6. Write the equation of the line in slope-intercept form that has the following points: (1, -3) (0, -5)

Answer:

9

Step-by-step explanation:

Based on the graph at 9 inches it first leveled out

Answer:

25x

Step-by-step explanation:

(-5) square × X square

=25x

Answer:

25x  

Step-by-step explanation:

you multiply -5x to itself

Answer:

Your total purchase was $9.40

Step-by-step explanation:

6 x 0.49 = 2.94

1 x 3.48 = 3.48

2 x 1.49 = 2.98

2.94 + 3.48 + 2.98 = 9.4

Answer:

option 4

Step-by-step explanation:

option 2 and 4 are the more closest options but option 2 is a graph function between the negative numbers, so the answer is option 4.

Answer:

(I am not sure what the question means but according to what I understood I got these as my answers)

33 pounds = 55%

114 = 240 %

95% = 57 pounds

The algebraic expression that represents the given phrase is b + 15 = 180.

The algebraic expression that represents the phrase "the sum of the number of boys and 15 girls is 180" is:b + 15 = 180Explanation:Let's assume that the number of boys is "b".

According to the question, the total number of students in the class can be calculated by adding the number of boys (b) to the number of girls (15) which is equal to 180 students.Using algebra,

we can write this statement as an equation: b + 15 = 180Hence, the algebraic expression that represents the given phrase is b + 15 = 180

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I guess you're asking about the probability density for the random variable \(|A-B|\) where \(A,B\) are independent and identically distributed uniformly on the interval (0, 15). The PDF of e.g. \(A\) is

\(\mathrm{Pr}(A=a) = \begin{cases}\dfrac1{15} & \text{if } 0 < a < 15 \\\\ 0 & \text{otherwise}\end{cases}\)

It's easy to see that the support of \(|A-B|\) is the same interval, (0, 15), since \(|x|\ge0\), and

• at most, if \(A=15\) and \(B=0\), or vice versa, then \(|A-B|=15\)

• at least, if \(A=B\), then \(|A-B|=0\)

Compute the CDF of \(C=|A-B|\) :

\(\mathrm{Pr}(C\le c) = \mathrm{Pr}(|A - B| \le c) = \mathrm{Pr}(-c \le A - B \le c)\)

This probability corresponds to the integral of the joint density of \(A,B\) over a subset of a square with side length 15 (see attached). Since \(A,B\) are independent, their joint density is

\(\mathrm{Pr}(A=a,B=b) = \begin{cases}\dfrac1{15^2} & \text{if } (a,b) \in (0,15) \times (0,15) \\ 0 &\text{otherwise}\end{cases}\)

The easiest way to compute this probability is by using the complementary region. The triangular corners are much easier to parameterize.

\(\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \mathrm{Pr}(|A-B| > c) \\\\ ~~~~~~~~ = 1 - \int_0^{15-c} \int_{a+c}^{15} \frac{db\,da}{15^2} - \int_c^{15} \int_0^{a-c} \frac{db\,da}{15^2} \\\\ ~~~~~~~~ = 1 - \frac1{225} \left(\int_0^{15-c} (15 - a - c) \, da + \int_c^{15} (a - c) \, da\right)\)

In the second integral, substitute \(a=15-a'\) and \(da=-da'\), so that

\(\displaystyle \int_c^{15} (a-c) \, da = \int_{15-c}^0 (15-a'-c) (-da') = \int_0^{15-c} (15 - a' - c) \, da'\)

which is the same as the first integral. This tells us the joint density is symmetric over the two triangular regions.

Then the CDF is

\(\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \frac2{225} \int_0^{15-c} (15 - a - c) \, da \\\\ ~~~~~~~~ = 1 - \frac2{225} \left((15-c) a - \frac12 a^2\right) \bigg|_{a=0}^{a=15-c} \\\\ ~~~~~~~~ = \begin{cases}0 & \text{if } c < 0 \\\\ 1 - \dfrac{(15-c)^2}{225} = \dfrac{2c}{15} - \dfrac{c^2}{225} & \text{if } 0 \le c < 15 \\\\ 1 & \text{if } c \ge 15\end{cases}\)

We recover the PDF by differentiating with respect to \(c\).

\(\mathrm{Pr}(|A-B| = c) = \begin{cases}\dfrac2{15} - \dfrac{2c}{225} & \text{if } 0 < c < 15 \\\\ 0 & \text{otherwise}\end{cases}\)

The proportionality constant between half-life and inverse rate constant affects the slope of a plot of ln(t1/2) vs. 1/t. Half-life is the amount of time it takes for half of the initial quantity of a substance to decay or react to form a product.

The rate constant is the proportionality constant in the rate law equation that relates the rate of the reaction to the concentration of reactants raised to some power. The rate constant is also used in the equation for the half-life of a reaction.The half-life of a reaction is proportional to the inverse of the rate constant. Mathematically, it can be represented a :t1/2 ∝ 1/kTherefore, as the rate constant increases, the half-life of the reaction decreases and vice versa. In other words, there is an inverse relationship between the half-life of a reaction and the rate constant.

This relationship can be observed in a plot of ln(t1/2) vs. 1/t. The slope of this plot is equal to the proportionality constant between half-life and inverse rate constant, which is given by the expression:k = ln(2)/t1/2Therefore, the slope of the plot is directly proportional to ln(2), which is a constant. This means that the proportionality constant between half-life and inverse rate constant does affect the slope of a plot of ln(t1/2) vs. 1/t.

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Answer:

a. y=7x or n=7t

b. 42

c. 19 years

d. The answer to that would be that the equation represent an average American uses 7 trees every single year.

Two possible solutions are: (4.5, 8.5) and (8, 6.3)

Let x represents hours dog walking and y represents hours washing cars.

So, we have:

Earnings = Rate of dog walk * x + Rate of car wash * y

This gives

Earnings = 7x + 11y

He wants to earn at least $125.

This means that:

7x + 11y ≥ 125

See attachment for the graph of the inequality

From the attached graph, two possible solutions are: (4.5, 8.5) and (8, 6.3)

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Answer:

\(x\)-intercepts at \((1,0)\) and \((0.8,0)\)

Step-by-step explanation:

A quadratic in the form \(ax^2+bx+c=y\) crosses the \(x\)-axis when \(y=0\).

The first step is replacing \(f(x)\) with \(y\).

So: \(y= -5x^2+9x-4\). With the information above, we can find the \(x\) intercepts by setting \(y = 0\).

Therefore \(0=-5x^2+9x-4\).

Now we can use the quadratic formula because it is in the form \(ax^2+bx+c=0\).

Note the quadratic formula: \(\frac{-b\frac{+}{-}\sqrt{b^2-4ac} }{2a } = x\).

To find the values of \(a\),\(b\) and \(c\) we can compare the equation to the general equation.

Therefore: \(a=-5\), \(b=9\) and \(c=-4\).

Now put these values into the quadratic formula:

\(\frac{-9\frac{+}{-}\sqrt{(9)^2-4(-5)(-4)} }{2(-5) } = x\)

And simplify:

\(\frac{-9\frac{+}{-}\sqrt{81-80} }{-10 }\) , \(\frac{-9\frac{+}{-}\sqrt{1} }{-10 }\).

\(\sqrt{1} =1\), therefore \(x = \frac{-9\frac{+}{-}1 }{-10 }\)

Now we can have two values for \(x\). One when we take away the discriminant (\(b^2-4ac\)) and one when we add it.

So \(x = \frac{-10}{-10} = 1\)

or

\(x = \frac{-8}{-10} = 0.8\)

Therefore \(x\)-intercepts at \((1,0)\) and \((0.8,0)\)

Answer:

(x + 7)(x - 4) = x2 - 4x + 7x - 28 = x2 + 3x - 28

To find the circumference of a circle with a radius of 7 cm, Erin should use 2×7×π. This is because the circumference of a circle is calculated by the formula 2×π×r.

The circumference is the measure of the perimeter of a circle. Since we know that from every point on the circle to its center has an equal length and is said to be its radius, the perimeter of the circle we can write as

Circumference = 2 × π × r units.

Here r is the radius of the circle.

It is given that Erin wants to find the circumference of a circle with a radius of 7 cm.

So, first, she has to know the formula for finding the circumference.

Here the radius is given as 7 cm i.e., r = 7 cm

Then, the circumference of the given circle is

= 2 × π × r

On substituting the radius value into the above formula, we get

Circumference = 2 × π × 7 cm

⇒ 14π cm

So, she needs to use option A. 2 × 7 × π  for finding the required circumference. This is the first step for finding the circumference of a circle.

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According to a recent study of 125 students at Texas Woman's University, the average student spends 6.7 nights per month in an alcohol-related event, with a standard deviation of 3.4 nights.

The study sample consisted of 125 students, and the average number of nights spent in an alcohol-related event was found to be 6.7, with a standard deviation of 3.4. With this information, we can calculate the margin of error for the confidence interval using the formula:

margin of error = (critical value) × (standard deviation / sqrt(sample size)). For a 95% confidence level, the critical value is approximately 1.96. Plugging in the values, we get the margin of error as \(\((1.96) \times \frac{3.4}{\sqrt{125}} \approx 0.61\)\).

To determine the confidence interval, we take the average (6.7) and subtract the margin of error (0.61) to get the lower bound: 6.7 - 0.61 = 6.1 nights. Similarly, we add the margin of error to the average to get the upper bound: 6.7 + 0.61 = 7.3 nights. Therefore, we can accurately report to the parents of potential/incoming freshman that the typical student at Texas Woman's University spends between 6.1 and 7.3 nights per month in an alcoholic environment, with 95% confidence.

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The length of the picture frame is less than 14.5 inches.

The perimeter of a rectangle is defined as the addition of the lengths of the rectangle's four sides.

The perimeter of a rectangle = 2(L+W)

Let the width of the picture frame is b

So length = b +3

The perimeter of the picture frame (p) as

⇒ 2 (b + b+3)

⇒ (2b+3)

⇒ 4b+6

Given that perimeter is less than 52 inches.

⇒ p < 52

⇒ 4b + 6 < 52

⇒ 4b < 46

⇒ b < 11.5

So, the length = b+3 =11.5 + 3 = 14.5

Hence,  the length of the picture frame is less than 14.5 inches.

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The number that will complete the square to solve equation is 1/36.

What is completing the square?

For some values of h and k, completing the square is an elementary algebraic method for changing a quadratic polynomial of the form

ax^2 + bx + c to the form a^2 + k. In other words, the quadratic expression is completed by inserting a perfect square trinomial.

Consider, the given polynomial

3w^2 - w - 24 = 0

Rewrite the polynomial in the form ax^2 + bx = c

Add 24 on both sides,

3w^2 - w - 24 + 24 = 24

⇒ 3w^2 - w = 24

Divide both sides by 3,

\(w^2-\frac{1}{3}w = 8\)

To complete the square x^2 + bx, we add \((\frac{b}{2} )^2\)

Here, b = -1/3

So, \((\frac{b}{2})^2 = (\frac{-1/3}{2})^2 = (-\frac{1}{6})^2 = \frac{1}{36}\)

Adding to both sides of the equation, we have

\(w^2-\frac{1}{3}w+\frac{1}{36} = 8+\frac{1}{36}\)

Hence, the number that will complete the square to solve an equation is 1/36.

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Answer:

623 dvds in one display case.

Step-by-step explanation:

9 display cases 5615 dvds

5615 divided by 9=623.8 repeated

you cant have .8 dvds so the answer would be 623 dvds in one display case.

The potential at the center of the square due to the four equal charges is (8k * q) / l.

To find the potential at the center of the square due to the four charges, we can calculate the electric potential at that point due to each charge individually and then sum them up.

Given:

Four equal charges, q, placed at the corners of a square of side length l.

The center of the square is at the center of the coordinate system.

The electric potential, V, at a point due to a point charge q is given by the formula:

V = k * q / r

Where:

k is the electrostatic constant (k = 9 * 10^9 Nm^2/C^2)

q is the magnitude of the charge

r is the distance from the charge to the point

Since the charges are placed at the corners of a square, the distance from each charge to the center of the square is l/2. Thus, the potential at the center due to each charge is:

V = k * q / (l/2) = 2k * q / l

Since all four charges are equal, the total potential at the center of the square is the sum of the potentials due to each charge:

V_total = 2k * q / l + 2k * q / l + 2k * q / l + 2k * q / l

V_total = (8k * q) / l

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The potential at the center of the square is the sum of the potentials due to each charge, which can be calculated using the formula V = k * q / r. Since the charges are placed at the corners of the square of side l, the distance from the center of the square to each charge is sqrt(2) * l/2. Therefore, the potential at the center is (4 * sqrt(2) * k * q) / l.

The potential at the center of the square can be found by considering the contributions from each of the four charges. Since the charges are equal and placed at the corners of the square, the electric field at the center will be zero. Therefore, the potential at the center of the square is simply the sum of the potentials due to each charge.

The potential due to a single charge can be calculated using the formula:

V = k * q / r

where V is the potential, k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance.

Since the charges are placed at the corners of the square of side l, the distance from the center of the square to each charge is sqrt(2) * l/2. Plugging in these values, we get:

V = 4 * (k * q / (sqrt(2) * l/2))

V = (4 * sqrt(2) * k * q) / l

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To find the value of the second Taylor polynomial P2(x) of the given function f(x) = 6cos(2x) - 2x^2 at x0 = 0, we can use the Taylor polynomial formula and evaluate it at x = 1.3.

The second Taylor polynomial P2(x) of a function f(x) at x0 is given by the formula:

P2(x) = f(x0) + f'(x0)(x - x0) + (f''(x0)/2!)(x - x0)^2

First, we need to find the first and second derivatives of the function f(x) = 6cos(2x) - 2x^2. The first derivative is f'(x) = -12x - 12sin(2x), and the second derivative is f''(x) = -24 - 24cos(2x).

Next, we evaluate these derivatives at x0 = 0:

f(0) = 6cos(0) - 2(0)^2 = 6

f'(0) = -12(0) - 12sin(0) = 0

f''(0) = -24 - 24cos(0) = -48

Now, we substitute these values into the Taylor polynomial formula:

P2(x) = f(0) + f'(0)(x - 0) + (f''(0)/2!)(x - 0)^2

      = 6 + 0(x) + (-48/2)(x^2)

      = 6 - 24x^2

Finally, we evaluate P2(1.3):

P2(1.3) = 6 - 24(1.3)^2

       = 6 - 24(1.69)

       = 6 - 40.56

       = -34.56

Therefore, P2(1.3) is approximately -34.56.

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The hypothesis test in which the significance level is a = 0.05 and the probability power of the test is (B) 0.82.

To find the power of the test, we subtract the probability of a Type II error from 1.

Given:

Significance level (α) = 0.05

Probability of Type II error (β) = 0.18

Power = 1 - β

Power = 1 - 0.18

Power = 0.82

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In scenario three, we explore additional options e and f, each with their own unique probabilities and potential outcomes.

Option e introduces a 50% chance of winning $0 or an unspecified amount, while option f offers a 50% chance of winning either $20 or $60.

These options introduce different potential outcomes and associated probabilities compared to the original scenarios. In option e, there is an equal chance of winning nothing or winning an unspecified amount of money.

The introduction of these options expands the range of possible outcomes and probabilities in scenario three, providing different risk-reward trade-offs for the decision-maker.

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The orthonormal basis of p is {N1, N2, N3} = {(1/3, -2/3, 2/3), (1/√15, 3/√15, -1/√15), (-2/√33, -1/√33, 4/√33)}.

Let {v1, v2, v3} be the given basis of p.

Apply Gram-Schmidt orthonormalization process to B = {(1, -2, 2), (2, 2, 1), (-2, 1, 3)} as follows:v1 = (1, -2, 2)N1 = v1/‖v1‖ = (1/3, -2/3, 2/3)v2 = (2, 2, 1) - (v2 ⋅ N1) N1= (2, 2, 1) - (5/3, -4/3, 4/3)= (1/3, 10/3, -1/3)N2 = v2/‖v2‖ = (1/√15, 3/√15, -1/√15)v3 = (-2, 1, 3) - (v3 ⋅ N1) N1 - (v3 ⋅ N2) N2= (-2, 1, 3) - (-4/3, 8/3, -4/3) - (-2/√15, -4/√15, 7/√15)= (-2/3, -2/3, 10/3)N3 = v3/‖v3‖ = (-2/√33, -1/√33, 4/√33)

Therefore the orthonormal basis of p is {N1, N2, N3} = {(1/3, -2/3, 2/3), (1/√15, 3/√15, -1/√15), (-2/√33, -1/√33, 4/√33)}.Answer: {(1/3, -2/3, 2/3), (1/√15, 3/√15, -1/√15), (-2/√33, -1/√33, 4/√33)}.

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The orthonormal basis for the given basis isB = {B₁, B₂, B₃} = {(1, -2, 2)/3, (1, 3, 0)/√10, (-1/√10)(1, 1, -3)}Given basis is B = {(1, -2, 2), (2, 2, 1), (-2, 1, -2)}

Let’s begin the Gram-Schmidt orthonormalization process for the given basis and transform it into an orthonormal basis.

Step 1: Normalize the first vector of the basis.B₁ = (1, -2, 2)

Step 2: Project the second vector of the basis onto the first vector and subtract it from the second vector of the basis.

B₂ = (2, 2, 1) - projB₁B₂= (2, 2, 1) - [(2+(-4)+2)/[(1+4+4)] B₁]B₂ = (2, 2, 1) - (0.5)(1, -2, 2)B₂ = (1, 3, 0)

Step 3: Normalize the vector obtained in step 2.B₂ = (1, 3, 0)/ √10

Step 4: Project the third vector of the basis onto the orthonormalized first and second vectors and subtract it from the third vector.

B₃ = (-2, 1, -2) - projB₁B₃ - projB₂B₃ = (-2, 1, -2) - [(2+(-4)+2)/[(1+4+4)] B₁] - [(1+9+0)/10 B₂]

B₃ = (-2, 1, -2) - (0.5)(1, -2, 2) - (1.0)(1/ √10)(1, 3, 0)B₃ = (-2, 1, -2) - (0.5)(1, -2, 2) - (1/√10)(1, 3, 0)

B₃ = (-1/√10)(1, 1, -3)

Therefore, the orthonormal basis for the given basis isB = {B₁, B₂, B₃} = {(1, -2, 2)/3, (1, 3, 0)/√10, (-1/√10)(1, 1, -3)}

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Answer:

g = f - e

Step-by-step explanation:

f - g = e

- g = e - f

g. = f - e

a. Yes, the columns of matrix a are linearly independent if the equation ax = 0 has the trivial solution.

This means that the only solution to the equation is x = 0, which implies that the vectors in the matrix are not linear combinations of each other.

b. No, if s is a linearly dependent set, then each vector is not necessarily a linear combination of the other vectors in s. A linear combination is a combination of two or more vectors where each vector is multiplied by a scalar and then added together.

A linearly dependent set means that one of the vectors can be expressed as a linear combination of the other vectors, but not necessarily all of them.c. No, the columns of any 4 5 matrix are not necessarily linearly dependent.

Linear dependence is determined by the rank of the matrix. If the rank of the matrix is equal to the number of columns, then the columns of the matrix are linearly independent.

However, if the rank of the matrix is less than the number of columns, then the columns of the matrix are linearly dependent.

d. Yes, if x and y are linearly independent, then any linear combination of x and y must equal to zero. This means that the coefficients of x and y must be equal to zero in order for the linear combination to equal zero.

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Complete question:

Determine if the columns of the matrix form a linearly independent set. Justify your answer. Choose the correct answer below. A. The columns of the matrix do form a linearly independent set because the set contains more vectors than there are entries in each vector. B. The columns of the matrix do form a linearly independent set because there are more entries in each vector than there are vectors in the set. C. The columns of the matrix do not form a linearly independent set because the set contains more vectors than there are entries in each vector. D. The columns of the matrix do not form a linearly independent set because there are more entries in each vector than there are vectors in the set.

Answer:

There are 200 base balls

Step-by-step explanation:

240 divided by 6 is 40, times 40 by the 5 and you get 200

This is a simple problem of ranges, we will find that the ranges for the main floor and the balcony are:

[-6ft, 8ft] or {x ∈ R | -6ft ≤ x ≤ 8ft}

[26ft, 37ft] or {x ∈ R | 26ft ≤ x ≤ 37ft}

respectively.

We know that:

The main floor of the auditorium ranges from 6ft below the stage  to 8 feet above the stage.

The floor of the balcony ranges from 26 to 37 feet above the stage

The ranges relative to the stage will be, in interval notation, the interval that contains the two extremes of the range.

For the main floor, it goes from -6ft to 8ft (the first one is negative because is 6ft below the stage) then this interval will be: [-6ft, 8ft]

While in set-builder notation we want to specify that the variable must be a real number, so we start with: {x ∈ R

And we need to specify that x can go from -6ft to 8ft, then we add the restriction:

{x ∈ R | -6ft ≤ x ≤ 8ft}

Notice that we use a close interval here.

While for the balcony, the minimum is 26ft and the maximum is 37ft, then the interval will be [26ft, 37ft]

For the set-builder notation we will have:

{x ∈ R | 26ft ≤ x ≤ 37ft}

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