What is the slope of the line through the points (-2, -1) and (4, 3)?


2/3


2/5


5/2

Answer:

i think h=17

Step-by-step explanation:

-18h = 17 = xh/6

Subtract xh/6 from both sides of the equation.

-18 - xh/6 = 17

Answer:

Step-by-step explanation:8

Answer:

y=3x+2

Step-by-step explanation:

y=mx+b

b=2

m=3

plug in

y=3x+2

Answer:

I. Length = 600 meters

II. Width = 30 meters

Step-by-step explanation:

Given the following data;

Length = 120m

Width = 6m

Drawing scale = 1:5

To find the length and width using the given drawing scale;

This ultimately implies that, with this drawing scale, the length and width of the rectangle would increase by a factor of 5 (multiplied by 5) i.e the rectangle is 5 times bigger in real-life than on the diagram.

For the length;

120 * 5 = 600 meters

For the width;

6 * 5 = 30 meters

Therefore, the dimensions of the rectangle using the given drawing scale is 600 meters by 30 meters.

the answer are 4kg ,12kg and 20 kg

the conditional probability pr[x>1000 | x>500] is P(x > 500) = 1 - CDF(500).

Given that x has an exponential likelihood thickness work with β = 500, we are able to compute the likelihood that x is more noteworthy than 500, i.e., P(x > 500).

For an exponential conveyance with parameter β, the likelihood thickness work (PDF) is given by:

f(x) = (1/β) * e^(-x/β), where x ≥ 0.

To discover P(x > 500), we got to coordinate the PDF from 500 to limitlessness:

P(x > 500) = ∫[500, ∞] (1/β) * e^(-x/β) dx.

Let's calculate this likelihood:

P(x > 500) = ∫[500, ∞] (1/500) * e^(-x/500) dx.

To calculate this indispensably, ready to utilize the truth that the necessity of the PDF over its whole extent is rise to 1. So, ready to revamp the likelihood as:

P(x > 500) = 1 - P(x ≤ 500).

Since the exponential conveyance is memoryless, P(x ≤ 500) is break even with the total conveyance work (CDF) at 500.

P(x > 500) = 1 - CDF(500).

The CDF of the exponential dissemination is given by:

CDF(x) = ∫[0, x] (1/β) * e^(-t/β) dt.

To calculate P(x > 500), we have to assess CDF(500) and subtract it from 1.

Presently, let's calculate P(x > 500):

P(x > 500) = 1 - CDF(500)

= 1 - ∫[0, 500] (1/500) * e^(-t/500) dt.

To calculate the conditional probability P(x > 1000 | x > 500), we have to consider the occasion that x > 500 is our modern test space. The conditional probability is at that point given by:

P(x > 1000 | x > 500) = P(x > 1000, x > 500) / P(x > 500).

Since x takes after an exponential conveyance, it is memoryless, which implies the likelihood of x > 1000 given x > 500 is the same as the likelihood of x > 500. Hence, we have:

P(x > 1000 | x > 500) = P(x > 500) = 1 - CDF(500).

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Answer:

x = 5

x = 2

Step-by-step explanation:

2(2x + 4) = 8x - 12

2x + 4 = 4x - 6

x + 2 = 2x - 3

x = 5

Question 17

8/4 = (x + 2)/(2x - 2)

2(2x - 2) = x + 2

4x - 4 = x + 2

3x = 6

x = 2

The value of the expression 6x(y^2 - z) when x = 0, y = -1, and z = 2 is 0.

To find the value of the algebraic expression 6x(y^2 - z) when x = 0, y = -1, and z = 2, we substitute the given values into the expression.

First, let's evaluate the inner expression (y^2 - z):

Substituting y = -1 and z = 2, we have (-1)^2 - 2 = 1 - 2 = -1.

Now, we substitute x = 0 and the result of the inner expression (-1) into the outer expression:

6x(y^2 - z) = 6(0)(-1) = 0.

Therefore, when x = 0, y = -1, and z = 2, the value of the expression 6x(y^2 - z) is 0.

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Answer:

The answer to the question provided is 14.

Step-by-step explanation:

》The Distance Formula:

[tex] d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } [/tex]

》Plug in.

[tex]d = \sqrt{( - 5 - 9)^{2} + (3 - 3) ^{2} } \\ d = \sqrt{( - 14)^{2} + (0) ^{2} } \\ d = \sqrt{196 + 0} \\ d = \sqrt{196} \\ d = 14[/tex]

Answer:

86 degrees

Step-by-step explanation:

180 - (74 + 20) = 86

Answer:

2

Step-by-step explanation:

14 / (7)

Answer:

D.) $100.36

Step-by-step explanation:

66.02/100 = 0.6602

0.6602 x 109.5 = 72.2919

72.2919/100 = 0.722919

0.722919 x 138.83 = 100.36

Based on the given data, a 95% confidence interval estimate of the mean number falls between 12.22 and 16.78.

To construct a confidence interval for the mean, we need to calculate the sample mean and the margin of error. The formula for the margin of error is:

Margin of Error = Z * [tex]\frac{standard deviation}{\sqrt{n} }[/tex]

where Z is the critical value corresponding to the desired confidence level (for 95% confidence level, Z ≈ 1.96), Standard Deviation is the sample standard deviation, and n is the sample size.

From the given data, we calculate the sample mean to be 14.33 and the sample standard deviation to be 1.91. Since the population is assumed to be normally distributed, we can use the Z-distribution.

Using the formula for the margin of error, we find:

Margin of Error = 1.96 * (1.91 / √9) ≈ 1.39

The confidence interval is calculated by subtracting and adding the margin of error to the sample mean:

Confidence Interval = (14.33 - 1.39, 14.33 + 1.39) = (12.94, 15.72)

Rounded to at least two decimal places, the 95% confidence interval estimate of the mean number is approximately (12.22, 16.78). This means that we can be 95% confident that the true mean number falls within this interval.

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Answer:

1.A,B,C,D

2. AB, CD

3. AC, BD

4. Line AD (don't take my word for this one)

Step-by-step explanation:

Answer:

Step-by-step explanation:

3x^2 + 24x + 48 + 43 - 48

3(x^2 + 8x + 16) - 5

3(x + 4)^2 - 5

vertex at (-4,-5)

Answer:

vertex = (- 4, - 5 )

Step-by-step explanation:

The equation of a quadratic in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Given

y = 3x² + 24x + 43 ← factor out 3 from the first 2 terms

  = 3(x² + 8x) + 43

To complete the square

add/subtract ( half the coefficient of the x- term)² to x² + 8x

y = 3(x² + 2(4)x + 16 - 16) + 43

y = 3(x + 4)² - 48 + 43

y = 3(x + 4)² - 5 ← in vertex form

with vertex = (- 4, - 5 )

Given:Standard deviation, s = $2000Confidence level = 96%Margin of error, E = $500We have to find the sample size, n.

Sample size formula is given as:\[n={\left(\frac{z\text{/}2\times s}  {E}\right)}^{2}\]Where, z/2 is the z-score at a 96% confidence level. Using the standard normal table, we can get the value of z/2 as follows:z/2 = 1.750Incorporating all the values in the formula, we get:\[n={\left(\frac{1.750\times 2000}{500}\right)}^{2}\] Simplifying,\[n=21\]Therefore, a sample size of 21 is required if we wish to be 96% confident that our mean will be within $500 of the true mean salary of college instructors.

To determine the sample size needed to be 96% confident that the mean salary will be within $500 of the true mean salary, we can use the formula for sample size in a confidence interval.

The formula is:

n = (Z * σ / E)^2

Where:

n is the required sample size

Z is the z-score corresponding to the desired confidence level (in this case, 96% confidence level)

σ is the standard deviation of the population (given as $2000)

E is the maximum error tolerance (given as $500)

First, we need to find the z-score corresponding to a 96% confidence level. The remaining 4% is split evenly between the two tails of the distribution, so we look up the z-score that corresponds to the upper tail of 2% (100% - 96% = 4% divided by 2).

Using a standard normal distribution table or a calculator, the z-score for a 2% upper tail is approximately 2.05.

Now we can substitute the values into the formula:

n = (Z * σ / E)^2

n = (2.05 * 2000 / 500)^2

Calculating this expression:

n = (4100 / 500)^2

n = 8.2^2

n = 67.24

Rounding up to the next whole number, the required sample size is approximately 68.

Therefore, a sample size of 68 is needed to be 96% confident that the mean salary will be within $500 of the true mean salary of college instructors.

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Given information: The starting salaries of college instructors have a standard deviation (SD) of $2000. The sample size that is needed if we want to be 96% confident that our mean will be within $500 of the true mean salary of college instructors is to be calculated.

Hence, 49 is the sample size that is needed if we want to be 96% confident that our mean will be within $500 of the true mean salary of college instructors.

The formula for the sample size required is as follows:

[tex]n = [(Z \times \sigma) / E]^{2}[/tex]

Here, Z is the value from the normal distribution for a given confidence level, σ is population standard deviation, E is the maximum error or the margin of error, which is [tex]\$500n = [(Z \times \sigma) / E]^2[/tex]

On substituting the given values, we get:

[tex]n = [(Z \times \sigma) / E]^2[/tex]

[tex]n= [(Z \times \$2000) / \$500]^2[/tex]

[tex]n = [(1.7507 \times \$2000) / \$500]^2[/tex]

[tex]n = (7.003 \times 7.003)[/tex]

n = 49 (rounded off to the next whole number)

Hence, 49 is the sample size that is needed if we want to be 96% confident that our mean will be within $500 of the true mean salary of college instructors.

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We know that the line integral of the curve C is equal to the difference between the anti-derivative at the final point B and the antiderivative at the initial point A.  Therefore, Vg⋅dr= F(B) - F(A)⇒ Vg⋅dr= [2(5)²(5) + 3(5)² + C] - [90 + C]⇒ Vg⋅dr= 184

The question states that the function g(x,y) = 2x² + 3y² and the curve C is the arc of a curve from point A(4,3) to point B(5,5). The task is to find the value of the line integral along curve C.

Therefore, we need to use the fundamental theorem for line integrals to evaluate the line integral. To use the fundamental theorem for line integrals, we must first evaluate the gradient vector field of the function. Then we need to find the antiderivative of the gradient vector field of the function. We can obtain the antiderivative by integrating the gradient vector field along the curve C using the initial and final points of the curve. The value of the line integral of the curve C is equal to the difference between the antiderivative at the initial point A and the antiderivative at the final point B, i.e., Vg⋅dr= F(B) - F(A).

Step-by-step solution: Given, the function g(x,y) = 2x² + 3y²Let us calculate the gradient vector of the function g(x,y).∇g(x,y) = [∂g/∂x, ∂g/∂y]⇒ ∇g(x,y) = [4x, 6y]Therefore, the gradient vector field of g(x,y) is V = [4x, 6y].

Now, we need to find the antiderivative of the gradient vector field of the function. Let us integrate V along the curve C from A(4,3) to B(5,5). The curve C is given by y = x + 1.We know that the line integral along curve C is given by the formula, Vg⋅dr= ∫C V . dr = F(B) - F(A)

Therefore, we need to find the antiderivative F of V.F(x,y) = ∫V dx⇒ F(x,y) = 2x²y + 3y² + C. Since we have two variables, we need to find the value of C using the initial point A(4,3).F(4, 3) = 2(4)²(3) + 3(3)² + C⇒ F(4, 3) = 90 + C

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Given that the function is G(x, y) = 2x² + 3y² and Arc of a curve C from point A(4, 3) to point B(5, 5). The value of the line integral [tex]\int _C[/tex] (2x² + 3y²) ds is 106.67.

Solution: In the given question, we have a function G(x, y) = 2x² + 3y² and an arc of a curve C from point A(4, 3) to point B(5, 5).

We are to use the fundamental theorem for line integrals to find the value of  [tex]\int _C[/tex] (2x² + 3y²) ds.

Step 1: First, we will find the parametric equations of the given curve.

The points A(4, 3) and B(5, 5) are given.

We can write the parametric equations of the curve C as: x = f(t) and y = g(t), where a ≤ t ≤ b, and f(a) = 4, g(a) = 3, f(b) = 5, g(b) = 5.

Here, the curve C is the straight line from A(4, 3) to B(5, 5), so we can choose any convenient parameterization.

A possible one is: t → r(t) = (4 + t, 3 + t), 0 ≤ t ≤ 1.

Step 2: Next, we will find dr/dt and ds/dt.

We have: r(t) = (4 + t)i + (3 + t)j

⇒ dr/dt = i + j.

Square of the magnitude of the tangent vector: |dr/dt|² = (1)² + (1)²

= 2.

Magnitude of the normal vector:

|n| = √(ds/dt)²

= √(2)

= √2.

Magnitude of the velocity vector:

|v| = √(dr/dt)²

= √2.

Step 3: Now, we will find the limits of integration and substitute the required values in the integral.

Given:  [tex]\int _C[/tex] (2x² + 3y²) ds.

We have: r(t) = (4 + t)i + (3 + t)j

⇒ r'(t) = i + j

⇒ |r'(t)| = √2.

We know that the length of the curve C from A to B is given by:

Length of the curve = [tex]\int _C[/tex] ds

= [tex]\int_a^b[/tex] |r'(t)| dt

= [tex]\int_0^1[/tex] √2 dt

= √2.

Now, we have the value of ds: ds = √2 dt.

Then, we can write the integral as follows:

[tex]\int _C[/tex] (2x² + 3y²) ds = [tex]\int_0^1[/tex] (2(4 + t)² + 3(3 + t)²) √2 dt

= [tex]\int_0^1[/tex] (32 + 32t + 10t²) √2 dt

= [32t + 16t² + (10/3)t³[tex]]_0^1[/tex]

= 32 + 16 + (10/3)

= 106.67.

Thus, the value of the line integral  [tex]\int _C[/tex] (2x² + 3y²) ds is 106.67.

The required answer is: 106.67.

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Answer:

its 62!

p-by-step explanation:

Answer:

62.

Step-by-step explanation:

A=2(wl+hl+hw)=2·(3·5+2·5+2·3)=62

y=-1x+4

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hope it helps

sorry I had no idea how to explain

Answer:

area = 175 cm^2

Step-by-step explanation:

area = base * height

area = 5 cm + 35 cm

area = 175 cm^2

Answer: D

Step-by-step explanation:

just got it correct.

Answer:

the answer is D

hope it help

Answer: D, x²-6x+9

Step-by-step explanation:

(x-3)² is the same as (x-3)(x-3).

So you would do:

x times x=x²

x times -3=-3x

-3 times x=-3x

-3 times -3=9

The equation would look like this:

x²-3x-3x+9

Then you would have to collect the like terms. Like terms are terms that have the same variables and powers. So it would look like this:

x²-6x+9

Hope this helps :)

Answer:

838.4

Step-by-step explanation:

From the question,

Total worth of the property = 10480.

Eugenia share: Her 10 brothers share = 1:4

Total ratio = 1+4 = 5.

Her 10 brother's share = (4/5)(10480)

Her 10 brother's share = 41920/5

Her 10 brother's share = 8384.

If the 10 brothers shared their portion equally,

Each brother's share = 8384/10

Each brother's share = 838.4

Answer:

i dont know

Step-by-step explanation:

Answer:

The answer is D

Step-by-step explanation:

They are using 135 times the weeks, but the car already had 100 miles on it to start.  Juan buys a car that has 40 miles originally on it and he drives it 150 miles per week. So, the word problem is D

There is no good answer for this.

You would do s/12, since there is 1 adult for every 12 kids. Then, round the number to the nearest 1. (if it is a decimal, there is a number and remainder of students.)

Answer:

There are 10 odd numbers: {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} There are 4 factors of 8: {1, 2, 4, 8} There is 1 number in both lists: {1} Probability of an odd or a factor of 8 = (10 + 4 - 1)/20 = 13/20

The PGF of the number of tosses until the sequence HTH appears is given by G(t) = (G × t / (1 - (1/2) × t).

To find the expected number of tosses until three consecutive heads appear, we can approach the problem using the concept of the probability generating function (PGF).

Let's define a random variable X as the number of tosses until three consecutive heads appear. We want to find E(X), the expected value of X.

To determine the PGF of X, we consider the possible outcomes at each toss. There are three possible outcomes: T (tails), H (heads), and the sequence HTH (three consecutive heads).

At the first toss, the possible outcomes are T and H. The PGF for this situation is given by:

[tex]G1(t) = Pr(X = 1) \times t^1 + Pr(X = 2) \times t^2[/tex]

Since we can have either T or H on the first toss, we have Pr(X = 1) = 1/2 and Pr(X = 2) = 1/2. Therefore:

[tex]G1(t) = (1/2) \times t + (1/2) \times t^2[/tex]

Now, let's consider the situation after the first toss:

If the first toss resulted in T, we are back to the starting point. Therefore, the PGF is G(t).

If the first toss resulted in H, we are one step closer to our goal (HTH). The PGF for this situation is G(t) × t.

Combining these two cases, we have:

[tex]G(t) = (1/2) \times t + (1/2) \times t^2 \times G(t)[/tex]

Simplifying the equation, we get:

[tex]G(t) = (1/2) \times t / (1 - (1/2) \times t^2)[/tex]

Next, let's consider the situation after the second toss:

If the second toss resulted in T, we are back to the starting point. Therefore, the PGF is G(t).

If the second toss resulted in H, we are still one step closer to our goal (HTH). The PGF for this situation is G(t) × t.

Combining these two cases, we have:

G(t) = (1/2) × t + (1/2) × t × G(t)

Simplifying the equation, we get:

G(t) = (1/2) × t / (1 - (1/2) × t)

Finally, we can calculate the expected value E(X) using the PGF:

E(X) = G'(1)

To find the derivative of G(t), we can use the quotient rule:

G'(t) = [(1 - t) × 1 - t × (-1/2)] / (1 - (1/2) × [tex]t)^2[/tex]

Simplifying the equation, we get:

G'(t) = 1 / (1 - (1/2) × [tex]t)^2[/tex]

Evaluating G'(1), we have:

[tex]E(X) = G'(1) = 1 / (1 - (1/2) \times 1)^2 = 1 / (1 - 1/2)^2 = 1 / (1/2)^2 = 1 / (1/4) = 4[/tex]

Therefore, the expected number of tosses until three consecutive heads appear is 4.

Additionally, the PGF of the number of tosses until the sequence HTH appears is given by:

G(t) = (G × t / (1 - (1/2) × t)

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Answer:

this is true but what do we have to do?

Step-by-step explanation:

Answer: x = 35, angle 1 = 85, angle 2 = 70, angle 3 = 25

Step-by-step explanation:

Finding x: 85+2x+x-10=180

3x= 105

x= 35

x-10 = angle 3 due to a property of parallels lines so put x in to get that angle 3 is 25 degrees

angle 2x = 70 degrees

70 + 25+angle 1 = 180

angle 1 = 85

angle 2 = 180-85-25 = 70