without detailed computation, give an argument that is time dependent

Answer:

Step-by-step explanation:

b. The value of x is 9

c. The probability that a student picked had just played two games = 11/20

A set is the mathematical model for a collection of different things.

If G represent Gaelic football

R represent Rugby

S represent soccer

therefore,

n(G and R) only = 16-4 = 12

n( G and S) only = 42-4 = 38

n( Sand R) only = x-4

n( G) only = 65-(38+12+4)

= 65-54

= 11

n( S) only = 57-(38+x-4+4)

= 57-38-x

= 19-x

n(R) only = 34-(16+x-4+4)

= 34-16-x

= 18-x

b. 100 = 12+38+x-4+11+19-x+18-x+4+6

100 = 12+38+11+19+18+4+7+x-x-x

100 = 109-x

x = 109-100 = 9

c. probability that a student picked played just two games;

sample space = 12+38+x-4

= 50+9-4

= 55

total outcome = 100

= 55/100 = 11/20

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Answer:

sorry, could you be a little more specific? like add an equation. i would love to answer this question, but i cant without more information. if you can add some more i will gladly answer the question for you.

Step-by-step explanation:

If a > 1, the graph of y = a|x| will be vertically stretched (or "taller") or if 0 < a < 1, the graph of y = a|x| will be vertically compressed (or "shorter") or If a is negative, the graph of y = a|x| will be a reflection of the graph of y = |x|  than the graph of y = |x| than the graph of y = |x|,

A graph is a visual representation of a set of objects, called vertices or nodes, that are connected by lines or edges. It is used to study relationships and patterns between these objects.

According to the given information :

The graph of the function y = |x| is a V-shaped graph that passes through the origin, with the arms of the V opening upward and downward at a slope of 1. When we introduce a coefficient 'a' to the function, the graph of y = a|x| is stretched or compressed vertically.

Specifically, if a > 1, the graph of y = a|x| will be vertically stretched (or "taller") than the graph of y = |x|, and the arms of the V will be steeper. On the other hand, if 0 < a < 1, the graph of y = a|x| will be vertically compressed (or "shorter") than the graph of y = |x|, and the arms of the V will be less steep.

If a is negative, the graph of y = a|x| will be a reflection of the graph of y = |x| about the x-axis, resulting in the same shape as y = |x| but flipped upside down

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Answer: the polygon sides is 15

If the measure of one exterior angle of a regular polygon is 24, Number of sides of polygon with each angle of 24 is 15 sides.

The argument provided is deductive and valid.

This is because deductive reasoning involves using general premises to arrive at a specific conclusion, and the argument here follows this pattern. The premise is that the French Revolution did not happen in 1771, and the conclusion is that it must have happened in 1988. This conclusion is logically valid because it necessarily follows from the given premise.

However, it is important to note that the argument does not provide any evidence or support for why the French Revolution would have happened in 1988, so the conclusion may not necessarily be true.

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We can be 99% confident that the true proportion of people with kids in the population falls within this interval.

To construct the confidence interval, we first need to calculate the sample proportion of people with kids:

p = 66/300 = 0.22

Next, we need to find the critical value for a 99% confidence interval. We can use a Z-table or calculator to find that value, which is 2.576.

Now we can use the formula for the confidence interval:

p ± Zα/2 * sqrt(p(1-p)/n)

Substituting in our values, we get:

0.22 ± 2.576 * sqrt(0.22(1-0.22)/300)

Simplifying this expression, we get:

0.22 ± 0.066

Therefore, the 99% confidence interval for the true population proportion of people with kids is:

(0.154, 0.286)

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Answer:

they will collide in 4 seconds

The eukaryotic Msc gene can be represented as follows: 5'-A

(exon)-intron 1-B (exon)-intron 2-C (exon)-intron 3-D (exon)-3'. The gene contains four exons (A, B, C, and D) and three introns.

b) The difference in amino acid sequences between the pancreatic cells and hepatocytes can be explained by alternative splicing. This process allows for different combinations of exons to be included in the mature mRNA, resulting in multiple protein isoforms with distinct sequences.

c) One possible mature mRNA for the Msc gene in hepatocytes could include exons A, B, and D, with exon C skipped due to alternative splicing: 5'-A (exon)-B (exon)-D (exon)-3'.

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False.

The range of the transformation T is not necessarily equal to R^m.

The range of a linear transformation T: R^n -> R^m is the set of all possible output vectors of T, i.e., the set of all vectors y in R^m such that there exists an input vector x in R^n such that T(x) = y.

The range of a transformation T can be thought of as the span of the columns of the matrix A that represents T, which is the set of all possible linear combinations of the columns of A.

Therefore, the range of the transformation T will depend on the column space of A, which is a subspace of R^m, and not necessarily equal to R^m. The dimension of the column space of A will give the rank of the matrix A, and the rank of A can be at most min(m, n).

Step-by-step explanation:

Let's break down the expression 17(4 + 15) to better understand the problem:

17(4 + 15) = 17(19)

= 323

So the expression simplifies to 323.

Student A evaluated the expression by adding the product of 17 and 4 to the product of 17 and 15.

So, Student A's evaluation is incorrect because they did not distribute the 17 to each term inside the parentheses. The correct process would be:

17(4 + 15) = 17*4 + 17*15

= 68 + 255

= 323

Student B evaluated the expression by determining the product of 17 and 19.

So, Student B's evaluation is correct because 17 times 19 gives us 323, which is the solution to the expression 17(4+15).

Therefore, Student A's evaluation is incorrect, while Student B's evaluation is correct.

Answer:

/2975

Step-by-step explanation:

Pythagorean theorem = A^2 + B^2 = C^2

We already know C^2 (60 ft) and B^2 (25ft)

We need to find A^2

C^2 - B^2 = A^2

60^2 - 25^2

3600 - 625 = 2975

Find the square root of 2975

There is no whole number squared that equals 2975

Height of the building is square root 2975

Check statement

/2975+ 25^2

2975 + 625 = 3600

The square root of 3600 is 60^2

Making the statement true

A^2 + B^2 = C^2

A^2 = 2975

B^2 = 25^2

C^2 = 60^2

2975 + 25^2 = 60^2

The proof of the identity cos²x(1 + tan²x) = 1 is complete using the mentioned rules.

Hi! I'd be happy to help you complete the proof of the identity cos²x(1 + tan²x) = 1 using the given terms.

1. Statement: cos²x(1 + tan²x) = cosx (sec²x)
  Rule: Identity (using the identity tan²x = sec²x - 1)

2. Statement: cosx (sec²x) = cosx (1 + cos²x)
  Rule: Identity (using the identity sec²x = 1/cos²x)

3. Statement: cosx (1 + cos²x) = cos²x + cos⁴x
  Rule: Distributive Property (cosx * 1 + cosx * cos²x)

4. Statement: cos²x + cos⁴x = 1
  Rule: Pythagorean Identity (since cos²x + sin²x = 1, we substitute sin²x with 1 - cos²x and simplify)

So, the proof of the identity cos²x(1 + tan²x) = 1 is complete using the mentioned rules.

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Answer:

sorry can't understand this language

Answer:4 feet

Step-by-step explanation:

2x8=16

64 divided by 16= 4 feet

The surface area of the given cylinder is 112π cm².

Given is a cylinder.

Radius of the base = 4 cm

Height of the cylinder = 10 cm

Here there are two circular bases and a lateral face.

Area of the bases = 2 × (πr²)

                             = 2 × π (4)²

                             = 32π cm²

Area of the lateral face = 2π rh

                                      = 2π (4)(10)

                                      = 80π

Total area = 112π cm²

Hence the total surface area of the cylinder is 112π cm².

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By combining like terms, we can simplify equations and expressions. This makes it easier to solve for a single variable, or to check the accuracy of a given equation.

15. 7x⁴ - 5x⁴= 2x⁴16. 32y + 5y = 37y17. 6b + 7b - 10 = 13b - 1018. 2x + 3x + 4 = 5x + 419. y + 4 + 3(y + 2) = 4y + 1020. 7a²- a²+ 16 = 6a² + 1621. 3y²+ 3(4y²- 2) = 15y² - 622. z² + z + 4z³+ 4z² = 5z² + 4z³23. 0.5(x⁴ - 3) + 12 = 0.5x⁴ + 924. 1/4 * (16 + 4p) = 4 + p

An equation is a statement that asserts the equality of two expressions, with each expression being composed of numbers, variables, and/or mathematical operations. Equations are used to solve problems in mathematics, science, engineering, economics, and other fields. Equations offer the opportunity to describe relationships between different variables and to develop models that can be used to predict the behavior of systems.

15. 7x⁴ - 5x⁴= 2x⁴

16. 32y + 5y = 37y

17. 6b + 7b - 10 = 13b - 10

18. 2x + 3x + 4 = 5x + 4

19. y + 4 + 3(y + 2) = 4y + 10

20. 7a²- a²+ 16 = 6a² + 16

21. 3y²+ 3(4y²- 2) = 15y² - 6

22. z² + z + 4z³+ 4z² = 5z² + 4z³

23. 0.5(x⁴ - 3) + 12 = 0.5x⁴ + 9

24. 1/4 * (16 + 4p) = 4 + p

Conclusion:
By combining like terms, we can simplify equations and expressions. This makes it easier to solve for a single variable, or to check the accuracy of a given equation. It is important to remember that like terms must have the same base and exponent to be combined.

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Any two disjoint closed subsets of Y can be separated by disjoint open subsets of Y, which implies that Y is a normal space.

Let X be a normal space and let Y be a closed subspace of X.

We want to show that Y is also normal.

To show that Y is normal, we need to show that for any two disjoint closed subsets A and B of Y, there exist disjoint open subsets U and V of Y such that A is a subset of U and B is a subset of V.

Since A and B are closed subsets of Y, they are also closed subsets of X. By the normality of X, there exist disjoint open subsets U' and V' of X such that A is a subset of U' and B is a subset of V'. Since Y is a closed subspace of X,

we can find closed subsets U and V of X such that U' is a subset of U and V' is a subset of V, and U ∩ Y = U' and V ∩ Y = V'.

Since A is a closed subset of Y and U ∩ Y = U',

we have A ∩ (X - U) = A ∩ (Y - U') = ∅.

Similarly, since B is a closed subset of Y and V ∩ Y = V',

we have B ∩ (X - V) = B ∩ (Y - V') = ∅.

Therefore, U and V are disjoint open subsets of Y such that A is a subset of U and B is a subset of V.

Therefore, we have shown that any two disjoint closed subsets of Y can be separated by disjoint open subsets of Y, which implies that Y is a normal space.

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The analytic solution is y(x) = (1/9)sin(3x) - (1/9)sin(3). This solution is unique since there are no arbitrary constants remaining after applying the boundary conditions.

The given differential equation is:

y''(x) = sin(3x)

We can solve this by first finding the general solution to the homogeneous equation y''(x) = 0, which is simply y(x) = Ax + B, where A and B are constants determined by the boundary conditions.

Next, find a particular solution to the non-homogeneous equation y''(x) = sin(3x).

Since sin(3x) is a trigonometric function, we can try a particular solution of the form y(x) = Csin(3x) + Dcos(3x), where C and D are constants to be determined.

Taking the first and second derivatives of this expression:

y'(x) = 3Ccos(3x) - 3Dsin(3x)

y''(x) = -9Csin(3x) - 9Dcos(3x)

Substituting these into the original equation:

-9Csin(3x) - 9Dcos(3x) = sin(3x)

Equating coefficients of sin(3x) and cos(3x):

-9C = 1 and -9D = 0

Solving for C and D:

C = -1/9 and D = 0

So, the particular solution is:

y(x) = (-1/9)sin(3x)

Therefore, the general solution to the non-homogeneous equation is:

y(x) = Ax + B - (1/9)sin(3x)

Using the boundary conditions y(0) = 0 and y() = 0:

0 = A + B

0 = A - (1/9)sin(3)

Solving for A and B:

A = (1/9)sin(3) and B = -(1/9)sin(3)

So, the final analytic solution is:

y(x) = (1/9)sin(3x) - (1/9)sin(3)

The solution is unique, as there are no arbitrary constants remaining after applying the boundary conditions.

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The equation of the tangent plane to the surface z=[tex]2x^2[/tex]+[tex]y^2[/tex]−5y  at the point (1, 2, -4) is 4x-y-z = 6, which is not one of the options given. Therefore, the correct option is (a) none of these.

To find the equation of the tangent plane to the surface z=[tex]2x^2[/tex]+[tex]y^2[/tex]−5y  at the point (1, 2, -4), we need to find the partial derivatives of the surface with respect to x and y at that point.

∂z/∂x = 4x

∂z/∂y = 2y - 5

At the point (1, 2, -4), these partial derivatives are:

∂z/∂x = 4(1) = 4

∂z/∂y = 2(2) - 5 = -1

So the normal vector to the tangent plane is <4, -1, 1>.

Using the point-normal form of the equation of a plane, we get:

4(x - 1) - 1(y - 2) = 1(z + 4)

Simplifying, we get:

4x-y-z = 6

Therefore, the equation of the tangent plane to the surface z=[tex]2x^2[/tex]+[tex]y^2[/tex]−5y at the point (1, 2, -4) is 4x-y-z = 6, which is not one of the options given. Therefore, the answer is (a) none of these.

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Each entry in the table is the probability of the corresponding outcome (e.g. Upper C 1 and Upper R 1) occurring.

a. The completed contingency table is:

Upper C 1   Upper C 2   Total
Upper R 1      4           8          12
Upper R 2     10           8          18
Total              14          16          30

b. To find ​P(Upper C 1​), we add up the values in the Upper C 1 column and divide by the total number of observations:

​P(Upper C 1​) =[tex]\frac{(4 + 10)} { 30} = 0.47[/tex]
To find ​P(Upper R 2​), we add up the values in the Upper R 2 row and divide by the total number of observations:

​P(Upper R 2​)[tex]= \frac{18} { 30} = 0.6[/tex]

To find ​P(Upper C 1 ​& Upper R 2​), we look at the intersection of the Upper C 1 column and the Upper R 2 row, which is 10. We then divide by the total number of observations:

​P(Upper C 1 ​& Upper R 2​) = 10 / 30 = 0.33

c. The joint probability distribution is:

Upper C 1   Upper C 2   Total
Upper R 1    0.13       0.27      0.4
Upper R 2   0.33       0.27      0.6
Total            0.47       0.53      1.0

Each entry in the table is the probability of the corresponding outcome (e.g. Upper C 1 and Upper R 1) occurring.

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A is not invertible when x = 0 or x = 1.

To determine when the given matrix A is not invertible, we need to find when its determinant is equal to zero. Therefore, we can compute the determinant of A by expanding it along any row or column. Expanding along the first column, we have:

|A| = x | 1 -1 1 |

-1 | 1 1 1 |

1 |-1 0 2x|

  (0 + 0 + 2x)

= x[(1)(0)-(1)(2x)] - (-1)(0-2x) + (1)[(1)(-1)-(1)(-1)]

= -2x^2 + 2x + 0

= 2x(-x + 1)

Therefore, A is not invertible when x = 0 or x = 1.

If x = 0, then the third row of A is equal to the sum of the first and second rows, so the rows of A are linearly dependent. Thus, A is not invertible in this case.

If x = 1, then the first and third columns of A are equal, so the columns of A are linearly dependent. Thus, A is not invertible in this case as well.

In summary, A is not invertible when x = 0 or x = 1.

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The mean of X1, ..., Xn is 121 * y, which indicates that the underlying data have a central tendency of 121 * y. Therefore, the mean and variance of X1, ..., Xn are 121 * y and 242, respectively.

The mean provides information about the central tendency of the underlying data, while the variance provides information about the spread or variability of the underlying data.

The mean and variance of X1, ..., Xn can be calculated as follows:

Mean:

E[X1, ..., Xn] = E[X1] + ... + E[Xn] = n * E[X1]

= n * y

= 121 * y

Variance:

E[(X1 - E[X1])2 + ... + (Xn - E[Xn])2] = n * E[(X1 - E[X1])2]

= n * 02

= 121 * 02

= 242

Therefore, the mean and variance of X1, ..., Xn are 121 * y and 242, respectively.

The mean and variance of a random variable are important parameters for describing the probability distribution of that variable.

In this case, the mean of X1, ..., Xn is 121 * y, which indicates that the underlying data have a central tendency of 121 * y.

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The mean of a given uniform probability distribution with minimum value as 5 and maximum value 15 is given by 10.

In the  uniform probability distribution,

Minimum value is equal to 5

Maximum value is equal to 15

Let us consider the minimum value represented by x.

And maximum value represented by y.

The mean can be calculated using formula.

Mean = ( Minimum value + Maximum value ) / 2

         = ( x + y ) / 2

         = ( 5 + 15 ) / 2

         = 10.

Therefore, the mean of the uniform probability distribution with given minimum and maximum value is equal to 10.

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If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect the size of the confidence interval to increase.

This is because a higher level of confidence requires a wider interval to encompass a larger range of possible values. The sample size does not necessarily need to change to adjust the confidence interval. Therefore, the correct answer is C. the size of the confidence interval to increase.
If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect C. the size of the confidence interval to increase. This is because a higher confidence level requires a larger range to ensure the true population parameter is captured with more certainty.

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The area under one arch of the cycloid x = r(t − sin t), y = r(1 − cos t) for 0 ≤ t ≤ 2π is 4πr².

To find the area under one arch of the cycloid x = r(t − sin t), y = r(1 − cos t) for 0 ≤ t ≤ 2π, we can use the formula for finding the area under a curve:

A = ∫[a,b] f(x) dx

In this case, we need to find the integral of y with respect to x:

A = ∫[0,2π] y dx

We can solve for y in terms of t by substituting x = r(t − sin t) into the equation for y:

y = r(1 − cos t)

dx = r(1 − cos t) dt

Substituting these into the formula for the area, we get:

A = ∫[0,2π] r(1 − cos t)(r(1 − cos t) dt)

Simplifying, we get:

A = r² ∫[0,2π] (1 − cos t)² dt

Using the trig identity (1 − cos 2t) = 2 sin² t, we can simplify the integrand:

A = r² ∫[0,2π] (1 − cos t)² dt

= r² ∫[0,2π] (1 − 2cos t + cos² t) dt

= r² ∫[0,2π] (1 − 2cos t + (1 − sin² t)) dt

= r² ∫[0,2π] 2(1 − cos t) dt

= r² [2t − 2sin t] from 0 to 2π

= 4πr²

Therefore, the area under one arch of the cycloid x = r(t − sin t), y = r(1 − cos t) for 0 ≤ t ≤ 2π is 4πr².

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Answer:

Step-by-step explanation:

You want a pair of transformations that will map ∆ABC to ∆A'B'C'.

We note that segment BC points downward, and segment B'C' points upward. This suggests a vertical reflection.

We also note that point A' is 2 units left of point A, suggesting a horizontal translation. It is as far below the x-axis as A is above the x-axis.

The two transformations that map ∆ABC to ∆A'B'C' are ...

These transformations are independent of each other, so may be applied in either order.

The complete formal proof of p->(q->(r->p)) from no premises, with an empty premise line:

1. |_
2. | |_ p    (Assumption)
3. | | |_ q  (Assumption)
4. | | | |_ r (Assumption)
5. | | | | p (Copy: 2)
6. | | | q->(r->p) (Implication Introduction: 4-5)
7. | | p->(q->(r->p)) (Implication Introduction: 2-6)
8. |_ p->(q->(r->p)) (Implication Introduction: 1-7)

In this proof,

we start with an empty premise line (line 1), and then assume p (line 2).

From there, we assume q (line 3) and r (line 4), and then use the copy rule to copy p from line 2 (line 5).

We then use implication introduction to conclude q->(r->p) (line 6), and then use implication introduction again to conclude p->(q->(r->p)) from lines 2-6 (line 7).

Finally, we use implication introduction one last time to conclude p->(q->(r->p)) from line 1 and line 7 (line 8).

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The value of the probability P(E or F) is 0.55.

In science, the probability of an event is a number that indicates how likely the event is to occur.

It is expressed as a number in the range from 0 and 1, or, using percentage notation, in the range from 0% to 100%. The more likely it is that the event will occur, the higher its probability.

To find the probability of the event E or F, we can use the formula:
P(E or F) = P(E) + P(F) - P(E and F)

We are given that P(E) = 0.20 and P(F) = 0.45, and we also know that P(E and F) = 0.10.

Substituting these values into the formula, we get:
P(E or F) = 0.20 + 0.45 - 0.10
P(E or F) = 0.55

Therefore, the probability of the event E or F is 0.55.

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