Answer:
If k = −1 then the system has no solutions.
If k = 2 then the system has infinitely many solutions.
The system cannot have unique solution.
Step-by-step explanation:
We have the following system of equations
[tex]x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2[/tex]
The augmented matrix is
[tex]\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right][/tex]
The reduction of this matrix to row-echelon form is outlined below.
[tex]R_2\rightarrow R_2-R_1[/tex]
[tex]\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right][/tex]
[tex]R_3\rightarrow R_3-2R_1[/tex]
[tex]\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right][/tex]
[tex]R_3\rightarrow R_3-R_2[/tex]
[tex]\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right][/tex]
The last row determines, if there are solutions or not. To be consistent, we must have k such that
[tex]k^2-k-2=0[/tex]
[tex]\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2[/tex]
Case k = −1:
[tex]\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right][/tex]
If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.
Case k = 2:
[tex]\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right][/tex]
This gives the infinite many solution.
We use matrix row reduction to determine the values of k that result in no solution, a unique solution, or infinitely many solutions in the system of equations.
Explanation:To determine the values of k for which the system of equations has no solution, a unique solution, or infinitely many solutions, we will use the concept of matrix row reduction. First, let's rewrite the system of equations in augmented matrix form:
[1 -2 3 2 | 0] [2 1 1 -1 | 0] [2 -1 4 -k^2 | 0]
Performing row reduction on this augmented matrix, we can find the values of k where each situation occurs. If there is a row of 0's followed by a non-zero constant (in the rightmost column), then the system has no solution. If the row reduction yields a matrix with a non-zero row followed by zeroes (except for the last row), then the system has infinitely many solutions. Otherwise, the system has a unique solution.
Learn more about Systems of Equations here:https://brainly.com/question/35467992
#SPJ3
Write an equation in slope-intercept form of the line perpendicular to y = - 1 5 x + 1 4 that passes through the point (3, 4).
The equation of the line is [tex]y=\frac{1}{15} x+\frac{19}{5}[/tex]
Explanation:
The equation is [tex]y=-15x+14[/tex] and passes through the point (3,4)
To find the equation of the line in slope intercept form, first we shall find the slope.
This equation is of the slope-intercept form [tex]y=m x+b[/tex], we shall find the value of slope.
Thus, slope m = -15
Since, the line is perpendicular, the negative slope is given by [tex]\frac{-1}{m}[/tex]
Thus, the new slope is [tex]m=\frac{1}{15}[/tex]
Now, we shall find the equation of the line perpendicular to the slope [tex]\frac{1}{15}[/tex] is
[tex]y-y_{1}=\frac{1}{15} \left(x-x_{1}\right)[/tex]
Let us substitute the points (3,4), we have,
[tex]y-4=\frac{1}{15} \left(x-3\right)[/tex]
Muliplying the term within the bracket, we get,
[tex]y-4=\frac{1}{15}x-\frac{1}{5}[/tex]
Adding both sides of the equation by 4, we get,
[tex]y=\frac{1}{15}x-\frac{1}{5}+4[/tex]
Adding the like terms, we have,
[tex]y=\frac{1}{15} x+\frac{19}{5}[/tex]
Thus, the equation in slope intercept form of the line is [tex]y=\frac{1}{15} x+\frac{19}{5}[/tex]
A group of five applicants for a pair of identical jobs consists of three men and two women. The employer is to select two of the five applicants for the jobs. Let S denote the set of all possible outcomes for the employer’s selection. Let A denote the subset of outcomes corresponding to the selection of two men and B the subset corresponding to the selection of at least one woman. List the outcomes in A, B, A ∪ B, A ∩ B, and A ∩ B. (Denote the different men and women by M1, M2, M3 and W1, W2, respectively.)
Answer:
A= {M1,M2},{M2,M3}, {M2,M3}
A U B = S
A n B = 0
A n B'= A
Step-by-step explanation:
A= ( Two males) = { (M1,M2), (M2,M3), (M2,M3)
B= (Atleast one female) = {M1,W1}, {M,W1}, {M3,W1}, {M1,W2} , {M2,W2}, {M3,W2}
Following are the solution to the required function:
Set function:Given that there are five applicants with three men and two women.
Let S be the subset of the set of all possible outcomes,
[tex]\{M_1, M_2\}, \{M_2, M_3\},\{M_3,M_1\},\{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}[/tex]
Let A denote the subset of outcomes corresponding to the selection of two men.
The possible outcomes of A are,
[tex]\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}[/tex]
Let B be the subset corresponding to the selection of at least one woman.
[tex]\{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}[/tex]
Then [tex]\bar{B} =[/tex]
[tex]\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}[/tex]
Find [tex]A\cup B\\\\[/tex]
[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\} \cup \{ \{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}, \{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\[/tex]
Find [tex]A\cap B\\\\[/tex]
[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\} \cap \{ \{W_1,M_1\},\{W_2,M_1\}, \{W_1, M_2\}, \\ \{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\ =\{\phi\}[/tex]
Find [tex]A\cap \bar{B}\\\\[/tex]
[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\cap\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\\\\=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\\\\[/tex]
Learn more about the set function here:
brainly.com/question/25009504
The manager of the local grocery store has determined that, on average, 4 customers use the service desk every half-hour. Assume that the number of customers using the service desk has a Poisson distribution. What is the probability that during a randomly selected half-hour period, exactly 2 customers use the service desk
Answer:
There is a 14.65% probability that during a randomly selected half-hour period, exactly 2 customers use the service desk.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
[tex]e = 2.71828[/tex] is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
The manager of the local grocery store has determined that, on average, 4 customers use the service desk every half-hour.
This means that [tex]\mu = 4[/tex]
What is the probability that during a randomly selected half-hour period, exactly 2 customers use the service desk?
This is P(X = 2). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-4}*(4)^{2}}{(2)!} = 0.1465[/tex]
There is a 14.65% probability that during a randomly selected half-hour period, exactly 2 customers use the service desk.
A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is used to measure the glucose concentration in the blood sample five times. The measured glucose concentrations are 104.5 , 96.2 , 102.2 , 98.3 , and 101.8 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor. A. 104.5 mg/dL absolute error = 2.5 mg / dL relative error = 0.025 B. 96.2 mg/dL absolute error = −5.8 mg / dL relative error = 0.057 C. 102.2 mg/dL absolute error = 0.2 mg / dL relative error = 0.020 D. 98.3 mg/dL absolute error = −3.7 mg / dL relative error = 0.036 E. 101.8 mg/dL absolute error = −0.2 mg / dL relative error =
Answer:
The Absolute Error is the difference between the actual and measured value.
[tex]Absolute \:error = |Actual \:value - Measured \:value|[/tex]
The Relative Error is the Absolute Error divided by the actual measurement.
[tex]Relative \:error = \frac{Absolute \:error}{Actual \:value}[/tex]
We know that the actual value is 102.0 mg/dL.
To find the absolute error and relative error for each measurement made by the glucose monitor you must use the above definitions.
a) For a concentration of 104.5 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102-104.5\right|\\Absolute \:error =\left|-2.5\right|\\Absolute \:error =2.5[/tex]
[tex]Relative \:error = \frac{2.5}{102.0}=0.0245[/tex]
b) For a concentration of 96.2 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-96.2\right|\\Absolute \:error =\left|5.8\right|\\Absolute \:error =5.8[/tex]
[tex]Relative \:error = \frac{5.8}{102.0}=0.0569[/tex]
c) For a concentration of 102.2 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-102.2\right|\\Absolute \:error =\left|-0.2\right|\\Absolute \:error =0.2[/tex]
[tex]Relative \:error = \frac{0.2}{102.0}=0.00196[/tex]
d) For a concentration of 98.3 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-98.3\right|\\Absolute \:error =\left|3.7\right|\\Absolute \:error =3.7[/tex]
[tex]Relative \:error = \frac{3.7}{102.0}=0.0363[/tex]
e) For a concentration of 101.8 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-101.8\right|\\Absolute \:error =\left|0.2\right|\\Absolute \:error =0.2[/tex]
[tex]Relative \:error = \frac{0.2}{102.0}=0.00196[/tex]
A typical incoming telephone call to your catalog sales force results in a mean order of $28.63 with a standard deviation of $13.91. You may assume that orders are received independently of one another. a. Based only on this information, can you find the probability that a single incoming call will result in an order of more than $40? Why or why not? b. An operator is expected to handle 110 incoming calls tomorrow. Find the mean and standard devi- ation of the resulting total order. c. What is the approximate probability distribution of the total order to be received by the operator in part b tomorrow? How do you know? d. Find the (approximate) probability that the operator in part b will generate a total order of more than $3,300 tomorrow. e. Find the (approximate) probability that the operator in part b will generate an average order between $27 and $29 tomorrow.
Answer:
Step-by-step explanation:
Hello!
The study variable for this exercise is:
X: Price of an order of a sales catalog placed per telephone.
You don't have information about the distribution of this variable, but you know that the mean is μ= $28.63 and the standard deviation is σ= $13.91
a. You need to calculate the probability of a single incoming call resulting in an order of more than $40, symbolically P(X>$40). To be able to calculate this probability you need to know what the distribution of the variable is. Without knowing the form of the distribution, i.e. the probability density function, you cannot tell what is the asked probability.
b. If the operator is expected to handle 110 calls in a day (tomorrow) this means that you have a sample of n= 110 calls, and in each call, you are going to take the information of the order placed by the customer.
Note:
If X₁, X₂, ..., Xₙ be the n random variables that constitute a sample, then any function of type θ = î (X₁, X₂, ..., Xₙ) that depends solely on the n variables and does not contain any parameters known, it is called the estimator of the parameter.
When the function i (.) It is applied to the set of the n numerical values of the respective random variables, a numerical value is generated, called parameter estimate θ.
This follows the concepts:
1) The function i (.) It is a function of random variables, so it is also a random variable, that is to say, that every estimator is a random variable.
2) From the above, it follows that Î has its probability distribution and therefore mathematical hope, E (î), and variance, V (î).
And:
The central limit theorem states that if a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
This means that if you have the study variable X with a certain distribution, then it's the sample mean X[bar], that is also an aleatory variable, will have the same distribution as it's origin variable. On the other hand, if the distribution of the said variable is unknown, so will be the distribution of its sample mean, but if the sample is large enough, then you can apply the central limit theorem and approximate the distribution of the sample mean to normal, symbolically:
X~?(μ;δ²) and n ≥ 30 then X[bar]≈N(μ;δ²/n )
The mean of said approximation is the same as the mean of the variable of origin.
μ= $28.63
And the standard deviation will be the same as the original variable but affected by the sample size:
δ/√n = $13.91/110= $0.126 ≅ $0.13
c. X[bar]≈N(μ;δ²/n )
d. Using the aproximation you can calculate the asked probabilities with the standard normal:
P(X[bar] > $3,300) = P(Z > [tex]\frac{3.300- 28.63}{13.91/\sqrt{110} }[/tex]) = P(Z > -19.098)= 0
e.
P(27 < X[bat] < 29) = P(X<29) - P(X<27) = P(Z<[tex]\frac{29-28.63}{0.13}[/tex]) - P(Z<[tex]\frac{27-28.63}{0.13}[/tex])
P(Z<0.278) - P(Z<-1.229)= 0.609 - 0.110= 0.499
I hope it helps!
The table shows the functions representing the height and base of a triangle for different values of x The area of the triangle when x= 2 is 14. Which equation can be used to represent the area of the triangle, A(x)?
Answer: option 2 is the correct answer.
Step-by-step explanation:
When x = 2, the area is 14.
It means that height = 2² + 3 = 7
It means that base = 2² = 4
Area = 1/2 × 7 × 4 = 14
Therefore, it is a right angle triangle, the formula for determining the area of the triangle is expressed as
Area = 1/2bh
Where
b represents the base of the triangle.
h represents the height of the triangle.
Since height is f(x) = x² + 3 and base is g(x) = 2x,
The equation that can be used to represent the area is
A(x) = 0.5(f.g)(x)
Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=1/x, y=0, x=1, and x=4 about the line y=−1.
The volume of the solid obtained by rotating the region about the line y=-1 is approximately 24.27π cubic units.
To find the volume of the solid formed by rotating the region about the line y=-1, we can use the washer method. Here's how:
1. Identify the washers:
Imagine rotating the shaded region between the curves y=1/x, y=0, x=1, and x=4 about the line y=-1. This will create a series of washers stacked on top of each other. Each washer will have a hole in the middle due to the rotation about the line y=-1.
2. Define the parameters for each washer:
The inner radius (r₁) of each washer is the distance from the line y=-1 to the curve y=1/x. This can be expressed as 1 + 1/x.
The outer radius (r₂) of each washer is the distance from the line y=-1 to the x-axis (y=0). This is simply 1.
The thickness (dx) of each washer is the infinitesimal change in x.
3. Set up the integral:
Since we are rotating about a horizontal axis, the volume of each washer can be calculated using the formula for the volume of a washer:
dV = π[(r₂)² - (r₁)²] dx
The total volume of the solid is then the sum of the volumes of all the washers, which can be represented by a definite integral:
V = ∫⁴ π[(1)² - (1 + 1/x)²] dx
4. Evaluate the integral:
This integral can be solved using the power rule and the sum rule for integration. Simplifying the result, you will get:
V = π[8x - 3ln(x + 1)] |⁴
Finally, evaluate the integral at the limits of integration (x = 1 and x = 4) and subtract the results to find the total volume of the solid:
V = π[(32 - 3ln(5)) - (8 - 3ln(2))] ≈ 24.27 π cubic units
Therefore, the volume of the solid obtained by rotating the region about the line y=-1 is approximately 24.27π cubic units.
Use the graph to fill in the blank with the correct number. f(−2) = ________ X, Y graph. Plotted points negative 3, 0, negative 2, 2, 0, 1, and 1, negative 2. Numerical Answers Expected! Answer for Blank 1:
The given points
[tex](-3, 0)[/tex]
[tex](-2, 2)[/tex]
[tex](0, 1)[/tex]
[tex](1, -2)[/tex]
Imply that
[tex]f(-3)=0[/tex]
[tex]f(-2)=2[/tex]
[tex]f(0)=1[/tex]
[tex]f(1)=-2[/tex]
Answer:
f(2) = -1.
Step-by-step explanation:
There are three workstations available having steady-state probabilities of 0.99, 0.95, 0.85 of being available on demand. What is the probability that at least two of the three will be available at any given time?
Answer:
99.065% probability that at least two of the three will be available at any given time.
Step-by-step explanation:
We have these following probabilities:
99% probability of the first workstation being available
95% probability of the second workstation being available
85% probability of the third workstation being avaiable
Two being available:
We can have three outcomes
First and second available, third not. So
0.99*0.95*0.15 = 0.141075
First and third available, second not. So
0.99*0.05*0.85 = 0.042075
Second and third available, first not. So
0.01*0.95*0.85 = 0.008075
Adding them all
P(2) = 0.141075 + 0.042075 + 0.008075 = 0.191225
Three being available:
P(3) = 0.99*0.95*0.85 = 0.799425
What is the probability that at least two of the three will be available at any given time?
P = P(2) + P(3) = 0.191225 + 0.799425 = 0.99065
99.065% probability that at least two of the three will be available at any given time.
The probability that at least two out of the three workstations are available is 0.967.
Explanation:We can find the probability that at least two out of the three workstations are available using the concept of complementary events. The probability of at least two workstations being available is equal to 1 minus the probability of none or only one workstation being available.
Let's calculate the probability of none or only one workstation being available:
Probability of none being available: 0.01 * 0.05 * 0.15 = 0.00075
Probability of only one being available: (0.99 * 0.05 * 0.15) + (0.01 * 0.95 * 0.15) + (0.01 * 0.05 * 0.85) = 0.03225
Now, subtracting this from 1:
1 - (0.00075 + 0.03225) = 0.967
Therefore, the probability that at least two out of the three workstations are available at any given time is 0.967.
Learn more about probability here:https://brainly.com/question/32117953
#SPJ3
A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (In other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 0.5 meters, its length is 5 meters, and its top is 2 meters under the ground, find the total amount of work needed to pump the gasoline out of the tank. (The density of gasoline is 673 kilograms per cubic meter; use g=9.8 m/s2.)
Final answer:
The work needed to pump the gasoline out of the underground tank is approximately 51.7 kJ, calculated using the density of gasoline, the volume of the cylindrical tank, and the gravitational energy required to lift the gasoline 2 meters up to ground level.
Explanation:
The total amount of work needed to pump the gasoline out of the tank can be determined using the concepts of physics specifically mechanical work and fluid dynamics. We know the density of gasoline is 673 kg/m3, the gravitational acceleration is 9.8 m/s2, the cylinder's radius is 0.5 meters, its length is 5 meters, and the top of the cylinder is 2 meters below ground level. The total volume of the cylinder is given by the formula for the volume of a cylinder V = πr2h, where r is the radius and h is the length of the cylinder. In this case, V = π(0.5)2(5) ≈ 3.927 m3. The total mass m of the gasoline can be calculated by multiplying the density of gasoline by the volume, m = density × volume = 673 kg/m3 × 3.927 m3 ≈ 2643.871 kg.
Since the gas tank is underground, the work done to lift the gasoline to ground level is W = mgh, where m is the mass of the gasoline, g is acceleration due to gravity, and h is the height the gasoline is lifted. We must lift the gasoline 2 meters to reach ground level, so the work done is W = 2643.871 kg × 9.8 m/s2 × 2 m ≈ 51738.7856 J or 51.7 kJ (since 1 J = 1 kg·m2/s2). Thus, the work required to pump the gasoline out of the tank would be approximately 51.7 kJ.
A whistle is made of a square tube with a notch cut in its edge, into which a baffle is brazed. Determine the dimensions d and θ for the baffle. Take b = 6.5 cm.
Answer:
d = 7.51 cm
θ = 60°
Step-by-step explanation:
The baffle used in the notch of a whistle is a triangular baffle (Isosceles triangle OAB).
For the isosceles triangle, the sides with equal dimension are OA and OB which is represented with d
d = the vertical component of the side of a triangle
It is given by
6.5cm = d sin60
d = 6.5/sin60
d = 7.505553499465134
d = 7.51 cm -------- Approximated
To calculate angle θ
Angle on a straight line is = 180
So, 60 + 60 + θ = 180
θ= 180 - 120
θ = 60°
(See attachments below)
There are five sales associates at Mid-Motors Ford. The five associates and the number of cars they sold last week are: Sales Associate Cars Sold Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6
a. How many different samples of size 2 are possible?
b. List all possible samples of size 2, and compute the mean of each sample.
c. Compare the mean of the sampling distribution of the sample mean with that of the
population.
Answer:
a) There are 10 different samples of size 2.
b) See the explanation section
c) See the explanation section
Step-by-step explanation:
a) We need to select a sample of size 2 from the given population of size 5. We use combination to get the number of difference sample.
[tex]\{ {{5} \atop {2}} \} = \frac{5!}{2!(5-2)!} \\= \frac{5!}{2!3!} \\= \frac{120}{2*6} \\= \frac{120}{12} \\=10[/tex]
b) Possible sample of size 2:
Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6
Peter Hankish and Connie Stallter ( Mean = (8 + 6)/2 = 14/2 = 7)Peter Hankish and Juan Lopez (Mean = (8 + 4)/2 = 12/2 = 6)Peter Hankish and Ted Barnes (Mean = (8 + 10)/2 = 18/2 = 9)Peter Hankish and Peggy Chu (Mean = (8 + 6)/2 = 14/2 = 7)Connie Stallter and Juan Lopez (Mean = (6 + 4)/2 = 10/2 = 5)Connie Stallter and Ted Barnes (Mean = (6 + 10)/2 = 16/2 = 8)Connie Stallter and PeggyChu (Mean = (6 + 6)/2 = 12/2 = 6)Juan Lopez and Ted Barnes (Mean = (4 + 10)/2 = 14/2 = 7)Juan Lopez and Peggy Chu (Mean = (4 + 6)/2 = 10/2 = 5)Ted Barnes and Peggy Chu (Mean = (10 + 6)/2 = 16/2 = 8)c) The mean of the population is:
[tex]mean = \frac{(8+6+4+10+6)}{5} \\= \frac{34}{5} \\= 6.8[/tex]
Comparing the mean of the population and the sample; we can say that most of the 2-size sample have their mean higher than that of the population sample. And the variation with the mean is not much. Some sample have their mean greater than population mean, while some sample have their mean greater than the population mean.
This question is based on the statistics. Therefore, the answers of all the questions are explained below.
Given:
There are five sales associates at Mid-Motors Ford. Sales Associate Cars Sold Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6.
(a) We have to find different samples of size 2 are possible.
Thus, we have to select sample of size 2 from given population of size 5.
So, by using combination,
[tex]5_{c_2} = \dfrac{5!}{2! (5-2)!} =\dfrac{120}{12} = 10[/tex]
Thus, 10 different samples of size 2 are possible.
(b) We have to find list all possible samples of size 2, and compute the mean of each sample.
Peter Hankish 8 ,Connie Stallter 6, Juan Lopez 4, Ted Barnes 10, Peggy Chu 6.
Peter Hankish and Connie Stallter ( Mean = [tex]\dfrac{8+6}{2} = 7[/tex] Peter Hankish and Juan Lopez (Mean =[tex]\dfrac{8+4}{2} = 6[/tex] Peter Hankish and Ted Barnes (Mean = [tex]\dfrac{8+10}{2} = 9[/tex] Peter Hankish and Peggy Chu (Mean = [tex]\dfrac{8+6}{2} = 7[/tex] Connie Stallter and Juan Lopez (Mean = [tex]\dfrac{4+6}{2} =5[/tex] Connie Stallter and Ted Barnes (Mean = [tex]\dfrac{10+6}{2} = 8[/tex] Connie Stallter and PeggyChu (Mean = [tex]\dfrac{6+6}{2} = 6[/tex] Juan Lopez and Ted Barnes (Mean = [tex]\dfrac{4+10}{2} = 7[/tex] Juan Lopez and Peggy Chu (Mean = [tex]\dfrac{4+6}{2} = 5[/tex] Ted Barnes and Peggy Chu (Mean = [tex]\dfrac{10+6}{2} = 8[/tex]
(c) The mean of the population is:
[tex]Mean = \dfrac{8+6+4+10+6}{5}\\\\Mean = \dfrac{34}{5}\\\\Mean = 6.8[/tex]
For more details, prefer this link;
https://brainly.com/question/19146184
Let p be the statement "There is no pollution in New Jersey." The statement "The whole world is polluted" is the negation of the statement "There is no pollution in New Jersey." Is the above statement true?
Answer:
No. It is not true.
Step-by-step explanation:
p = There is no pollution in New Jersey
¬p = There is pollution in New Jersey.
Removing the 'no' in the statement yield the negation.
The given statement "The whole world is polluted" is not correct because it has gone beyond it context/domain. Statement p is about New Jersey, so the negation should be about New Jersey.
The negation can also be written as: "New Jersey is polluted".
In logic, the negation of a statement is its direct contradiction. In this case, 'The whole world is polluted' is not the negation of 'There is no pollution in New Jersey'.
Explanation:In the field of logic and reasoning, the negation of a statement is a statement which contradicts or denies the original one. If the statement 'p' is 'There is no pollution in New Jersey', its negation would be 'There is pollution in New Jersey' because it is the exact opposite of the original statement. However, the statement 'The whole world is polluted' is not the direct negation of 'There is no pollution in New Jersey'. While it implies that there is pollution in New Jersey, it goes beyond that by including every other part of the world.
Learn more about Logic and Reasoning here:https://brainly.com/question/22813859
#SPJ3
Find g prime left parenthesis x right parenthesis for the given function. Then find g prime left parenthesis negative 3 right parenthesis, g prime left parenthesis 0 right parenthesis, and g prime left parenthesis 2 right parenthesis. g left parenthesis x right parenthesis equals StartRoot 4 x EndRoot
Answer: For x = 0, -3, our expression is undefined and for x = 2, we have 0.707.
Step-by-step explanation: From the question, we have
g(x) = \sqrt{4x}
Simplifying the right-hand side, we have:
g(x) = 2x^{1/2}
Differentiating with respect to $x$ using the second principle, we have,
g'(x) = 2 * \frac{1}{2} * x^{\frac{1}{2} - 1}
= x^{-1/2}
So from the indical laws, g'(x) =x^\frac{-1}{2} = 1/\sqrt{x}
For values of g'(x) when x = -3, we have
g(x) = 1/\sqrt{-3}
g(x) is undefined for values of x when x is -3 since the square root of a negative number is not defined. However, using complex solution we have
g(x) = 1/\sqrt{-3}
But \sqrt{-1} = i; then \sqrt{-3} = \sqrt(-1 * 3)
This is same as \sqrt(-1) * \sqrt(3)
And then we have 1.732i
For x = 2, we have
g’(2) = 1/\sqrt (x)
= 1/\sqrt(2) = 0.707
For x = 0, we have
g’(0) = 1/\sqrt (0)
= 1/0
Here again for x = 0, our expression is undefined.
If the New England Patriots get home-field advantage, you believe there is a 60% probability they will make it to the Super Bowl. If not, this probability is only 30%. Assuming a 70% probability that the Patriots get home-field advantage, what is the probability they will make it to the Super Bowl?
Answer:
There is a 51% probability they will make it to the super bowl.
Step-by-step explanation:
We have these following probabilities:
A 70% probability that the Patriots get homefield advantage.
A 30% probability that the Patriots does not get homefield advantage.
If they get homefield advantage, a 60% probability of making the Super Bowl.
If they do not get homefield advantage, a 30% probability of making the Super Bowl.
What is the probability they will make it to the Super Bowl?
This is 60% of 70%(when they get homefield advantage and make the super bowl), and 30% of 30%(no homefield, no super bowl). So
[tex]P = 0.6*0.7 + 0.3*0.3[/tex]
[tex]P = 0.51[/tex]
There is a 51% probability they will make it to the super bowl.
Determine the truth values of these statements:
(a) The product of x2 and x3 is x6 for any real number x.
(b) x2 > 0 for any real number x.
(c) The number 315 − 8 is even.
(d) The sum of two odd integers is even.
Final answer:
The product of x^2 and x^3 equals x^6 for any real number x is true. The statement that x^2 > 0 for any real number x is false, as it should state x^2 >= 0. 315 - 8 being even is false since it results in an odd number. The sum of two odd integers being even is true.
Explanation:
Determine the truth values of these statements:
The product of x2 and x3 is x6 for any real number x. This statement is true because according to the laws of exponents, when multiplying powers with the same base, you add the exponents. Therefore, x2 * x3 = x2+3 = x6.
x2 > 0 for any real number x. This statement is false because when x = 0, x2 = 0, not greater than 0. The correct statement should be x2 >= 0 for any real number x.
The number 315 − 8 is even. This statement is true because 315 − 8 = 307, and any number ending in 7 is odd. Therefore, the statement is false.
The sum of two odd integers is even. This statement is true because when you add two odd numbers, the sum is always even. For example, 3 + 5 = 8.
The newly elected president needs to decide the remaining 7 spots available in the cabinet he/she is appointing. If there are 13 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?
Answer: 8648640 ways
Step-by-step explanation:
Number of positions = 7
Number of eligible candidates = 13
This can be done by solving the question using the combination Formula for selection in which we use the combination formula to choose 7 candidates amomg the possible 13.
The combination Formula is denoted as:
nCr = n! / (n-r)! * r!
Where n = total number of possible options.
r = number of options to be selected.
Hence, selecting 7 candidates from 13 becomes:
13C7 = 13! / (13-7)! * 7!
13C7 = 1716.
Considering the order they can come in, they can come in 7! Orders. We multiply this order by the earlier answer we calculated. This give: 1716 * 7! = 8648640
Binomial Distribution. Surveys repeatedly show that about 40% of adults in the U.S. indicate that if they only had one child, they would prefer it to be a boy. Suppose we took a random sample of 15 adults and the number who indicated they preferred a boy was 8. This would be considered a rare event because the probability of 8 or more is so low.
True/False
Answer:
False
Step-by-step explanation:
We are given the following information:
We treat adult who prefer one child to be a boy as a success.
P(prefer one child to be a boy) = 40% = 0.4
Then the number of adults follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 15 and x = 8
We have to evaluate:
[tex]P(x \geq 8)\\= P(x = 8) + P(x = 9)+...+ P(x = 14) + P(x =15)\\\\= \binom{15}{8}(0.4)^{8}(1-0.4)^{7} +\binom{15}{9}(0.4)^{9}(1-0.4)^{6}+...\\\\...+\binom{15}{14}(0.4)^{14}(1-0.4)^{1} +\binom{15}{8}(0.4)^{15}(1-0.4)^{0}\\\\= 0.2131[/tex]
Since the probability of 8 or more is 0.2131 is not very small, thus, it is not a rare event.
Thus, the given statement is false.
Suppose we know that the functions r and s are are everywhere differentiable and that u(3)=0. Suppose we also know that for 1 ≤ x ≤ 3, the area between the x-axis and the non negative functions h(x)=u(x)dv/dx is 15, and that on the same interval, the area between the x-axis and the non negative function k(x) = v(x)du/dx is 20. Determine u(1)v(1).
Integrating by parts, we have
[tex]\displaystyle\int_1^3\underbrace{u(x)\dfrac{\mathrm dv}{\mathrm dx}}_{h(x)}\,\mathrm dx=u(3)v(3)-u(1)v(1)-\int_1^3\underbrace{v(x)\dfrac{\mathrm du}{\mathrm dx}}_{k(x)}\,\mathrm dx[/tex]
We're given [tex]u(3)=0[/tex] and [tex]\int_1^3h(x)\,\mathrm dx=15[/tex] and [tex]\int_1^3k(x)\,\mathrm dx=20[/tex]. So we have
[tex]15=-u(1)v(1)-20\implies\boxed{u(1)v(1)=-35}[/tex]
In a multicriteria decision problem: a. It is impossible to select a single decision alternative. b. The decision maker must evaluate each alternative with respect to each criterion. c. Successive decisions must be made over time. d. Each of these choices are true.
Answer:
in a multicriteria decision problem the decision maker must evaluate each alternative with respect to each criterion.
Step-by-step explanation:
Multiple-criteria decision-making (MCDM) is a sub-discipline of operations research that explicitly evaluates multiple conflicting criteria in decision making (both in daily life and in settings such as business, government and medicine e.t.c)
In multicriteria decision problems, each alternative must be evaluated separately based on each criterion. It's not impossible to choose a single decision path, but it can be complex and might require multiple stages of decision-making over time. All statements in the question are relatively true.
Explanation:In multicriteria decision problems, various factors or criteria come into play. It's important to note that all the provided statements have some truth in them. From a decision-maker's perspective, one must evaluate each alternative against each criterion. This ensures that the pros and cons of each option are meticulously considered. Contrary to option a, it's not impossible to select a single decision alternative. However, the decision-making process could become complicated due to differing priorities and preferences, ultimately delaying the selection of a single alternative. On the other hand, statement c is partially true; depending on the complexity and scale of the decision, it might require several rounds of decision-making over time. Hence, all of these statements are indeed relatively accurate.
Learn more about multicriteria decision here:https://brainly.com/question/36597508
#SPJ3
The half-life of Sodium-24 is 15 hours. If you start with 600 grams of sodium-24 how much would be left after 4 days? Which of the following equations could you use to solve for the amount of grams left after 4 days?
Step-by-step explanation:
A = A₀ ½^(t / T)
where A is the amount left, A₀ is the original amount, t is time, and T is the half life.
4 days is 96 hours, so the amount left is:
A = 600 ½^(96 / 15)
To find how much Sodium-24 will be left after 4 days, we calculate the number of half-lives in 96 hours (which is 6.4) and use the formula remaining amount = initial amount × (1/2)n, resulting in approximately 8.79 grams remaining.
Explanation:The question asks how much Sodium-24 would be left after 4 days, given its half-life of 15 hours. Firstly, we convert 4 days into hours, which is 4 days × 24 hours/day = 96 hours. Next, we divide 96 hours by the half-life of Sodium-24, which is 15 hours, to find out how many half-lives have passed. The result is 96/15 = 6.4 half-lives.
Using the half-life decay formula, which is remaining amount = initial amount × (1/2)n, where 'n' is the number of half-lives, we can substitute the given values to find the number of grams left.
Remaining sodium-24 = 600 g × (1/2)6.4 ≈ 600 g × 0.01465 ≈ 8.79 g
Therefore, approximately 8.79 grams of Sodium-24 would remain after 4 days.
A ball is kicked upward with an initial velocity of 32 feet per second. The ball's height, h (in feet), from the ground is modeled by where t is measured in seconds. How much time does the ball take to reach its highest point? What is its height at this point?
Answer:
1. t = 0.995 s
2. h = 15.92 ft
Step-by-step explanation:
First we have to look at the following formula
Vf = Vo + gt
then we work it to clear what we want
Vo + gt = Vf
gt = Vf - Vo
t = (Vf-Vo)/g
Now we have to complete the formula with the real data
Vo = 32 ft/s as the statement says
Vf = 0 because when it reaches its maximum point it will stop before starting to lower
g = -32,16 ft/s² it is a known constant, that we use it with the negative sign because it is in the opposite direction to ours
t = (0 ft/s - 32 ft/s) / -32,16 ft/s²
we solve and ...
t = 0.995 s
Now we will implement this result in the following formula to get the height at that time
h = (Vo - Vf) *t /2
h = (32 ft/s - 0 ft/s) * 0.995 s / 2
h = 32 ft/s * 0.995 s/2
h = 31.84 ft / 2
h = 15.92 ft
The ball takes 2 seconds to reach its highest point and the height at this point is 32 feet.
Explanation:To find the time it takes for the ball to reach its highest point, we can use the equation h = -16t^2 + 32t, where h is the height and t is the time. The maximum height is reached when the ball is at its highest point, which occurs when the ball is not moving vertically.
At this point, the velocity of the ball is 0, so the equation v = -16t + 32 can be used to find the time. Setting v = 0 and solving for t, we get t = 2 seconds.
Substituting this time value into the equation for height, we can find the height at this point. h = -16(2)^2 + 32(2) = 32 feet.
Therefore, the ball takes 2 seconds to reach its highest point and the height at this point is 32 feet.
A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately normally distributed with variance of 1000 (psi)2. A random sample of 12 specimens has a mean compressive strength of 3250 psi. (a) Construct a 95% confidence interval on mean compressive strength. (b) Suppose that the manufacturer of the concrete claims the average compressive strength is 3270 psi. Based on your answer in part (a), what would you say about this claim
Answer:
a) [tex]3250-1.96\frac{31.623}{\sqrt{18}}=3232.108[/tex]
[tex]3250+1.96\frac{31.623}{\sqrt{18}}=3267.892[/tex]
So on this case the 95% confidence interval would be given by (3232.108;3267.892)
b) For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X= 3250[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma^2 =1000[/tex]represent the sample standard variance
[tex] s = \sqrt{1000}[/tex] represent the sample deviation
n=12 represent the sample size
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]3250-1.96\frac{31.623}{\sqrt{18}}=3232.108[/tex]
[tex]3250+1.96\frac{31.623}{\sqrt{18}}=3267.892[/tex]
So on this case the 95% confidence interval would be given by (3232.108;3267.892)
Part b
For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.
A house worth $180,000 has a coinsurance clause of 75 percent. The owners insure the property for $101,250. They then have a loss that results in a $50,000 claim. They will receive $ ____.00 from insurance.
Answer:
$37,500
Step-by-step explanation:
We have been given that a house worth $180,000 has a coinsurance clause of 75 percent. The owners insure the property for $101,250. They then have a loss that results in a $50,000 claim.
We will use loss settlement formula to solve our given problem.
[tex]\text{Loss settlement}=\frac{\text{Loss}\times\text{Limit of insurance}}{\text{Actual cash value}\times \text{Coinsurance}\%}[/tex]
Upon substituting our given values, we will get:
[tex]\text{Loss settlement}=\frac{\$50,000\times\$101,250}{\$180,000\times 75\%}[/tex]
[tex]\text{Loss settlement}=\frac{\$50,000\times\$101,250}{\$180,000\times 0.75}[/tex]
[tex]\text{Loss settlement}=\frac{\$5,062,500,000}{\$135,000}[/tex]
[tex]\text{Loss settlement}=\$37,500[/tex]
Therefore, they will receive $37,500 from insurance.
The correct answer is $37,500. A house worth $180,000 has a coinsurance clause of 75 percent. The owners insure the property for $101,250. They then have a loss that results in a $50,000 claim. They will receive $37,500.00 from insurance.
A house worth $180,000 has a coinsurance clause of 75 percent. This means the owners must insure the house for at least 75% of its value to receive full coverage on claims. The required coverage amount is calculated as follows:
Required Insurance Coverage = 75% of $180,000 = 0.75 * $180,000 = $135,000
The owners insured the property for only $101,250. When a loss occurs, the amount received will be proportionate to the actual coverage relative to the required coverage:
[tex]Payout Ratio = \frac{Actual\ Insurance}{Required\ Insurance}[/tex]
[tex]Payout\ Ratio = \frac{\$ 101,250}{\$135,000} \approx 0.75[/tex]
Since the claim amount is $50,000, the actual payout from the insurance will be:
Insurance Payout = Payout Ratio * Claim Amount
Insurance Payout [tex]\approx[/tex] 0.75 * $50,000 = $37,500
Therefore, the owners will receive $37,500.00 from the insurance.
A short quiz has two true-false questions and one multiple-choice question with four possible answers. A student guesses at each question. Assuming the choices are all equally likely and the questions are independent of each other, the following is the probability distribution of the number of answers guessed correctly. What is the Probability of getting less than all three right
Final answer:
This probability question in Mathematics aims to calculate the chances of guessing answers correctly on quizzes or exams with true-false and multiple-choice questions.
The probability of getting less than all three right on the quiz with true-false and multiple-choice questions can be calculated by summing the probabilities of getting 0, 1, or 2 correct answers.
Explanation:
Probability of Getting Less Than All Three Right:
For the quiz described, the probability distribution of the number of correct answers is as follows:
0 correct: 1/8
1 correct: 3/8
2 correct: 3/8
3 correct: 1/8
To find the probability of getting less than all three right, you would add the probabilities of getting 0, 1, or 2 correct, which is 1/8 + 3/8 + 3/8 = 7/8.
Clay on the deep seafloor accumulates at a rate of about 1 millimeter per 1,000 years. How long would it take to accumulate 5 centimeters of clay?
Answer:
It will take 50,000 years to accumulate 5 centimeters of clay.
Step-by-step explanation:
The relationship between millimeter and centimer is that:
1ml = 0.1cm
So
How many ml are 5 cm?
1 ml - 0.1cm
x ml - 5cm
[tex]0.1x = 5[/tex]
[tex]x = \frac{5}{0.1}[/tex]
[tex]x = 50[/tex] ml
Clay on the deep seafloor accumulates at a rate of about 1 millimeter per 1,000 years. How long would it take to accumulate 5 centimeters of clay?
5cm is 50 ml.
1 ml per 1000 years.
So
1 ml - 1000 years
50 ml - x years
[tex]x = 50*1000[/tex]
[tex]x = 50,000[/tex]
It will take 50,000 years to accumulate 5 centimeters of clay.
Answer:
It will take 50000 years to accumulate 5 centimeters of clay.
Step-by-step explanation:
Clay on the deep seafloor accumulates at a rate of about 1 millimeter per 1,000 years. To determine the amount of time it will take to accumulate 5 centimeters of clay, we would convert 5 centimeters to millimeters.
1 centimeter = 10 millimeters
5 centimeters = 5 × 10 = 50 millimeters
Therefore,
If 1 millimeter = 1000 years,
Then, 50 millimeters = 50 × 1000 =
50000 years.
In a data set with mean of 12 and standard deviation of 4, at least what percent of data falls between 4 and 20?
Answer:
At least 95% of data falls between 4 and 20.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 12
Standard deviation = 4
At least what percent of data falls between 4 and 20?
4 = 12 - 2*4
So 4 is two standard deviations below the mean
20 = 12 + 2*4
So 20 is two standard deviations above the mean
By the Empirical Rule, at least 95% of data falls between 4 and 20.
Production managers on an assembly line must monitor the output to be sure that the level of defective products remains small. They periodically inspect a random sample of the items produced. If they find a significant increase in the proportion of items that must be rejected, they will halt the assembly process until the problem can be identified and repaired.
a. In this context, what is a Type I error?
b. In this context, what is a Type II error?
c. Which type of error would the factory owner consider more serious?
d. Which type of error might customers consider more serious?
Answer:
(a) Type I error in our context is that our test indicates that the proportion of defective products has increased after inspecting but in actual the proportion of defective products was small.
(b) Type II error in our context is that our test indicates that the proportion of defective products has remained small after inspecting but in actual the proportion of defective products was increased.
(c) Factory owner would consider Type 1 error more serious.
(d) Customers will consider Type II error more serious.
Step-by-step explanation:
Let [tex]H_0[/tex] = Proportion of defective products remains small
[tex]H_1[/tex] = Proportion of defective products increases
(a) Type I error represents that we have rejected our null hypothesis given the fact that null hypothesis is True.
Interpretation of this Type I error in our context is that our test indicates that the proportion of defective products has increased after inspecting but in actual the proportion of defective products was small.
(b) Type II error represents that we have accepted our null hypothesis given the fact that null hypothesis is False.
Interpretation of this Type II error in our context is that our test indicates that the proportion of defective products has remained small after inspecting but in actual the proportion of defective products was increased.
(c) Factory owner would consider Type 1 error more serious because after inspecting and testing he assumed that the proportion of defective products has increased due to which he will halt the assembly process till the time the problem is identified and is repaired but in actual he should continue his assembly process as in actual the proportion of defective products was small.
(d) Customers will consider Type II error more serious because after inspecting and testing factory owner assumed that the proportion of defective products is small and he will keeps on producing products and assembly process will keeps on going but in actual the proportion of defective products was increased and due to which customers will not get good quality products and they will not be able to purchase the products further.
a. Type I Error: Incorrectly concluding there's a significant increase in defective items when there isn't, leading to unnecessary halting of the assembly line.
b. Type II Error: Failing to detect a real increase in defective items, allowing the assembly to continue with actual defects.
c. The factory owner would consider a Type II error more serious.
d. Customers might find a Type I error more serious due to potential delays and disruptions in product availability.
In the context of the assembly line production process:
a. Type I error: Rejecting a null hypothesis (assuming there is no significant increase in defective products) when it is actually true.
This means that the production line is falsely halted due to the mistaken belief that there is a problem when there actually isn't. This can lead to unnecessary downtime, lost productivity, and increased costs.
b. Type II error: Failing to reject a null hypothesis (assuming there is no significant increase in defective products) when it is actually false.
This means that the production line continues to operate despite the presence of a problem that is causing an increase in defective products. This can lead to subpar products being shipped to customers, damaging the company's reputation and potentially leading to recalls or lawsuits.
c. The factory owner would consider a Type II error to be more serious.
A Type II error allows defective products to reach customers, which can damage the company's reputation, lead to recalls or lawsuits, and erode customer trust. While a Type I error can cause some inconvenience and expense, it is ultimately better to err on the side of caution and halt production if there is any suspicion of a problem.
d. Customers might consider a Type I error to be more serious.
Customers would prefer to receive products that are free of defects, even if it means that production is occasionally halted unnecessarily. A Type I error ensures that defective products are not shipped to customers, while a Type II error allows defective products to reach customers, which can cause inconvenience, frustration, and even safety hazards.
You draw five cards at random from a standard deck of 52 playing cards. What is the probability that the hand drawn is a full house? (A full house is a hand that consists of two of one kind and three of another kind.)
Answer:
The probability that the hand drawn is a full house is 0.00144.
Step-by-step explanation:
In a full house we have a hand that consists of two of one kind and three of another kind, i.e 5 cards are selected.
The number of ways of selecting 5 cards from 52 cards is:
[tex]{52\choose 5} = \frac{52!}{5!(52-5)!} \\=\frac{52!}{5!\times47!} \\=2598960[/tex]
In a deck of 52 cards there are 13 kind of cards, namely{K, Q, J, A, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Two kinds can be selected in, [tex]{13\choose 2}=\frac{13!}{2!\times(13-2)!} =\frac{13!}{2!\times11!} =78[/tex] ways
One of the two kinds can be selected for 3 cards combination in [tex]{2\choose 1} = 2[/tex] ways.
There are 4 cards of each kind.
So 3 cards combination can be selected from any of the two kinds in [tex]{4\choose 3} =\frac{4!}{3!(4-3)!} =4[/tex] ways.
And 2 cards combination can be selected from any of the two kinds in [tex]{4\choose 2} =\frac{4!}{2!(4-2)!} =6[/tex] ways.
Thus, total number of ways to select a full house is:
[tex]{13\choose 2}\times{2\choose 1}\times{4\choose 3}\times{4\choose 2}\\=78\times2\times4\times6\\=3744[/tex]
The probability that the hand drawn is a full house is:
[tex]\frac{Number\ of\ ways\ of\ Drawing\ a\ Full\ house)}{Number\ of\ ways\ of\ Selecting\ 5\ cards } =\frac{3744}{2598960} =0.00144[/tex]
Thus, the probability of playing a full house is 0.00144.
is 0.963 close to 3/4
Answer:
I would say No
Step-by-step explanation:
3/4 = .75
like 3 quarters
.75 about .963 no its not its about .2 away
There's no such concept as "close" in mathematics. Or at least, you have to specify when you consider two numbers to be "close".
All we can say is that, since 3/4=0.75, the two numbers are
[tex]0.963-0.75=0.213[/tex]
units apart. Is this small enough to consider them as "close"? Is this big enough to consider them not to be "close"?
You should clarify more what you mean so that a definitive answer can be given.