Answer:
336 different predictions.
1/336 probability of making the correct prediction by chance
Step-by-step explanation:
The order is important.
For example, Team A, B and C is a different outcome than team B, A, C.
So we use the permutations formula to solve this problem:
Permutations formula:
The number of possible permutations of x elements from a set of n elements is given by the following formula:
[tex]P_{(n,x)} = \frac{n!}{(n-x)!)}[/tex]
In this problem, we have that:
Permutations of 3 from a set of 8. So
[tex]P_{(8,3)} = \frac{8!}{(8-3)!} = 336[/tex]
What is the probability of making the correct prediction by chance?
There are 336 possible outcomes.
By chance, you predict 1.
So there is a 1/336 probability of making the correct prediction by chance
Final answer:
You can make 336 different predictions for the top 3 soccer teams out of 8, and the chance of making the correct prediction by chance is approximately 0.298%.
Explanation:
To calculate the number of different predictions you could make for the top 3 teams out of 8, without considering ties, you use permutations since the order matters. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of teams and r is the number of positions to fill.
In this case, n = 8 teams and r = 3 positions. Therefore, the calculation is P(8, 3) = 8! / (8-3)! = 8 x 7 x 6 = 336 different predictions.
To find the probability of making the correct prediction by chance, since there is only one correct prediction out of all possible predictions, the probability is 1 / 336. Thus, the probability is approximately 0.00298, or 0.298%.
Suppose that operators A^ and B^ are both Hermitian, i.e, A^` = A^ and B^` = B^.
Answer the following and show your work:
(a) Is A^² Hermitian?
(b) Is A^B^ Hermitian?
(c) Is A^B^+ B^A^ Hermitian?
(d) Is it possible for A^ to have complex eigenvalues, or must they be real?
Answer:
a) A^² is a Hermitian operator
b) A^B^ is not a Hermitian operator
c) A^B^+ B^A^ is a Hermitian operator
d) It is not possible to be complex it must be a real number
Step-by-step explanation:
In order to understand this solution we need to define the concept Hermitian
HERMITIAN
This can be defined as a matrix whose elements are real and symmetrical i.e. it a square matrix that is equal to its own conjugate, or we can simply put that its a matrix in which those pairs of element that are symmetrically placed with respect to the principal diagonal are complex conjugates.i.e the diagonal elements( Hermitian operators) are real numbers while others are complex numbers.
The solution to the question above are on the first and second uploaded image.
Which measurement is most accurate to describe the amount that a teacup can hold? 6 fl. oz. 2 cups 1 pint 1.5 quarts
Answer:
6 fl. oz
Step-by-step explanation:
The most accurate measurement for the volume a teacup can hold is 6 fluid ounces.
To determine the most accurate measurement for the amount that a teacup can hold, we need to consider the typical volume of a teacup and understand the relationship between different units of capacity. A standard teacup holds about 6 fluid ounces. Given the unit conversion factors, we know that:
1 cup = 8 fluid ounces
1 pint = 2 cups = 16 fluid ounces
1 quart = 2 pints = 32 fluid ounces
1.5 quarts = 48 fluid ounces
Considering these units:
6 fluid ounces is less than 1 cup (8 fl oz).
2 cups equal 16 fluid ounces, which is more than a typical teacup can hold.
1 pint (16 fluid ounces) is also more than what a teacup can hold.
1.5 quarts (48 fluid ounces) is significantly more than a teacup's capacity.
Therefore, the most accurate measurement to describe the amount a teacup can hold is 6 fluid ounces.
How to find the area of a square ABC D
Answer:
The answer to your question is 13 u²
Step-by-step explanation:
We know that the small triangle is surrounded by right triangles so we can use the Pythagorean theorem to find the lengths of the small triangle
AD² = 3² + 2²
Simplify
AD² = 9 + 4
AD² = 13
AD = [tex]\sqrt{13}[/tex]
Find the area of the square
Area = side x side
Area = AD x AD
Area = [tex]\sqrt{13} x \sqrt{13}[/tex]
Area = 13 u²
A multiple choice test has 10 questions with 3 choices each. a. How many ways are there to answer the test? b. What is the probability that two papers have the same answers?
Answer:
a. 59049 ways
b. [tex]1.69\times10^{-5}[/tex]
Step-by-step explanation:
If the multiple choice test has 10 questions with 3 choices each, for the each question there are 3 ways to answer the test. Then for n question there are [tex]3^n[/tex] ways to answer the test
The total number of ways to answer 10 questions at 3 choices each is
[tex]3^{10} = 59049[/tex]ways
b. There are 59049 ways to answer the test but if one test must match one other test then the probability for that to happen is
[tex]\frac{1}{59049} = 1.69\times10^{-5}[/tex]
There are 59,049 ways to answer a 10-question test with 3 choices per question. The probability of two tests having the same answers, with all answer patterns being equally likely, is 1/59,049.
Explanation:This problem belongs to the domain of combinatorics. (a) As there are 3 choices for each question and 10 questions, there are 3^10 or 59,049 ways to answer the test. This assumes you are answering every question. (b) This is essentially the question of the probability that two randomly selected tests are identically answered. Assuming all ways to answer a test are equally likely, there are 59,049 different ways to answer, so the probability that two randomly answered tests are the same is 1/59,049 or approximately 0.000017.
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Suppose that you are hiking on a terrain modeled by z=xy+y^3−x^2. You are at the point (2,1,−1).
(a) Determine the slope you would encounter if you headed due West from your position. What angle of inclination does this correspond to?
(b) Determine the slope you would encounter if you headed due North-West from your position. What angle of inclination does this correspond to?
(c) Determine the slope you would encounter if you headed due South-West from your position. What angle of inclination does this correspond to?
(d) Determine the steepest slope you could encounter from your position and the direction of that slope (as a unit vector).
Answer:
a) D_u . ∀ f ( 2 , 1 ) = 3 , Q = 72 degrees
b) D_u . ∀ f ( 2 , 1 ) = 4*sqrt(2) , Q = 80 degrees
c) D_u . ∀ f ( 2 , 1 ) = - sqrt(2) , Q = -54.74 degrees
d) u = 1 / sqrt(34) *(-3i +5j) , D_u . ∀ f ( 2 , 1 ) = +/- sqrt (34) , Q = 80.27 degrees
Step-by-step explanation:
Given:
- The terrain is modeled as a surface:
f(x , y) = x*y + y^3 - x^2
At point P( 2 , 1 , -1 ) is the current position:
Find:
(a) Determine the slope you would encounter if you headed due West from your position. What angle of inclination does this correspond to?
(b) Determine the slope you would encounter if you headed due North-West from your position. What angle of inclination does this correspond to?
(c) Determine the slope you would encounter if you headed due South-West from your position. What angle of inclination does this correspond to?
(d) Determine the steepest slope you could encounter from your position and the direction of that slope (as a unit vector).
Solution:
- We will compute the gradient of the vector ∀ f @ point P( 2 , 1 , -1 ) as follows:
∀ f ( x , y ) = (y - 2x ) i + ( x + 3y^2) j
- Evaluate at point P ( 2 , 1 ):
∀ f ( 2 , 1 ) = (1 - 2*2 ) i + ( 2 + 3*1^2) j
∀ f ( 2 , 1 ) = -3 i + 5 j
- We will use the result of ∀ f @ point P( 2 , 1 , -1 ) for the all the parts.
a)
- The direction due west can be written as a unit vector u_w = - i .
- Now compute the directional derivative in the direction of u_w = - i
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . - i
D_u . ∀ f ( 2 , 1 ) = 3
- Now compute the angle of inclination Q for the following direction:
Q = arctan(3) = 71.57 = 72 degrees
Hence, along the direction due west from position we ascend with an inclination of approximately 72 degrees.
b)
- The direction due north-west can be written as a unit vector u_w = - i + j .
- Now compute the directional derivative in the direction of u_w = - i + j
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i + j) / sqrt(2)
D_u . ∀ f ( 2 , 1 ) = 8 / sqrt(2) = 4*sqrt(2)
- Now compute the angle of inclination Q for the following direction:
Q = arctan(4*sqrt(2)) = 79.98 = 80 degrees
Hence, along the direction due north-west from position we ascend with an inclination of approximately 80 degrees.
c)
- The direction due south-west can be written as a unit vector u_w = - i - j .
- Now compute the directional derivative in the direction of u_w = - i - j
D_u . ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i - j) / sqrt(2)
D_u . ∀ f ( 2 , 1 ) = -2 / sqrt(2) = - sqrt(2)
- Now compute the angle of inclination Q for the following direction:
Q = arctan(-sqrt(2)) = -54.74 degrees
Hence, along the direction due south-west from position we descend with an inclination of approximately 55 degrees.
d)
- We know that ∀ f ( 2 , 1 ) gradient points in the direction greatest increase, hence, So from P the direction of greatest increase is ∇f(P) = -3i +5j. The unit vector pointing in this direction is:
u = 1 / sqrt(34) *(-3i +5j)
- so we have:
D_u . ∀ f ( 2 , 1 ) = u . ∀ f ( 2 , 1 ) = (-3i +5j) . (-3i +5j) / sqrt(34)
= +/- sqrt (34)
- Hence, the steepest ascent or decent is of :
Q = arctan ( sqrt(34)) = 80.27 degrees
A chemical plant has an emergency alarm system. When an emergency situation exists, the alarm sounds with probability 0.95. When an emergency situation does not exist, the alarm sounds with probability 0.02. A real emergency situation is a rare event, with probability 0.004. Given that the alarm has just sounded, what is the probability that a real emergency situation exists?
Answer:
6.56% probability that a real emergency situation exists.
Step-by-step explanation:
We have these following probabilities:
A 0.4% probability that a real emergency situation exists.
A 99.6% probability that a real emergency situation does not exist.
If an emergency situation exists, a 95% probability that the alarm sounds.
If an emergency situation does not exist, a 2% probability that the alarm sounds.
The problem can be formulated as the following question:
What is the probability of B happening, knowing that A has happened.
It can be calculated by the following formula
[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
In this problem:
What is the probability of a real emergency situation existing, given that the alarm has sounded.
P(B) is the probability of there being a real emergency situation. So [tex]P(B) = 0.004[/tex].
P(A/B) is the probability of the alarm sounding when there is a real emergency situation. So P(A/B) = 0.95.
P(A) is the probability of the alarm sounding. This is 95% of 0.4%(real emergency situation) and 2% of 99.6%(no real emergency situation). So
P(A) = 0.95*0.04 + 0.02*0.996 = 0.05792
Given that the alarm has just sounded, what is the probability that a real emergency situation exists?
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.004*0.95}{0.05792} = 0.0656[/tex]
6.56% probability that a real emergency situation exists.
A player of a video game is confronted with a series of 3 opponents and a(n) 77% probability of defeating each opponent. Assume that the results from opponents are independent (and that when the player is defeated by an opponent the game ends).
Round your answers to 4 decimal places.
a. What is the probability that a player defeats all 3 opponents in a game?
b. What is the probability that a player defeats at least two opponents in a game?
c. If the game is played 2 times, what is the probability that the player defeats all 3 opponents at least once?
Answer:
a.) 0.4565
b.) 0.8656
c.) 0.4615
Step-by-step explanation:
We solve this using the probability distribution formula of combination.
nCr * p^r * q^n-r
Where
n = number of trials
r = successful trials
probability of success = p = 77% =0.77
Probability of failure= q = 1-0.77 = 0.23
a.) When exactly 3 opponents are defeated, When n = 3 and r = 3, probability becomes:
= 3C3 * 0.77³ * 0.23^0
= 1 * 0.456533 * 1
= 0.456533 = 0.4565 (4.d.p)
b.) When at least 2 opponents are defeated, that is when r = 2 and when r = 3,
When r = 2, probability becomes:
= 3C2 * 0.77² * 0.23¹
= 3 * 0.5929 * 0.23
= 0.409101
When 3 opponents are defeated, we calculated it earlier to be 0.456533
Hence, probability that at least 2 opponents are defeated
= 0.409101 + 0.456533
= 0.865634 = 0.8656(2.d.p)
c.) If 2 games are played, probability he defeat all 3 at least once in the game will be the sum (probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game) + (probability of defeating all three opponents in both games)
Probability of defeating all three opponents in the first game = 0.456533
Probability of not defeating all three opponents in the second game = 1 - 0.456533 = 0.543467
Hence ,
probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game = 0.465633 * 0.543467 = 0.253056
probability of defeating all three opponents in both games
= 0.456533 * 0.456533
=0.208422
Probability he defeats all three opponents at least once in 2games
= 0.253056 + 0.208422
=0.461478 = 0.4615(4.d.p)
A researcher wants to determine if socioeconomic status (low, moderate, high) is related to smoking (yes or no). The Chi-Square null hypothesis for this study is that socioeconomic status is related to smoking behavior.A. TrueB. False
Answer:
The chi-square null hypothesis for the study that "socioeconomic status is related to smoking behavior" is False
Step-by-step explanation:
The chi-square null hypothesis is false because the chi-square null hypothesis states that no relationship exists on the categorical variables in a population, they are all independent of each other.
A large consumer goods company ran a television advertisement for one of its soap products.
On the basis of a survey that was conducted, probabilities were assigned to the following
events.
B = individual purchased the product S = individual recalls seeing the advertisement B∩S = individual purchased the product and recalls seeing the advertisement
The probabilities assigned were P(B)=.20,P(S)=.40, and P(B∩S)=.12
a. What is the probability of an individual’s purchasing the product given that the individual
recalls seeing the advertisement? Does seeing the advertisement increase
the probability that the individual will purchase the product? As a decision maker,
would you recommend continuing the advertisement (assuming that the cost is
reasonable)?
b. Assume that individuals who do not purchase the company’s soap product buy from
its competitors. What would be your estimate of the company’s market share? Would
you expect that continuing the advertisement will increase the company’s market
share? Why or why not?
"c. The company also tested another advertisement and assigned it values of P(S)=.30
and P(B∩S)=.10. What is P(B|S) for this other advertisement? Which advertise-
ment seems to have had the bigger effect on customer purchases?"
a. The probability is 0.30. The advertisement seems to have a positive effect on customer purchases.
b. The market share is 0.20. The advertisement could potentially increase the company's market share
c. The conditional purchase probability of 0.333. The second advertisement seems to have had a slightly bigger effect on customer purchases.
a. We are asked to find the probability of an individual purchasing the product given that the individual recalls seeing the advertisement, i.e., P(B|S).
Using the formula for conditional probability:
P(B|S) = P(B∩S) / P(S)
Given:
P(B∩S) = 0.12
P(S) = 0.40
So, P(B|S)
= 0.12 / 0.40
= 0.30
Seeing the advertisement increases the probability that an individual will purchase the product from 0.20 (P(B)) to 0.30 (P(B|S)). Therefore, the advertisement seems to have a positive effect on customer purchases.
As a decision maker, if the cost of the advertisement is reasonable, it would be recommended to continue the advertisement since it increases the likelihood of product purchases.
b. The company's market share can be estimated by considering the probability of individuals purchasing the company's soap product and the probability of individuals purchasing from competitors.
The probability of an individual purchasing from competitors
= 1 - P(B)
= 1 - 0.20
= 0.80
Therefore, the company's market share is:
Market Share = P(B)
= 0.20
Continuing the advertisement could potentially increase the company's market share since the advertisement has a positive effect on customer purchases, as shown in part (a).
c. For the other advertisement:
Given:
P(S) = 0.30
P(B∩S) = 0.10
Using the formula for conditional probability:
P(B|S) = P(B∩S) / P(S)
So, P(B|S)
= 0.10 / 0.30
= 0.333
Comparing the two advertisements, the first advertisement had a conditional purchase probability of 0.30, while the second advertisement had a conditional purchase probability of 0.333. Therefore, the second advertisement seems to have had a slightly bigger effect on customer purchases.
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The advertisements increase the chance for an individual to purchase the product. The second advertisement yields a higher likelihood of purchase, implying it's more effective.
Explanation:The problem is related to conditional probability. To solve the question:
The probability of an individual purchasing the product given that the individual recalls seeing the advertisement is calculated by P(B|S) = P(B∩S) / P(S) = .12 / .40 = .30 or 30%. This is greater than the overall probability of purchasing the product, P(B) = .20 or 20%. Meaning the advertisement does increase the probability for an individual to purchase the product.The estimated market share of the company is simply P(B) = .20 or 20%. As advertisement increases the probability of purchase, it would likely increase the market share as long as the cost of advertising does not outweigh the added revenue.For the other advertisement, P(B|S) = P(B∩S) / P(S) = .10 / .30 = .33 or 33%, which is higher than the first advertisement. Therefore, the other advertisement has a larger effect on customer purchases.Learn more about Conditional Probability here:https://brainly.com/question/32171649
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A rain gutter is to be made of aluminum sheets that are 12 inches wide by turning up the edges 90 degrees.What depth will provide maximum cross-sectional area and hence allow the most water to flow?
Answer:
18 in^2
Step-by-step explanation:
1 )The three sides of the gutter add up to 12
2x+ y = 12
2) Subtract 2x from both sides.
y = 12 — 2x
3 )Find the area of the rectangle in terms of x and simplify.
Area = xy = x(12 — 2x) = -2x^2+12x = f(x)
4 ) x=-b/2a
x co-ordinate of the vertex= -12/2(-2)=3
5 )Plug in 3 for x into they equation.
y co-ordinate of the vertex= 12 — 2(3) = 6
6 ) Plug in 3 for x and 6 for y.
Area= xy = 3(6) = 18
RESULT
18 in^2
Whats an explicit rule for this? 14, 20, 26, 32, etc. Write an explicit formula for the nth term an.
Answer:
a(n)=14+6(n-1)
Step-by-step explanation:
The first term is 14. This would be an arithmetic sequence, so you will add 6 to every term: the common difference is 6.
14+6= 20
20+6=26
You have the formula a(n)= a(1)+d(n-1)
a(n)= 14+6(n-1)
14 for the first term, 6 for the common difference.
A punch recipe requires 2/5 of a cup of pineapple juice for every 2 1/2 cups of soda. What is the unit rate of soda to pineapple juice in the punch?
Answer:
The unit rate is 6 1/4 cups of soda per cup of pineapple juice
Step-by-step explanation:
we know that
To find out the unit rate of soda to pineapple juice in the punch, divide the cups of soda by the cups of pineapple juice
so
[tex]2\frac{1}{2} :\frac{2}{5}[/tex]
Convert mixed number to an improper fraction
[tex]2\frac{1}{2}=2+\frac{1}{2}=\frac{2*2+1}{2}=\frac{5}{2}[/tex]
substitute
[tex]\frac{5}{2} :\frac{2}{5}[/tex]
Multiply in cross
[tex]\frac{25}{4}= 6.25[/tex]
Convert to mixed number
[tex]6.25=6+0.25=6+\frac{1}{4}= 6\frac{1}{4}[/tex]
That means
The unit rate is 6 1/4 cups of soda per cup of pineapple juice
Answer:
6 1/4
Step-by-step explanation:
If an angle of 96 degrees is rotated 90 degrees clockwise. what the measure?
Answer:
The measure is 6°.
Step-by-step explanation:
If an angle of 96 degrees is rotated 90 degrees clockwise then the measure of the new angle will be given by
= 96° - 90° = 6°
Someone please help me!
Answer:
The answer to your question is below
Step-by-step explanation:
43.-
x = [tex]\sqrt{27^{2} + 22^{2}- 2(27)(22)cos 73}[/tex]
x = [tex]\sqrt{865.66}[/tex]
x = 29.42
44.-
x = [tex]\sqrt{10^{2} + 14^{2} -2(10)(14)cos 66}[/tex]
x = [tex]\sqrt{182.11}[/tex]
x = 13.49
45.-
cos x = [tex]\frac{11^{2} - 17^{2} - 10^{2}}{-2(11)(17)}[/tex]
cos x = 0.7166
x = 44.22°
46.-
x = [tex]\sqrt{16^{2} + 12^{2} -2(16)(12)cos75}[/tex]
x = [tex]\sqrt{300.61}[/tex]
x = 17.34
47.-
cos P = [tex]\frac{6^{2} - 13^{2} -11^{2}}{-2(13)(11)}[/tex]
cos P = 0.888
P = 27.36°
sinR/13 = sinP/6
sin R = 13sin27.36/6
sinR = 0.996
R = 84.71°
Q = 180 - 84.71 - 27.36
Q = 67.93°
48.-
D = 180 - 25 - 113
D = 42°
CD = 9Sin113/sin42
CD = 12.38
ED = 9sin25/sin42
ED = 5.68
NEED HELP ASAP!!!!!! What is another way to say "to the third power"?
Answer:
A number to the third power would be a number cubed, so the answer is "Cubed".
List the probability value for each possibility in the binomial experiment calculated at the beginning of this lab, which was calculated with the probability of a success being ½. (Complete sentence not necessary; round your answers to three decimal places)P(x=0) P(x=6)
P(x=1) P(x=7)
P(x=2) P(x=8)
P(x=3) P(x=9)
P(x=4) P(x=10)
P(x=5)
Answer:
a. P(X = 0)= 0.001
b. P(X = 1)= 0.001
c. P(X=2)= 0.044
d. P(X=3)= 0.117
e. P(X=4)= 0.205
f. P(X=5)= 0.246
g. P(X=6)= 0.205
h. P(X=7)= 0.117
i. P(X=8)= 0.044
j. P(X=9)= 0.001
k. P(X=10)= 0.001
Step-by-step explanation:
Hello!
You have the variable X with binomial distribution, the probability of success is 0.5 and the sample size is n= 10 (I suppose)
If the probability of success p=0.5 then the probability of failure is q= 1 - p= 1 - 0.5 ⇒ q= 0.5
You are asked to calculate the probabilities for each observed value of the variable. In this case is a discrete variable with definition between 0 and 10.
You have two ways of solving this excersice
1) Using the formula
[tex]P(X)= \frac{n!}{(n-X)!X!} * (p)^X * (q)^{n-X}[/tex]
2) Using a table of cummulative probabilities of the binomial distribution.
a. P(X = 0)
Formula:
[tex]P(X=0)= \frac{10!}{(10-0)!0!} * (0.5)^0 * (0.5)^{10-0}[/tex]
P(X = 0) = 0.00097 ≅ 0.001
Using the table:
P(X = 0) = P(X ≤ 0) = 0.0010
b. P(X = 1)
Formula
[tex]P(X=1)= \frac{10!}{(10-1)!1!} * (0.5)^1 * (0.5)^{10-1}[/tex]
P(X = 1) = 0.0097 ≅ 0.001
Using table:
P(X = 1) = P(X ≤ 1) - P(X ≤ 0) = 0.0107-0.0010= 0.0097 ≅ 0.001
c. P(X=2)
Formula
[tex]P(X=2)= \frac{10!}{(10-2)!2!} * (0.5)^2 * (0.5)^{10-2}[/tex]
P(X = 2) = 0.0439 ≅ 0.044
Using table:
P(X = 2) = P(X ≤ 2) - P(X ≤ 1) = 0.0547 - 0.0107= 0.044
d. P(X = 3)
Formula
[tex]P(X = 3)= \frac{10!}{(10-3)!3!} * (0.5)^3 * (0.5)^{10-3}[/tex]
P(X = 3)= 0.11718 ≅ 0.1172
Using table:
P(X = 3) = P(X ≤ 3) - P(X ≤ 2) = 0.1719 - 0.0547= 0.1172
e. P(X = 4)
Formula
[tex]P(X = 4)= \frac{10!}{(10-4)!4!} * (0.5)^4 * (0.5)^{10-4}[/tex]
P(X = 4)= 0.2051
Using table:
P(X = 4) = P(X ≤ 4) - P(X ≤ 3) = 0.3770 - 0.1719= 0.2051
f. P(X = 5)
Formula
[tex]P(X = 5)= \frac{10!}{(10-5)!5!} * (0.5)^5 * (0.5)^{10-5}[/tex]
P(X = 5)= 0.2461 ≅ 0.246
Using table:
P(X = 5) = P(X ≤ 5) - P(X ≤ 4) = 0.6230 - 0.3770= 0.246
g. P(X = 6)
Formula
[tex]P(X = 6)= \frac{10!}{(10-6)!6!} * (0.5)^6 * (0.5)^{10-6}[/tex]
P(X = 6)= 0.2051
Using table:
P(X = 6) = P(X ≤ 6) - P(X ≤ 5) = 0.8281 - 0.6230 = 0.2051
h. P(X = 7)
Formula
[tex]P(X = 7)= \frac{10!}{(10-7)!7!} * (0.5)^7 * (0.5)^{10-7}[/tex]
P(X = 7)= 0.11718 ≅ 0.1172
Using table:
P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.9453 - 0.8281= 0.1172
i. P(X = 8)
Formula
[tex]P(X = 8)= \frac{10!}{(10-8)!8!} * (0.5)^8 * (0.5)^{10-8}[/tex]
P(X = 8)= 0.0437 ≅ 0.044
Using table:
P(X = 8) = P(X ≤ 8) - P(X ≤ 7) = 0.9893 - 0.9453= 0.044
j. P(X = 9)
Formula
[tex]P(X = 9)= \frac{10!}{(10-9)!9!} * (0.5)^9 * (0.5)^{10-9}[/tex]
P(X = 9)=0.0097 ≅ 0.001
Using table:
P(X = 9) = P(X ≤ 9) - P(X ≤ 8) = 0.999 - 0.9893= 0.001
k. P(X = 10)
Formula
[tex]P(X = 10)= \frac{10!}{(10-10)!10!} * (0.5)^{10} * (0.5)^{10-10}[/tex]
P(X = 10)= 0.00097 ≅ 0.001
Using table:
P(X = 10) = P(X ≤ 10) - P(X ≤ 9) = 1 - 0.9990= 0.001
Note: since 10 is the max number this variable can take, the cummulated probability until it is 1.
I hope it helps!
In this problem, y = c1ex + c2e−x is a two-parameter family of solutions of the second-order DE y'' − y = 0. Find c1 and c2 given the following initial conditions. (Your answers will not contain a variable.) y(1) = 0, y'(1) = e c1 = Incorrect: Your answer is incorrect. c2 = Incorrect: Your answer is incorrect. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. y = Incorrect: Your answer is incorrect.
Answer:
c₁ = 1/2
c₂ = - e²/2
y = (1/2)*(eˣ - e²⁻ˣ)
Step-by-step explanation:
Given
y = c₁eˣ + c₂e⁻ˣ
y(1) = 0
y'(1) = e
We get y' :
y' = (c₁eˣ + c₂e⁻ˣ)' ⇒ y' = c₁eˣ - c₂e⁻ˣ
then we find y(1) :
y(1) = c₁e¹ + c₂e⁻¹ = 0
⇒ c₁ = - c₂/e² (I)
then we obtain y'(1):
y'(1) = c₁e¹ - c₂e⁻¹ = e (II)
⇒ (- c₂/e²)*e - c₂e⁻¹ = e
⇒ - c₂e⁻¹ - c₂e⁻¹ = - 2c₂e⁻¹ = e
⇒ c₂ = - e²/2
and
c₁ = - c₂/e² = - (- e²/2) / e²
⇒ c₁ = 1/2
Finally, the equation will be
y = (1/2)*eˣ - (e²/2)*e⁻ˣ = (1/2)*(eˣ - e²⁻ˣ)
Applying the initial conditions, it is found that the solution is:
[tex]y = \frac{1}{2}e^{x} - \frac{e^2}{2}e^{-x}[/tex]
------------------------
The solution for the PVI is given by:
[tex]y = c_1e^{x} + c_2e^{-x}[/tex]
------------------------
The condition [tex]y(1) = 0[/tex] means that when [tex]x = 0, y = 1[/tex], and thus, we get:
[tex]c_1e + c_2e^{-1} = 0[/tex]
[tex]c_1e+ \frac{c_2}{e} = 0[/tex]
[tex]c_1e^{2} + c_2 = 0[/tex]
[tex]c_2 = -c_1e^{2}[/tex]
------------------------
The derivative is:
[tex]y^{\prime}(x) = c_1e^{x} - c_2e^{-x}[/tex]
Applying the condition [tex]y^{\prime}(1) = e[/tex], we get:
[tex]c_1e - \frac{c_2}{e} = e[/tex]
Considering [tex]c_2 = -c_1e^{2}[/tex]:
[tex]c_1e + c_1\frac{e^2}{e} = e[/tex]
[tex]c_1e + c_1e = e[/tex]
[tex]2c_1e = e[/tex]
[tex]2c_1 = 1[/tex]
[tex]c_1 = \frac{1}{2}[/tex]
------------------------
The second constant is:
[tex]c_2 = -c_1e^{2} = -\frac{e^2}{2}[/tex]
And the solution is:
[tex]y = \frac{1}{2}e^{x} - \frac{e^2}{2}e^{-x}[/tex]
A similar problem is given at https://brainly.com/question/13244107
The last digit of the heights of 66 statistics students were obtained as part of an experiment conducted for a class. Use the frequency distribution to construct a histogram.
Digit Frequency
0 16
1 3
2 5
3 4
4 5
5 14
6 5
7 5
8 4
9 5
What can be concluded from the distribution of the digits?
The histogram is shown below.
You'll have 10 bars. Under each bar is a label from 0 to 9. The height of each bar represents the frequency of each units digit.
The histogram shows two bars that are relatively large compared to the rest. These bars have frequency of 16 and 14 (for units digits of 0 and 5 respectively). The distribution is nearly bimodal. If the two frequencies mentioned were the same, say both 16, then it would be exactly bimodal. As you can probably guess, bimodal means "two modes".
The rest of the bar heights are nearly the same, so the remaining portion of the distribution is nearly uniform. A uniform distribution has every bar the same height. Collectively, all the bars of a uniform distribution forms a larger rectangle.
When analyzing data on the number of employees in small companies in one town, a researcher took square roots of the counts. Some of the resulting values, which are reasonably symmetric, were 4, 5, 5, 7, 7, 8, and 11.
What were the original values, and how are they distributed?
Answer:
The original values are : 16, 25, 25, 49, 49, 64, 121.
Step-by-step explanation:
We know that a researcher took square roots of the counts. Some of the resulting values, which are reasonably symmetric, were 4, 5, 5, 7, 7, 8, and 11. We calculate the original values:
[tex]4^2=16\\5^2=25\\5^2=25\\7^2=49\\7^2=49\\8^2=64\\11^2=121\\[/tex]
The original values are : 16, 25, 25, 49, 49, 64, 121.
We conclude that the original data is not simmetric.
The original values obtained by reversing the square root transformation are 16, 25, 25, 49, 49, 64, and 121. These values show variability in the number of employees across different small companies in the town. The transformed dataset was made more symmetrical for statistical analysis.
When a researcher applies a square root transformation to a dataset, the purpose is often to make the data more symmetrical and easier to analyze using certain statistical methods.
Given the transformed values 4, 5, 5, 7, 7, 8, and 11, we can reverse the transformation to find the original values.
The square of each transformed value yields the original data points:
4² = 165² = 255² = 257² = 497² = 498² = 6411² = 121Thus, the original values are 16, 25, 25, 49, 49, 64, and 121. These values are distributed with some repeated data points and a range from 16 to 121.
This distribution indicates variability in the number of employees across the small companies studied.
Use the roster method to write each of the given sets. (Enter EMPTY for the empty set.)
(a) The set of natural numbers x that satisfy x + 4 = 1.
(b) Use set-builder notation to write the following set.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Answer:
a) Empty set
b) [tex]\{x : x \in N \text{ and } x < 13\}[/tex]
Step-by-step explanation:
Roster form is a comma separated list form of set.
a) The set of natural numbers x that satisfy x + 4 = 1.
[tex]x + 4 = 1\\x = -3 \notin N[/tex]
Thus, x is an empty set.
b) set-builder notation for the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.
We use x to represent this set. Now x belongs to natural number and is less than equal to 12.
Thus, it can be written as:
[tex]\{x : x \in N \text{ and } x < 13\}[/tex]
You toss a fair coin 5 times. What is the probability of at least one head? Round to the nearest ten- thousandth.
Answer:
0.9688
Step-by-step explanation:
For each time the coin is tossed, there are only two possible outcomes. Either it is heads, or it is tails. The probabilities for each coin toss are independent from each other. 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:
For each time the coin is tossed, heads or tails are equally as likely, since the coin is fair. So [tex]p = \frac{1}{2} = 0.5[/tex]
You toss a fair coin 5 times. What is the probability of at least one head?
Either there are no heads, or there is at least one head. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We want to find [tex]P(X \geq 1)[/tex], when [tex]n = 5[/tex].
So
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.03125 = 0.96875[/tex]
Rounding to the nearest ten-thousandth(four decimal places), this probability is 0.9688.
The probability of getting at least one head when tossing a fair coin 5 times is 31/32 or 0.96875 when rounded to the nearest ten-thousandth.
Explanation:The subject of this question is probability, a field within Mathematics. To answer your question: the probability of getting at least one head when tossing a fair coin 5 times can be found by calculating the probability of not getting a head (which is tossing tails 5 times in a row) and then subtracting that from 1 (representing certainty). Each toss of the fair coin has two outcomes, heads or tails, with equal probability of 1/2. If you toss the coin 5 times, the total number outs comes is 2^5, or 32. The chance of getting all tails is (1/2)^5, which is 1/32.
So, the probability of not getting a head (only tails) in 5 tosses is 1/32. Subtract this from 1 to find the probability of getting at least one head: This equals 1 - 1/32 = 31/32 = 0.96875, when rounded to the nearest ten-thousandth.
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An education researcher collects data on how many hours students study at various local colleges. The researcher calculates an average to summarize the data. The researcher is using ______.A)measure of central tendency
B) descriptive statistical method
C) intuitive statistical method
D) inferential statistical method
Answer:
Correct option is (B) descriptive statistical method
Step-by-step explanation:
Descriptive statistics branch in statistics deals with the representation of the data using distinct brief coefficients. These coefficients are used as either the representative of the sample or the population.
The descriptive statistics branch is divided into two sub branches:
Measure of central tendencyMeasure of dispersion.The three measures of central tendency are:
Mean (or Average)MedianMode.The measures of dispersion are:
VarianceStandard deviationRangeKurtosisSkewnessThe education researcher computes the average number of hours student study at various local colleges.
The average of a data is the mean value which is the measure of central tendency.
Thus, the researcher is using descriptive statistical method to summarize the data.
Final answer:
The education researcher is using descriptive statistical methods by calculating an average of study hours, which is a measure of central tendency, a fundamental aspect of descriptive statistics.
Explanation:
An education researcher who collects data on how many hours students study at various local colleges and then calculates an average to summarize this data is using descriptive statistical methods. Descriptive statistics involve organizing and summarizing data to provide a clear overview of its characteristics. Examples of descriptive statistics include measures of central tendency (mean, median, mode), which indicate the typical value within a data set, and measures of variability (range, variance, standard deviation), which show how spread out the data points are. The calculation of an average, or mean, falls under the measure of central tendency, making it a key component of descriptive statistics.
For the next two questions, let the null and alternative hypotheses be LaTeX: H_0H 0: LaTeX: \mu=\:8μ = 8 and LaTeX: H_aH a : LaTeX: \mu>8μ > 8. Assume that the population standard deviation LaTeX: \sigmaσ is not known. Becca collects a sample of size LaTeX: n=9n = 9 and computes LaTeX: \overline{x}=11x ¯ = 11 and LaTeX: s=6s = 6. Is LaTeX: \sigmaσ known?
Answer:
[tex]t=\frac{11-8}{\frac{6}{\sqrt{9}}}=1.5[/tex]
[tex]p_v =P(t_{(8)}>1.5)=0.086[/tex]
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, we can conclude that the mean is higher than 8 at 5% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=11[/tex] represent the mean height for the sample
[tex]s=6[/tex] represent the sample standard deviation for the sample
[tex]n=9[/tex] sample size
[tex]\mu_o =8[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test. (assumed)
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 8, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 8[/tex]
Alternative hypothesis:[tex]\mu > 8[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{11-8}{\frac{6}{\sqrt{9}}}=1.5[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=9-1=8[/tex]
Since is a one side upper test the p value would be:
[tex]p_v =P(t_{(8)}>1.5)=0.086[/tex]
Conclusion
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, we can conclude that the mean is higher than 8 at 5% of signficance.
To determine if the population standard deviation is known, we can use the formula for the standard error of the mean (SEM) and use the t-distribution for a sample size of 9.
Explanation:To determine if the population standard deviation LaTeX: \sigma\sigma is known, we can use the formula for the standard error of the mean (SEM):
SE = \frac{s}{\sqrt{n}}
If the sample size is less than or equal to 30, we can use the t-distribution to find the critical value for a given level of significance. If the sample size is greater than 30, we can use the z-distribution. In this case, since the sample size is 9, the t-distribution should be used.
Thus, with a sample size of 9 and the population standard deviation not known, \sigma\sigma is not known.
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the average age of men who had walked on the moon was 39 years, 11months, 15days. Is the value aparameter or a statistic?
Answer:
Parameter
Step-by-step explanation:
We are given the following in the question:
The average age of men who had walked on the moon was 39 years, 11 months, 15 days.
Population and sample:
Population is the collection of all observation for variable of interest or individual of interest.Sample is a subset for population.Parameter and statistic:
Any variable or value describing a population is known as parameter.Any value describing a sample is known as statistic.Population of interest:
men who had walked on the moon
Value:
average age of men who had walked on the moon
Thus, the give value describes a population and hence, it is a parameter.
Binomial Distribution. Research shows that in the U.S. federal courts, about 90% of defendants are found guilty in criminal trials. Suppose we take a random sample of 25 trials. (For this problem it is best to use the Binomial Tables).Based on a proportion of .90, what is the variance of this distribution?
Answer:
The variance of this distribution is 0.0036.
Step-by-step explanation:
The variance of n binomial distribution trials with p proportion is given by the following formula:
[tex]Var(X) = \frac{p(1-p)}{n}[/tex]
In this problem, we have that:
About 90% of defendants are found guilty in criminal trials. This means that [tex]p = 0.9[/tex]
Suppose we take a random sample of 25 trials. This means that [tex]n = 25[/tex]
Based on a proportion of .90, what is the variance of this distribution?
[tex]Var(X) = \frac{p(1-p)}{n}[/tex]
[tex]Var(X) = \frac{0.9*0.1}{25} = 0.0036[/tex]
The variance of this distribution is 0.0036.
what is the solution to the equation A/2= -5
Answer:
A = -10
Step-by-step explanation:
A/2 = -5
Multiply both sides by the denominator of the fraction
We have A/2 x2 = -5 x 2
A = -10
Answer:
-10
Step-by-step explanation:
It is the easiest equation.
A/2= -5
At first, we have to multiply both the sides by 2. Therefore, we can get,
[tex]\frac{A*2}{2}[/tex] = (-5 × 2)
or, A = -10
Therefore, the value of A is -10. It remains negative because we cannot multiply both the sides by -1. If we do that, we cannot determine the constant.
Answer: A = -10
If a customer at a particular grocery store uses coupons, there is a 50% probability that the customer will pay with a debit card. Thirty percent of customers use coupons and 35% of customers pay with debit cards. Given that a customer does not pay with a debit card, the probability that the same customer does not use coupons is ________. A) 0.52 B) 0.60 C) 0.77 D) 0.85
Answer:
A. 0.52
Step-by-step explanation:
Let D be the event that person used Debit card and C b the event that person used coupon.
We have to find the probability of customer does not use coupons given that a customer does not pay with a debit card,
P(C'/D')=P(C')P(D'/C')/[P(C')P(D'/C')+P(D')P(D'/C')]
We are given that P(D)=0.35, P(C)=0.30 and P(D/C)=0.5.
P(D')=1-0.35=0.65
P(C')=1-0.3=0.7
P(D'/C')=0.5.
P(C'/D')=0.7(0.5)/[0.7(0.5)+0.65(0.5)]
P(C'/D')=0.35/[0.35+0.325]
P(C'/D')=0.35/[0.35+0.325]
P(C'/D')=0.35/0.675
P(C'/D')=0.5185=0.52
Thus, the probability of customer does not use coupons given that a customer does not pay with a debit card is 0.52.
The curves r1(t) = 3t, t2, t4 and r2(t) = sin(t), sin(2t), 5t intersect at the origin. Find their angle of intersection, θ, correct to the nearest degree.
To find the angle of intersection (θ) between the curves r1(t) and r2(t) at the origin, we calculate the dot product of their tangent vectors and use the arccosine formula. θ ≈ 79 degrees.
To find the angle of intersection (θ) between the curves r1(t) = (3t, [tex]t^2[/tex], [tex]t^4[/tex]) and r2(t) = (sin(t), sin(2t), 5t) at the origin, we can use the dot product formula for angles between vectors.
First, we need to calculate the tangent vectors at the origin for both curves. The tangent vector for r1(t) is (3, 2t, [tex]4t^3[/tex]), and for r2(t), it is (cos(t), 2cos(2t), 5).
Next, evaluate these vectors at t = 0 (the origin) to get the tangent vectors at the point of intersection: r1'(0) = (3, 0, 0) and r2'(0) = (1, 2, 5).
Now, calculate the dot product of these vectors:
r1'(0) · r2'(0) = (3 × 1) + (0 × 2) + (0 × 5) = 3.
The magnitude of r1'(0) is [tex]\sqrt{ (3^2 + 0^2 + 0^2)[/tex] = 3, and the magnitude of r2'(0) is [tex]\sqrt{(1^2 + 2^2 + 5^2[/tex]) = [tex]\sqrt{(1 + 4 + 25)[/tex] = √30.
Now, use the dot product formula for angles:
cos(θ) = (r1'(0) · r2'(0)) / (|r1'(0)| ×|r2'(0)|)
cos(θ) = 3 / (3 × [tex]\sqrt{30}[/tex]) = 1 / [tex]\sqrt30}[/tex]
Now, find θ:
θ = arc cos(1 / [tex]\sqrt{30[/tex])
Using a calculator, θ ≈ 79 degrees (rounded to the nearest degree).
So, the angle of intersection θ is approximately 79 degrees.
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There are 39 members on the Central High School student government council. When a vote took place on a certain proposal, all of the seniors and none of the freshmen voted for it. Some of the juniors and some of the sophomores voted for the proposal and some voted against it.
If a simple majority of the votes cast is required for the proposal to be adopted, which of the following statements, if true, would enable you to determine whether the proposal was adopted?
a. There are more seniors than freshmen on the council.
b. A majority of the freshmen and a majority of the sophomores voted for the proposal.
c. There are 18 seniors on the council.
d. There are the same number of seniors and freshmen combined as there are sophomores and juniors combined.
e. There are more juniors than sophomores and freshmen combined, and more than 90% of the juniors voted against the proposal.
Answer:
Option a
Step-by-step explanation:
Given that all of the seniors and none of the freshmen voted for it.
Some of the juniors and some of the sophomores voted for the proposal and some voted against it.
If the proposal was to be adopted then a simple of majority of votes should have been cast.
Since all seniors voted for it, and no freshmen number of freshmen has to be less than seniors then only majority would have voted.
Regarding juniors and sophomores we assume some voted and some against thus approximately nullifying their votes.
So the correct option would be
a. There are more seniors than freshmen on the council.
Option b wrong since only some voted for it.
C is wrong because actual seniors were not given.
D is wrong because while seniors voted for sophomores voted against.
e is wrong since sophomores and freshmen nullified each other
Match each shape on the left to every name that describes it on the right. Some answer options on the right will be used more than once.
The student is undertaking an English language assignment designed to strengthen their understanding of vocabulary, word formation, and spelling through a variety of exercises including word scrambles, pattern recognition, spelling reviews, and word categorization.
Explanation:The student appears to be working on a language arts activity related to vocabulary and word structure. The question likely requires them to engage with various linguistic exercises such as word scrambles, identifying patterns in spelling or pronunciation, reviewing correct spellings, and categorizing words. Such tasks are designed to help students learn about word formation, synonyms, antonyms, and the nuances of English spelling.
Word Scrambles and Patterns
For word scrambles, students are expected to rearrange the letters to form meaningful words. In doing this, they might uncover a hidden word that pertains to the lesson's focus. Identifying patterns in words might involve recognizing prefixes, suffixes, or roots that appear consistently across different words.
Reviewing Spelling
The student is also asked to choose the word with the correct spelling. This likely involves comparing similar words and identifying the correctly spelled one, perhaps with the aid of a dictionary for verification.
Categorizing Words
In the task of sorting words into groups, students might have to classify words based on different criteria such as part of speech, phonetic features, or spelling patterns. This reinforces their understanding of language structure and proper spelling conventions.