex]
So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.
Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.
The convenience sampling is not too useful since is not a probabilistic method.
Step-by-step explanation:
For this case we can clasify this study as an enumerative study and inferential since they want to identify the average distance between the hometowns of students and their campuses.
For this case the sampling frame represent all the 23 campuses, and is known for the researcher,
For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected [tex]p=\frac{1}{N}[/tex]
So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.
Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.
The convenience sampling is not too useful since is not a probabilistic method.
Different sampling methods that can be employed to estimate the average distance between CSU student hometowns and their campuses include Simple Random Sampling, Stratified Random Sampling, Cluster Sampling, and Systematic Sampling. Each has its own advantages and potential drawbacks.
Explanation:There are several sampling methods that can be employed to gauge the average distance between student hometowns and their respective California State University (CSU) campuses.
A Simple Random Sampling method could be applied, where every student from every campuse has the exact same chance of being selected. However, this method may not give a representative sample if certain campuses have more or less students than others.
Another method is Stratified Random Sampling, where students are first divided into groups or 'strata' based on their campus, and then random samples are taken from each stratum. This ensures a balanced sample from all campuses.
Then comes the Cluster Sampling, it divides the student population into 'clusters' based on their hometown, and then randomly selected clusters are surveyed. It would be useful for large-scale surveys.
Last is Systematic Sampling. In this method, every nth student on a list would be selected. This approach ensures evenly distributed selection of students across the whole population but requires a complete listing of all students at the CSU campuses.
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Problem 3 in your textbook presents three different sample spaces for a horse race with five horses running. Let the horses be designated by the letters A, B, C, D, and E. A more complete characterization of an outcome of the horse race would be to designate which horse finishes in first through fifth places
(a) Consider the sample space for the set of outcomes characterized in this way. How many such outcomes are in the sample space?
b) How many outcomes are in the event that horse A finishes first?
(c) If G is the event that horse A finishes first and H is the event that horse B does not finish second, describe in words the event GnH. How many outcomes are in this event?
Answer:
a) 120
b) 24
c) 18
Step-by-step explanation:
part a
The sample space is defined by the rank of the horses.
Hence, 5 ranks would permute to = 5! = 120 outcomes
part b
Fixing the first position for horse A we are left with four horses and 4 positions, the position are permutated to 4! = 24 outcomes
part c
Fixing the first position for horse A we are left with four horses and 4 positions, and horse B can not finish second hence:
A _ B _ _ the rest can permute hence, 3! = 6 outcomes
A _ _ B _ the rest can permute hence, 3! = 6 outcomes
A _ _ _ B the rest can permute hence, 3! = 6 outcomes
Total outcomes is sum of 3 cases above = 18 outcomes
jacob wants to buy packets of fruit juice. on tuesday, he buys 8 packets of orange and 6 packets of Apple the packets make a total of 34 cups of fruit juice on Wednesday he buys 14 packets of orange and 6 packets of Apple this allows him to make 46 cups of fruit juice the packets of orange can make blank cups of juice the packets of Apple can make blank cups of juice
Answer:
Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.
Step-by-step explanation:
Let number of cups of juice in each packet of orange be 'x'.
Let number of cups of juice in each pack of apples be 'y'.
Given:
On Tuesday;
Number of packets of oranges = 8
Number of packets of apples = 6 packets.
Amount of fruit juice = 34 cups
So we can say that;
Amount of fruit juice is equal to sum of Number of packets of oranges multiplied by number of cups of juice in each packet of orange and Number of packets of apples multiplied by number of cups of juice in each pack of apples.
framing in equation form we get;
[tex]8x+6y=34 \ \ \ \ equation \ 1[/tex]
On Wednesday;
Number of packets of oranges = 14
Number of packets of apples = 6
Amount of fruit juice = 46 cups
So we can say that;
Amount of fruit juice is equal to sum of Number of packets of oranges multiplied by number of cups of juice in each packet of orange and Number of packets of apples multiplied by number of cups of juice in each pack of apples.
framing in equation form we get;
[tex]14x+6y=46 \ \ \ \ equation \ 2[/tex]
Now Subtracting equation 1 from equation we get;
[tex]14x+6y-(8x+6y)=46-34\\\\14x+6y-8x-6y=12\\\\6x=12[/tex]
Now dividing both side by 6 we get;
[tex]\frac{6x}{6}=\frac{12}{6}\\\\x=2\ cups[/tex]
Now we will substitute the value of 'x' in equation 1 we get;
[tex]8x+6y=34\\\\8\times2+6y=34\\\\16+6y=34[/tex]
Subtracting both side by 16 we get;
[tex]16+6y-16=34-16\\\\6y= 18[/tex]
Now dividing both side by 6 we get;
[tex]\frac{6y}{6}=\frac{18}{6}\\\\y=3[/tex]
Hence Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.
The cost, C (in dollars) to produce g gallons of a chemical can be expressed as C=f(g). Using units, explain the meaning of the following statements in terms of the chemical. (a) f(400)=700 The statement f(400)=700 means
Answer:
To produce 400 gallons of a chemical, it costs 700 dollars.
Step-by-step explanation:
We have that:
The cost, C (in dollars) to produce g gallons of a chemical can be expressed as C=f(g).
The interpretation is:
To produce g gallons of a chemical it costs C dollars.
So
f(400)=700
This means that to produce 400 gallons of a chemical, it costs 700 dollars.
Final answer:
The statement f(400)=700 means that the cost to produce 400 gallons of a chemical is 700 dollars. This cost function indicates the relationship between the number of gallons produced and the total cost, facilitating the calculation of expenses for chemical production.
Explanation:
The statement f(400)=700 means that to produce 400 gallons of a chemical, the cost is 700 dollars. The function f(g) represents the cost C, in dollars, to produce g gallons of the chemical. Therefore, when we input 400 into the function, it outputs the cost associated with producing that quantity, which in this case is 700 dollars.
Understanding the function as a rate of change, the cost to produce an additional gallon can be determined through ordinary algebra. If it were the case that this chemical firm is in a constant cost industry, where the cost per additional unit produced remains the same (e.g., 10 USD/oz), then this would simplify the exercise of calculating the cost for any number of gallons produced.
Conversion factors in chemistry are often used to relate amounts of substances in reactions, similarly in economics, conversion factors can tell us how much of a good can be bought or produced per unit of currency, like dollars per gallon, which is USD/unit.
A book is randomly chosen from a library shelf. For each of the following characteristics of the book, decide whether the characteristic is a continuous or a discrete random variable: a. Weight of the book (e.g., 2.3 pounds) b. Number of chapters in the book (e.g., 10 chapters) c. Width of the book (e.g., 8 inches) d. Type of book (0=hardback or 1=paperback) e. Number of typographical errors in the book (e.g., 4 errors
Answer:
a) Continuous since the weight can take decimals for example 2.3 pounds.
b) Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters
c) Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous
d) Discrete since that's a binary variable and only take integers.
e) Discrete since the number of errors can't be 4.3 for example, and can't take decimals.
Step-by-step explanation:
Previous concepts
A continuous random variable by definition is a "random variable where the data can take infinitely many values" defined on a interval.
And a discrete random variable is a random variable that only can takes integers and is defined over a domain
Solution to the problem
a. Weight of the book (e.g., 2.3 pounds)
Continuous since the weight can take decimals for example 2.3 pounds.
b. Number of chapters in the book (e.g., 10 chapters)
Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters
c. Width of the book (e.g., 8 inches)
Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous
d. Type of book (0=hardback or 1=paperback)
Discrete since that's a binary variable and only take integers.
e. Number of typographical errors in the book (e.g., 4 errors)
Discrete since the number of errors can't be 4.3 for example, and can't take decimals.
sin(a+b)*sin(a-b)=cos^2b-cos^2a
Answer:
Step-by-step explanation:
To prove sin(a+b)*sin(a-b)=cos^2b-cos^2a
we simplify the left side sin(a+b)*sin(a-b) first
sin(a+b) = sin a cos b + cos a sin b
sin(a-b) = sin a cos b - cos a sin b
sin(a+b)*sin(a-b) = (sin a cos b + cos a sin b) x (sin a cos b -cos a sin b)
sin a cos b((sin a cos b + cos a sin b) - cos a sin b (sin a cos b + cos a sin b)
open the bracket
sin a cos b(sin a cos b) + sin a cos b(cos a sin b) -cos a sin b (sin a cos b)+ cos a sin b ( cos a sin b)
sin²a cos²b + sin a cos b cos a sin b - cos a sin b sin a cos b + cos²a sin²b
sin²a cos²b + 0 + cos²a sin²b
sin²a cos²b + cos²a sin²b
sin²a = 1-cos² asin²b = 1-cos² b(1-cos² a)cos² b - cos² a(1-cos² b)
= cos² b - cos² a cos² b - cos² a +cos² a cos² b
choose like termscos² b - cos² a - cos² a cos² b + cos² a cos² b = cos² b - cos² a + 0
cos² b - cos² a
left hand side equals right hand side
A sock drawer contains eight navy blue socks and five black socks with no other socks. If you reach in the drawer and take two socks without looking and without replacement, what is the probability that: a) you will pick a navy sock and a black sock? b) the colors of the two socks will match? c) at least one navy sock will be selected?
Answer:
a. the probability of picking a navy sock and a black sock = P (A & B)
= (8/13 ) * (5/12) = 40/156 = 0.256
b. the probability of picking two navy or two black is
= 56/156 + 20/156 = 76/156 = 0.487
c. the probability of either 2 navy socks is picked or one black & one navy socks.
= 40/156 + 56/156 = 96/156 = 0.615
Step-by-step explanation:
A sock drawer contains 8 navy blue socks and 5 black socks with no other socks.
If you reach in the drawer and take two socks without looking and without replacement, what is the probability that:
Solution:
total socks = N = 8 + 5 + 0 = 13
a) you will pick a navy sock and a black sock?
Let A be the probability of picking a navy socks first.
Then P (A) = 8/13
without replacing the navy sock, will pick the black sock, total number of socks left is 12.
Let B be the probability of picking a black sock again.
P (B) = 5/12.
Then, the probability of picking a navy sock and a black sock = P (A & B)
= (8/13 ) * (5/12) = 40/156 = 0.256
b) the colors of the two socks will match?
Let A be the probability of picking a navy socks first.
Then P (A) = 8/13
without replacing the navy sock, will pick another navy sock, total number of socks left is 12.
Let B be the probability of another navy sock again.
P (B) = 7/12.
Then, the probability of picking 2 navy sock = P (A & B)
= (8/13 ) * (7/12) = 56/156 = 0.359
Let D be the probability of picking a black socks first.
Then P (D) = 5/13
without replacing the black sock, will pick another black sock, total number of socks left is 12.
Let E be the probability of another black sock again.
P (E) = 4/12.
Then, the probability of picking 2 black sock = P (D & E)
= (5/13 ) * (4/12) = 5/39 = 0.128
Now, the probability of picking two navy or two black is
= 56/156 + 20/156 = 76/156 = 0.487
c) at least one navy sock will be selected?
this means, is either you pick one navy sock and one black or two navy socks.
so, if you will pick a navy sock and a black sock, the probability of picking a navy sock and a black sock = P (A & B)
= (8/13 ) * (5/12) = 40/156 = 0.256
also, if you will pick 2 navy sock, Then, the probability of picking 2 navy sock = P (A & B)
= (8/13 ) * (7/12) = 56/156 = 0.359
now either 2 navy socks is picked or one black one navy socks.
= 40/156 + 56/156 = 96/156 = 0.615
In general, ___ % of the values in a data set lie at or below the median. ___ % of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 1600 test scores, __ of them would be at or below the median. If a sample consists of 1600 test scores, ____ of them would be at or above the first quartile (Q1)
Answer:
In general, 50 % of the values in a data set lie at or below the median. 75 % of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 1600 test scores, 0.5*1600 = 800 of them would be at or below the median. If a sample consists of 1600 test scores, 0.75*800 = 1200 of them would be at or above the first quartile (Q1)
Step-by-step explanation:
The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.
The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.
The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.
The answer is:
In general, 50 % of the values in a data set lie at or below the median. 75 % of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 1600 test scores, 0.5*1600 = 800 of them would be at or below the median. If a sample consists of 1600 test scores, 0.75*800 = 1200 of them would be at or above the first quartile (Q1)
Using the principle of quartile distribution, the percentage amount of data at or below the median and third quartile are 50% and 75% respectively. Number of scores below the median and at or above the first quartile is 800 and 1200 respectively.
The median(Q2) refers to the second quartile of the the distribution which is 50% of the data or below. The Lower quartile (Q1) is 25% or below of the distribution while 75% of the distribution refers to the upper quartile (Q3)The median of 1600 can be calculated thus :
50% × 1600 = 0.5 × 1600 = 800Above the first quartile :
Below first quartile = 25% × 1600 = 400Above first quartile = 1600 - 400 = 1200Therefore, the score above the first quartile is 1200.
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Consider an economy with a flat rate tax system. Each dollar of income over $5000 is taxed at 20%. (Income below $5000 is tax free.) In general, T2(Y-5000) = -1000+ .2Y, where T is taxes and Y is income. Suppose that the population mean income is $20,000 and that the population standard deviation of incomes is $8000. All families have at least $5000 of income. a. find the mean of T b. Find the standard deviation of T. c. if the population contains 20 million families, what is the government's total tax revenue.
Answer:
a) [tex] E(T) = 0.2 E(Y) -1000= 0.2*20000 -1000=3000[/tex]
b) [tex] Sd(T) = \sqrt{0.2^2 Var(Y)}=\sqrt{0.2^2 8000^2}= 1600[/tex]
c) Assuming 20 million of families and each one with a mean of income of 20000 for each family approximately then total income would be:
[tex] E(T) = 20000000*20000= 40000 millions[/tex]
And if we replace into the formula of T we have:
[tex] T = 0.2*400000x10^6 -1000= 790000 millions[/tex]
Approximately.
Step-by-step explanation:
For this case we knwo that Y represenet the random variable "Income" and we have the following properties:
[tex] E(Y) = 20000, Sd(Y) = 8000[/tex]
We define a new random variable T "who represent the taxes"
[tex] T = 0.2(Y-5000) = 0.2Y -1000[/tex]
Part a
For this case we need to apply properties of expected value and we have this:
[tex] E(T) = E(0.2 Y -1000)[/tex]
We can distribute the expected value like this:
[tex] E(T) = E(0.2 Y) -E(1000)[/tex]
We can take the 0.2 as a factor since is a constant and the expected value of a constant is the same constant.
[tex] E(T) = 0.2 E(Y) -1000= 0.2*20000 -1000=3000[/tex]
Part b
For this case we need to first find the variance of T we need to remember that if a is a constant and X a random variable [tex] Var(aX) = a^2 Var(X)[/tex]
[tex] Var(T) = Var (0.2Y -1000)[/tex]
[tex] Var(T)= Var(0.2Y) -Var(1000) + 2 Cov(0.2Y, -1000)[/tex]
The covariance between a random variable and a constant is 0 and a constant not have variance so then we have this:
[tex] Var(T) =0.2^2 Var(Y)[/tex]
And the deviation would be:
[tex] Sd(T) = \sqrt{0.2^2 Var(Y)}=\sqrt{0.2^2 8000^2}= 1600[/tex]
Part c
Assuming 20 million of families and each one with a mean of income of 20000 for each family approximately then total income would be:
[tex] E(T) = 20000000*20000= 40000 millions[/tex]
And if we replace into the formula of T we have:
[tex] T = 0.2*400000x10^6 -1000= 790000 millions[/tex]
Approximately.
For each random variable, state whether the random variable should be modeled with a Binomial distribution or a Poisson distribution. Explain your reasoning. State the parameter values that describe the distribution and give the probability mass function.
Random Variable 1. A quality measurement for cabinet manufacturers is whether a drawer slides open and shut easily. Historically, 2% of drawers fail the easy slide test. A manufacturer samples 10 drawers from a batch. Assuming the chance of failure is independent between drawers, what type of distribution could be used to model the number of failed drawers from the sample of 10?
Random Variable 2. The warranty for a particular system on a new car is 2 years. During which there is no limit to the number of warranty claims per car. Historically, the average number of claims per car during the period is 0.8 claims. What type of distribution could be used to model the number of warranty claims per car?
Answer:
1) This random variable should be modelled using a binomial distribution since we have independence between the events and a bernoulli trial each time when the experiment is conducted, a fixd value for the sample size n and for the probability of success.
Let X the random variable of interest, on this case th distribution would be given by:
[tex]X \sim Binom(n=10, p=0.02)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x} = (10Cx) (0.02)^x (1-0.02)^{10-x}[/tex]
2) For this case we don't have a sample size provided and we just have an average rate for a given period, so then we can assume that the best distribution for this case is the Poisson distribution.
Let X the random variable that represent the number of claims per car. We know that [tex]X \sim Poisson(\lambda)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
Where [tex]\lambda=0.2[/tex] represent the mean of occurrences in the interval of 2 years provided.
And f(x)=0 for other case.
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Random variable 1
This random variable should be modelled using a binomial distribution since we have independence between the events and a bernoulli trial each time when the experiment is conducted, a fixd value for the sample size n and for the probability of success.
Let X the random variable "number of failed drawers", on this case th distribution would be given by:
[tex]X \sim Binom(n=10, p=0.02)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x} = (10Cx) (0.02)^x (1-0.02)^{10-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Random variable 2
For this case we don't have a sample size provided and we just have an average rate for a given period, so then we can assume that the best distribution for this case is the Poisson distribution.
Let X the random variable that represent the number of claims per car. We know that [tex]X \sim Poisson(\lambda)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
Where [tex]\lambda=0.2[/tex] represent the mean of occurrences in the interval of 2 years provided.
And f(x)=0 for other case.