The data in the table describes the preferred type of exercise of 9th graders.




Find the marginal relative frequency for students who prefer swimming as their preferred type of exercise.

39%

35%

19%

16%

Answer:13.33

Step-by-step explanation:

The value of x in the diagram to the right is equal to 58°.

In Mathematics and Geometry, a supplementary angle simply refers to two (2) angles or arc whose sum is equal to 180 degrees.

Additionally, the sum of all of the angles on a straight line is always equal to 180 degrees. In this scenario, we can reasonably infer and logically deduce that the sum of the given angles are supplementary angles:

x + 6 + 116° = 180°

By rearranging and collecting like-terms, the value of x is given by:

x + 122° = 180°

x = 180° - 122°

y = 58°.

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The correct answer is: 2. y = 26

To solve for the dependent variable (y), we need to use the given information that the individual wants to have two hot dogs for each guest and 6 extra hot dogs for potential friends.

If the independent variable is 10, then the total number of guests would be 10.

So, the equation to find the number of hot dogs needed (y) would be:

y = (2 hot dogs per guest) x 10 guests + 6 extra hot dogs

y = 20 + 6

y = 26

.
An individual is hosting a cookout for the kickball team and wants to have two hot dogs for each guest (x), and 6 extra hot dogs in case some teammates bring friends. The independent variable (x) is 10. To solve for the dependent variable (y), we use the equation:

y = 2x + 6

Now, substitute the value of x:

y = 2(10) + 6

y = 20 + 6

y = 26

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The derivative of y with respect to t (dy/dt) for the curve with parametric equations y = ln(t) and x = 4t^5 is dy/dt = 1/t.

To find dy/dt, we differentiate y = Int with respect to t:

dy/dt = d/dt (Int)

Recall that the derivative of an integral with respect to its upper limit is equal to the integrand evaluated at the upper limit. Therefore, we have:

dy/dt = 1/t

Given parametric equations:
y = ln(t)
x = 4t^5

(i) To find dy/dt, we need to differentiate y with respect to t.

y = ln(t)

Differentiating with respect to t:

dy/dt = d(ln(t))/dt

Using the chain rule, we know that the derivative of ln(t) with respect to t is 1/t:

dy/dt = 1/t

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The vector projection of x onto y is p = [4 - sin^2(t)] / 10 [cos(t), -sin(t), 3].

(a) To find the vector projection of x onto y, we use the formula:

p = (x ⋅ y / ||y||^2) y

where ⋅ denotes the dot product and ||y|| is the magnitude of y.

First, we compute the dot product:

x ⋅ y = (-5)(3) + (4)(-5) + (5)(3) = -15 - 20 + 15 = -20

Next, we compute the magnitude of y:

||y|| = √(3^2 + (-5)^2 + 3^2) = √34

Now we can plug these values into the formula:

p = (-20 / 34) [3, -5, 3] = [-1.41, 2.35, -1.41]

Therefore, the vector projection of x onto y is p = [-1.41, 2.35, -1.41].

(b) To find the vector projection of x onto y, we use the same formula:

p = (x ⋅ y / ||y||^2) y

where ⋅ denotes the dot product and ||y|| is the magnitude of y.

First, we compute the dot product:

x ⋅ y = cos(t)cos(t) + sin(t)(-sin(t)) + 1(3) = cos^2(t) - sin^2(t) + 3

Next, we compute the magnitude of y:

||y|| = √(cos^2(t) + (-sin^2(t)) + 3^2) = √(cos^2(t) + sin^2(t) + 9) = √10

Now we can plug these values into the formula:

p = [cos^2(t) - sin^2(t) + 3] / 10 [cos(t), -sin(t), 3]

Simplifying the numerator, we get:

p = [(cos^2(t) + 3) - (sin^2(t))] / 10 [cos(t), -sin(t), 3]

Using the identity cos^2(t) + sin^2(t) = 1, we can simplify further:

p = [(1 + 3) - (sin^2(t))] / 10 [cos(t), -sin(t), 3]

p = [4 - sin^2(t)] / 10 [cos(t), -sin(t), 3]

Therefore, the vector projection of x onto y is p = [4 - sin^2(t)] / 10 [cos(t), -sin(t), 3].

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The evaluate value of integral

[tex]\int_{0}^{\infty}7xe^{-x²} dx[/tex], is equals to the [tex] \frac{ 7 }{2}[/tex]

and limit of integral is finite so, this integral converges.

Integral test is used to check the Integral convergence. Integral is converge whose limit exists and is finite, and integral divergence is defined as an integral whose limit is either ±∞ , or nonexistent. When evaluating an integral with one boundary at infinity, that is [tex]\int_{a}^{\infty} f(x) dx = \lim_{A→ ∞ }\int_{a}^{A} f(x) dx [/tex]. We have an integral say [tex]I =\int_{0}^{+ \infty}7xe^{- x²} dx [/tex]

[tex] =\int_{0}^{\infty} 7xe^{- x²} dx [/tex]

We have to evaluate it and check it converges or not. Now, put x² = z

=> 2xdx = dz

when x = 0 => z = 0 and x = ∞=> z = ∞

[tex]\int_{0}^{\infty}7xe^{-x²} dx = \int_{0}^{\infty}\frac{ 7 }{2}e^{ - z} dz [/tex]

[tex]= \frac{ 7 }{2}\int_{0}^{\infty}e^{ - z} dz [/tex]

Now, consider the limits of integral, [tex]= \frac{ 7 }{2}\lim_{ε → ∞}\int_{0}^{ε}e^{ - z} dz \\ [/tex]

[tex]= \frac{ 7 }{2}\lim_{ε → ∞}[ -e^{ - z} ]_{0}^{ε} \\ [/tex]

[tex]= \frac{ 7 }{2}\lim_{ε → ∞}( 1 -e^{ -ε} ) \\ [/tex]

[tex]= \frac{ 7 }{2}( 1 -e^{ - \infty} )[/tex]

[tex]= \frac{ 7 }{2}( 1 - 0 ) = \frac{ 7 }{2}[/tex]

which is a finite number. Hence, integral is converges.

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Complete question:

Evaluate the integrals that converge enter 'DNE' if integral Does Not Converge

[tex]I =\int_{0}^{ + \infty} 7xe^{- x²} dx [/tex]

Jump to Problem: [ 1 2 3 4 5 ,]

The emergency department of the hospital can be considered as a rare event occurring independently and with a constant rate (on average 7 per hour), which makes the Poisson distribution an appropriate choice.

The most appropriate distribution to use for calculating probabilities, expected value, and variance of the number of patients that arrive between 6 PM and 7 PM for a given day would be the Poisson distribution. The Poisson distribution is commonly used to model the number of occurrences of a rare event in a fixed period of time, where the events occur independently and with a constant rate. In this case, the number of patients arriving in the emergency department of the hospital can be considered as a rare event occurring independently and with a constant rate (on average 7 per hour), which makes the Poisson distribution an appropriate choice.

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An equation of the line in fully simplified slope-intercept form include the following: y = -3x/2 + 8.

In Mathematics and Geometry, the point-slope form of a straight line can be calculated by using the following mathematical expression:

y - y₁ = m(x - x₁)

Where:

First of all, we would determine the slope of this line;

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Slope (m) = (2 - 5)/(4 - 2)

Slope (m) = -3/2

At data point (2, 5) and a slope of -3/2, a linear equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - 5 = -3/2(x - 2)  

y = -3x/2 + 3 + 5

y = -3x/2 + 8

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The exact values of the cosecant, secant, and cotangent ratios of -7pi/4 radians are -√(2), -√(2), and 1.

Here are the exact values of the cosecant, secant, and cotangent ratios of -7π/4 radians:

The cosecant of an angle is equal to the length of the hypotenuse of a right triangle with that angle as its opposite side, divided by the length of the opposite side. The formula for cosecant is cosec(θ) = 1/sin(θ).

In this case, the sine of -7π/4 radians is -√(2)/2, so the cosecant is -2/√(2), which simplifies to -√(2).

The secant of an angle is equal to the length of the hypotenuse of a right triangle with that angle as its adjacent side, divided by the length of the adjacent side. The formula for secant is sec(θ) = 1/cos(θ).

In this case, the cosine of -7π/4 radians is -√(2)/2, so the secant is -2/√(2), which simplifies to -√(2).

The cotangent of an angle is equal to the length of the adjacent side of a right triangle with that angle as its opposite side, divided by the length of the opposite side.

The formula for cotangent is cot(θ) = 1/tan(θ). In this case, the tangent of -7π/4 radians is 1, so the cotangent is 1.

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Answer:

converges.

Step-by-step explanation:

We can use the ratio test to determine whether the series converges or diverges.

The ratio test states that for a series ∑aₙ, if the limit of the absolute value of the ratio of successive terms is less than 1, then the series converges absolutely. If the limit is greater than 1 or does not exist, then the series diverges.

Let's apply the ratio test to the given series:

lim k→∞ |(k^(2+1)−k+2) / (3(k+1)^(4)+2(k+1)^(2)+1) * (3k^(4)+2k^(2)+1) / (k^(2)−k+1)|

= lim k→∞ |(3k^(6) + 8k^(5) - 5k^(4) - 6k^(3) + 9k^(2) + 2k + 1) / (3k^(6) + 12k^(5) + 23k^(4) + 22k^(3) + 13k^(2) + 4k + 1)|

= 3/3 = 1

Since the limit of the absolute value of the ratio of successive terms is 1, the ratio test is inconclusive. We need to use another test.

Let's try the limit comparison test, where we compare the given series to another series whose convergence or divergence is known.

We can choose the series ∑[infinity]->k=1 1/k^(2). This series converges by the p-series test since p=2>1.

Now, let's find the limit of the ratio of the two series:

lim k→∞ [(k^(2)−k+1) / (3k^(4)+2k^(2)+1)] / (1/k^(2))

= lim k→∞ k^(4)(k^(2)-k+1)/(3k^(4)+2k^(2)+1)

= 1/3

Since the limit is a finite positive number, both series have the same convergence behavior. Therefore, the given series converges by comparison to the convergent series ∑[infinity]->k=1 1/k^(2).

Therefore, the given series converges.

The given series can be determined to be a convergent series using the Limit Comparison Test.

To apply the Limit Comparison Test, we need to find another series whose behavior is known. We can do this by simplifying the given series by dividing both the numerator and denominator by k^4. This gives us:

[(k^2/k^4) - (k/k^4) + (1/k^4)] / [3 + (2/k^2) + (1/k^4)]

Now, as k approaches infinity, all the terms containing k will approach zero, leaving us with:

[0 - 0 + 1/k^4] / [3 + 0 + 0]

Simplifying this expression further gives us:

1 / 3k^4

Now, we can compare this to the known convergent p-series, 1/k^2, by taking the limit of the ratio of their terms as k approaches infinity:

lim as k -> infinity of [(1/3k^4)/(1/k^2)] = lim as k -> infinity of (k^(-2))/3 = 0

Since the limit is a finite value, we can conclude that the given series is convergent by the Limit Comparison Test. Therefore, the series converges.

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a. Null hypothesis

There is no significant difference in need for achievement scores between individuals who fit the Type A behavior pattern and those who fit the Type B behavior pattern. Research hypothesis: Individuals who fit the Type A behavior pattern have significantly higher need for achievement scores than individuals who fit the Type B behavior pattern.
b. The proper statistical test to use in this case is an independent samples t-test.

c. An independent samples t-test was conducted to compare the mean need for achievement scores of Type A and Type B individuals. The results indicated that the mean need for achievement score for Type A individuals (M = 11.4, SD = 2.2) was significantly higher than the mean score for Type B individuals (M = 7.2, SD = 1.9), t(18) = 4.28, p < .001. Therefore, the research hypothesis was supported, indicating that individuals who fit the Type A behavior pattern have significantly higher levels of need for achievement than individuals who fit the Type B behavior pattern.

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To find a polynomial function f(x) of least degree having only real coefficients and zeros of 5 and 2-i, we know that the complex conjugate of 2-i, which is 2+i, must also be a zero. This is because complex zeros of polynomials always come in conjugate pairs.

So, we can start by using the factored form of a polynomial:

f(x) = a(x - r1)(x - r2)(x - r3)...

where a is a constant and r1, r2, r3, etc. are the zeros of the polynomial. In this case, we have:

f(x) = a(x - 5)(x - (2-i))(x - (2+i))

Multiplying out the factors, we get:

f(x) = a(x - 5)((x - 2) - i)((x - 2) + i)
f(x) = a(x - 5)((x - 2)^2 - i^2)
f(x) = a(x - 5)((x - 2)^2 + 1)

To make sure that f(x) only has real coefficients, we need to get rid of the complex i term. We can do this by multiplying out the squared term and using the fact that i^2 = -1:

f(x) = a(x - 5)((x^2 - 4x + 4) + 1)
f(x) = a(x - 5)(x^2 - 4x + 5)

Now, we just need to find the value of a that makes the degree of f(x) as small as possible. We know that the degree of a polynomial is determined by the highest power of x that appears, so we need to expand the expression and simplify to find the degree:

f(x) = a(x^3 - 9x^2 + 24x - 25)
Degree of f(x) = 3

Since we want the least degree possible, we want the coefficient of the x^3 term to be 1. So, we can choose a = 1:

f(x) = (x - 5)(x^2 - 4x + 5)
Degree of f(x) = 3

Therefore, the polynomial function f(x) of least degree having only real coefficients and zeros of 5 and 2-i is:

f(x) = (x - 5)(x^2 - 4x + 5)
To find a polynomial function f(x) of least degree with real coefficients and zeros of 5 and 2-i, we need to remember that if a polynomial has real coefficients and has a complex zero (in this case, 2-i), its conjugate (2+i) is also a zero.

Step 1: Identify the zeros
Zeros are: 5, 2-i, and 2+i (including the conjugate)

Step 2: Create factors from zeros
Factors are: (x-5), (x-(2-i)), and (x-(2+i))

Step 3: Simplify the factors
Simplified factors are: (x-5), (x-2+i), and (x-2-i)

Step 4: Multiply the factors together
f(x) = (x-5) * (x-2+i) * (x-2-i)

Step 5: Expand the polynomial
f(x) = (x-5) * [(x-2)^2 - (i)^2] (by using (a+b)(a-b) = a^2 - b^2 formula)
f(x) = (x-5) * [(x-2)^2 - (-1)] (since i^2 = -1)
f(x) = (x-5) * [(x-2)^2 + 1]

Now we have a polynomial function f(x) of least degree with real coefficients and zeros of 5, 2-i, and 2+i:
f(x) = (x-5) * [(x-2)^2 + 1]

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Answer:

Step-by-step explanation:

I knew this because i got this wright on my assignment

"There is enough evidence to conclude that the mean price of homes are not all the same for these three areas."

(a) The estimate of the common standard deviation can be found by taking the square root of the mean square within groups: sqrt(0.0464) ≈ 0.215.

(b) The F-statistic is given as 1.41/0.0464 ≈ 30.43.

(c) The critical value for the F-distribution with 3 and 30 degrees of freedom at the 0.025 significance level is approximately 3.12 (obtained from a statistical table or calculator). Since the calculated F-statistic of 30.43 is greater than the critical value of 3.12, we reject the null hypothesis and conclude that there is enough evidence to conclude that the mean price of homes are not all the same for these three areas. Therefore, the valid conclusion for this hypothesis test at the significance level of 0.025 is: "There is enough evidence to conclude that the mean price of homes are not all the same for these three areas."

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To write the equation of a line with zero slope that passes through the point (3, 28)= y = 28

we first need to understand what a zero slope means. A zero slope indicates that the line is horizontal, meaning it doesn't rise or fall as it moves horizontally. This means that the y-value of every point on the line is constant.

Since the line passes through the point (3, 28), we know that the constant y-value is 28. Thus, the equation of the line with zero slope passing through (3, 28) is simply:

y = 28

This equation represents a horizontal line that goes through all points with a y-coordinate of 28, including the given point (3, 28).

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61% of all males in the US between the ages of 40 and 49 have a fat consumption of less than 104.47 g per day

a) rv X = the fat consumption of a randomly selected male in the US between the ages of 40 and 49

b) [tex]P(X ≥ 91.94) = P(Z ≥ \frac{(91.94 - 103.1)}{4.32} /) = P(Z ≥ -2.57) = 0.0051[/tex]

c) [tex]P(X ≥ 93.64) = P(Z ≥ \frac{(93.64 - 103.1)}{4.32} ) = P(Z ≥ -2.19) = 0.0143[/tex]

d) [tex]P(91.94 ≤ X ≤ 93.64) = P(Z ≤ (\frac{93.64 - 103.1}{4.32} ) - P(Z ≤ (\frac{91.94 - 103.1)}{4.32} ) = P(Z ≤ -2.19) - P(Z ≤ -2.57) = 0.0143 - 0.0051 = 0.0092[/tex]

e) [tex]P(X ≥ 118.22) = P(Z ≥ (\frac{118.22 - 103.1}{4.32} ) = P(Z ≥ 3.50) = 0.0002[/tex]

f) no, since the probability of having a value of fat consumption at least that high is less than or equal to 0.05

g) Using the standard normal table, we find the z-score corresponding to the 61st percentile to be approximately 0.28. Therefore, we have:

[tex]0.28 = \frac{x-103.1}{4.32}[/tex]

x = 104.47

So 61% of all males in the US between the ages of 40 and 49 have a fat consumption of less than 104.47 g per day. The units are grams per day.

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The inequality that projects Chandra's finish times, x, for any 100 meter sprint is x < 18 seconds. This is due to the reason of her finish times were under 18 seconds this season.

The inequality for finish times in a 100 meter sprint is applied to differentiate the performance of two or more athletes.
t1 - t2 > k

Here
t1 and t2 = finish times of two athletes
k = constant that depends on the level of competition and other factors. Inequality refers to the topic of an order relationship that is considered to be greater than,or equal to, less than, under two numbers or algebraic expressions.

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To find the percentage of 84 out of 113, we need to first calculate what percentage of the total score 84 represents.
If a score of 113 is 40%, then we can set up a proportion:
113 / 100 = 40 / x
where x represents the percentage we are trying to find.

Cross-multiplying, we get:
113x = 4000
Dividing both sides by 113, we get:
x = 35.4
So, 84 represents approximately 35.4% of the total score of 113.

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P value of 0.039 rejects the assumption of independence of Part Quality and Supplier. Further supplier evaluation is recommended.

To answer this question, we need to perform a chi-square test of independence to determine if Part Quality depends on Supplier. The given data is:

Supplier     Good      Minor Defect    Major Defect
A                100              5                     8
B                160              4                     7
C                150              27                   11

Step 1: Calculate the expected values for each cell.

Step 2: Apply the chi-square test formula: χ² = Σ[(O - E)² / E], where O is the observed value and E is the expected value.

Step 3: Calculate the p-value using the chi-square distribution with the appropriate degrees of freedom. In this case, df = (number of rows - 1) * (number of columns - 1) = (3 - 1) * (3 - 1) = 4.

Step 4: Compare the p-value to the given significance level (α = 0.05). If the p-value is less than α, reject the null hypothesis and conclude that Part Quality depends on Supplier.

Based on the given data, the correct answer is b. P value of 0.039 rejects the assumption of independence of Part Quality and Supplier. Further supplier evaluation is recommended.

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Based on the information given, we know that the roots of the equation 5x + 13x + k = 0 are reciprocal of each other. This means that if one root is represented by r, the other root can be represented by 1/r.

Using the sum and product of roots formula, we can find that the sum of the roots is: r + 1/r = -13/5
Multiplying both sides by r, we get: r^2 + 1 = -13/5r
Multiplying both sides by 5r, we get: 5r^3 + 5r = -13
Simplifying, we get: 5r^3 + 5r + 13 = 0
This is a cubic equation that can be solved using the cubic formula. However, we do not need to solve for r to find the value of k.
We know that the product of the roots is: r * 1/r = 1
Using the product of roots formula, we can find that the product of the roots is: k/5 = 1
Multiplying both sides by 5, we get: k = 5
Therefore, the value of k is 5.

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(a) The marginal density functions fX and fY is FY(y) = 3y(2y+1)

(b)The probability that X + Y ≤ 1 is P(X + Y ≤ 1) = 5/16

(a) To discover the negligible thickness work of X, we coordinated the joint thickness work with regard to y over the extent of conceivable values of y:    

fX(x) = ∫ f(x,y) dy = ∫ 3(2−x)y dy,   0<x<2

Assessing the necessary, we get:

fX(x) = (3/2)*(2-x)²,   0<x<2

To discover the negligible thickness work of Y, we coordinated the joint thickness work with regard to x over the extent of conceivable values of x:

FY(y) = ∫ f(x,y) dx = ∫ 3(2−x)y dx,   0<y<1

Assessing the necessary, we get:

FY(y) = 3y(2y+1),   0<y<1

(b) To calculate the likelihood that X + Y ≤ 1, we got to coordinate the joint thickness work over the locale of the (x,y) plane where X + Y ≤ 1:

P(X + Y ≤ 1) = ∫∫ f(x,y) dA,   where A is the locale X + Y ≤ 1

We will modify the condition X + Y ≤ 1 as y ≤ 1−x. So the limits of integration for y are to 1−x, and the limits of integration for x are to 1:

P(X + Y ≤ 1) = [tex]∫0^1 ∫0^(1−x)[/tex] 3(2−x)y dy dx

Evaluating the inner integral, we get:

[tex]∫0^(1−x)[/tex] 3(2−x)y dy = (3/2)*(2−x)*(1−x)²

Substituting this into the external indispensably, we get:

P(X + Y ≤ 1) = ∫0^(3/2)*(2−x)*(1−x)²dx

Assessing this necessarily, we get:

P(X + Y ≤ 1) = 5/16

Hence, the likelihood that X + Y ≤ 1 is 5/16.

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please see attached...

ignore 8/52 in the top right hand corner

The equation for a line parallel to f(x) = -3x - 5 and passing through the point (2, -6) is y = -3x.

An equation for a line parallel to f(x) = -3x - 5 and passing through the point (2, -6). Here are the steps:

Step 1: Identify the slope of the given line, f(x) = -3x - 5. Since it's in the form y = mx + b, where m is the slope, we see that the slope of the given line is -3.

Step 2: Since we want a line parallel to the given line, the slope of our new line will be the same, which is -3.

Step 3: Use the point-slope form of a linear equation, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point the line passes through. In this case, m = -3 and the point is (2, -6), so x1 = 2 and y1 = -6.

Step 4: Plug the values into the point-slope form equation: y - (-6) = -3(x - 2)

Step 5: Simplify the equation. First, change y - (-6) to y + 6, then distribute -3: y + 6 = -3x + 6

Step 6: Write the equation in slope-intercept form (y = mx + b) by subtracting 6 from both sides: y = -3x

So, the equation for a line parallel to f(x) = -3x - 5 and passing through the point (2, -6) is y = -3x.

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The solutions to the equation 5x²+14x=x+6 are x = 4/5 or x = -3 we solved by using quadratic formula

The given equation is 5x²+14x=x+6

We have to solve for x

Subtract x from both sides

5x²+13x=6

Subtract 6 from both sides

5x²+13x-6=0

Now we can use the quadratic formula to solve for x:

x = (-b ± √(b²- 4ac)) / 2a

where a = 5, b = 13, and c = -6.

Substituting these values and simplifying:

x = (-13 ±√(13²- 4(5)(-6))) / (2 × 5)

x = (-13 ± √289)) / 10

x = (-13 ± 17) / 10

So we get two solutions:

x = 4/5 or x = -3

Therefore, the solutions to the equation 5x^2 + 14x = x + 6 are x = 4/5 or x = -3.

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The statement "2(y – 2) for some y ∈ Z} = 2z for some z ∈ Z}" means that there exists an integer y such that when you multiply 2 by y-2, you get an even integer that is equal to 2 times some other integer z. In other words, there exists some even integer that can be expressed as 2 times some other integer z, and that even integer can also be expressed as 2 multiplied by the difference of an integer y and 2.

To solve the equation 2(y - 2) for some y ∈ Z} = 2z for some z ∈ Z}, follow these steps:

Step 1: Start with the given equation, 2(y - 2) = 2z.

Step 2: Distribute the 2 on the left side of the equation: 2y - 4 = 2z.

Step 3: Solve for y in terms of z: 2y = 2z + 4.

Step 4: Divide both sides of the equation by 2: y = z + 2. Now, the equation is in the form y = z + 2, where both y and z are integers (y, z ∈ Z}).

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Answer: [tex]10\sqrt[3]{210}[/tex] units, (about 59.4)

Step-by-step explanation:

a hemisphere is half a sphere.

the volume of a sphere is [tex]\frac{4}{3} \pi r^3[/tex]

since we need half of this, the volume of a hemisphere would be: [tex]\frac{4}{6} \pi r^3[/tex]

this simplified nicely to: [tex]\frac{2}{3} \pi r^3[/tex]

next, we want to find the radius, given the volume. So lets set up the equation.

[tex]140000\pi = \frac{2}{3} \pi r^3[/tex]

[tex]140000 = \frac{2}{3} r^3[/tex]    --- cancel a pi from both sides.

[tex]210000 = r^3[/tex] ---- multiply both sides by 3/2 to cancel the 2/3.

[tex]\sqrt[3]{210000 }= r[/tex] ---- take the cube root of both sides to find r

[tex]10\sqrt[3]{210} = r[/tex]

Thats the exact answer: the radius is [tex]10\sqrt[3]{210}[/tex] units.

a decimal approximation is about 59.4 units.

The required monthly payment for this compound-interest loan is approximately $1,199.10.

Calculating the monthly payment for a compound-interest loan, you will need to use the following formula:

Monthly Payment = [tex]P (r  (1 + r)^n) / ((1 + r)^n - 1)[/tex]

Where:
P, principal amount = $200,000
r, monthly interest rate = annual rate / 12
n, total number of payments = 30 years × 12 payments per year

For this loan:
P = $200,000
Annual Rate = 6% = 0.06
Monthly Rate (r) = 0.06 / 12 = 0.005
Number of Payments (n) = 30 * 12 = 360

Now, putting in the values into the formula:

Monthly Payment = 200,000 × (0.005 × [tex](1 + 0.005)^{360}) / ((1 + 0.005)^{360} - 1)[/tex]
Calculating this, you get:

Monthly Payment ≈ $1,199.10

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The probability of rolling a sum of 12 with these dice is  1/36.

The likelihood of rolling an entirety of 12 with two standard dice can be found utilizing the equation:

P(D1 + D2 = 12) = number of ways to induce an entirety of 12 / total possible results

There's as it were one way to roll a whole of 12: rolling a 6 on both dice.

The whole conceivable results can be found by noticing that there are 6 conceivable results for each dice roll, since each kick the bucket has 6 sides. In this manner, the whole number of conceivable results is:

6 x 6 = 36

So the likelihood of rolling a whole of 12 with two standard dice is:

P(D1 + D2 = 12) = 1/6 * 1/6 = 1/36

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The ratio of the trigonometric function of the right triangle, cos A is 3/5.

Given that,

In △ABC , ∠C is a right angle.

Then the opposite side to the right angle will be the hypotenuse.

So AB is the hypotenuse.

Sin A = BC / AB [ Since sine of an angle is opposite side / hypotenuse]

BC / AB = 4/5

BC = 4 and AB = 5

Using the Pythagoras theorem,

Third side, AC = √(5² - 4²) = 3

Cos of an angle is the ratio of adjacent side to the hypotenuse.

Cos A = 3/5

Alternatively, we can use the identity,

sin²A + cos²A = 1

to find the value of cos A.

Hence the value of cos A is 3/5.

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The randomly select 500 students and observe that 85 of them smoke. Estimate the probability that a randomly selected student smokes , the correct answer is 27.
To estimate the probability that a randomly selected student smokes, we use the proportion of students who smoke in our sample of 500. We observed that 85 out of 500 students smoke, so the proportion is: 85/500 = 0.17
To convert this proportion to a probability, we simply round to two decimal places: 0.17 ≈ 0.27
Therefore, the estimated probability that a randomly selected student smokes is approximately 0.27, which is answer choice a.

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