2.
The graph of a quadratic function is shown on the grid. What ordered pair best represents the vertex of the
graph?
Two ch ractor
-10-
-9-
-8+
-9-
-10-

a) The probability that three tubes are defective is approximately 0.0146, or 1.46%.

b) The probability that at least four tubes are defective is 0.4686 or 46.86%.

To solve this problem, we can use the binomial probability formula:

P(X = k) = (n choose k) * p^k * (1-p)^(n-k)

where X is the number of defective tubes, n is the total number of tubes inspected, p is the probability that a tube is defective, and (n choose k) is the binomial coefficient, which represents the number of ways to choose k items out of n.

(a) To find the probability that three tubes are defective out of six, we can plug in n = 6, k = 3, and p = 0.1 into the formula:

P(X = 3) = (6 choose 3) * 0.1^3 * 0.9^3
          = 20 * 0.001 * 0.729
          = 0.01458

Therefore, the probability that three tubes are defective is approximately 0.0146, or 1.46%.

(b) To find the probability that at least four tubes are defective out of six, we can use the complementary probability:

P(X >= 4) = 1 - P(X < 4)

To find P(X < 4), we can add up the probabilities of having zero, one, two, or three defective tubes:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
               = (6 choose 0) * 0.1^0 * 0.9^6 + (6 choose 1) * 0.1^1 * 0.9^5 + (6 choose 2) * 0.1^2 * 0.9^4 + (6 choose 3) * 0.1^3 * 0.9^3
               = 0.53144

Therefore, P(X >= 4) = 1 - 0.53144 = 0.46856, or approximately 46.86%.

So the probability that at least four tubes are defective is 0.4686 or 46.86%.

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To find the total number of different routes from town A to town C, we can first find the number of different routes from A to B and then multiply it by the number of different routes from B to C. There are 7 different roads between A and B and 4 different roads between B and C. Therefore, the total number of different routes from A to C is 7 x 4 = 28.

(b) To find the total number of different routes from town A to town C and back, we can use the product rule. There are 28 different routes from A to C (as calculated in part a) and 28 different routes from C to A (since we can use any road once in each direction). Therefore, the total number of different routes from A to C and back is 28 x 28 = 784.

(c) To find the total number of different routes from town A to town C and back in part (b) that visit town B at least once, we can use the principle of inclusion-exclusion. There are 28 different routes from A to C and 28 different routes from C to A. However, we need to subtract the routes that do not visit B at all. To find this number, we can use the product rule again, since there are 5 different roads between A and C that do not go through B (2 from A to C and 3 from C to A). Therefore, the number of routes that do not visit B at all is 2 x 3 = 6. So, the total number of different routes from A to C and back in part (b) that visit B at least once is 28 x 28 - 6 = 784 - 6 = 778.

(d) To find the total number of different routes from town A to town C and back in part (b) that do not use any road twice, we can use the principle of permutations. Since we cannot use any road twice, we need to find the number of permutations of the roads. There are 7 roads between A and B, 4 roads between B and C, and 2 roads between A and C. Therefore, the total number of different routes from A to C and back in part (b) that do not use any road twice is 7P2 x 4P2 x 2P2 = 126 x 12 x 2 = 3024.

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The remaining eigenvalue is λ = -0.007. Thus, the eigenvalues of A are:
X1 = 0
X2 = 0
X3 = -0.007
Arranging them in ascending order, we get:
X1 = -0.007
X2 = 0
X3 = 0

To find the characteristic polynomial of A, we first need to compute the determinant of (A - λI), where I is the identity matrix and λ is a scalar variable:

|0-λ  0.00 0.007|
|0.00 0-λ  0.00 |
|0.00 0.00 0-λ  |

Expanding along the first row, we get:

(0-λ) |0-λ  0.00| - 0.00 |0.00 0-λ | + 0.007 |0.0 0.00|
       |0.00 0-λ |            |0.0 0.00|                 |0.00 0-λ |

Simplifying, we obtain:

-λ³ + (-0.007)λ² = 0

Factoring out λ², we get:

λ²(-λ - 0.007) = 0

Therefore, the characteristic polynomial of A is:

p(A) = λ³ + 0.007λ²

To find the eigenvalues of A, we need to solve the equation p(A) = 0. We can see that one of the roots is λ = 0, which has multiplicity 2 (since it appears as a factor of λ² in the characteristic polynomial). To find the third eigenvalue, we need to solve:

λ³ + 0.007λ² = 0

Factoring out λ² again, we obtain:

λ²(λ + 0.007) = 0

Therefore, the remaining eigenvalue is λ = -0.007. Thus, the eigenvalues of A are:

X1 = 0
X2 = 0
X3 = -0.007

Arranging them in ascending order, we get:

X1 = -0.007
X2 = 0
X3 = 0

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The rate of Change of Function A is 1/2 and function B is 3/2.

We have to the average rate of change of each function from x=2 to x=4.

For Function A:

Here, f(2)= 1 and f(4) = 2

So, the rate of change

= f(4)- f(2)/ (4-2)

= (2-1)/ 2

= 1/2

Function B:

Here, f(2)= 4 and f(4) = 7

So, the rate of change

= f(4)- f(2)/ (4-2)

= (7- 4)/ 2

= 3/2

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Answer:

To find the degree of the product, we need to multiply the highest degree terms of the two factors.

In this case, the two factors are -2x^2 and (4x^3 - 5x^2).

The highest degree term in -2x^2 is -2x^2 itself, which has a degree of 2.

The highest degree term in (4x^3 - 5x^2) is 4x^3, which has a degree of 3.

When we multiply these terms, we get:

-2x^2 * 4x^3 = -8x^(2+3) = -8x^5

Therefore, the degree of the product is 5.

The answer is D) 5.

Step-by-step explanation:

highest exponent number

Step-by-step explanation:

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Ha: difference in Apgar score between first-born and second-born twins is not equal to zero.

The most appropriate test for this scenario would be a paired t-test, as the researchers collected data from the same set of twins and are comparing the differences in Apgar score between the first-born and second-born twins.

The appropriate alternative hypothesis for this test would be "Ha: difference in Apgar score between first-born and second-born twins is not equal to zero."

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The sum of the coefficients in the simplified polynomial is -54.

Adding two integers always results in an integer, if the two integers are positive, their sum will be positive, if two integers are negative, they will yield a negative sum)

To find the sum of the coefficients of the simplified polynomial, first, distribute the constants and then combine like terms.

The given polynomial is:

[tex]$3(x^{10} - x^7 2x^3 - x 7) 4(x^3 - 2x^2 - 5)$[/tex]

Distribute the constants:

[tex]$3x^{10} - 3x^7 - 6x^3 - 3x - 21 + 4x^3 - 8x^2 - 20$[/tex]
Combine like terms:

[tex]$3x^{10} - 3x^7 + (-6x^3 + 4x^3) + (-8x^2) + (-3x) + (-21 - 20)$[/tex]

Which simplifies to:

[tex]$3x^{10} - 3x^7 - 2x^3 - 8x^2 - 3x - 41$[/tex]

Now, sum the coefficients:

[tex]$3 - 3 - 2 - 8 - 3 - 41 = -54$[/tex]

So, the sum of the coefficients in the simplified polynomial is -54.

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100°

Supplementary angle pairs sum to 180°.

Supplementary Angles

Supplementary angle pairs form a straight line. Since straight lines have a measure of 180°, the sum of supplementary angles is always 180°. Supplementary angles do not necessarily have to be adjacent, but the angles above are. Since the angles above create a straight line together, they must be supplementary angles.

Solving for j

Now that we know that the sum must be 180°, we can create an equation to find j.

To solve this, all we need to do is subtract 80 from both sides.

Angle j must have a measure of 100°.

To find all vectors in R3 orthogonal to ū = (-1,1,2) which are a linear combination of vectors ū1 = (1,0,1) and ū2 = (2,2,1), we can use the cross product of ū1 and ū2 to get a vector that is orthogonal to both ū1 and ū2. Then, we can use the dot product to find the scalar multiple of that vector that is orthogonal to ū.

First, we find the cross product of ū1 and ū2:

ū1 x ū2 = (2,-1,-2)

This vector is orthogonal to both ū1 and ū2. To find the scalar multiple of this vector that is orthogonal to ū, we take the dot product:

(2,-1,-2) · (-1,1,2) = 0

This tells us that any scalar multiple of (2,-1,-2) is orthogonal to ū. Therefore, any linear combination of ū1 and ū2 that is a scalar multiple of (2,-1,-2) will also be orthogonal to ū.

To find the 2-norm of these vectors, we can use the formula:

||x|| = sqrt(x1^2 + x2^2 + x3^2)

Let's call the scalar multiple of (2,-1,-2) k:

k(2,-1,-2) = (2k, -k, -2k)

To find the value of k that gives a 2-norm of 5, we set ||k(2,-1,-2)|| = 5:

sqrt((2k)^2 + (-k)^2 + (-2k)^2) = 5

Simplifying this equation, we get:

sqrt(9k^2) = 5

3k = 5

k = 5/3

Therefore, the vector that is a linear combination of ū1 and ū2 and is orthogonal to ū and has a 2-norm of 5 is:

(2/3, -5/3, -10/3)

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Therefore, the standard form polynomial expression that represents the area of the triangle is: [tex](3/2)g^2h - 3gh + h[/tex].

The expression [tex]3g^2 - 6g + 2[/tex] does not represent the area of a triangle because it is not in the form of a polynomial expression that represents the area of a triangle. The area of a triangle is given by the formula:

A = (1/2)bh

Here A is the area, b is the base of the triangle, and h is the height of the triangle.

To write a polynomial expression in standard form that represents the area of a triangle, we need to simplify the formula for A using algebra. Let's assume that [tex]3g^2 - 6g + 2[/tex] represents the base of the triangle and h represents the height of the triangle. Then, we have:

A =[tex](1/2)(3g^2 - 6g + 2)h[/tex]

A =  [tex](3/2)g^2h - 3gh + h[/tex].

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An example of a tiered observation when measuring temperature would be measurements rounded off to the nearest degree.

This means that when you observe and measure the temperature, you would round the values to the nearest whole degree, providing a concise and uniform set of data points for analysis or comparison.

One example of a tiered observation when measuring temperature could be only considering those temperatures in a certain range, such as only observing temperatures that fall within the range of 60-70 degrees Fahrenheit. This would be a tiered observation because it is limiting the range of data being observed.

However, it's important to note that measurements rounded off to the nearest degree could also be considered a tiered observation because it's grouping data into discrete categories based on the rounding method used.
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The form of the following quadratic include the following: D. not a quadratic.

In Mathematics and Geometry, the standard or general form of a quadratic function can be modeled and represented by using the following quadratic equation;

y = ax² + bx + c

Where:

Mathematically, the vertex form of a quadratic equation is given by this formula:

f(x) = a(x - h)² + k

Where:

Additionally, the intercept form of a quadratic equation is given by this formula:

f(x) = a(x - p)(x - q)

In conclusion, we can logically deduce that the given expression is a polynomial function.

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No, It is not a function because the x's (input) do repeat.

We know that;

A relation between a set of inputs having one output each is called a function.

Now, We have to given that;

In the given relation,  

The x's (input) do repeat.

Hence, It is not a function.

Thus, Correct statement is,

No, It is not a function because the x's (input) do repeat.

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The statement "the heights of the bars represent relative class frequencies" is true. The correct answer is C.

A histogram is a graphical representation of the distribution of numerical data. It is commonly used to display the frequency distribution of continuous data in a graphical form.

The horizontal axis of the histogram represents the range of values of the variable being measured, and this range is divided into equal intervals called bins. The vertical axis represents the frequency, or the number of times a value appears in each bin.

The statement "the heights of the bars represent relative class frequencies" is also true. In a histogram, the height of each bar represents the frequency or count of data points that fall within each bin. This height is proportional to the frequency of data points within each bin, and it is often normalized to show the relative frequency of each bin.

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Answer:

[tex]f(x)=\frac{4x+3}{2x}[/tex]

Step-by-step explanation:

We are given the following function: [tex]f(x)=\frac{8x^2+2x-3}{4x^2-2x}[/tex], and we want to simplify it.

Starting with the numerator, we can factor 8x² + 2x - 3 to become (2x-1)(4x+3).

We can also pull out 2x from the denominator to get 2x(2x-1).

Now, our function will look like:

[tex]f(x)=\frac{(2x-1)(4x+3)}{2x(2x-1)}[/tex]

We can cancel 2x-1 from both the numerator and denominator.

We are left with:

[tex]f(x)=\frac{4x+3}{2x}[/tex]

The humidity should be approximately 68%.

a)

When you run the command -

> xtabs(water ~ temp+humid, snail)/4

you get the following output -

Now, we see that the humidity of 60% lies exactly in between the humidity of 45% and 75%. And also the temperature of 25oC lies exactly in between the temperature of 20oC and 30oC.

So, we can proceed by taking the average values to estimate the water content.

Create a colum for the humidity of 60% in between humidity of 45% and 75% by taking the mean of humidity of 45% and 75% as shown -

Humidity

45% 60% 75%

Temp 20 72.5 77 81.5

30 69.5 73.875 78.25

Now, similarly create a row for the temperature of 25oC by taking the average of rows for the temperature of 20oC and 30oC as shown -

Humidity

45% 60% 75%

Temp 20 72.5 77 81.5

25 71 75.4375 79.875

30 69.5 73.875 78.25

So, we can see that the estimated water content for 60% humidity and temperature of 25oC is = 75.4375.

--------------------------------------

b)

Use the following code to fit the regression model for 'water' with predictors 'temperature' and 'humidity'.

> model <- lm(water ~ temp+humid, snail)

Now, you can view the parameters using the code -

> coefficients(model)

This will give you the following output -

> coefficients (model)\n(Intercept)\nhumid\n52.6108059-0.1833333 0.4734890\ntemp\n

So, the estimated regression model is -

Water = 52.6108 - 0.1833(temp) + 0.4735(humid)

We can now use this model to predict the water content for humidity of 60% and temperature of 25oC using following code -

First define your new data using code -

> newdata = data.frame(temp = 25, humid = 60)

And now use -

> predict(model, newdata)

to get the predicted value. You will get the output as -

76.43681

-------------------------------------

c)

Again, define your new parameters as -

> newdata2 = data.frame(temp = 30, humid = 75)

And use the model to predict the water content using -

> predict(model, newdata2)

you will get the output as -

82.62248

So, the predicted water content for 75% humidity and 300C temperature is = 82.62248.

From part (a), we get that the average water content for given condition is 78.25%. The average method used in part (a) is straight forward and doesn't involve much mathematics while the linear regression method uses complex algorithm to predict the value but has much more accuracy than the simple average method because its not necessary that data is always changing with constant rate.

-------------------------------

d)

For a predicted response of 52.6%, we would have -

Water = 52.6108 - 0.1833(temp) + 0.4735(humid) = 52.6

=> 0.4735 (humid) = 0.1833(temp)

Or temp \approx 2.6 (humid)

So, any pair of values satisfying the above relation would give the predicted value same as the intercept value.

For example, humidity = 60% and temperature = 156oC

or, humidity = 45% and temperature = 117oC

But note that the regression model has been trained on values of temperature ranging between 20 to 30 while we are using the temperature of more than 100oC to get the predicted value same as intercept value.

So, this doesn't represent a reasonable prediction.

----------------------------------------------------------

e)

For, predicted value of water = 80%, and temperature of 25oC, the humidity would be -

Water = 52.6108 - 0.1833(temp) + 0.4735(humid) = 80

=> 52.6108 - 0.1833(25)+ 0.4735(humid) = 80

=> humid = 67.52%

So, humidity should be approximately 68%.

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a) This state-diagram corresponds to a birth-death process because the transitions only depend on the current state and not on any previous history of the system. b) We can sum over all values of mp(n) = ∑p(n,m). c. This system can be modeled as a birth-death process, where the states represent the number of taxis and the number of people waiting in line.

Steady-state probabilities of waiting passengers and taxis can be found using balance equations and summing probabilities for the respective cases

a. To draw the state-diagram for this system, we need to identify the different states of the system. In this case, the states are the number of taxis and the number of people waiting in line. Let's denote the number of taxis by n and the number of people waiting in line by m. The states can be represented as (n,m).

For each state, there are two possible transitions: a taxi can arrive, or a passenger can board a taxi. If a taxi arrives, the system moves to state (n+1,m) with probability μ, if there is room for the taxi. If there is no room, the taxi departs immediately and the system moves to state (n,m) with probability λ. If a passenger boards a taxi, the system moves to state (n,m-1) with probability μ. If there are no passengers waiting, the taxi joins the queue and the system moves to state (n+1,m) with probability λ.

This state-diagram corresponds to a birth-death process because the transitions only depend on the current state and not on any previous history of the system.

b. To find the steady-state probability of having n persons waiting in line, we need to use the balance equations. Let p(n,m) be the steady-state probability of being in state (n,m). Then, the balance equations are:

λp(n-1,m) + μp(n,m-1) = (λ+p)m(n,m) + μ(n+1)p(n+1,m)

for n >= 0 and m >= 0. We also have the normalization condition:

∑p(n,m) = 1.

We can solve these equations to find the steady-state probabilities. In this case, we are interested in the probabilities of having n persons waiting in line, so we can sum over all values of m:

p(n) = ∑p(n,m).

c. To find the steady-state probability of having m taxis waiting in the taxi stand, we can use a similar approach. The balance equations are:

λp(n-1,m) + μp(n,m-1) = λ(n+1)p(n+1,m) + (μ+p)m(n,m)

for n >= 0 and m >= 0. We can solve these equations to find the steady-state probabilities. In this case, we are interested in the probabilities of having m taxis waiting in the stand, so we can sum over all values of n:

p(m) = ∑p(n,m).

Overall, this system can be modeled as a birth-death process, where the states represent the number of taxis and the number of people waiting in line. We can use the balance equations to find the steady-state probabilities of having n persons waiting in line or m taxis waiting in the stand.

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Approximately 63.7 grams of radium will be present in 430 years, given that 80 grams are present now.

The half-life of radium is 1690 years, which means that after 1690 years, half of the initial amount will remain. We can use this information to calculate the amount of radium that will be present in 430 years, given that 80 grams are present now.

Let A(t) be the amount of radium present at time t, measured in grams. Then, the formula for the amount of radium after time t, given the initial amount A0, is:

[tex]A(t) = A0 * (1/2)^(t/1690)[/tex]

We can use this formula to find the amount of radium that will be present in 430 years, by setting t = 430 and A0 = 80:

[tex]A(430) = 80 * (1/2)^(430/1690)[/tex]

A(430) ≈ 63.7 grams

Therefore, approximately 63.7 grams of radium will be present in 430 years, given that 80 grams are present now.

The reason for this decrease in the amount of radium over time is due to the process of radioactive decay. Radium atoms are unstable and undergo radioactive decay, which results in the emission of alpha particles and the transformation of the radium atom into a different element. The half-life of radium is the time it takes for half of the initial amount of radium to decay. As the radium atoms continue to decay over time, the amount of radium present decreases exponentially, following the formula above.

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The given relation:

Input    output

-5             7

1                4

6              1

7               4

Is a function.

A relation maps elements (inputs) from one set into elements (outputs)of another set, and a relation is called a function if every element of the first set is mapped into only one element of the second set.

Here the first set is:

Input

-5

1

6

7

And the correspondent pairings are:

7

4

1

4

Notice that every one of the inputs appears only once, then this is a function.

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The solution of the quadratic equations are shown below.

There are various methods that we could use when we want to solve a quadratic equation and these include;

1) Formula method

2) Graphical method

3) Completing the square method

4) Factor method

We have solved the following quadratic equations by factoring.

1)  3x^2 + 13x +10 = 0

x = - 1 and -10/3

2)  12n²-11n +2=0

n = 2/3 and 1/4

3) 5x^2 + 8x +3=0

x = -3/5 and -1

4)  10a²+ a -2=0

a = 2/5 and -1/2

5) 8x^2 + 10x + 3 = 0

x = -1/2 and -3/4

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The amount of a restaurant charge if the ingredients cost $10 is,

⇒ $35

We have to given that;

A restaurant should charge its customers about 3. 5 times the cost of the ingredients.

Hence, We get;

The amount of a restaurant charge if the ingredients cost $10 is,

⇒ 3.5 x $10

⇒ $35

Thus, The amount of a restaurant charge if the ingredients cost $10 is,

⇒ $35

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The probability that (-3) < Y < (-2) for a sample size of 36 is approximately 0.0385.

Given, Yi, i=1, 2, ,…,n, are i.i.d. random variables, each distributed N(-3,81),

i.e., Yi ~ N(-3,81)

We need to find Pr[(-3) < Y < (-2)] for a sample size of 36.

First, we need to standardize the variable Y as follows:

Z = (Y - μ) / σ

where μ is the mean of Y, and σ is the standard deviation of Y.

Here, μ = -3 and σ = 9 (since the standard deviation is the square root of the variance, which is given as 81).

So,

Z = (Y - (-3)) / 9 = (Y + 3) / 9

Now, we need to find Pr[(-3) < Y < (-2)] in terms of Z:

Pr[(-3) < Y < (-2)] = Pr[(-3 + 3)/9 < Z < (-2 + 3)/9]

= Pr[0 < Z < 1/9]

We can use the standard normal distribution table or calculator to find the probability of Z lying between 0 and 1/9.

Using a standard normal distribution table or calculator, we get:

Pr[0 < Z < 1/9] ≈ 0.0385

Therefore, the probability that (-3) < Y < (-2) for a sample size of 36 is approximately 0.0385.

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The number of people at the party who play basketball and are under six feet tall, |B∩U] = 31 . The probability that a randomly chosen party-goer plays basketball and is under six feet tall, P(BU) = 0.732 .

Using the formula: |B∩U| = |B| + |U| - |B∪U|

where, |B| = 30 and |U| = 15 .

|B∪U| = |B| + |U| - |B∩U| + |(not B)∩(not U)|

where, |(not B)∩(not U)| = 9

|B∪U| = 30 + 15 - |B∩U| + 9

|B∪U| = 54 - |B∩U|

So, |B∩U| = 30 + 15 - |B∪U|

|B∪U| = 30 + 15 - |B∩U| + 9

|B∩U| = 36 - |B∪U|

Substituting |B∪U| into the earlier equation:

|B∩U| = 30 + 15 - (36 - |B∪U|)

|B∩U| = 9 + |B∪U|

Using the equation above:

|B∪U| = |B| + |U| - |B∩U| + |(not B)∩(not U)|

Substituting this into the earlier equation:

|B∩U| = 9 + (54 - |B∩U|)

2|B∩U| = 63

|B∩U| = 31.5

Therefore, the number of people at the party who play basketball and are under six feet tall, |B∩U|, is approximately 31.

To find the probability, P(B∩U),

P(B∩U) = |B∩U|/|S|

where |S| is the size of the sample space = 43

Substituting the value of |B∩U|:

P(B∩U) = 31.5/43

P(B∩U) ≈ 0.732

Therefore, the probability that a randomly chosen party-goer plays basketball and is under six feet tall, P(B∩U), is 0.732.

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The exact values of the variables are;

x = 9

y =14.5

To determine the value of the variables, we have that the trigonometric identities are;

From the diagram shown, we can se that the triangle is an isosceles triangle

An  isosceles triangle has two of its sides and angles equal to each other.

Then, the value of the variable x would be;

x = 18/2 = 9

Using the Pythagorean theorem;

15²- 9² = y²

find the square value

y² = 225 - 16

y² = 209

Find square root

y = 14. 5

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Answer:

The volume of a cone is given by the formula:

V = (1/3)πr^2h

where r is the radius of the base, h is the height, and π is the constant pi (approximately 3.14).

Plugging in r = 6 and h = 15, we get:

V = (1/3)π(6^2)(15) = 540π cubic feet

Rounding to the nearest hundredth, we get:

V ≈ 1696.63 cubic feet

Therefore, the volume of the cone is approximately 1696.63 cubic feet.

To check, we can use the formula for the volume of a cone to calculate the volume using different methods. For example, we can use the fact that the cone is one-third the volume of a cylinder with the same base and height. The cylinder has radius 6 feet and height 15 feet, so its volume is:

V_cylinder = π(6^2)(15) = 540π cubic feet

Dividing by 3, we get:

V_cone = (1/3)V_cylinder = (1/3)(540π) = 180π cubic feet

Rounding this to the nearest hundredth, we get:

V_cone ≈ 565.49 cubic feet

This is reasonably close to our previous answer of 1696.63 cubic feet, so we can be confident that our calculation is correct.

Step-by-step explanation:

The solution to the inequality 15 ≤ 5x + 20 < 35 is -1 ≤ x < 3.

Option C is the correct answer.

We have,

To solve the inequality 15 ≤ 5x + 20 < 35,

We need to isolate the variable x by performing the same operation on all three parts of the inequality.

15 ≤ 5x + 20 < 35

Subtract 20 from all three parts:

-5 ≤ 5x < 15

Divide all three parts by 5:

-1 ≤ x < 3

Therefore,

The solution to the inequality 15 ≤ 5x + 20 < 35 is -1 ≤ x < 3.

This means that any value of x between -1 (inclusive) and 3 (exclusive) will satisfy the inequality

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Using the first thirteen rules of inference to derive the conclusions of the given symbolized arguments.

A. Conclusion: v-X

B. Conclusion: L

C. Conclusion: Z

In the first argument, the disjunctive syllogism and conditional proof rules were used to derive the conclusion. In the second argument, the material implication, double negation, commutation, modus ponens, and disjunctive syllogism rules were used to derive the conclusion. In the third argument, the material implication, De Morgan's Law, disjunctive syllogism, addition, and negation elimination rules were used to derive the conclusion.

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Using the rules of inference to derive the conclusions of the given symbolized arguments.

A. Conclusion: v-X

X-M (Premise)

M (Assumption for Conditional Proof)

|-(v-M) (Premise)

v (Disjunctive Syllogism: 3, M)

v-X (Conditional Proof: 2-4)

B. Conclusion: L

L» (B v 0) (Premise)

-(-0.-B) (Premise)

-0 v -(-B) (Material Implication: 2)

-0 v B (Double Negation: 3)

B v 0 (Commutation: 1)

-B»0 (Material Implication: 5)

-B (Disjunctive Syllogism: 4,6)

0 (Modus Ponens: 6,7)

L (Disjunctive Syllogism: 1,8)

C. Conclusion: Z

Ko (Y.Z) (Premise)

(K.C) v ( MK) (Premise)

-Z (Assumption for Conditional Proof)

-(Y.Z) (Material Implication: 1)

-Y v -Z (De Morgan's Law: 4)

Y (Disjunctive Syllogism: 2, MK)

-Z v -Z (Addition: 3)

Z (Negation Elimination: 7)

The disjunctive syllogism and conditional proof procedures were employed to reach the conclusion in the first argument.

The material implication, double negation, commutation, modus ponens, and disjunctive syllogism rules were employed to reach the conclusion in the second argument. The material implication, De Morgan's Law, disjunctive syllogism, addition, and negation elimination rules were utilized to obtain the conclusion in the third argument.

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Answer:

54734431

Step-by-step explanation:

54734431

Answer:

a) a = 1 ; b = 2 ;c =-1

c) -1 + √2  ; -1 - √2

Step-by-step explanation:

        x² + 2x = 1

a)      x² + 2x - 1 = 0

Compare with ax² + bx + c = 0

a = 1 ; b = 2 and c = -1

b)  

      [tex]\boxed{x=\dfrac{-b \± \sqrt{b^2-4ac}}{2a}}[/tex]

           [tex]= \dfrac{-2 \± \sqrt{2^2-4*1*(-1)}}{2*1}\\\\\\\\C) \ =\dfrac{-2 \± \sqrt{4+4}}{2}\\\\=\dfrac{-2 \± \sqrt{8}}{2}\\\\=\dfrac{-2 \±2\sqrt{2}}{2}\\\\=\dfrac{2(-1 \± \sqrt{2})}{2}\\\\= -1 \± \sqrt{2}[/tex]

          x  = -1 + √2   or x = -1 -√2