b.
If it's a (+,+) it's the 1st quadrant. If it's a (-,+) it's the 2nd quadrant. If its a (-,-) it's the 3rd quadrant. And if it's a (+,-) it's the 4th quadrant
Answer:
B
Step-by-step explanation:
(1 point) Write the given second order equation as its equivalent system of first order equations. u′′+6.5u′−1.5u=8sin(3t),u(1)=−3,u′(1)=−1.5 Use v to represent the "velocity function", i.e. v=u′(t). Use u and v for the two functions, rather than u(t) and v(t). (The latter confuses webwork. Functions like sin(t) are OK.)
Answer:
The equivalent system of first-order differential equations is:
[tex]\[ \begin{cases}u' = v \\v' = 8\sin(3t) - 6.5v + 1.5u\end{cases} \][/tex]
with the initial conditions:
[tex]\[ \begin{cases}u(1) = -3 \\v(1) = -1.5\end{cases} \][/tex]
To convert the given second-order differential equation into an equivalent system of first-order equations, we follow these steps:
1. Identifying the original second-order differential equation:
u'' + 6.5u' - 1.5u = 8sin(3t)
2. Then we introduce a new variable v to represent the first derivative of u:
v = u'
3. Now ,we write the second-order equation as a system of first-order equations:
The original equation is:
u'' + 6.5u' - 1.5u = 8sin(3t)
Using the new variable v = u' , the second derivative u'' can be written as v'. Therefore, we have:
v' + 6.5v - 1.5u = 8sin(3t)
4. Formulating the system of first-order equations:
The first equation comes directly from the definition of v:
u' = v
The second equation comes from the rewritten second-order equation:
v' = 8sin(3t) - 6.5v + 1.5u
5. Let us include the initial conditions:
The initial conditions provided are:
u(1) = -3
u'(1) = -1.5
Since v = u', this translates to:
v(1) = -1.5
6. Writing the system with initial conditions:
The system of first-order equations is:
[tex]\[ \begin{cases} u' = v \\ v' = 8\sin(3t) - 6.5v + 1.5u \end{cases} \][/tex]
With the initial conditions:
[tex]\[ \begin{cases} u(1) = -3 \\ v(1) = -1.5 \end{cases} \][/tex]
Yoselin has 7 boxes of coins. Each box has 28 coins. How many coins does 1 point
Yoselin have in all? Choose best equation. *
28-7=21
28x7=196
28+7=35
28/7=4
Answer:
28x7=196 (Pls give Brainliest)
Assume that random guesses are made for ninenine multiple choice questions on an SAT test, so that there are nequals=99 trials, each with probability of success (correct) given by pequals=0.50.5. Find the indicated probability for the number of correct answers. Find the probability that the number x of correct answers is fewer than 44.
Answer:
The probability that the number of correct answers is 4 is 0.2461.
Step-by-step explanation:
Let X = number of correct answers.
The probability that an answer is correct is,P (X) = p = 0.50.
The total number of questions is, n = 9.
The event of an answer being correct is independent of the other answers.
The success of each trial is defined as a correct answer with equal probability of success for each trial, i.e. 0.50.
The random variable X follows a Binomial distribution with parameter n = 9 and p = 0.50.
The probability mass function of X is:
[tex]P(X=x)={9\choose x}\times0.50^{x}\times (1-0.50)^{9-x};\ x=0,1,2,3...[/tex]
Compute the value of P (X = 4) as follows:
[tex]P(X=4)={9\choose 4}\times(0.50)^{4}\times (1-0.50)^{9-4}[/tex]
[tex]=126\times 0.0625\times 0.03125\\=0.24609375\\\approx 0.2461[/tex]
Thus, the probability that the number of correct answers is 4 is 0.2461.
A six-sided die is rolled, and the number N on the uppermost face is recorded. Then a fair coin is tossed N times, and the total number Z of heads to appear is observed. Determine the mean and variance of Z by viewing Z as a random sum of N Bernoulli random variables. Determine the probability mass function of Z, and use it to find the mean and variance of Z.
Answer:
1. Mean is 1.75
2. The variance is 1.6042
3.
The distribution function is:
Z Z/K
0 21/128
1 5/16
2 33/128
3 1/6
4 29/384
5 1/48
6 1/384
Step-by-step explanation:
The mean of Z is given as:
E(Z) =Σ6, k=0 Kp (Z = k)
Σ6,k=0 K 1/6 Σ6, n=k (n k) (1/2)^n
=( 0(21/128) + 1(5/16) + 2( 33/128) + 3 (1/6) + 4 (29/384) + 5 (1/48) + 6 (1/384))
=7/4
=1.75
Thus, the mean Z is 1.75
The variance of Z is given as:
Var (Z) = E (Z^2) - (E (Z)) ^2
Therefore,
E(Z^2) = Σ 6, k=0 K^2P ( Z=K)
= ( 0(21/128 + 1(5/16) + 4(33/128) + 9(1/6) + 16(29/384) + 25(1/48) + 36(1/384))
=14/3
Var (Z) = 14/7 - (7/4)^2
= 14/7 - 49/16
=77/48
=1.6042
Thus, the variance is 1.6042
The probability of mass function is given as:
P(Z=k) = 1/6 Σ 6, n=k (n k) (1/2)^n
The distribution function is
Z Z/K
0 21/128
1 5/16
2 33/128
3 1/6
4 29/384
5 1/48
6 1/384
An amusement park has a diameter of 975 feet and has a circular walking path around the entire park. the maintenance worker has to walk around the park 3 times a day how far does he walk a day
Answer: 2925 feet a day
Step-by-step explanation:
you gotta multiply 975 x 3 and you gonna 2925.
Find the inner product for (-4,9,8) . (3,2,-2) and state whether the vectors are perpendicular.
a
-10; no
c.
10; no
b.
-10; yes
d.
10; yes
Answer:
-10; no
Step-by-step explanation:
-4*3 + 9*2 + 8*-2 = -10-10 does not equal 0 so it is not perpendicular
Final answer:
The inner product of the vectors (-4,9,8) and (3,2,-2) is -10. Since the inner product is not zero, the vectors are not perpendicular. Therefore, the correct answer is: -10; no.
Explanation:
The inner product (also known as the dot product) of two vectors (-4,9,8) and (3,2,-2) is calculated by multiplying the corresponding components of the two vectors and summing the result:
Inner product = (-4)×3 + 9×2 + 8×(-2)
= -12 + 18 - 16
= -10
To determine if the vectors are perpendicular, we check if their inner product is zero. Since the inner product in this case is -10, not zero, the vectors are not perpendicular.
A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. Suppose that this chart is being used in phase I and the averages from a set of m samples or subgroups from this process are plotted on this chart. What is the probability that at least one of the averages will plot outside the control limits when m
Answer:
Check the explanation
Step-by-step explanation:
Ans=
A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134
M = 10: 1 – 0.9973^10 = 0.0267
M = 20: 1 – 0.9973^20 = 0.0526
M = 30: 1 – 0.9973^30 = 0.0779
M = 50: 1 – 0.9973^50 = 0.126
18)
Ans=
Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values
Jenny buys a television for the sale
price of $72.59. The television
normally costs $145.18. What percent is
being saved?
What -3 as a decimal
Answer:
-0.03
Step-by-step explanation:
Answer:
-.03 hope this helps
A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 49.049.0 and 54.054.0 minutes. Find the probability that a given class period runs between 51.2551.25 and 51.7551.75 minutes. Find the probability of selecting a class that runs between 51.2551.25 and 51.7551.75 minutes.
Answer:
10% probability that a given class period runs between 51.25 and 51.75 minutes.
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability of finding a value X between c and d, d greater than c, is given by the following formula:
[tex]P(c \leq X \leq d) = \frac{d-c}{b-a}[/tex]
Uniformly distributed between 49 and 54 minutes
This means that [tex]b = 54, a = 49[/tex]
Find the probability that a given class period runs between 51.25 and 51.75 minutes.
[tex]P(51.25 \leq X \leq 51.75) = \frac{51.75 - 51.25}{54 - 49} = 0.1[/tex]
10% probability that a given class period runs between 51.25 and 51.75 minutes.
this uses pythagorean theorem
Step-by-step explanation:
x²=a²+b²
x=√6²+12²
x=√180
x=3√2v
y²=16²+12²
y=√400
y=20
Answer: the answer for rafter 1 is 13.4 and the answer for rafter 2 is 20
Step-by-step explanation: I just know
State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a 95% confidence level to a 90% confidence level. b. Changing from a sample size of 25 to a sample size of 250. c. Changing from a standard deviation of 20 pounds to a standard deviation of 30 pounds.
Answer:
A) Confidence Interval will become narrower. B) Confidence Interval will become narrower. C) Confidence Interval will become broader.
Step-by-step explanation:
Confidence Interval is the probable range around sample statistic, in which the population parameter is expected to lie.
Confidence Level shows the average percentage level of confidence interval, expected to contain population parameter. Lower confidence level implies narrower Confidence Interval
Bigger sample size reduces margin error (sample statistic, population parameter difference). Parameter-statistic proximity implies: narrower confidence interval around statistic, expected to contain parameter.
Standard Deviation is a measure of dispersion, spread. So, higher standard deviation implies more spread & broader confidence interval.
A company purchases a small metal bracket in containers of 5,000 each. Ten containers have arrived at the unloading facility, and 250 brackets are selected at random from each container. The fraction nonconforming in each sample are 0, 0, 0, 0.004, 0.008, 0.020, 0.004, 0, 0, and 0.008. Do the data from this shipment indicate statistical control
Answer:
Do the data from this shipment indicate statistical control: No
Step-by-step explanation:
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
= (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10
= 0.044/10
= 0.0044
Calculating the lower control limit (LCL) using the formula;
LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n
= 0.0044 - 3*√(0.0044(1-0.0044))
= 0.0044- (3*0.0042)
= 0.0044 - 0.01256
= -0.00816 ∠ 0
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
= 0.0044 + 3*√(0.0044(1-0.0044))
= 0.0044+ (3*0.0042)
= 0.0044 + 0.01256
=0.01696∠ 0
Do the data from this shipment indicate statistical control: No
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment do not indicate statistical control.
The data from this shipment does not indicate statistical control.
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
[tex]\frac{(0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)}{10}\\=\frac{0.044}{10}\\=0.0044[/tex]
Calculating the lower control limit (LCL) using the formula;
LCL
= (x-bar) - 3*√(x-bar(1-x-bar))/n
[tex]= 0.0044 - 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044- (3*0.0042)\\= 0.0044 - 0.01256\\= -0.00816[/tex]
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
[tex]= 0.0044 + 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044+ (3*0.0042)\\= 0.0044 + 0.01256\\=0.01696[/tex]
Do the data from this shipment indicate statistical control:
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment does not indicate statistical control.
Learn more about mean: https://brainly.com/question/12892403
the experimental probability that an SUV will pass by andis store is 0.4. If 500 cars pass by andis store, how many can she expect to be SUVs?
Answer:
50
Step-by-step explanation:
Answer:
the answer is 200
Step-by-step explanation:
A circle has a radius of 6. An arc in this circle has a central angle of 48 degrees. What is the length of the arc
Write an exact, simplified answer in terms of pi.
Answer:
[tex] \huge \pi \: units [/tex]
Step-by-step explanation:
[tex]l = \frac{48 \degree}{360 \degree} \times 2\pi \: r \\ \\ = \frac{2}{24} \times 2 \times \pi \times 6 \\ \\ = \frac{24}{24} \times \pi \\ \\ = \pi \: units \\ [/tex]
During a 12-hour period, the temperature in a city dropped from a high of 66°F to a low of −29°F. What was the range of the temperatures during this period?
Answer:
95
Step-by-step explanation:
66-66=0
0-29=-29
66+29=95
The range of the temperatures during this period is 95.
Given the following data:
Time period = 12 hoursHighest temperature = 66°FLowest temperature = −29°FTo determine the range of the temperatures during this period:
Range is simply calculated by taking the difference between the highest number and the lowest number in a sample.
Mathematically, range is given by the formula;
[tex]Range = highest \;number -lowest \;number[/tex]
Substituting the given parameters into the formula, we have;
[tex]Range = 66-(-29)\\\\Range =66+29[/tex]
Range = 95
Therefore, the range of the temperatures during this period is 95.
Read more: https://brainly.com/question/16205323
How many faces does the shape have
Answer:
5 faces
Step-by-step explanation:
4 triangular, 1 square
Researchers are investigating the effectiveness of leg-strength training on cycling performance. A sample of 7 men will be selected to participate in a training program that lasts for one month. Peak power during cycling will be recorded for each man both before training and after training. The mean difference in times will be used to construct a 95 percent confidence interval for the mean difference in the population.
Answer:
A. The interval will be narrower if 15 men are used in the sample.
Step-by-step explanation:
Hello!
When all other things remain the same, which of the following statements about the width of the interval is correct?
A. The interval will be narrower if 15 men are used in the sample.
B. The interval will be wider if 15 men are used in the sample.
C. The interval will be narrower if 5 men are used in the sample.
D. The interval will be narrower if the level is increased to 99% confidence.
E. The interval will be wider if the level is decreased to 90% confidence.
Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.
The general structure for the CI is "point estimate"±" margin fo error"
Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)
The margin of error of the interval is:
d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
1) The sample size changes, all other terms of the interval stay the same.
As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.
↓d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↑n)
↑d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↓n)
Correct option: A. The interval will be narrower if 15 men are used in the sample.
2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:
↓d= ↓[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
↑d= ↑[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)
I hope it helps!
Answer:
The interval will be narrower if 15 men are used in the sample.
Step-by-step explanation:
Find the equation of the line that Contains the point (4,-2) and is perpendicular to the line y = -2x+5
5 people want to evenly share a 1/3 pound bag of peanuts. How many pounds should each person get?
Answer:
1 / 15
Step-by-step explanation:
1/3 / 5 =
1/ 15
Final answer:
Each person will get 1/15 of a pound of peanuts when a 1/3 pound bag is shared evenly among 5 people.
Explanation:
To figure out how much each of the 5 people should get from a 1/3 pound bag of peanuts, we need to divide the total weight of the peanuts by the number of people. This gives us:
1/3 pound ÷ 5 = 1/15 pound per person.
This means each person will get 1/15 of a pound of peanuts. In other calculations such as the candy survey, determining percent uncertainty, or unit conversions as in Michaela's party scenario, a similar process of division or unit conversion is applied to find the answer.
27.) What shape do you create if you cut a square in
half horizontally or vertically?
Answer:
A rectangle!!
Step-by-step explanation:
When you cut a square horizontally or vertically, you get two smaller triangles that are half the area of the square. When you split the rectangle again, you get a square 1/4 the size of the regular square.
The following is the (edited) output for the test: A Two-sample T-Test and CI for the data. Sample 1(M) has N=112, Mean=7.38, StDev=6.95, SE Mean=0.66. Sample 2 (F) has N=101, Mean=7.15, StDev=6.31, SE Mean = 0.63. The difference is mu (1) - mu (2) and its estimate is 0.230000. The 95% lower bound for difference is -1.271079. The T-Test of difference: T-Value = 0.25, P- Value =0.400, DF=210. From the output we learn that: (i) The data provide sufficient evidence to reject H0 and to conclude that the mean depression score for male teens is larger than that of female teens. (ii) The data provide sufficient evidence to conclude that male and female teens do not differ in mean depression score. (iii) The data do not provide sufficient evidence to conclude that the mean depression score of male teens is larger than that of female teens. (iv) The data do not provide sufficient evidence to reject H0, so we accept it, and conclude that male and female teens do not differ in mean depression score.
Answer:
Step-by-step explanation:
Hello!
You have the output:
Two-Sample T-Test and Cl
Sample N Mean StDev SE Mean
1(M) 112 7.38 6.95 0.66
2(F) 101 7.15 6.31 0.63
Difference = mu (1) - mu (2)
Estimate for difference: 0.230000
95% lower bound for difference: -1.271079
T-Test of difference: T-value = 0.25 P-Value = 0.400 DF= 210
This output summarizes the information of the two samples and indicates the order the populations where studied.
It also informs you of the value of the statistic under the null hypothesis and the p-value.
Unfortunately, there is no information on the type of hypotheses that were tested, i.e. if they where two-tailed or one-tailed, in the latter case, there is no information if it was left-tailed or right-tailed). Likewise is not specified if the test was done for a specific value of the parameter. (for example μ₁ - μ₂ = 0 or μ₁ - μ₂ = θ₀)
For these reasons, the data provided by the output isn't enough to conclude any hypothesis.
From all the provided answers the one more likely to be correct is:
(iii) The data do not provide sufficient evidence to conclude that the mean depression score of male teens is larger than that of female teens.
I hope this helps!
Answer:
iv
Step-by-step explanation:
Since the p-value (0.4) is greater than the significance level (0.05), we can conclude that the result is not significant. This means that there is no enough statistical evidence to reject the null hypothesis H0. Therefore, we must accept it and conclude that the mean depression score for male and female teens do not differ.
Show that y=sin(t) is a solution to (dydt)2=1−y2. Enter your answers below in terms of the independent variable t in the order in which the terms were given. Be sure you can justify your answer.
Answer:
y = sin(t) is a solution to the differential equation
(dy/dt)² = 1 - y²
Step-by-step explanation:
Given (dy/dt)² = 1 - y²
Suppose y = sin(t) is a solution, then it satisfies the differential equation.
That is
[d(sin(t))]² = 1 - y²
Let y = sin(t)
dy/dt = d(sin(t)) = cos(t)
(dy/dt)² = cos²t
But cos²t + sin²t = 1
=> 1 - sin²t = cos²t
So
(dy/dt)² = 1 - sin²t
Since sin²t = (sint)² = y²,
we have
(dy/dt)² = 1 - y²
as required.
The differential equation becomes [tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]
Given the function;
[tex]y = sint[/tex]Take the differential of the function
[tex]\frac{dy}{dt} = cost[/tex]Square both sides of the equation to have:
[tex](\frac{dy}{dx} )^2 = (cost)^2[/tex]
Recall from trigonometry identity that [tex]sin^2t + cos^2t = 1[/tex]
Hence, [tex]cos^2t = 1- sin^2t[/tex]
Replace into the differential expression to have:
[tex](\frac{dy}{dx} )^2 = 1-sin^2t[/tex]
Recall that y = sin(t). On replacing, the differential equation becomes:
[tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]
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PLEASE HELP
Find the volume and surface area of the 3-dimensional figure below.
Volume=
Surface Area=
Answer:
volume: 18 ft³surface area: 42 ft²Step-by-step explanation:
The volume of a cuboid is the product of its dimensions:
V = LWH = (3 ft)(3 ft)(2 ft)
V = 18 ft³
The area is the sum of the areas of the faces. Since opposite faces have the same area, we can figure the area from ...
A = 2(LW +H(L+W)) = 2((3 ft)(3 ft) +(2 ft)(3 ft +3 ft)) = 2(9 ft² +12 ft²)
A = 42 ft²
Based on a survey, assume that 2525% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when fivefive consumers are randomly selected, exactly threethree of them are comfortable with delivery by drones. Identify the values of n, x, p, and q.
Answer:
0.0879 is the probability that out of 5 randomly selected consumers, three are comfortable with delivery by drones.
Step-by-step explanation:
We are given the following information:
We treat drone deliveries as a success.
P(consumers comfortable having drones deliver) = 25% = 0.25
Then the number of consumers 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.
We have to evaluate:
P(Exactly 3 customers out of 5 are comfortable with delivery by drones)
Here,
[tex]n = 5\\x = 3\\p = 0.25\\q = 1 - p = 1-0.25=0.75[/tex]
Putting values, we get,
[tex]P(x =3)\\\\= \binom{5}{3}(0.25)^3(1-0.25)^2\\\\= 0.0879[/tex]
0.0879 is the probability that out of 5 randomly selected consumers, three are comfortable with delivery by drones.
How many 1 sixths are in 2 and why !!!! PLeASE HURRY
2 x (c^2 -5) for c=4
Answer:
22
Step-by-step explanation:
PEMDAS
4^2 = 16
16 - 5 = 11
11 x 2 = 22
Answer:
22
Step-by-step explanation:
2 (c^2 -5)
Let c=4
2 (4^2 -5)
PEMDAS
Parentheses first,
(4^2 -5)
Exponents
16-5 =11
Replace into expression
2(11)
22
Which of the hypotheses below would be suited for testing with a one-variable chi-square test? It was hypothesized that more people would choose the number 7 as their 'lucky' number than any other number. People who choose the number 7 as their 'lucky' number are significantly more superstitious than people who choose the number 13 as their 'lucky' number. Choice of 'lucky' number is directly related to measures superstition. All of these.
Answer:
It was hypothesized that more people would choose the number 7 as their 'lucky' number than any other number.
Step-by-step explanation:
Given that one variable chi-square is used to test whether a single categorical variable follows a hypothesized population distribution. The Chi Square statistic compares the tallies or counts of categorical responses between two (or more) independent groups
The null hypothesis (H0) for the test is that all proportions are equal.
The alternate hypothesis (H1) is given condition in the question.
A. It was hypothesized that more people would choose the number 7 as their 'lucky' number than any other number.
This is suited for testing with a one-variable chi-square test because we are testing if the proportion of people who choose number 7 is greater than the proportion of any other numbers. So, we are therefore comparing more than 2 proportions.
B. People who choose the number 7 as their 'lucky' number are significantly more superstitious than people who choose the number 13 as their 'lucky' number.
This is not suited for testing with a one-variable chi-square test. A z test is more preferable in this instance because we are testing just two proportions.
C. Choice of 'lucky' number is directly related to measures superstition.
This is not suited for testing with a one-variable chi-square test because chi square test is not used for showing relationship between variables.
D. All of these. Since option A is correct, this option can not be correct.
g Concerning 5 card poker hands from a 52 card deck- how many hands contain 2 pairs(that is 2 pairs of 2 different kinds plus a fifth card of some third kind- Example would be 2 jacks, 2 kings and a 5. Remember a 52 card deck contains 4 suits(hearts, Diaminds, Spades and club)of which has 13 kinds of cards having increasing values of 2 through 10, jack,queen,king and ace(having the value of 1 or a value higher than the king.))
Answer:
attached
Step-by-step explanation:
attached
How do I find the complement?
Let U={1,2,3,6,10,13,14,16,17} . Determine the complement of the set {3,10,16}
Let A be some subset of a universal set U. The "complement of A" is the set of elements in U that do not belong to A.
For example, if U is the set of all integers {..., -2, -1, 0, 1, 2, ...} and A is the set of all positive integers {1, 2, 3, ...}, then the complement of A is the set {..., -2, -1, 0}.
Notice that the union of A and its complement make up the universal set U.
In this case,
U = {1, 2, 3, 6, 10, 13, 14, 16, 17}
The set {3, 10, 16} is a subset of U, since all three of its elements belong to U.
Then the complement of this set is all the elements of U that aren't in this set:
{1, 2, 6, 13, 14, 17}
The complement of a set A with respect to a set B includes elements in B that are not in A. The complement of the set {3,10,16} within {1,2,3,6,10,13,14,16,17} is {1,2,6,13,14,17}. The process involves removing elements in A from B.
Explanation:In mathematics, specifically in set theory, the complement of a set A, with respect to a set B, refers to the elements in set B that are not in set A. Let's use this definition to find the complement of {3,10,16} in the universal set U={1,2,3,6,10,13,14,16,17}.
First, list all the elements in U. Next, remove those elements which appear in the set {3,10,16}. The remaining elements constitute the complement of {3,10,16} given U. With this procedure, we find that the complement of {3,10,16} with respect to U is {1,2,6,13,14,17}. This method can be applied to any sets within a given universal set to find their complements.
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