Final answer:
The probability of correctly choosing five random numbers from balls numbered 1 to 36 is calculated by multiplying the probability of choosing each number correctly, which is (1/36)^5, rounded to 0.0000 to four decimal places.
Explanation:
The student is asking about the probability of choosing five numbers that match five randomly selected balls when the balls are numbered 1 through 36. This is a question of combinatorial probability, where we are interested in the probability of one specific outcome in a set of possibilities.
To solve this, we need to calculate the probability of choosing each ball correctly. The probability of choosing the first number correctly is 1/36, since there is only one correct number out of 36. Likewise, the probability of choosing the second number correctly is also 1/36, and the same logic applies for the third, fourth, and fifth numbers. As these events are all independent (choosing one number does not affect the others), we can find the total probability by multiplying the individual probabilities together:
P(choosing all five numbers correctly) = P(choosing 1st number correctly) × P(choosing 2nd number correctly) × ... × P(choosing 5th number correctly) = (1/36)^5.
The exact value of this probability is quite small, and one would usually leave it as a fraction to avoid rounding errors. However, the instructions specify to round to four decimal places, so let's calculate:
(1/36)^5 = 1/60466176, which is a very small likelihood and as a decimal, it's approximately 0.0000000165, but you can rounded to 0.0000 when expressing it to four decimal places as per instruction.
What is the term associated with scores that are at the extreme ends of the distribution?
Answer:outliers
Step-by-step explanation:
Determine whether the data described are qualitative or quantitative. The maximum speed limit on interstate highways.
a. Qualitative
b. Quantitative
The data of the maximum speed limit on interstate highways is quantitative because it is expressible as a numerical value, which is an amount or quantity.
Explanation:The data described, 'the maximum speed limit on interstate highways', is quantitative. This is because data is quantitative when it is expressible as an amount or quantity. In this case, a speed limit is typically represented as a numerical value in miles or kilometers per hour, which is an amount or quantity. Therefore, the data of the maximum speed limit on interstate highways is quantitative.
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Help ? Pls ???????? Don’t know
Answer:
y=-3x+7
Step-by-step explanation:
y=mx+b is the formula you are going to end up once completed.
This, by given two points:
(4,-5) (3,-2)
[tex]\frac{-2-5}{3-4}[/tex] = Point-Slope Intercept Form
[tex]\frac{3}{-1}[/tex] = After subtracting both from above
-3 = is your slope, decreasing.
Substitute -3 for y=mx+b
y=-3x+b
Use one of the given points (any) to find what b equals.
(4,-5)
x y
-5=-3(4)+b
-5= -12+b
7=b
Final Equation:
y=-3x+7
Answer: y = -3x + 7
Step-by-step explanation:
The formula for calculating equation of line that passes through two points is given by :
[tex]\frac{y-y_{1}}{x-x_{1}}[/tex] = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]x_{1}[/tex] = 4
[tex]x_{2}[/tex] = 3
[tex]y_{1}[/tex] = -5
[tex]y_{2}[/tex] = -2
substituting the values into the equation , we have :
[tex]\frac{y-(-5)}{x-4}[/tex] = [tex]\frac{-2-(-5)}{3-4}[/tex]
[tex]\frac{y+5}{x-4}[/tex] = [tex]\frac{-2+5}{3-4}[/tex]
[tex]\frac{y+5}{x-4}[/tex] = [tex]\frac{3}{-1}[/tex]
[tex]\frac{y+5}{x-4}[/tex] = -3
y + 5 = -3( x - 4 )
y + 5 = -3x + 12
y = -3x + 12 - 5
y = -3x + 7
Compute the line integral of the vector field F=⟨3zy−1,4x,−y⟩F=⟨3zy−1,4x,−y⟩ over the path c(t)=(et,et,t)c(t)=(et,et,t) for −9≤t≤9
Answer:
Step-by-step explanation:
10 granola bars and twelve bottles of water cost $23. 5 granola bars and 4 water bottles of water of water coat $10. how much does one granola and one water bottle cost
Answer:
One granola costs $1.40.
One water bottle costs $0.75.
Step-by-step explanation:
This question can be solved by a simple system of equations.
I am going to say that
x is the cost of each granola bar.
y is the cost of each bottle of water.
The first step is building the system:
10 granola bars and twelve bottles of water cost $23.
This means that:
10x + 12y = 23.
5 granola bars and 4 water bottles of water of water cost $10
This means that:
5x + 4y = 10
So we have to solve the following system of equations:
10x + 12y = 23
5x + 4y = 10
I am going to multiply the second equation by -2, and use the addition method. So:
10x + 12y = 23
-10x - 8y = -20
10x - 10x + 12y - 8y = 23 - 20
4y = 3
y = 0.75
y = 0.75 means that each water bottle costs 75 cents.
5x + 4y = 10
5x = 10 - 4y
5x = 10 - 4*0.75
5x = 7
x = 1.4.
x = 1.4 means that each granola costs $1.40.
Answer:one granola costs $1.4
One bottle of water costs $0.75
Step-by-step explanation:
Let x represent the cost of one granola.
Let y represent the cost of one water bottle.
10 granola bars and twelve bottles of water cost $23. It means that
10x + 12y = 23 - - - - - - - - - - - 1
5 granola bars and 4 water bottles of water of water cost $10. It means that
5x + 4y = 10 - - - - - - - - - - - 2
Multiplying equation 1 by 1 and equation 2 by 2, it becomes
10x + 12y = 23
10x + 8y = 20
Subtracting, it becomes
4y = 3
y = 3/4 = 0.75
Substituting y = 0.75 into equation 1, it becomes
10x + 12 × 0.75 = 23
10x + 9 = 23
10x = 23 - 9 = 14
x = 14/10 = 1.4
Let X be the time in minutes between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ = 1, compute the following: (If necessary, round your answer to three decimal places.)
(a) The expected time between two successive arrivals is minutes.
(b) The standard deviation of the time between successive arrivals is minutes.
Answer:
a) 1
b) 1
Step-by-step explanation:
Data provided in the question:
X = The time in minutes between two successive arrivals
X has an exponential distribution with λ = 1
Now,
a) The expected time between two successive arrivals is minutes i.e
E(X) = [tex]\frac{1}{\lambda}[/tex]
or
E(X) = [tex]\frac{1}{1}[/tex]
or
E(X) = 1
b) The standard deviation of the time between successive arrivals is minutes
i.e
σₓ = [tex]\sqrt{\frac{1}{\lambda^2}}[/tex]
or
σₓ = [tex]\sqrt{\frac{1}{1^2}}[/tex]
or
σₓ = 1
If an angle of 145 is divided into 5 angles with ratios 1: 3: 5: 8: 12, what is the measure
of the 4th largest angle?
Answer:
15 °
Step-by-step explanation:
The five angles are splitted in the ratio 1:3:5:8:12
The 1st biggest angle is at ratio 12
The 2nd biggest angle is at ratio 8
The 3rd biggest angle is at ratio 5
The 4th biggest angle is at ratio 3.
Time solve for the 4th biggest angle, we will find the sum of the ratios:
Therefore, 1 + 3 + 5 + 8 + 12 = 29
Ratio of 4th biggest angle:
= (3)/(29). * 145
= (3 * 45) / 29
= 3 * 5
= 15 °
NOTE: If either of the products is not defined, type UNDEFINED for you answer. If the product is defined, type the dimension in the form mxn with NO spaces in betwee.
Answer:
a) Dimension of AB is DEFINED
b) Dimension of BA is UNDEFINED
Step-by-step explanation:
A matrix is always represented with (mxn) rows and columns. the rows are the elements in the horizontal line while the columns make up the elements in the vertical line. however, there are rules in multiplication of matrix.
To multiply matrix, multiply elements in the rows of the first matrix by the elements in the columns of the second matrix. for example if you're multiplying a 3by2 matrix by a 2by3 matrix, the resulting matrix will be a 3by3 matrix from (mxn) -rows and columns.
from the question, matrix A is a 3by5 (3x5) matrix i.e it has 3rows and 5columns.
matrix B is a 5by2 (5x2) matrix i.e it has 5rows and 2columns. multiplying AB = (3x5) X (5x2), hence the resultant matrix will be a 3by2 (3x2) and this shows that multiplication or dimension of AB is DEFINED.
As for the multiplication of BA = (5x2) X (3x5), from this multiplication, it is not possible as such we can't determine any resultant matrix, this makes multiplication or dimension of BA to be UNDEFINED.
Assume that in the absence of immigration and emigration, the growth of a country's population P(t) satisfies dP/dt = kP for some constant k > 0 (see Equation (1) of Section 1.3). Determine a differential equation governing the growing population P(t) of the country when individuals are allowed to immigrate into the country at a constant rate r > 0. (Use P for P(t).)
dP dt =
What is the differential equation for the population P(t) of the country when individuals are allowed to emigrate at a constant rate r > 0?
dP dt =
Answer:
A) Differential equation for population growth in case of individual immigration is:
[tex]\frac{dP}{dt}=kP+r[/tex]
B) Differential equation for population growth in case of individual emigration is:
[tex]\frac{dP}{dt}=kP-r[/tex]
Step-by-step explanation:
Population growth rate in the absence of immigration and emigration is given as:
[tex]\frac{dP}{dt}=kP--(1)[/tex]
A) When individuals are allowed to immigrate:
Let r be the constant rate of individual immigration given that r >0.
Differential equation for population growth in this case is:
[tex]\frac{dP}{dt}=kP+r[/tex]
B) In case of individual emigration:
Let r be the constant rate of individual emigration given that r >0.
Differential equation for population growth in this case is:
[tex]\frac{dP}{dt}=kP-r[/tex]
Given {(x, f(x)):(x, f(x)) = (-2, 0), (3, 4), (7, 12), (9, -13)}, find the set of ordered pairs (x, 4[f(x)]).
Answer:
{(x, 4f(x)) = (-2,0), (3, 16), (7, 48), (9, -52)}
Step-by-step explanation:
We have the following set of ordered pairs:
(x, f(x))
(-2,0). So f(-2) = 0. That is when x = -2, y = 0.
(3,4). So f(3) = 4. That is, when x = 3, y = 4.
(7,12). So f(7) = 12. That is, when x = 7, y = 12.
(9, -13). So f(9) = -13. That is when x = 9, y = -13.
find the set of ordered pairs (x, 4[f(x)]).
We just need to multiply the values of y by 4. So the set is
{(x, 4f(x)) = (-2,0), (3, 16), (7, 48), (9, -52)}
Answer the following question using the appropriate counting technique, which may be either arrangement with repetition, permutations, or combinations. Be sure to explain why this counting technique applies to the problem.
How many possible birth orders with respect to gender are possible in a family with five children? (For example, BBBGG and BGBGGBBBGG and BGBGG are different orders.)
A. Permutations because the selections come from a single group of items, no item can be selected more than once and the order of the arrangement matters.
B. Arrangements with repetitions because there are r selections from a group of n choices and choices can be repeated.
C. Combinations because the selections come from a single group of items, no item can be selected more than once, and the order of the arrangement does not matter.
D. Arrangements with repetitions because the selections come from a single group of items, and the order of the arrangement matters.
Final answer:
The total number of possible birth orders with respect to gender in a family with five children is 32.
Explanation:
The appropriate counting technique to answer this question is A. permutations. Permutations are used when the selections come from a single group of items, no item can be selected more than once, and the order of the arrangement matters. In this case, the birth orders with respect to gender can be represented by arranging the genders of the children in different orders.
Since there are 5 children, there are 5 positions to fill. The first position can be filled with either a boy or a girl, so there are 2 options. The second position can also be filled with either a boy or a girl, so there are 2 options again. This goes on for all 5 positions.
Therefore, the total number of possible birth orders with respect to gender is 2 * 2 * 2 * 2 * 2 = 32.
It can be helpful to classify a differential equation, so that we can predict the techniques that might help us to find a function which solves the equation. Two classifications are the order of the equation -- (what is the highest number of derivatives involved) and whether or not the equation is linear .
Linearity is important because the structure of the the family of solutions to a linear equation is fairly simple. Linear equations can usually be solved completely and explicitly. Determine whether or not each equation is linear:
1. (1+y2)(d2y/dt2)+t(dy/dt)+y=et
2. t2(d2y/dt2)+t(dy/dt)+2y=sin t
3. (d3y/dt3)+t(dy/dt)+(cos2(t))y=t3
4. y''-y+y2=0
Answer:
Second-order nonlinear ordinary differential equation. Second-order linear ordinary differential equation. Third-order linear ordinary differential equation. Second-order nonlinear ordinary differential equationStep-by-step explanation:
The objective is to determine whether or not each of the following equation is linear:
[tex](1+y^2)\frac{d^2y}{dt^2} + t \frac{dy}{dt} = e^t[/tex] [tex]t^2\frac{d^2y}{dt^2} + t \frac{dy}{dt} +2y= \sin t[/tex] [tex]\frac{d^3y}{dt^3} + t \frac{dy}{dt} + \cos (2t) y = t^3[/tex] [tex]y''-y+y^2 = 0[/tex][tex](1)[/tex]
We can rewrite this equation in the form
[tex](1+y^2)y''(t) +ty'(t) = e^t[/tex].
As we can see, this is an second-order nonlinear differential equation, because of the term [tex]1+y^2[/tex] next to [tex]y''(t)[/tex].
[tex](2)[/tex]
We can rewrite this equation in the form
[tex]t^2y''(t) + t y'(t) +2y= \sin t[/tex].
This is an second-order linear ordinary differential equation.
[tex](3)[/tex]
We can rewrite this equation in the form
[tex]y'''(t)+ t y' + \cos (2t) y = t^3[/tex]
This is an third-order linear ordinary differential equation.
[tex](4)[/tex]
This is an second-order nonlinear ordinary differential equation, because of the term [tex]y^2[/tex].
.... is a process where ordered pairs of points that solve an equation are found. The points are plotted on a grid and then connected with a smooth curve.
Answer: The answer is Point-by-point graphing
Step-by-step explanation:
Point-by-point graphing is a process where ordered pairs of points that solve an equation are found. The points are plotted on a grid and then connected with a smooth curve.
The quantity demanded x for a certain brand of MP3 players is 100 units when the unit price p is set at $80. The quantity demanded is 1100 units when the unit price is $40. Find the demand equation.
p =
Answer:p = (2100 - x)/25
Step-by-step explanation:
According to the law of demand, when the price,p of the brand of MP3 players is is high, the quantity,x of the brand of MP3 players demanded would be low and when the price,p of the brand of MP3 players is is low, the quantity,x of the brand of MP3 players demanded would be high.
To derive the demand equation, we would apply the slope intercept form of equation which is expressed as
y = mx + c
Where
m = slope
c = intercept
The slope, m would be
(y2 - y1)/(x2 - x1)
Slope = (1100 - 100)/(40 - 80) = 1000/-40
Slope = - 25
To find the y intercept, we would substitute m = - 25, y = 1100 and x = 40 into y = mx + c. It becomes
1100 = - 25 × 40 + c
1100 = - 1000 + c
c = 1100 + 1000 = 2100
y = - 25x + 2100
Therefore, the demand equation is
x = - 25p + 2100
25p = 2100 - x
p = (2100 - x)/25
Calls are repeatedly placed to a busy phone line until a connect is achieved. Let c = connect and b = busy, select the correct sample space, S, below: A. S = {b, c} B. S = {bbbbbb, ...} C. S = {cb, cbb, cbbb, ...} D. S = {c, bc, bbc, bbbc, ...}
Answer:
(D)
Step-by-step explanation:
[tex](A)[/tex]
Let's have a look at the first sample space, the set
[tex]S = \{b,c\}[/tex]
As we can see, it consists of two elements. Element [tex]b[/tex] implies that a call was made, but the line was busy, and the element [tex]c[/tex] implies that the call was connected. However, this sample space doesn't correspond to the event in which the calls are repeatedly made until a connect is achieved. The element [tex]c[/tex] is suitable, because it is possible that the line wasn't busy at the moment of calling, but the element [tex]b[/tex] implies that after receiving the busy signal, there were no further attempts to connect to the line.
[tex](B)[/tex]
The second sample space is
[tex]S =\{bbbbbb, \ldots \}[/tex]
It corresponds to the case in which after a few attempts to connect to the line, the person gave up. Therefore, it doesn't correspond to our problem.
[tex](C)[/tex]
The third sample space is
[tex]S = \{ cb, cbb, cbbb, \ldots \}[/tex]
It corresponds to the case in which after one successful attempt, the person tried again for a few more times, and also gave up before connecting again. Therefore, it doesn't correspond to our problem either.
[tex](D)[/tex]
The last sample space is
[tex]S = \{ c, bc, bbc, bbbc, \ldots\}[/tex]
The first element corresponds to the case that the line wasn't busy in the first attempt. The second element, [tex]bc[/tex], corresponds to the case in which the line was busy in the first attempt but it wasn't in the second. The third is the one in which the line was busy for three attempts, but wasn't in the fourth, and so on. As we can see, the common thing about all this elements is that the there were no further attempts after a connection was achieved, which corresponds to our case. Therefore, this is the correct sample space.
The correct option is D. S = {c, bc, bbc, bbbc, ...}.
The sample space for this scenario should include all possible outcomes of the calling process until a connection is made.
Since calls are placed repeatedly until the line is no longer busy, the sequence of events will consist of a series of 'b's followed by a 'c'.
The 'c' represents the successful connection, and it must be the last event in the sequence.
Option A, S = {b, c}, is incorrect because it only includes one busy signal or one connection without considering the sequence of events. Option B, S = {bbbbbb, ...}, is incorrect because it only includes sequences of busy signals without ever achieving a connection. Option C, S = {cb, cbb, cbbb, ...}, is incorrect because it starts with a connection 'c', which is not possible since the connection can only occur after a sequence of busy signals. Option D, S = {c, bc, bbc, bbbc, ...}, is the correct sample space. It represents all possible sequences of busy signals followed by one successful connection.A school psychologist wants to test the effectiveness of a new method of teaching logic. She recruits 200 sixth-grade students and randomly divides them into two groups. Group 1 is taught by means of the new method, while group 2 is taught by traditional methods. The same teacher is assigned to teach both groups. At the end of the year, an achievement test (graded on a scale from 1-10) is administered and the results of the two groups are compared. Complete parts (A) through (D) below.
A. What is the response variable in this experiment?
a. The score on the achievement test.
b. The students' ability in Health.
c. The score of Group 2 on the achievement test.
d. The score of Group 1 on the achievement test.
B. Which of the following explanatory variables is manipulated?
a. Intelligence
b. Teacher
c. Method of teaching
d. Grade level
C. How many levels of the treatment are there?
D. What type of experimental design is this?
a. Survey
b. Randomized block design
c. Completely randomized assignment
d. Matched Pair
In a statistical experiment, a response variable is something that “responds” to the changes made in the experiment. The changes are made to the independent variable (also called the manipulated variable).
For the experiment in question:
The response variable is the score on the achievement test.
The explanatory variable that is manipulated is the method of teaching
There are two levels of treatment ( the new teaching method and the traditional treating method)
The type of experimental design used is: Completely randomized assignment
The school psychology experiment's response variable is the score on the achievement test. The manipulated explanatory variable is the method of teaching. There are two treatment levels - the new method and the traditional method of teaching. The experiment follows a completely randomized assignment design.
Explanation:In this experiment conducted by the school psychologist, various aspects come into play.
A. The response variable in this experiment is option a: The score on the achievement test. The response variable is what you measure in the experiment and what is affected during the experiment. In this case, it's the score that the students get on the achievement test at the end of the year.
B. In this experiment, the explanatory variable that is manipulated is option c: Method of teaching. This is because it is the factor that the experimenter is changing in order to measure the effect on the response variable.
C. There are Two levels of the treatment in this experiment. The two modes of teaching – the new method and the traditional method constitute the two levels of treatment.
D. The type of experimental design used here is option c: Completely randomized assignment. The students were randomly divided into two groups for the purpose of the study which fits under this type of experimental design.
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A music school has budgeted to purchase 3 musical instruments. They plan to purchase a piano costing $4,000, a guitar costing $600, and a drum set costing $750. The mean cost for a piano is $4,500 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $850 with a standard deviation of $100. (Enter your answers to two decimal places.)
1.) How many standard deviations above or below the average piano cost is the piano?
2.) How many standard deviations above or below the average guitar cost is the guitar?
3.) How many standard deviations above or below the average drum set cost is the drum set?
4.) Which cost is the lowest, when compared to other instruments of the same type?
5.) Which cost is the highest when compared to other instruments of the same type?
The piano is 0.2 standard deviations below the average piano cost, while the guitar is 0.5 standard deviations above the average guitar cost. The drum set is 1 standard deviation below the average drum set cost. The drums have the lowest cost compared to other instruments of the same type, and the piano has the highest cost.
Explanation:1.) To find how many standard deviations above or below the average piano cost is the piano, we need to calculate the z-score. The formula for calculating the z-score is
z = (x - mean) / standard deviation
Let's plug in the values for the piano:
z = ($4,000 - $4,500) / $2,500
z = -0.2
So, the piano is 0.2 standard deviations below the average piano cost.
2.) Following the same formula, let's calculate the z-score for the guitar:
z = ($600 - $500) / $200
z = 0.5
Thus, the guitar is 0.5 standard deviations above the average guitar cost.
3.) Using the formula for the z-score again, let's calculate the z-score for the drum set:
z = ($750 - $850) / $100
z = -1
The drum set is 1 standard deviation below the average drum set cost.
4.) The cost of the drums is the lowest when compared to other instruments of the same type.
5.) The cost of the piano is the highest when compared to other instruments of the same type.
Which expression is equivalent to 6 minus (negative 8)?
Answer:
6 - (-8)
is 6 + 8
Step-by-step explanation:
Answer:
b
Step-by-step explanation:
Use De Morgan's law to select the statement that is equivalent to: "It is not true that the employee received a large bonus and has a big office."
a. The employee received a big bonus or has a big office.
b. The employee did not receive a big bonus and does not have a big office.
c. The employee did not receivę a big bonus or does not have a big office.
d. The employee received a big bonus and has a big office
Answer:
c. The employee did not receive a big bonus or does not have a big office.
Step-by-step explanation:
From the original statement, "It is not true that the employee received a large bonus and has a big office.", it can be concluded that the employee either has a big office, received a large bonus or neither. That is, either one, the other, or neither conditions are met.
Applying De Morgan's law:
a. This leaves out the possibility that both conditions are not met
b. This only considers that both conditions are not met
c. This statement means that one of the conditions are not met, which is correct.
d. This assumes both conditions are true, which is incorrect.
The statement equivalent to the negation of both an employee receiving a large bonus and having a big office is that the employee did not receive a big bonus or does not have a big office. Hence the correct
To apply De Morgan's laws to the statement "It is not true that the employee received a large bonus and has a big office," we need to negate the conjunction and change the 'and' to 'or' while negating each of the individual statements within the conjunction. De Morgan's laws tell us that the negation of a conjunction is equivalent to the disjunction of the negations. So the statement becomes:
The employee did not receive a large bonus or does not have a big office.
Therefore, the correct answer is (c) The employee did not receive a big bonus or does not have a big office.
The population of a region is growing exponentially. There were 40 million people in 1980 (when ????=0) and 80 million people in 1990. Find an exponential model for the population (in millions of people) at any time ????, in years after 1980.
Answer:
[tex]P(t) = 40e^{0.06931t}[/tex]
Step-by-step explanation:
The population exponential equation is as follows.
[tex]P(t) = P(0)e^{rt}[/tex]
In which P(t) is the population in t years from now, P(0) is the population in the current year and r(decimal) is the growth rate.e = 2.71 is the Euler number.
Find an exponential model for the population (in millions of people) at any time ????, in years after 1980.
There were 40 million people in 1980 (when ????=0).
This means tht P(0) = 40.
So
[tex]P(t) = P(0)e^{rt}[/tex]
[tex]P(t) = 40e^{rt}[/tex]
80 million people in 1990.
1990 is 10 years after 1980. So P(10) = 80. We use this to find the value of r.
So
[tex]P(t) = 40e^{rt}[/tex]
[tex]80 = 40e^{10r}[/tex]
[tex]e^{10r} = 2[/tex]
Applying ln to both sides, since [tex]\ln{e^{a}} = a[/tex]
[tex]\ln{e^{10r}} = \ln{2}[/tex]
[tex]10r = 0.6931[/tex]
[tex]r = \frac{0.6931}{10}[/tex]
[tex]r = 0.06931[/tex]
So the exponential model for the population is:
[tex]P(t) = 40e^{0.06931t}[/tex]
The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development is the Space-Based Infrared System (SBIRS), which uses satellite imagery to detect and track missiles (Chance, Summer 2005) The probability that an intruding object (e.g., a missile) will be detected on a flight track by SBIRS is .8. Consider a sample of 20 simulated tracks, each with an intruding object. Let x equal the number of these tracks where SBIRS detects the object.
a. Demonstrate that x is (approximately) a binomial random variable.
b. Give the values of p and n for the binomial distribution. .8.20
c. Find P(x = 15), the probability that SBIRS will detect the object on exactly 15 tracks. .17456
d. Find P(x lessthanorequalto 15), the probability that SBIRS will detect the object on at least 15 tracks. .804208
e. Find E(x) and interpret the result. 16
Answer:
a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:
1) Independence between the trials (satisfied)
2) A value of n fixed , for this case is 20 (satisfied)
3) Probability of success p =0.2 fixed (Satisfied)
So then we have all the conditions and we can assume that:
[tex] X \sim Bin(n =20, p=0.8)[/tex]
b) [tex] X \sim Bin(n =20, p=0.8)[/tex]
c) [tex]P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456[/tex]
d) [tex] P(X \geq 15) = P(X=15)+ .....+P(X=20) [/tex]
[tex]P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456[/tex]
[tex]P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218[/tex]
[tex]P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205[/tex]
[tex]P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137[/tex]
[tex]P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058[/tex]
[tex]P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012[/tex]
[tex] P(X\geq 15)=0.804208 [/tex]
e) [tex] E(X) = np = 20*0.8 = 16[/tex]
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".
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:
1) Independence between the trials (satisfied)
2) A value of n fixed , for this case is 20 (satisfied)
3) Probability of success p =0.2 fixed (Satisfied)
So then we have all the conditions and we can assume that:
[tex] X \sim Bin(n =20, p=0.8)[/tex]
Part b
[tex] X \sim Bin(n =20, p=0.8)[/tex]
Part c
For this case we just need to replace into the mass function and we got:
[tex]P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456[/tex]
Part d
For this case we want this probability: [tex] P(X\geq 15) [/tex]
And we can solve this using the complement rule:
[tex] P(X \geq 15) = P(X=15)+ .....+P(X=20) [/tex]
[tex]P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456[/tex]
[tex]P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218[/tex]
[tex]P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205[/tex]
[tex]P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137[/tex]
[tex]P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058[/tex]
[tex]P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012[/tex]
[tex] P(X\geq 15)=0.804208 [/tex]
Part e
The expected value is given by:
[tex] E(X) = np = 20*0.8 = 16[/tex]
A number is equal to 3 times a smaller number. Also, the sum of the smaller number and 4 is the larger number. The situation is graphed on the coordinate plane below, where x represents the smaller number and y represents the larger number. On a coordinate plane, a line goes through (0, 4) and (2, 6) and another line goes through (1, 3) and (2, 6). Which two equations represent the situation? y = one-third x and y = x minus 4 y = one-third x and y = x + 4 y = 3 x and y = x + 4 y = 3 x and y = x minus 4.
Answer: two equations represent the situation are
y = 3x and y = x + 4
Step-by-step explanation:
The smaller number was represented by x and its values on the x coordinate.
The larger number was represented by y and its values on the y coordinate.
The larger number is equal to 3 times a smaller number. This means that
y = 3x
Also, the sum of the smaller number and 4 is the larger number. This means that
y = x + 4
Answer:
two equations represent the situation are
y = 3x and y = x + 4
Step-by-step explanation:
Inferential statistics involves using population data to make inferences about a sample. Group of answer choices True False
Answer:
False. Is the inverse process.
See explanation below.
Step-by-step explanation:
We need to remember two important concepts:
A parameter, is a quantity or value who describe a population desired, for example the population mean [tex]\mu[/tex] or the population standard deviation [tex]\sigma[/tex]
A statistic, is a quantity or value who represent the information of the sample data, for example the sample mean [tex] \bar X[/tex] or the sample deviation [tex] s[/tex]
Based on this we can analyze the statement:
"Inferential statistics involves using population data to make inferences about a sample"
False. Is the inverse process.
If we know the population data then we indeed have parameters and we don't need to do any type of inference in order to estimate these parameters with the statistics.
What we do generally is use the information from the sample in order to obtain statistics representative of the population with the aim to estimate the parameters unknown of the population
a farmer looks over a field and sees 28 heads and 78 feet. some are goats, some are ducks. how many of each animal are there?
Answer:there are 11 goats and 17 ducks.
Step-by-step explanation:
Let x represent the number of goats in the field.
Let y represent the number of ducks in the field.
A farmer looks over a field and sees 28 heads and 78 feet. some are goats, some are ducks. A goat has one head and a duck also has one head. It means that
x + y = 28
A goat has 4 feets and a duck has two feets. It means that
4x + 2y = 78 - - - - - - - - - - - - - 1
Substituting x = 28 - y into equation 1, it becomes
4(28 - y) + 2y = 78
112 - 4y + 2y = 78
- 4y + 2y = 78 - 112
- 2y = - 34
y = - 34/- 2
y = 17
x = 28 - y = 28 - 17
x = 11
To find the number of goats and ducks in the field, set up a system of equations using the given information. Solve the system of equations using elimination or substitution method. After solving, you will find that there are 11 goats and 17 ducks in the field.
Explanation:Let's solve this problem by setting up a system of equations. Let's assume that the number of goats is G and the number of ducks is D.
From the given information, we can set up two equations:
1) G + D = 28 (equation 1) - since the total number of heads is 28
2) 4G + 2D = 78 (equation 2) - since each goat has 4 feet and each duck has 2 feet
We can solve these equations using substitution or elimination. Let's solve by elimination method:
Multiplying equation 1 by 2, we get:
2G + 2D = 56 (equation 3)
Subtracting equation 3 from equation 2, we get:
2G = 22
G = 11
Substituting the value of G into equation 1, we get:
11 + D = 28
D = 17
Therefore, there are 11 goats and 17 ducks in the field.
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Assume that the random variable X is normally distributed, with mean μ = 110 and standard deviation σ = 20. Compute the probability P(X > 126).
Answer:
P(X > 126) = 0.2119
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 110, \sigma = 20[/tex]
P(X > 126) is the 1 subtracted by the pvalue of Z when X = 126. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{126 - 110}{20}[/tex]
[tez]Z = 0.8[/tex]
[tez]Z = 0.8[/tex] has a pvalue of 0.7881.
P(X > 126) = 1 - 0.7881 = 0.2119
The probability P(X > 126) for a normally distributed random variable X with mean μ = 110 and standard deviation σ = 20 is approximately 21.19%.
Explanation:This question is about computing the probability that a normally distributed random variable X with a specified mean and standard deviation exceeds a certain value. In our case, mean μ = 110 and standard deviation σ = 20, and we want to find the probability P(X > 126).
To solve this problem, we first need to compute the Z-score, which is a measure of how many standard deviations a given data point is from the mean, using the formula: Z = (X - μ) / σ. To compute P(X > 126), therefore, we first compute Z = (126 - 110) / 20 = 0.8.
Next, we look up the probability associated with Z = 0.8 in a standard normal distribution table. However, standard normal tables usually give the probability P(Z < z). Since we are looking for P(X > 126), or equivalently P(Z > 0.8), we need to subtract the value we get from the table from 1, since the total probability under a normal curve is one. If we look up Z = 0.8 in the table, we get approximately 0.7881. Hence, P(Z > 0.8) = 1 - P(Z ≤ 0.8) = 1 - 0.7881 = 0.2119 or about 21.19%.
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Viola is a collegiate volleyball player whose protein needs have been determined to be 1.4 g/kg body weight. Viola is 6'2" tall and weighs 170 lbs. Based on this information, Viola should consume approximately _______ g protein daily.
Convert pounds to kilograms.
1 pound = 0.4536 kg.
170 pounds x 0.4536 = 77.11 kg.
Multiply weight by grams of protein per kg.
77.11 kg x 1.4 =107.95 grams. Round answer as needed.
Viola should consume approximately 108.18 grams of protein daily based on her weight and the recommended intake for collegiate volleyball players.
To calculate the amount of protein Viola should consume daily, we need to follow these steps:
1. Convert Viola's weight from pounds to kilograms:
170 lbs / 2.2 = 77.27 kg (rounded to 2 decimal places).
2. Determine her protein needs:
77.27 kg * 1.4 g/kg = 108.18 g of protein per day (rounded to 2 decimal places).
Therefore, Viola should consume approximately 108.18 grams of protein daily based on her weight and the recommended intake for collegiate volleyball players.
Four fair coins are tossed at once. What is the probability of obtaining 3 heads and one tail? 2
Answer:
There are 16 possible outcomes, all equally likely. There’s only one way to get four heads and only one way to get no heads. That leaves 14 ways to get two or three heads.
There’s four ways to get three heads. The first coin can be the tail, the second can be the tail, the third can be the tail, or the fourth can be the tail.
There’s also four ways to get one head. The first coin can be the head, the second can be the head, the third can be the head, or the fourth can be the head.
That leaves six ways left to get exactly two heads. So 6/16 or 3/8.
Step-by-step explanation:
Geologists have an interest in the structure and the history of the earth. A geologist can go back in time by drilling deep into the ground, retrieving a core sample, estimating the ages of the various layers, and examining the composition. A timeline can be built of the entire area from where the core sample was drilled. A geologist may retrieve several core samples to confirm the history of the earth's structure in that sampled area. Mountains, lakes, and unstable ground can easily impede a simple random sampling of a desired geographical area, therefore what is the most realistic sampling method that represents the actual drillings, comparisons, and scientific examinations of several core samples within the same geographical area? Stratified sampling Systematic sampling Cluster sampling Simple random sampling
Answer:
The answer is cluster sampling
Step-by-step explanation:
Going back to the basic types of sampling and how they are acconplished, cluster sampling is accomplished by dividing the population into groups (usually geographical) which in the above question is referred to as the geographical area. These groups are called clusters or blocks. The clusters are randomly selected and each element in the selected clusters are used. Therefore the most realistic sampling method that represents the actual drillings, comparisons, and scientific examinations of several core samples within the same geographical area is cluster sampling.
Final answer:
Stratified sampling is the most realistic sampling method for geologists to represent the actual drillings, comparisons, and scientific examinations of several core samples within the same geographical area.
Explanation:
Stratified sampling is the most realistic sampling method for geologists to represent the actual drillings, comparisons, and scientific examinations of several core samples within the same geographical area. This method involves dividing the population into distinct subgroups based on certain characteristics and then randomly selecting samples from each subgroup.
For example, in geology, geologists may stratify the geographical area based on different rock types or layers to ensure a representative sample of the earth's structure. By using stratified sampling, geologists can account for the diverse composition and ages of various layers within the sampled area, leading to more accurate interpretations of the history and structure of the earth.
Stratified sampling allows geologists to focus their analyses on specific layers or formations within the earth's structure, providing a comprehensive understanding of the geological history of a particular area. This method enhances the reliability and validity of the findings obtained from multiple core samples taken within the same geographical region.
Renee attached a 6ft6in extension cord to her computer’s 3ft8in power cord. What was the total length of the cords?
Answer:
36+24= 60
Step-by-step explanation:
just do 6x6 and add it to 3x8
Answer: 10ft 2in
Step-by-step explanation:
The correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
What fraction of the variability in fuel economy is accounted for by the engine size?
A. 27.5%
B. 63.8%
C. 71.8%
D. 84.1%
Answer: 59.91%.
Step-by-step explanation:
We know that the fraction of the variability in data values accounted by a model is given by [tex]r^2[/tex] , where r is the coefficient of correlation.We are given , that the correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
Then, the fraction of the variability in fuel economy is accounted for by the engine size would be [tex]r^2=( - 0.774)^2=0.599076\approx59.91\%[/tex]
[Multiply 100 to convert a decimal into percent]
Hence, the fraction of the variability in fuel economy is accounted for by the engine size is 59.91%.
None of the options are correct.
To solve such problems we must know about the fraction of the variability in data values or R-squared.
R- SquaredThe fraction by which the variance of the dependent variable is greater than the variance of the errors is known as R-squared. It is called so because it is the square of the correlation between the dependent and independent variables, which is commonly denoted by “r” in a simple regression model.fraction of the variability in data values = (coefficient of correlation)²= r²
The fraction of the variability in fuel economy is accounted for by the engine size is 59.91%.
Given to usThe correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.SolutionVariability in fuel economyAs it is given that the correlation between a car’s engine size and its fuel economy (in mpg) is r = - 0.774.
the variability in fuel economy = r²
= (-0.774)²
= 0.599076
= 59.91%
Hence, the fraction of the variability in fuel economy is accounted for by the engine size is 59.91%.
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