Right triangle. Find the exact values of x and y.​

We first calculate D⁴:

D⁴ = |1⁴ 0 0 |

|0 2⁴ 0 |

|0 0 4⁴|

Substituting into the formula, we get:

A⁴ =

Question 1:

To find the eigenvalues and eigenvectors of matrix A, we solve the characteristic equation:

|A - λI| = 0

where I is the identity matrix and λ is the eigenvalue.

Substituting A, we get:

|1-λ 1 3 |

|1 5-λ 1 | = 0

|3 1 1-λ|

Expanding the determinant, we get:

(1-λ) [(5-λ)(1-λ) - 1] - (1)[(1)(1-λ) - (3)(1)] + (3)[(1)(1) - (5-λ)(3)] = 0

Simplifying, we get:

-λ³ + 7λ² - 14λ + 8 = 0

This equation can be factored as:

-(λ-1)(λ-2)(λ-4) = 0

Therefore, the eigenvalues of A are λ1 = 1, λ2 = 2, and λ3 = 4.

To find the eigenvectors, we solve the equation (A-λI)x = 0 for each eigenvalue.

For λ1 = 1, we get:

|0 1 3 | |x1| |0|

|1 4 1 | |x2| = |0|

|3 1 -0 | |x3| |0|

Simplifying, we get the system of equations:

x2 + 3x3 = 0

x1 + 4x2 + x3 = 0

3x1 + x2 = 0

Solving this system, we get:

x1 = -3x3

x2 = x3

x3 = x3

So, the eigenvector corresponding to λ1 = 1 is:

v1 = (-3, 1, 1)

Similarly, for λ2 = 2, we get:

v2 = (-1, 1, -1)

And for λ3 = 4, we get:

v3 = (1, 1, -3)

Therefore, the eigenvalues of A are 1, 2, and 4, and the corresponding eigenvectors are (-3, 1, 1), (-1, 1, -1), and (1, 1, -3).

Question 2:

To find the matrix P that transforms A to diagonal form, we need to find the eigenvectors of A and use them as columns of P. That is:

P = [v1 v2 v3]

where v1, v2, and v3 are the eigenvectors of A.

From Question 1, we have:

v1 = (-3, 1, 1)

v2 = (-1, 1, -1)

v3 = (1, 1, -3)

So, the matrix P is:

P = |-3 -1 1|

| 1 1 1|

| 1 -1 -3|

To calculate A⁴, we use the formula:

Aⁿ = PDⁿP⁻¹

where Dⁿ is the diagonal matrix with the eigenvalues raised to the nth power.

So, we first calculate D⁴:

D⁴ = |1⁴ 0 0 |

|0 2⁴ 0 |

|0 0 4⁴|

Substituting into the formula, we get:

A⁴ =

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The conditions that must hold for inferential procedures to be valid for this scenario are:

(b) the two sample groups must be independent of each other, (c) the data distribution for both men's and women's hemoglobin levels must be normally distributed or each sample size must be larger than 30, and (d) the observations within each sample must be independent.

(b) The two sample groups must be independent of each other because the samples should not be related in any way that could influence the results.

(c) The data distribution for both men's and women's hemoglobin levels must be normally distributed or each sample size must be larger than 30. If the data is not normally distributed, then the Central Limit Theorem can be applied if the sample size is greater than 30.

(d) The observations within each sample must be independent to avoid bias in the results. This means that each observation should not be influenced by any other observation.

(a) The expected count for each level of the categorical variable must be at least 5. This condition is relevant for Chi-square tests of independence, which are not being used in this scenario.

(e) There must be 10 successes and 10 failures in each sample. This condition is relevant for tests of proportions, which are not being used in this scenario.

So b,c and d are correct options.

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G. I and III only. Thus, integrals I and III necessarily have the same value, and the correct answer is G.

I. ∫[a, b] f(x)dx: This integral compute the area under the curve of f(x) from x=a to x=b.

II. ∫[a/2, b/2] f(2x)dx: This integral computes the area under the curve of f(2x) from x=a/2 to x=b/2. The function f(2x) represents a horizontal compression of the original function f(x) by a factor of 2, and the limits of integration are also halved. So, this integral doesn't necessarily have the same value as integral I.

III. ∫[a+c, b+c] f(x−c)dx: This integral computes the area under the curve of f(x−c) from x=a+c to x=b+c. The function f(x−c) represents a horizontal shift of the original function f(x) by c units, but it does not change the shape of the curve. Since the limits of integration are also shifted by c units, this integral has the same value as integral I.

Thus, integrals I and III necessarily have the same value, and the correct answer is G.

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A. The domain is N and the image is the set of all natural numbers that are products of two natural numbers

B. The image is all values of y in Q such that (1 - 2x)/3 is defined.

C. we note that y² + 1 is always odd, so the image is the set of all odd natural numbers greater than or equal to 2.

What is function?

In mathematics, a function is a relationship between two sets of elements, called the domain and the range, such that each element in the domain is associated with a unique element in the range.

(a) R is a function because for each x in N, there exists a unique y in N such that xy. The domain is N and the image is the set of all natural numbers that are products of two natural numbers.

(b) R is a function because for each x in Q, there exists a unique y in Q such that 2x+3y=1. To find the domain, we solve for x in terms of y: 2x = 1 - 3y, x = (1 - 3y)/2. The domain is all values of y in Q such that (1 - 3y)/2 is defined. Simplifying, we get y ≠ 1/3. Therefore, the domain is Q - {1/3}. To find the image, we solve for y in terms of x: 3y = 1 - 2x, y = (1 - 2x)/3. The image is all values of y in Q such that (1 - 2x)/3 is defined. Therefore, the image is Q.

(c) R is a function because for each x in N, there exists a unique y in N such that y²+1=x. To find the domain, we solve for x in terms of y: y² = x - 1, y = ±√(x - 1). Since we are given that x and y are natural numbers, the domain is the set of all natural numbers greater than or equal to 2. To find the image, we note that y² + 1 is always odd, so the image is the set of all odd natural numbers greater than or equal to 2.

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Answer:

v would be 28 degrees

Step-by-step explanation:

You do the total angle take away by 43.

71 - 43 = 28

Answer:

The kinetic energy (KE) of an object with mass (m) moving at a velocity (v) can be calculated using the formula:

KE = 1/2 * m * v^2

Substituting the given values:

KE = 1/2 * 1200 kg * (8.33 m/s)^2

KE = 41,147.5 J

Therefore, the kinetic energy of the object is 41,147.5 Joules.

Step-by-step explanation:

We would need a sample size of at least 17 mice per group to detect a difference in proportions with a power of 0.8 and a significance level of 0.05.

To determine the sample size needed for a study, we can use power analysis. In this case, we want to detect a difference in proportions between two groups with a significance level of 0.05 and a power of 0.8.

We can use the following formula to calculate the sample size per group:

n = (Z_1-α/2 + Z_1-β)² × (p_1(1-p_1) + p_2(1-p_2)) / (p_1 - p_2)²

where:

Z_1-α/2 is the z-score corresponding to the chosen significance level (0.05/2 = 0.025 for a two-tailed test)

Z_1-β is the z-score corresponding to the chosen power (0.8 in this case)

p_1 is the proportion of wild type mice that develop cancer (0.25)

p_2 is the proportion of mutant mice that develop cancer (0.75)

Plugging in the values, we get:

n = (1.96 + 0.84)² × (0.25(1-0.25) + 0.75(1-0.75)) / (0.75 - 0.25)²

n = 16.48

Therefore, we would need a sample size of at least 17 mice per group to detect a difference in proportions with a power of 0.8 and a significance level of 0.05.

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Means that the derivative of A at a, dA(a), is equal to L. Hence, A is differentiable at a with dA(a) = L.

To prove that the function A is differentiable at a with dA(a) = L, we need to show that:

lim(x→a) [A(x) - A(a) - L(a)(x-a)] / ||x-a|| = 0

We know that A(x) = c + L(x) for all x in RP, where c is a constant and L is a linear transformation from RP to R9.

Then, we have:

A(a) = c + L(a)

L(a)(x-a) = L(x-a) + L(a-a) = L(x-a)

Substituting these into the limit expression, we get:

lim(x→a) [c + L(x) - c - L(a) - L(x-a)] / ||x-a||

= lim(x→a) [L(x) - L(a)] / ||x-a||

Since L is a linear transformation, it is continuous. Therefore, we can write:

lim(x→a) [L(x) - L(a)] / ||x-a|| = L( lim(x→a) [x-a] / ||x-a|| )

But lim(x→a) [x-a] / ||x-a|| = u, a unit vector in the direction of x-a.

Therefore, we have:

lim(x→a) [L(x) - L(a)] / ||x-a|| = Lu

This means that the derivative of A at a, dA(a), is equal to L. Hence, A is differentiable at a with dA(a) = L.

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The predicted number of guests at the festival in 2022 is 1846.15, rounded to two decimal places.

To calculate the 2-year Weighted Moving Average (WMA) to predict the number of guests at the festival in 2022, you will need to follow these steps:

1. First, you need to calculate the weighted average for the most recent two years. To do this, you multiply the number of guests in 2021 (2968) by the first weight (0.7), and you multiply the number of guests in 2020 (1657) by the second weight (0.3). Then you add the two products together to get the weighted average for 2020-2021:

Weighted Average = (2968 x 0.7) + (1657 x 0.3) = 2077.9

2. Next, you need to calculate the weighted average for the previous two years. To do this, you multiply the number of guests in 2019 (1559) by the first weight (0.7), and you multiply the number of guests in 2018 (1771) by the second weight (0.3). Then you add the two products together to get the weighted average for 2018-2019:

Weighted Average = (1559 x 0.7) + (1771 x 0.3) = 1614.4

3. Finally, you take the average of the two weighted averages calculated in steps 1 and 2 to get the 2-year Weighted Moving Average for 2022:

2-year WMA for 2022 = (2077.9 + 1614.4) / 2 = 1846.15

Therefore, the predicted number of guests at the festival in 2022 is 1846.15, rounded to two decimal places.

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The initial mortality rate in 1998 per 100,000 people is :

88.8 deaths

To find the initial mortality rate in 1998, you'll need to use the given relation :

M = 88.8(0.9418)^t, where M is the number of deaths per 100,000 people, and t is the number of years since 1998.

Identify the value of t for 1998. Since 1998 is the starting year, t = 0.

Substitute the value of t into the equation. M = 88.8(0.9418)^0

Calculate M. Since any number raised to the power of 0 is 1, the equation becomes M = 88.8(1), which simplifies to M = 88.8.

So, the initial mortality rate in 1998 is 88.8 deaths per 100,000 people.

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Answer:

First, we need to calculate the total exemptions for Kareem, his wife, and their child:

Total exemptions = 3 x 4,050 = 12,150

Next, we subtract the standard deduction and exemptions from their adjusted gross income to find their taxable income:

Taxable income = 92,600 - 12,600 - 12,150 = 67,850

Therefore, the correct answer is (B) 67,850.

Step-by-step explanation:

Branliest please

If Charlotte will take out the garbage for 15 weeks,  Charlotte will earn 327.67 dollars on the 15th week.

To find how much Charlotte will earn on the 15th week, we can use the formula for the sum of a geometric series:

Sₙ = a(1 - rⁿ) / (1 - r)

where Sₙ is the sum of the first n terms of the series, a is the first term, r is the common ratio, and n is the number of terms.

In this case, a = 1 cent, r = 2 (since each week Charlotte earns twice as much as she did the week before), and n = 15. Substituting these values into the formula gives:

S₁₅ = 1(1 - 2¹⁵) / (1 - 2)

S₁₅ = (1 - 32768) / (-1)

S₁₅ = 32767 cents

Therefore, Charlotte will earn 327.67 dollars on the 15th week.

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Using the moment generating function technique, we can determine the limiting distribution of X as n approaches infinity, but the exact form of the distribution will depend on the value of t and may require additional approximation methods to evaluate.

To find the limiting distribution of the random variable X with a chi-square distribution, we can use the moment generating function (MGF) technique. The moment generating function of X is defined as M_X(t) = E(e^(tX)).

For a chi-square distribution with n degrees of freedom, the probability density function (pdf) is given by:

f(x) = (1/(2^(n/2) * Γ(n/2))) * x^((n/2)-1) * e^(-x/2)

To find the moment generating function, we evaluate the integral:

M_X(t) = ∫[0 to ∞] e^(tx) * f(x) dx

Substituting the pdf into the MGF expression, we have:

M_X(t) = ∫[0 to ∞] e^(tx) * (1/(2^(n/2) * Γ(n/2))) * x^((n/2)-1) * e^(-x/2) dx

Simplifying, we get:

M_X(t) = (1/(2^(n/2) * Γ(n/2))) * ∫[0 to ∞] x^((n/2)-1) * e^((t-1/2)x) dx

To find the limiting distribution, we take the limit of the MGF as n approaches infinity. Using the property of the gamma function, we have:

lim(n->∞) (1/(2^(n/2) * Γ(n/2))) = 1

So, the limiting moment generating function becomes:

lim(n->∞) M_X(t) = ∫[0 to ∞] x^((n/2)-1) * e^((t-1/2)x) dx

To evaluate this integral, we need to use techniques such as Laplace's method or the saddlepoint approximation. The exact form of the limiting distribution depends on the specific value of t and may not have a closed-form expression.

Therefore, using the moment generating function technique, we can determine the limiting distribution of X as n approaches infinity, but the exact form of the distribution will depend on the value of t and may require additional approximation methods to evaluate.

The result obtained for the limiting distribution of the random variable X with a chi-square distribution as n approaches infinity has practical implications in various areas. Here are a few examples:

Approximation of chi-square distributions: The limiting distribution can be used as an approximation for chi-square distributions with large degrees of freedom. When the degrees of freedom are sufficiently large, the limiting distribution can provide a good approximation to the chi-square distribution. This can be useful in statistical analysis and hypothesis testing, where chi-square distributions are commonly used.

Central Limit Theorem: The result is related to the Central Limit Theorem, which states that the sum or average of a large number of independent and identically distributed random variables tends to follow a normal distribution. Since the chi-square distribution arises in various statistical contexts, the limiting distribution can help in approximating the distribution of sums or averages involving chi-square random variables.

Statistical inference: The limiting distribution can have implications for statistical inference procedures. For example, in hypothesis testing or confidence interval estimation involving chi-square statistics, knowledge of the limiting distribution can aid in determining critical values or constructing confidence intervals. It can also be used to assess the asymptotic properties of estimators based on chi-square distributions.

Simulation studies: The limiting distribution can be used in simulation studies to generate random samples that mimic chi-square distributions with large degrees of freedom. This can be helpful in situations where directly simulating from the chi-square distribution is computationally expensive or difficult.

Overall, understanding the limiting distribution of the chi-square distribution as n approaches infinity provides insights into the behavior of chi-square random variables and can be used as a practical tool in various statistical applications, such as approximation, inference, and simulation studies.

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We can see here that the vertices will be:

The vertex, in geometry, is the intersection of two or more lines, curves, or edges. It can also refer to the vertex of a parabola, which is where a function reaches its highest or lowest value.

We can see here that the equation of the hyperbola is seen in standard form. It is known that the center of the hyperbola is at (h, k) is (2, 1).

The distance between the center and vertices = a

where a² = coefficient of the positive term

So we see that  a² = 16

a = 4.

Also, the distance between the center and co-vertices = b

where b² = 4

b = 2.

Thus,

Vertex 1 = (2 + 4, 1) = (6, 1)

Vertex 2 = (2 - 4, 1) = (-2, 1).

Therefore, the vertices are:

(6, 1) and (-2, 1).

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The final amount after 20 years, compounded quarterly is $6,353.98.

The final amount after 5 years, compounded continuously, is  $1,145.10.

we can use the formula for compound interest:

[tex]A = P(1 + r/n)^(^n^\times^t^)[/tex]

where A is the final amount,

P is the principal (starting amount),

r is the annual interest rate, t is the time in years, and n is the number of times compounded per year.

Plugging in the given values, we get:

[tex]A = 3750(1 + 0.035/4)^(^4^\times^2^0^)[/tex]

A = $6,353.98

Therefore, the final amount after 20 years, compounded quarterly, is approximately $6,353.98.

P= $1,000; r=2.8%, t= 5yrs compounded continuously ​

For the second problem, we can use the formula for continuous compounding:

[tex]A = Pe^(^r^t^)[/tex]

[tex]A = 1000e^(^0^.^0^2^8^\times^5^)[/tex]

A = $1,145.10

Therefore, the final amount after 5 years, compounded continuously, is  $1,145.10.

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If the corners of the kitchen are (8, 4), (−3, 4), (8, −8), and (−3, −8), the area of the kitchen is 132 square feet. So, the correct option is C.

To find the area of the rectangular kitchen, we need to use the formula for the area of a rectangle, which is A = L x W, where A is the area, L is the length, and W is the width.

From the given ordered pairs, we can determine the length and width of the rectangle. The length is the distance between the points (8,4) and (-3,4), which is 8 - (-3) = 11 feet. The width is the distance between the points (8,4) and (8,-8), which is 4 - (-8) = 12 feet.

Now that we know the length and width, we can find the area by multiplying them together:

A = L x W = 11 x 12 = 132 square feet

Therefore, the correct answer is C.

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Answer    C. 132 fT2  

Step-by-step explanation:

a) To calculate the probability that the sample proportion of defective cups from Supplier A is less than 10%, we need to find the probability that the sample proportion is less than 0.10. Thus, we need to find P(p < 0.10).

We can use the central limit theorem to approximate the distribution of the sample proportion as a normal distribution, with mean μ = 0.08 and standard deviation [tex]σ = \sqrt{0.08 (\frac{1-0.08)}{200} )}= 0.024[/tex]. Then, we can standardize the distribution and use a standard normal table or calculator to find the probability:

[tex]P (p < 0.10)=P(\frac{p-u}{σ} < \frac{0.10-0.08}{0.024} = P(z < 0.83)=0.7977[/tex]

Therefore, the probability that the sample proportion of defective cups from Supplier A is less than 10% is approximately 0.7977.

b) To calculate the probability that the sample proportion of defective cups from Supplier A is at least 5%, we need to find the probability that the sample proportion is greater than or equal to 0.05. Thus, we need to find P(p≥ 0.05).

Using the same approach as in part (a), we can find that                                                             P(p < 0.05)=-0.0207. Therefore, P(p ≥ 0.05) = 1 - P(p< 0.05) =0.9793.

Therefore, the probability that the sample proportion of defective cups from Supplier A is at least 5% is approximately 0.9793.

c) To calculate the probability that the sample proportion of defective cups from Supplier A is between 5% and 10%, we need to find the probability that 0.05 ≤ p < 0.10. We can use the same approach as in part (a) to find that P(p < 0.05) = 0.0207 and P(p < 0.10) = 0.7977. Therefore, P(0.05 ≤ p < 0.10) = P(p < 0.10) - P(p < 0.05) = 0.7770.

Therefore, the probability that the sample proportion of defective cups from Supplier A is between 5% and 10% is approximately 0.7770.

d) To calculate the probability that the sample proportion of defective cups from Supplier A is exactly 8%, we need to find P(p = 0.08). Since the sample proportion is a discrete random variable, we can use the binomial distribution to find the probability:

[tex]P(p=0.08)=(200 choose 16) (0.080)^{16} (1-0.08)^{184} = 0.1567[/tex]

Therefore, the probability that the sample proportion of defective cups from Supplier A is exactly 8% is approximately 0.1567.

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if you add up 2 interior degrees the answer is the exterior degree

so the answer is 145

The value of expression is,

⇒ x ÷ 12

We have to given that;

The algebraic expression is,

⇒ x divided by 12

Hence, We can formulate;

The value of correct expression is,

⇒ x ÷ 12

⇒ x / 12

Thus, The value of expression is,

⇒ x ÷ 12

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The actual length and width of the kitchen in the scale drawing is: 22.5 ft x 17.5 ft

A scale drawing is defined as an enlargement of an object. An enlargement changes the size of an object by multiplying each of the lengths by a scale factor to make it larger or smaller. The scale of a drawing is usually stated as a ratio.

Now, the scale factor of the given drawing is seen as 2 in : 5 ft

From the drawing the dimensions of the kitchen are:

9" x 7"

Thus:

True length = (9 * 5)/2 = 22.5 ft

True width = (7 * 5)/2 = 17.5 ft

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The solution to the equation ex + x = 7 using Newton's method with an initial guess of Xo = 2 is approximately 1.4449.

(a) To determine the equation y = a + mx of the least square line that best fits the given data points (2,1), (1,1), (3,2), we first need to calculate the mean of x and y, and then calculate the slope m and y-intercept a of the line.

Mean of x: (2 + 1 + 3)/3 = 2

Mean of y: (1 + 1 + 2)/3 = 4/3

To calculate the slope m, we need to find the sum of (xi - x-bar)(yi - y-bar) and the sum of (xi - x-bar)^2 for each data point:

(2-2)(1-4/3) + (1-2)(1-4/3) + (3-2)(2-4/3) = -1/3

(2-2)^2 + (1-2)^2 + (3-2)^2 = 6

So, m = (-1/3) / 6 = -1/18

To calculate the y-intercept a, we can use the formula a = y-bar - m(x-bar):

a = (4/3) - (-1/18)(2) = 5/9

Therefore, the equation of the least square line is y = (5/9) - (1/18)x.

(b) We want to solve the equation ex + x = 7 using Newton's method with an initial guess of Xo = 2.

First, we need to find the derivative of the function f(x) = ex + x:

f'(x) = ex + 1

Then, we can use the formula for Newton's method:

Xn+1 = Xn - f(Xn) / f'(Xn)

Plugging in X0 = 2 and using four significant digits:

X1 = 2 - (e^2 + 2) / (e^2 + 1) = 1.574

X2 = 1.574 - (e^1.574 + 1.574) / (e^1.574 + 1) = 1.4633

X3 = 1.4633 - (e^1.4633 + 1.4633) / (e^1.4633 + 1) = 1.445

X4 = 1.445 - (e^1.445 + 1.445) / (e^1.445 + 1) = 1.4449

Therefore, the solution to the equation ex + x = 7 using Newton's method with an initial guess of Xo = 2 is approximately 1.4449.

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Answer:

replace

y

with

0

and solve for

x

.

Step-by-step explanation:

x=-1/3,2

Roots of a quadratic equation using the factorisation method, we need to find two numbers that multiply to the constant term of the equation and add up to the coefficient of the linear term. Then, we can use these two numbers to factor the quadratic expression and solve for the roots.

a) For the quadratic equation x^2 - 10x + 16 = 0, we need to find two numbers that multiply to 16 and add up to -10. These numbers are -2 and -8, so we can write the quadratic as (x - 2)(x - 8) = 0. Setting each factor equal to zero, we get x - 2 = 0 and x - 8 = 0, which give us the roots x = 2 and x = 8.

b) For the quadratic equation 2x^2 + 3x - 9 = 0, we need to find two numbers that multiply to -18 (since 2*(-9) = -18) and add up to 3. These numbers are 6 and -3, so we can write the quadratic as 2x^2 + 6x - 9x - 9 = 0. Factoring by grouping, we get 2x(x + 3) - 9(x + 3) = 0, which simplifies to (2x - 9)(x + 3) = 0. Setting each factor equal to zero, we get 2x - 9 = 0 and x + 3 = 0, which give us the roots x = 9/2 and x = -3.

c) For the quadratic equation x^2 - 5x = 0, we can factor out an x to get x(x - 5) = 0. Setting each factor equal to zero, we get x = 0 and x - 5 = 0, which give us the roots x = 0 and x = 5.

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We can expect about 84 watermelons to weigh less than 17.36 pounds.

We have,

To answer this question, we need to use the concept of standard normal distribution.

First, we need to calculate the z-score of 17.36 using the formula:
z = (x - μ) / σ
where x is the weight we're interested in, μ is the mean weight, and σ is the standard deviation. Substituting the values given in the question, we get:
z = (17.36 - 20) / 2.9
z = -0.9655

Now, we can look up the area under the standard normal curve to the left of z = -0.9655 using a z-table or a calculator. The result is 0.1675.

This means that about 16.75% of the watermelons will weigh less than 17.36 pounds.

To find the actual number of watermelons, we can multiply this percentage by the total number of watermelons produced:
500 x 0.1675 = 83.75

Therefore,

We can expect about 84 watermelons to weigh less than 17.36 pounds.

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To determine the value of "r" between age and salary, we would need to conduct a statistical analysis, such as a correlation coefficient calculation. Without this information, it is impossible to determine the direction or strength of the relationship between age and salary.

Similarly, without the results of a regression analysis, it is not possible to determine the value of R², which represents the proportion of variance in the dependent variable (salary) that can be explained by the independent variable (age). Once this value is known, we can use the rubric to determine the strength of the predictor.

However, based on the rubrics provided, if the correlation coefficient (r) is close to -1 or 1, the relationship between age and salary would be considered very strong, either negatively or positively correlated. If the coefficient is closer to 0, the correlation would be considered weak or somewhat weak.

Similarly, R² measures the proportion of variance in the dependent variable (salary) that is explained by the independent variable (age). A value of 1 would indicate a perfect predictor, while a value of 0 would indicate no relationship between the variables. Values between 0 and 1 would indicate varying degrees of predictive power.

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Answer:

PART 1: K 5375.66

PART 2: 38.1051177665 or 38  210235533/2000000000

Step-by-step explanation:

1. (i) Compounded Semi Annually: A = P × [1 + r/n]nt A = K4,000 × [1 + 6%/2]2×5 A = K4,000 × [1 + 0.03]10 A = K4,000 × [1.03]10 A = K4,000 × [1.344] A = K 5375.66

2. √(243) + (3√ (75) - √(12)= 38.1051177665

38.1051177665 as a decimal: 38.1051177665

38.1051177665 as a a fraction: 38  210235533/2000000000

K5376.48 will be in the account after 5 years compounded semi-annually. K5396.32 will be in the account after 5 years compounded weekly. The value of simplification is  22√3.

Compounded semi-annually

The interest rate per period is r = 6% / 2 = 0.03

The number of periods is n = 5 x 2 = 10

The amount A after n periods is given by

A = K(1 + r)ⁿ

A = 4000(1 + 0.03)¹⁰

A = 4000 x 1.34412

A = K5376.48

Compounded weekly

The interest rate per period is r = 6% / 52 = 0.001153846

The number of periods is n = 5 x 52 = 260

The amount A after n periods is given by

A = K(1 + r)ⁿ

A = 4000(1 + 0.001153846)²⁶⁰

A = 4000 x 1.34908

A = K5396.32

√243 + 3√75 - √12

= √(81 x 3) + 3√(25 x 3) - √(4 x 3)

= 9√3 + 15√3 - 2√3

= 22√3

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Answer:

To find out how many minutes Valeria would practice in 4 weeks, we need to first find out how many minutes she practices per week.

Divide the total number of minutes she practices by the number of weeks she practices:

910 minutes ÷ 5 weeks = 182 minutes per week

Valeria practices 182 minutes per week.

To find out how many minutes she would practice in 4 weeks, we can multiply the minutes per week by the number of weeks:

182 minutes/week x 4 weeks = 728 minutes in 4 weeks

Valeria would practice 728 minutes in 4 weeks.

If a candy bar cost 95 cents, then I'll cost 3.8$ to buy the four candy's

To Find How much will it cost to buy 4 candy bars which comes in price of 95 cents per candy bar so therefore calculation will be:

One candy bar cost  = 95 cent

We will use multiplication to multiply the cost of each candy.

4 candy bar costs  = 4 x  price of one candy

4 candy bar costs  = 4 x  95 cents

4 candy bar costs  = 380 cents

After that we get:

100 cents = 1 $

Now we will divide:

4 candy bar costs  = 380/100$

4 candy bar costs  = 3.8 $

Therefore, If a candy bar is 95 cents, it will cost me 3.8$ to purchase the four candies together.

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