ratio of 3 boys and 4 girls there are now 12 boys

Answer:

Common ratio = r

r = [tex]\frac{a_2}{a_1} =\frac{3}{4}=0.75[/tex]

Answer:3.99

Step-by-step explanation:

i just kniow

The given equation for the domain is C=5.5b, where b represents the number of dollar bills produced and C represents the total cost of producing those dollar bills.

We are told that the mint produces at least 420 dollar bills but not more than 425 dollar bills. Therefore, the domain of the function C=5.5b for this situation is the set of values of b that satisfy this condition.

In interval notation, we can represent this domain as follows:

Domain: 420 ≤ b ≤ 425

Therefore, the domain of the function C=5.5b for this situation is 420 ≤ b ≤ 425.

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A polynomial that satisfies the given conditions is f(x) = a(x + 2)^2(x - 2)^2x^3, where a is a constant.

To find the polynomial f(x) that meets the given requirements, we can start by noting that since -2 and 2 are zeros of multiplicity 2, the factors (x + 2)^2 and (x - 2)^2 must be included in the polynomial. Additionally, since 0 is a zero of multiplicity 3, the factor x^3 must also be included.

So far, we have the polynomial in the form f(x) = a(x + 2)^2(x - 2)^2x^3, where a is a constant that we need to determine.

To find the value of a, we can use the fact that f(-1) = 45. Plugging in x = -1 into the polynomial, we get:

f(-1) = a(-1 + 2)^2(-1 - 2)^2(-1)^3

= a(1)^2(-3)^2(-1)

= 9a

Setting 9a equal to 45, we can solve for a:

9a = 45

a = 5

So the polynomial f(x) that satisfies the given conditions is:

f(x) = 5(x + 2)^2(x - 2)^2x^3.

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The probability that a student has a dog, given that they do not have a cat is 5/6.

The probability that a student has a dog given that they do not have a cat is :

P ( Has a dog | Does not have a cat ) = P ( Has a dog and does not have a cat) / P ( Does not have a cat )

Total number of students = 11 + 10 + 5 + 2 = 28

P ( Has a dog | Does not have a cat ):

= (10 / 28) / (12 / 28)

= 10 / 12

= 5 / 6

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We can say with 90% confidence that the true mean level of polyunsaturated fatty acid in this brand of diet margarine is between 16.95 and 17.23. The answer is 90%.

Using the t-distribution with 5 degrees of freedom (n-1), we can calculate the t-value for a 90% confidence interval. We use a one-tailed test because we want to find the confidence interval for values greater than 16.65:

t-value = t(0.90,5) = 1.476

Now we can calculate the margin of error (E) for a 90% confidence interval:

E = t-value * (s / √n) = 1.476 * (0.31 / √6) = 0.28

Finally, we can calculate the confidence interval:

16.95 + E = 16.95 + 0.28 = 17.23

Therefore, we can say with 90% confidence that the true mean level of polyunsaturated fatty acid in this brand of diet margarine is between 16.95 and 17.23. Since the range of values between 16.65 and 17.32 falls within this confidence interval, we can also say that we are 90% confident that the true mean level of polyunsaturated fatty acid falls within this range.

So, the answer is 90%.

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a) y = -0.75x + 25 approximate equation of the line of best fit for the data.

b) The prediction for time spent doing homework for someone who spends 12 hours watching TV is 16 hours.

a) I added the graph to Desmos, online graphic calculator and superimposed a line which I think is a good/decent fit to the given data. Image for reference:

Thus, we get the line of best fit (approximately) equation as

y = -0.75x + 25 (in equation)

b)  From the given equation, the prediction for time spent doing homework for someone who spends 12 hours watching TV is

y = -0.75x + 25.

y = -0.75(12) + 25

y = -9 + 25

y = 16

Thus, the prediction for time spent doing homework for someone who spends 12 hours watching TV is 16 hours.

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The phasor form (in the rms sense) of the given expressions are:

a. 20∠(-180°) V

b. 6 x 10⁻⁶∠90° A

c. 3.6 x 10⁻⁶∠(-20°) A

a. The given expression is in the form of 20 sin (ωt - φ), where ω is the angular frequency and φ is the phase angle in degrees. To convert it to phasor form, we need to express it as a complex number in the form of Vrms∠θ, where Vrms is the root mean square (rms) value of the voltage and θ is the phase angle in radians. In this case, the rms value is 20 V and the phase angle is -180° (since it is given as -180° in the expression). The phasor form can be represented as 20∠(-180°) V.

b. The given expression is in the form of 6 x 10⁻⁶ cos(ωt), where ω is the angular frequency. To convert it to phasor form, we need to express it as a complex number in the form of Irms∠θ, where Irms is the rms value of the current and θ is the phase angle in radians. In this case, the rms value is 6 x 10^(-6) A and the phase angle is 90° (since it is cos(ωt)). The phasor form can be represented as 6 x 10⁻⁶∠90° A.

c. The given expression is in the form of 3.6 x 10⁻⁶ cos(ωt - φ), where ω is the angular frequency and φ is the phase angle in degrees. To convert it to phasor form, we need to express it as a complex number in the form of Irms∠θ, where Irms is the rms value of the current and θ is the phase angle in radians. In this case, the rms value is 3.6 x 10⁻⁶ A and the phase angle is -20° (since it is given as -20° in the expression). The phasor form can be represented as 3.6 x 10⁻⁶∠(-20°) A.

THEREFORE, the phasor form (in the rms sense) of the given expressions are:

a. 20∠(-180°) V

b. 6 x 10⁻⁶∠90° A

c. 3.6 x 10⁻⁶∠(-20°) A.

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The limit of the sequence is ln(7) and the sequence converges to ln(7).

To determine the convergence of the sequence, we need to investigate the behavior of its terms as n approaches infinity.

We have:

[tex]an = ln(7n^2 + 3) − ln(n^2 + 3)[/tex]

To simplify this expression, we can use the property of logarithms that states [tex]ln(a) - ln(b) = ln(a/b)[/tex]:

[tex]an = ln[(7n^2 + 3)/(n^2 + 3)][/tex]

Now, let's investigate the behavior of the fraction inside the natural logarithm as n approaches infinity. We can use the fact that the leading term in the numerator and denominator dominates as n gets large:

[tex](7n^2 + 3)/(n^2 + 3) ≈ 7[/tex]

Therefore, as n approaches infinity, an approaches ln(7), which is a finite number. Thus, the sequence converges to ln(7).

Therefore, the limit of the sequence as n approaches infinity is:

[tex]lim n→∞ an = ln(7)[/tex]

Therefore, the limit of the sequence is ln(7) and the sequence converges to ln(7).

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The CDF of X for a constant parameter a > 0, a rayleigh random variable x is [tex]F_{X}(x) = 1 - e^{(-a^2x^2/2)[/tex] for x > 0, and 0 otherwise.

To find the CDF (Cumulative Distribution Function) of a Rayleigh random variable X with the given [tex]PDF f_X(x) = {a^2xe^{(-a^2x^2/2)[/tex] for x > 0, 0 otherwise}, we need to integrate the PDF from 0 to x. Here's the solution:

[tex]CDF F_X(x)[/tex] = ∫[tex][a^2xe^{(-a^2x^2/2)]}dx[/tex] from 0 to x

Let's denote u = [tex]a^2x^2/2[/tex]. Then, du = [tex]a^2xdx[/tex]. So the integral becomes:

[tex]F_X(x)[/tex] = ∫[tex][e^{(-u)}du][/tex] from 0 to [tex]a^2x^2/2[/tex]

Now, integrate [tex]e^{(-u)}[/tex] with respect to u:

[tex]F_X(x)[/tex] = [tex]-e^{(-u)[/tex] | from 0 to [tex]a^2x^2/2[/tex]

Evaluate the definite integral:

[tex]F_X(x)[/tex] = [tex]-e^{(-a^2x^2/2)} + e^{(0)} = 1 - e^{(-a^2x^2/2)[/tex]

Thus, the CDF of X is [tex]F_X(x)[/tex] = [tex]1 - e^{(-a^2x^2/2)[/tex] for x > 0, and 0 otherwise.

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Answer:

60

Step-by-step explanation:

The arc cosine of 0.5 can be written as cos⁻¹(0.5). To find its value in degrees, we can use a calculator or reference table. Specifically, we have:

cos⁻¹(0.5) ≈ 60 degrees

Without knowledge of the distribution of x, we cannot make any further calculations.

Without additional information about the distribution of variable x, we cannot determine p(2.5 < x < 6.5).

If we know the distribution, we could use the probability density function (PDF) or cumulative distribution function (CDF) to calculate the probability. For example, if x is a normally distributed variable with mean 5 and standard deviation 1, we could use the standard normal distribution to find:

p(2.5 < x < 6.5) = p((2.5-5)/1 < (x-5)/1 < (6.5-5)/1)

= p(-2.5 < z < 1.5)

= Φ(1.5) - Φ(-2.5)

≈ 0.7745 - 0.0062

≈ 0.7683

But without knowledge of the distribution of x, we cannot make any further calculations.

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The measurement of angle A is approximately 38.8 degrees and the measurement of angle B is approximately 51.2 degrees.

Triangles and the connections between their sides and angles are studied in the branch of mathematics known as trigonometry. Trigonometric functions like sine, cosine, and tangent are used to solve problems involving right triangles and other geometric shapes in a variety of disciplines, including science, engineering, and physics.

We can use trigonometry to solve for the angle A.

First, we can find the length of the hypotenuse AB using the Pythagorean theorem:

AB² = BC² + CA²

AB² = 19² + 22²

AB² = 905

AB = √(905)

AB = 30.1

Next, we can use the sine function to find the measure of angle A:

sin(A) = BC / AB

sin(A) = 19 / 30.1

A = sin⁻¹(19 / 30.1)

A = 38.8

Finally, we can use the fact that the sum of the angles in a triangle is 180 degrees to find the measure of angle B:

B = 90 - x

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Using Leibniz's rule, the final answer is xd/dx ∫ f(t) dt = x f(x) + ∫ f(t) dt + x f'(x)

Using Leibniz's rule, we have:

x d/dx ∫ f(t) dt = x f(x) + ∫ x d/dx f(t) dt

The first term x f(x) comes from differentiating the upper limit of integration with respect to x, while the second term involves differentiating under the integral sign.

If we assume that f(x) is a differentiable function, then by the chain rule, we have:

d/dx f(x) = d/dx [f(t)] evaluated at t = x

Therefore, we can rewrite the second term as:

∫ x d/dx f(t) dt = ∫ x d/dt f(t) dt evaluated at t = x

= ∫ f(t) dt + x f'(x)

Substituting this into the original equation, we obtain:

xd/dx ∫ f(t) dt = x f(x) + ∫ f(t) dt + x f'(x)

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Mr Tan is 69 years old now.

Let's denote Mr. Tan's age as T and Peiling's age as P.

We are given two pieces of information:
5 years ago, Mr Tan was 4 times as old as Peiling.
Peiling is 48 years younger than Mr Tan now.
Now let's translate these into equations:
T - 5 = 4 * (P - 5)
P = T - 48
Next, we'll solve for P in equation 1 and then substitute it into equation 2:
T - 5 = 4 * (P - 5)
T - 5 = 4P - 20
4P = T + 15
Now, substitute P from equation 2 into this equation:
4 * (T - 48) = T + 15
4T - 192 = T + 15
3T = 207
T = 69.

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Answer:

0.00446

Step-by-step explanation:

1 decailiters = 10,000 milliliters.

44.6 / 10,000= 0.00446

The f-statistic use to test the hypothesis.is 34.719264.

To calculate the F-statistic for testing the overall significance of the linear regression model,

we need to compare the regression sum of squares (SSR) to the residual sum of squares (SSE) and the degrees of freedom associated with each.

The formula for the F-statistic is:

F = (SSR / k) / (SSE / (n - k - 1))

where k is the number of predictor variables in the model, and n is the sample size.

Since the question does not provide the values of k and n, I will assume that k = 1 (simple linear regression) and use the information given in the previous question to find n.

From the previous question, we have:

SSE = 24.074572

MSE = SSE / (n - 2) = 2.674952

SSTO = SSR + SSE = 83.820408

R-squared = SSR / SSTO = 0.7136

We can use R-squared to find SSTO:

SSTO = SSR / R-squared = 83.820408 / 0.7136 = 117.539337

Then, we can use SSTO and MSE to find n:

SSTO / MSE = n - 2

117.539337 / 2.674952 = n - 2

n = 45

Now we can substitute the values of k, n, SSR, and SSE into the formula for the F-statistic:

F = (SSR / k) / (SSE / (n - k - 1))

F = ((117.539337 - 83.820408) / 1) / (24.074572 / (45 - 1 - 1))

F = 34.719264

Therefore, the F-statistic is 34.719264.

This value can be used to test the hypothesis that the slope coefficient is equal to zero, with a significance level determined by the degrees of freedom and the chosen alpha level.

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The general solution to the differential equation 6y'' + 6y = 2tan(6) - 1/2e^{3t} is y(t) = c1*cos(t) + c2*sin(t) + (1/6)tan(6) - (1/36)e^{3t}.

To find this solution, first, solve the homogeneous equation 6y'' + 6y = 0. The characteristic equation is 6r^2 + 6 = 0. Solving for r gives r = ±i.

The homogeneous solution is y_h(t) = c1*cos(t) + c2*sin(t), where c1 and c2 are constants. Next, find a particular solution y_p(t) for the non-homogeneous equation by using an ansatz. For the tan(6) term, use A*tan(6), and for the e^{3t} term, use B*e^{3t}.

After substituting the ansatz into the original equation and simplifying, we find that A = 1/6 and B = -1/36. Thus, y_p(t) = (1/6)tan(6) - (1/36)e^{3t}. Finally, combine the homogeneous and particular solutions to get the general solution: y(t) = c1*cos(t) + c2*sin(t) + (1/6)tan(6) - (1/36)e^{3t}.

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The vertex form of the quadratic function y = x² - 6x - 7 is y = (x - 3)² - 16

Given the quadratic function in the question:

y = x² - 6x - 7

The vertex form of a quadratic function is expressed as:

y = a(x - h)² + k

Where (h, k) is the vertex of the parabola and "a" is a coefficient that determines the shape of the parabola.

To write y = x² - 6x - 7 in vertex form, we need to complete the square.

We can do this by adding and subtracting the square of half the coefficient of x:

y = x² - 6x - 7

y = (x² - 6x + 9) - 9 - 7 (adding and subtracting 9)

y = (x - 3)² - 16

Hence, the vertex form is:

y = (x - 3)² - 16

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Let's find the domain and codomain for each of the given matrices.

(a) The given matrix product is: [ 6 3 ] [x1] [-1 7 ] [x2]

The domain of a transformation is the set of all possible input vectors. In this case, the input vector is [x1, x2]. Since there are no restrictions on the values of x1 and x2, the domain is all real numbers for both components.

Mathematically, the domain is R^2, where R represents the set of all real numbers. The codomain of a transformation is the set of all possible output vectors. The given transformation is a 2x2 matrix, which means it maps R^2 to R^2.

Thus, the codomain is also R^2.

(b) The given matrix product is: [ 2 1 -6 ] [x1] [ 3 7 -4 ] [x2] [ 1 0 3 ] [x3]

The domain of this transformation is the set of all possible input vectors. In this case, the input vector is [x1, x2, x3]. Since there are no restrictions on the values of x1, x2, and x3, the domain is all real numbers for each component.

Mathematically, the domain is R^3. The codomain of this transformation is the set of all possible output vectors. The given transformation is a 3x3 matrix, which means it maps R^3 to R^3.

Thus, the codomain is also R^3.

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The derivate f'(a) when the point (a, b) is a local minimum. In this case, the correct answer is: d. It's zero

When a point (a, b) is a local minimum, the derivative f'(a) will be zero. This is because, at a local minimum, the function changes its direction from decreasing to increasing, and the slope of the tangent line is zero.

If the point (a,b) is a local minimum of a differentiable function f, then f'(a) = 0.

This is because at a local minimum, the slope of the tangent line to the graph of f at point (a,b) is zero (since the derivative f'(x) gives the slope of the tangent line at point x). If the slope of the tangent line at point (a,b) is zero, then the derivative f'(a) must also be zero.

Therefore, the correct answer is (d) it's zero

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The value of p is 20/19, 19/16.

What is the distance formula?

Using their coordinates, an algebraic expression provides the distances between two points (see coordinate system). The distance formula is a formula used to determine how far apart two places are from one another. The dimensions of these points are unlimited.

Here, we have

Given: The distance between the points (1, 2p) and (1- p, 1) is 11-9p.

We have to find the value of p.

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

11 - 9p = [tex]\sqrt{(1-p-1)^2 +(1-2p)^2}[/tex]

(11 - 9p)² = 5p² - 4p + 1

121 + 81p² - 198p =  5p² - 4p + 1

121 + 81p² - 198p -  5p² + 4p - 1 = 0

120 + 76p² - 194p = 0

76p² - 194p + 120 = 0

38p² - 97p + 60 = 0

p = 20/19, 19/16

Hence, the value of p is 20/19, 19/16.

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The value of p is 20/19, 19/16.

What is the distance formula?

Using their coordinates, an algebraic expression provides the distances between two points (see coordinate system). The distance formula is a formula used to determine how far apart two places are from one another. The dimensions of these points are unlimited.

Here, we have

Given: The distance between the points (1, 2p) and (1- p, 1) is 11-9p.

We have to find the value of p.

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

11 - 9p = [tex]\sqrt{(1-p-1)^2 +(1-2p)^2}[/tex]

(11 - 9p)² = 5p² - 4p + 1

121 + 81p² - 198p =  5p² - 4p + 1

121 + 81p² - 198p -  5p² + 4p - 1 = 0

120 + 76p² - 194p = 0

76p² - 194p + 120 = 0

38p² - 97p + 60 = 0

p = 20/19, 19/16

Hence, the value of p is 20/19, 19/16.

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Answer:

Let's represent the number of stones each bag holds by the variable x.

Then the total number of stones in the 6 large bags is 6x, and the total number of stones in the 4 small bags is 4x.

Therefore, the total number of stones can be represented by the polynomial:

6x + 4x = 10x

So the polynomial is 10x, which represents the total number of stones.

Answer:

Step-by-step explanation:

large bags = x

small bags = y

6x + 4y

The amount of fencing, dirt and sod for the baseball field are: length of Fencing & 1410.5 ft. Area of the sod ≈ 118017.13ft² Area of the field covered with dist ≈ 7049.6ft²

Area of a circle = πr²

Circumference of a circle = 2πr

where r is the radius of the circle

The area of a Quarter of a circle is therefore;

Area of a circle/ 4

The perimeter of a Quarter of a Circle is;

The perimeter of a circle/4

Fencing = ¼ x 2 x π x 380 + 2 x 15 +2 x 380 + ¼ x 2 x π x 15

Fencing = 197.5π + 190π = 1410.5 feet.

Grass =

π/4 x (380 - 6)² + 87 ² - π/4 × (87 + 30)² + 2 x 380 x 15 + π/4 x 15² - (3/4) x π x 10² - 25π

= 31528π + 18969 = 118017.13

The area Covered by the sod is about 118017.13Sq ft.

Dirt = π/4 x 380 ² - π/4 x (380 - 6)² + π/4 (87 + 30)² - 87² + π100 = (18613π - 30276)/4

= 7049.6

The area occupied by the dirt is about 7049.6 Sq feet

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If a sample contains 12.5% parent, 3 half lives are passed and if the radioactive pair is K-40 and Ar-40,  the actual age of the sample is 3.75 billion years.

If a sample contains 12.5% parent, it means that 3 half-lives have passed. To calculate the actual age of the sample with the radioactive pair K-40 and Ar-40, you need to know the half-life of K-40, which is 1.25 billion years. The actual age of the sample can be found by multiplying the half-life by the number of half-lives passed: 1.25 billion years * 3 = 3.75 billion years. Therefore, the actual age of the sample is 3.75 billion years.

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Answer:

2 hours 30 mins

Step-by-step explanation:

We Know

Caleb and his friends went to see a movie at 7:35 p.m.

They left at 10:05

How long was the movie?

We Take

10:05 - 7:35 = 2 hours 30 mins

So, the movie is 2 hours 30 mins long.

Answer:

A) Slope=1

Step-by-step explanation:

To find the slope of the line passing through the points (2, 3) and (4, 5), we can use the slope formula:

slope = (y2 - y1) / (x2 - x1)

where (x1, y1) = (2, 3) and (x2, y2) = (4, 5).

Plugging in the values, we get:

slope = (5 - 3) / (4 - 2)

= 2 / 2

= 1

Therefore, the slope of the line is 1. Answer: A) 1.

Answer

m = 1

In-depth Explanation

To calculate slope, we use the formula

Where m is the slope and (y2, y1)( x2, x1) are points on the line.

Calculating :

Therefore, the slope is m = 1

The diameter of the tire based on the amplitude of the sinusoidal graph of the height of the tire valve, y inches above or below the center of the tire is 20 inches.

A sinusoidal function is a periodic function that is based on the sine or cosine function.

The shape of the graph is the shape of a sinusoidal function graph.

The y-coordinates of the peak and the through are; y = 10 and y = -10

The shape of the tire is a circle

The amplitude, A, of the sinusoidal function, which has a magnitude equivalent to the length of the radius of the tire is therefore;

A = (10 - (-10))/2 = 10

The radius of the tire, r = A = 10 inches

The diameter of the tire = 2 × The radius of the tire

The diameter of the tire = 2 × 10 inches = 20 inches

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+) Case 1: a = 0

=> f(x) = 0×x²+c = c

=> for all values of x, f(x) always = c (does not satisfy the requirement)

+) Case 2: a≠0

=> f(x) = ax²+c

=> for every non-zero a, f(x) has only one solution x

Ans: a≠0, c ∈ R

P/s: c can be any value (in case you don't know the symbols above)

Ok done. Thank to me >:333