Darius' pool can hold 13,650 gallons of water. After draining the pool completely so that it could be resurfaced, Darius is now refilling the pool. The pool is filling at a rate of 640 gallons per hour. There are currently 5, 810 gallons in the pool. How much longer does Darius need to fill his pool before it is full?

The following parts can be answered by the concept of differential equation. Options A, B, C, and F are first-order linear differential equations.

Based on the given terms, here is the classification of each equation as first-order linear differential equations or not:

A. X dy/dx - 4y = x²: This is a first-order linear differential equation, as it has the form (X dy/dx) - 4y = f(x).

B. dP/dt + 2tP = P + 4t - 2: This is a first-order linear differential equation, as it has the form (dP/dt) + g(t)P = h(t).

C. dy/dx + cos(x)y = 5: This is a first-order linear differential equation, as it has the form (dy/dx) + p(x)y = q(x).

D. de = y² - 3y: This is not a first-order linear differential equation, as it lacks the dy/dx term and does not have the standard form.

E. 2 + sin(x) = cos(x): This is not a differential equation, as there are no derivatives present.

F. sin(x) dy/dx - 3y = 0: This is a first-order linear differential equation, as it has the form (sin(x) dy/dx) - 3y = 0.

So, options A, B, C, and F are first-order linear differential equations.

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The length of the side of the cube whose surface area is 302. 46inches. is approximately 7.09 inches.

To find the length of the side of the cube, we need to use the formula for the surface area of a cube

Surface Area = 6s^2

Where s is the length of the side of the cube.

We are given that the surface area is 302.46 in, so we can set up the equation

302.46 = 6s^2

Dividing both sides by 6, we get:

50.41 = s^2

Taking the square root of both sides, we get

s = 7.09

So the length of the side of the cube is approximately 7.09 inches.

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Answer:

a. Mean:

(12 + 13 + 14(3) + 15(3) + 16(2) + 17(3)

+ 18(5) + 19 + 20 + 21 + 22(2))/23

= 389/23 = 16.91

b. Median: There are 23 observations, so the median is the 12th observation when the data are arranged in order. That observation is 17.

c. Mode: The mode is the age that appears the most times. That age is 18, which appears 5 times.

d. Range = largest value - smallest value

= 22 - 12 = 10

e. When the data are arranged in order, the first quartile is the 6th observation, 15, and the third quartile is the 18th observation, 18. So IQR = Q3 - Q1 =

18 - 15 = 3

f. 16.875 × 24 = 405

405 - 389 = 16

g. If the 24th member of this data is a 17-year-old, the median will remain 17.

Since the second derivative of f(x) is a constant positive number, there are no inflection points for [tex]f(x) = 3x^2 - 2[/tex].

The point of inflection or inflection point is a point in which the concavity of the function changes. It means that the function changes from concave down to concave up or vice versa. In other words, the point in which the rate of change of slope from increasing to decreasing manner or vice versa is known as an inflection point. Those points are certainly not local maxima or minima. They are stationary points.

To find the critical points of [tex]f(x) = 4x^2,[/tex] we need to find the values of x where the derivative of f(x) equals zero.

f'(x) = 8x

Setting f'(x) = 0, we get:

8x = 0

x = 0

Therefore, the critical point of[tex]f(x) = 4x^2[/tex] is at x = 0.

To determine if[tex]f(x) = 3x^2 - 2[/tex]has any inflection points, we need to find the second derivative of f(x) and check its sign.

f''(x) = 6

Since the second derivative of f(x) is a constant positive number, there are no inflection points for [tex]f(x) = 3x^2 - 2[/tex].

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The estimate for ln(0.99) is ln(2) - 0.01.

To estimate ln(0.99) using linearization, we first find the linear approximation of f(x) near x=1. We have:

f(1) = ln(2)

f'(x) = 1/x (by differentiating ln(x))

So, the equation of the tangent line at x=1 is:

y - ln(2) = 1/1 (x - 1)

y - ln(2) = x - 1

y = x - 1 + ln(2)

Now, we can use this linear approximation to estimate ln(0.99) as follows:

ln(0.99) ≈ 0.99 - 1 + ln(2) = ln(2) - 0.01


This estimate is an underestimate because ln(x) is a decreasing function for x in (0,1), which means that the tangent line at x=1 lies below the graph of ln(x) for x in (0,1). Therefore, the linear approximation underestimates the value of ln(0.99).

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The total surface area of the prism is 48.52 mm² and  the volume of the triangular prism is 17.70 mm³.

A triangular prism is a three-dimensional geometric shape that consists of two parallel triangular bases and three rectangular faces that connect the corresponding sides of the bases. It has a total of six faces, nine edges, and six vertices. The height of the prism is the perpendicular distance between the two bases, and the lateral edges are the three edges that connect the corresponding vertices of the bases. The volume of a triangular prism can be found by multiplying the area of one of the triangular bases by the height of the prism, and the surface area can be found by adding up the areas of each six faces. Triangular prisms are commonly used in architecture, engineering, and geometry.

To find the surface area of the triangular prism, we first need to find the area of the triangular base, which is an equilateral triangle with side length 2.7 mm.

Area of triangular base = (√3 / 4) x (side length)²

= (√3 / 4) x (2.7 mm)²

= 3.16 mm^2 (rounded to the nearest hundredth)

Since the base is an equilateral triangle, the perimeter is 3 times the side length:

Perimeter of triangular base = 3 x 2.7 mm

= 8.1 mm

Lateral area of prism = Perimeter of the triangle x Height

= 8.1 mm x 5.6 mm

= 45.36 mm²

The total surface area of the prism will be the sum of the area of the base and the lateral area:

Surface area = Area of triangular base + Lateral area of prism

= 3.16 mm² + 45.36 mm²

= 48.52 mm² (rounded to the nearest hundredth)

To find the volume of given triangular prism, we can use the formula:

Volume = Area of triangular base x Height of prism

= 3.16 mm² x 2.3 mm

= 17.696 mm³ = 17.70 mm³ (rounded to the nearest hundredth)

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Some inferential procedures have conditions that must be met, but others do not is false.

Deducible procedures,  similar as  thesis testing and confidence intervals, are statistical  styles used to make conclusions about a population grounded on a sample of data. These procedures calculate on the  supposition that the sample is representative of the population and that the data satisfy certain  hypotheticals.  

Some  exemplifications of  deducible procedures and their corresponding  hypotheticals include   t- tests Assumes that the data are  typically distributed and have equal  dissonances between groups.  ANOVA Assumes that the data are  typically distributed and have equal  dissonances between groups.  Linear retrogression Assumes that the relationship between the dependent and independent variables is direct, the  crimes are  typically distributed, and the  friction of the  crimes is constant.

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The projected online ad market share (as a percentage) at the beginning of 2005 is 26.7%.

(a) The projected online ad market share at any time t can be found by integrating the rate function R(t) with respect to t:
S(t) = ∫(−0.021t^2 + 0.3004t + 0.06) dt
S(t) = −0.007t^3 + 0.1502t^2 + 0.06t + C
where C is the constant of integration. We can find the value of C by using the fact that the online ad market share at the beginning of 2000 was 1.7%:
S(0) = −0.007(0)^3 + 0.1502(0)^2 + 0.06(0) + C = 0.017
C = 0.017
So the projected online ad market share at any time t is:
S(t) = −0.007t^3 + 0.1502t^2 + 0.06t + 0.017
(b) The beginning of 2005 corresponds to t = 5, so we can use the function S(t) to find the projected online ad market share at that time:
S(5) = −0.007(5)^3 + 0.1502(5)^2 + 0.06(5) + 0.017 = 0.267

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First, find dy/dt using the chain rule. Then, use dt/dr = 1 and the chain rule to find dy/dr. dy/dr = (12r / √(11k² + 12r²)), using chain rule and assuming r = k*t, where k is a constant.

Let [tex]u = 11 + 12t^2[/tex]. Then, we have y = √u, and we can utilize the chain rule to track down dy/dt:

[tex]dy/dt = (1/2)(u)^(- 1/2)(du/dt)[/tex]

[tex]= (1/2)(11 + 12t^2)^(- 1/2)(24t)[/tex]

Presently, we need to track down dy/dr. We know that dr/dt = 1, since r isn't a component of t. In this manner, by the chain rule,

dy/dr = dy/dt * dt/dr

We can settle for dt/dr by taking the equal of dr/dt:

dt/dr = 1/dr/dt = 1/1 = 1

Subbing the two qualities, we get:

dy/dr = dy/dt * dt/dr = [tex](1/2)(11 + 12t^2)^(- 1/2)(24t) * 1[/tex]

= 12t/√(11 + [tex]12t^2[/tex])

In this manner, dy/dr = 12t/√(11 + [tex]12t^2[/tex]). Therefore, the final answer for dy/dr is: dy/dr = (12r / √(11k² + 12r²))

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The critical points of the  f(x) = x - 5tan^(-1)(x) are x = -2 and x = 2.

To find the critical points of the function f(x) = x - 5tan^(-1)(x), you need to calculate the first derivative and then determine where it is equal to zero or undefined. Here are the steps:

Find the first derivative of f(x):
f'(x) = 1 - 5/(1 + x^2) (due to the derivative of tan^(-1)(x) = 1/(1 + x^2))

Set the derivative equal to zero and solve for x:
1 - 5/(1 + x^2) = 0

Solve the equation for x:
5/(1 + x^2) = 1
5 = 1 + x^2
x^2 = 4
x = ±2
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According to the given function f(x,y) = 22 - xy + y² - 5x + 5y and the domain D = {(x,y) | 0 < x < 4, -1 < y < 3}, we can find the maximum and minimum values of the function within the given domain.
To find the critical points, we need to take the partial derivatives of the function with respect to x and y, set them equal to zero, and solve for x and y.
f_x = -y - 5 = 0
f_y = -x + 2y + 5 = 0
Solving these equations simultaneously, we get the critical point (x,y) = (3,2).
To determine whether this critical point is a maximum or a minimum, we need to find the second partial derivatives of f(x,y) with respect to x and y.
f_xx = 0, f_yy = -2
Since f_yy is negative at the critical point, we conclude that (3,2) is a local maximum.
Next, we need to check the boundary of the domain to see if there are any maximum or minimum values. We can parameterize the boundary as follows:
1. x = 0, -1 ≤ y ≤ 3
2. x = 4, -1 ≤ y ≤ 3
3. 0 ≤ x ≤ 4, y = -1
4. 0 ≤ x ≤ 4, y = 3
We can then plug these values into the original function f(x,y) and compare the results to find the maximum and minimum values.
On the line x = 0, we have f(0,y) = 22 + y² + 5y, which has a maximum value of 33 when y = -5/2 and a minimum value of 11 when y = 1.
On the line x = 4, we have f(4,y) = 6 + y² + 5y, which has a maximum value of 33 when y = -5/2 and a minimum value of 11 when y = 1.
On the line y = -1, we have f(x,-1) = 28 - x - 5, which has a maximum value of 22 when x = 0 and a minimum value of 10 when x = 4.
On the line y = 3, we have f(x,3) = 10 - x + 15, which has a maximum value of 22 when x = 0 and a minimum value of 10 when x = 4.
Therefore, the maximum value of f(x,y) within the domain D is 33, which occurs at the points (0,-5/2), (3,2), and (4,-5/2), and the minimum value is 10, which occurs at the points (4,1) and (0,1).

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The evaluate value of a definite integral [tex] I = \int_{0}^{1} ( 2x + x^{\frac{1}{3}}) dx[/tex] is equals to the [tex] \frac{ 7}{4} [/tex] .

An important factor in mathematics is the sum over a period of the area under the graph of a function or a new function whose result is the original function that is called integral. Two types of integral definite or indefinite. When limits of integral is known, it is called definite integral. We have a definite integral, [tex] I = \int_{0}^{1} ( 2x + x^{\frac{1}{3}}) dx[/tex]

We have to evaluate this integral value.

Using linear property of an integral,

[tex] = \int_{0}^{1} 2x dx + \int_{0}^{1} x^{\frac{1}{3}} dx[/tex]

Using the rule of integration, [tex]=[\frac{ 2x²}{2}]_{0}^{1} + \frac{x^{\frac{1}{3} + 1}}{ \frac{1}{3} + 1}]_{0}^{1}[/tex]

[tex] = [\frac{ 2× 1²}{2}] + \frac{1^{\frac{4}{3}}}{ \frac{4}{3}}][/tex]

[tex] = (\frac{ 3}{4}] + 1 )[/tex]

[tex] = \frac{ 7}{4} [/tex]

Hence, required value is [tex] \frac{ 7}{4} [/tex] .

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Complete question:

Evaluate the integral [tex] I = \int_{0}^{1} ( 2x + x^{\frac{1}{3}}) dx[/tex].

The measure of the inscribed angle is equal to 90 degrees

The inscribed angle theorem mentions that the angle inscribed inside a circle is always half the measure of the central angle or the intercepted arc that shares the endpoints of the inscribed angle's sides. In a circle, the angle formed by two chords with the common endpoints of a circle is called an inscribed angle and the common endpoint is considered as the vertex of the angle.

In this problem, the side length of the square is 5 which forms 90 degrees to all the other sides.

The measure of the circumscribed angle is 90 degree

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The probability of landing on heads on the coin is 1/2. The probability of landing on a number less than 7 on the spinner is 6/8 or 3/4. Since these two events are independent, the probability of both events happening is the product of their individual probabilities:

[tex](1/2) \times (3/4) = 3/8[/tex]

Note:- I'm sorry to bother you but can you please mark me BRAINLEIST if this ans is helpfull

The answer is 24 .

A mathematical expression or equation may be simplified by being reduced to its most basic form. To do this, complex statements or equations must be simplified using mathematical operations including addition, subtraction, multiplication, division, and exponentiation. As it reduces errors, makes problems simpler to answer, and helps people understand mathematical concepts, simplification is a crucial mathematical talent. It is widely used in calculus, algebra, and other areas of mathematics.

According to the question,

8x8x8 / 2 = 256

=256+81-1

=336

=6x6x6 -4x5 = 196 =14²

=336/14

=24

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The equation for line B is y = (2/3)x + 7/3.

What is the slope?

The slope is a measure of how steep a line is, and it describes the rate at which the line is changing. It is defined as the ratio of the vertical change (rise) to the horizontal change (run) between any two points on a line.

We can start by using the fact that lines A and B are perpendicular.

The slopes of two perpendicular lines are negative reciprocals of each other, so we can find the slope of line B by taking the negative reciprocal of the slope of line A:

The slope of a line can be calculated by choosing any two points on the line and using the formula:

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

the slope of line A = -3/2

slope of line B = 2/3 (negative reciprocal of -3/2)

Now we can use the point-slope form of the equation of a line to write an equation for line B. The point-slope form is:

y - y1 = m(x - x1)

where m is the slope of the line and (x1, y1) is a point on the line.

We know that the point (-2, 1) lies on line B, so we can use that as our (x1, y1) values.

We also know that the slope of line B is 2/3. Plugging these values into the point-slope form, we get:

y - 1 = (2/3)(x + 2)

We can simplify this equation by distributing the 2/3:

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

y = (2/3)x + 4/3 + 1

y = (2/3)x + 7/3

Therefore, the equation for line B is y = (2/3)x + 7/3.

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The First and the second function are decreasing only over the interval(-1,1)

a function is a relation between a set of inputs (called the domain) and a set of outputs (called the range) such that each input is associated with exactly one output. The output value depends on the input value, and this relationship is often represented by a formula or a graph.

the first

when x=-1 f(-1)=3

         x=0 f(0)=0(from the table to know)

         x=1 f(1)=-3

3>0>-3

so,the first function is decreasing

the second

f(-1)=2 f(0)=0 f(1)=-8

8>0>-8

so,the second function is decreasing

the third and fourth are

f(-1)<f(a)<f(-1)

so,the function is increasing

the first and second function are decreasing only over the interval(-1,1)

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Therefore , f(x) = 4000  (0.925)ˣ is the function

Firewood loses 7.5% of its weight each week. In other words, 92.5% of the original weight of the firewood is still present after one week.

The function I gave is an illustration of a function that denotes the quantity of firewood that is still available x weeks after the start of the winter season.

A mathematical item called a function accepts an input and creates an output. The number of weeks since the start of the winter season is the input in this scenario, and the output is the weight of firewood still available.

The function can be used to express the weight of firewood left over x weeks after the start of winter:

f(x) = 4000(1 - 0.075)ˣ

Firewood loses 7.5% of its weight each week. In other words, 92.5% of the original weight of the firewood is still present after one week

where x represents how many weeks have passed since the start of the winter season.

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The marginal PDF of X is: fX(x) = (90 - 6x)/160, for 0 ≤ x ≤ 3, And the marginal PDF of Y is: fY(y) = (120 - 6y)/160, for 0 ≤ y ≤ 4

To find the value of the constant C, we integrate the joint probability density function over the entire range of X and Y, and set the result equal to 1, since the total probability over the entire range of the two variables must be equal to 1:

∫∫ f(x,y) dxdy = 1

∫∫ C(30 - 2x - 3y) dxdy = 1

We can evaluate this double integral by integrating over Y first and then X:

∫∫ C(30 - 2x - 3y) dxdy = C∫[0,4] ∫[0,3-2/3y] (30 - 2x - 3y) dxdy

= C∫[0,4] [30x -[tex]x^2[/tex] - 3xy] evaluated from 0 to 3-2/3y dy

= C∫[0,4] (90 - 36y + 4[tex]y^2[/tex])/3 dy

= C[(30y^2 - 36[tex]y^3/2[/tex] + 4[tex]y^3[/tex]/3)/3] evaluated from 0 to 4

= C(160/3)

Setting this equal to 1, we get:

C(160/3) = 1

C = 3/160

Therefore, the constant C is 3/160.

Now, we can find the marginal probability density functions of X and Y by integrating the joint probability density function over the range of the other variable. For the marginal PDF of X:

fX(x) = ∫ f(x,y) dy

fX(x) = ∫ 3/160 (30 - 2x - 3y) dy, for 0 ≤ x ≤ 3

fX(x) = (90 - 6x)/160, for 0 ≤ x ≤ 3

And for the marginal PDF of Y:

fY(y) = ∫ f(x,y) dx

fY(y) = ∫ 3/160 (30 - 2x - 3y) dx, for 0 ≤ y ≤ 4

fY(y) = (120 - 6y)/160, for 0 ≤ y ≤ 4

Therefore, the marginal PDF of X is:

fX(x) = (90 - 6x)/160, for 0 ≤ x ≤ 3

And the marginal PDF of Y is:

fY(y) = (120 - 6y)/160, for 0 ≤ y ≤ 4

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1) With the given alpha of .05 and the appropriate degrees of freedom (based on the number of groups and the sample size).
2) This can be done in Excel, using the ANOVA function, or with other statistical software.
3) The calculated P value will be given as part of the ANOVA test output.

4 SS error is the sum of squares within groups. These values will also be available in the ANOVA test output.

5) Once you have the data from the Excel file, you can perform these calculations and interpret the results.

Once you have the data from the Excel file, you can perform these calculations and interpret the results.

he critical F value, use an F-distribution table or an online calculator, with the given alpha of .05 and the appropriate degrees of freedom (based on the number of groups and the sample size).

Based on the information provided, this study involves a one-way between-groups ANOVA test. This is because the consumer psychologist is comparing the grease levels of burgers from three different fast-food restaurants (In-n-Out, Jack in the Box, and Whataburger), and the measurements are taken by four different people.

As for the critical F value, calculated F value, calculated P value, and partial eta squared, I am unable to access the "Burgers Grease Levels" Excel file on Blackboard. However, I can provide guidance on how to calculate these values:

1. To find the critical F value, use an F-distribution table or an online calculator, with the given alpha of .05 and the appropriate degrees of freedom (based on the number of groups and the sample size).

2. To calculate the F value, you will need to perform the one-way between-groups ANOVA test on the grease levels data. This can be done in Excel, using the ANOVA function, or with other statistical software.

3. The calculated P value will be given as part of the ANOVA test output.

4. To calculate partial eta squared, use the formula: partial[tex]{\eta}^2=SS_{effect} / (SS_{effect} + SS_{error}),[/tex] where SS_effect is the sum of squares between groups, and SS error is the sum of squares within groups. These values will also be available in the ANOVA test output.

Once you have the data from the Excel file, you can perform these calculations and interpret the results.

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Answer:

Step-by-step explanation:

so since he pours 295 grams and need to remove 12 spoonfuls the answer will me 1.7 milligrams

The constants value of  α = -4 and β = 0.

A differential equation is an equation which contains one or more terms and the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable) dy/dx = f(x) Here “x” is an independent variable and “y” is a dependent variable. For example, dy/dx = 5x.

The function is :

[tex]y"+\alpha y'+\beta y=t+e^(^4^t^)[/tex]

the form of the particular solution to this differential equation as prescribed by the method of undetermined coefficients is

[tex]yp(t)=A1t^2+A0t+B0te^(^4^t^).[/tex]

=> [tex]y ' = 2A1t + A0 + B0 [e^(^4^t^) +4 te^(^4^t^) ][/tex]

    [tex]y ' = 2A1t + A0 + B0 e^(^4^t^) +4B0 te^(^4^t^)[/tex]

=> [tex]y '' = 2A1 + 4B0e^(^4^t^) + 4B0 [ e^(^4^t^) + 4te^(^4^t^)[/tex]

    [tex]y '' = 2A1 + 4B0e^(^4^t^) + 4B0e^(^4^t^) + 16B0te^(^4^t^)[/tex]

Now substitute the values of y ' and y '' in the differential equation:

[tex]y"+\alpha y'+\beta y=t+e^(^4^t^)[/tex]

[tex]2A1 + 4B0e^(^4^t^) + 4B0e^(^4^t^) + 16B0te^(^4^t^) + \alpha {2A1t + A0 + B0e^(^4^t^) + 4B0 te^(^4^t^) } + \beta {A1 t^2 + A0 t + B0 t e6(^4^t^)} = t + e^(^4^t^)[/tex]

Next, we equate coefficients

1) Constant terms of the left side = constant terms of the right side:

[tex]2A1+ 2\alpha A0 = 0[/tex] ..... eq (1)

2) Coefficients of [tex]e^(^4^t^)[/tex] on both sides

8B0 + αB0 = 1 => B0 (8 + α) = 1 .... eq (2)

3) Coefficients on t

2αA1 + βA0 = 1 .... eq (3)

4) Coefficients on [tex]t^2[/tex]

βA1 = 0 ....eq (4)

A1 ≠ 0 => β =0

5) terms on [tex]te^(^4^t^)[/tex]

16B0 + 4αB0 + βB0 = 0 => B0 (16 + 4α + β) = 0 ... eq (5)

B0 ≠ 0 => 16 + 4α + β = 0

Use the value of β = 0 found previously

16 + 4α = 0 => α = - 16 / 4 = - 4.

α = - 4 and β = 0

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The  calculated difference between the sample means tracks a normal distribution with mean m₁ - m₂  and standard deviation √(5500.33).

Then the lifetimes of both brands of tires come from the same typical dissemination, the distinction in comparison to their test implies that it is ordinary dissemination.

Precisely, on the off chance that we let X and Y.

This projects  the test which implies the primary and moment brands, separately, and let s indicate the common standard deviation (given as the square root of 33002), at that point the conveyance of the distinction X-Y is additionally typical.

Now the  mean m₁ - m₂ and the standard deviation is given by the square root of the whole of the fluctuations, which in this case is the square root of 2 times the fluctuation of each test cruel (since the test sizes and changes are broken even with):
√(2) × √(33002/12) = √(5500.33)

Therefore, the difference between the sample means follows a normal distribution with mean m₁ - m₂

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The best test to use in this case is a two sample t-test of means

The most suitable test to use in this case is a two sample t-test of means since this study compares the mean test scores of two independent groups.

The 2-sample t-test use the sample data provided from two groups and gives the t-value. The process is somewhat close to the usual t-test and we can use the concept of the signal to noise ratio. However, the two-sampled t-test requires independent variable.

In a two-sample t-test, the numerator is the signal which is the difference between the two means.

The default null hypothesis of a 2-sample t-test can be said to be of two groups that are equal

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a) the estimated amount of dirt in the pile after 3 hours is 53,000 cubic feet. b) our estimate in Part A (53,000) was an underestimate. c) the specific solution that satisfies the initial condition P(0) = 35,000 is:

[tex]P = 7000 + 25,000e^{(t/5)[/tex]

Part A:

To estimate the amount of dirt in the pile after 3 hours, we will use the tangent line to the graph of P at t = 0.

First, we need to find P(0) and P'(0) to determine the equation of the tangent line.

P(0) is given as 35,000 cubic feet, which is the initial amount of dirt in the pile.

To find P'(0), we plug in t = 0 into the differential equation:

dP/dt = 1/5 (P - 7000)

dP/dt = 1/5 (35,000 - 7000)

dP/dt = 6,000

Therefore, P'(0) = 6,000.

Now, we can use the point-slope form of the equation of a line to find the tangent line:

y - y1 = m(x - x1)

P - 35,000 = 6,000(t - 0)

P = 6,000t + 35,000

To estimate the amount of dirt in the pile after 3 hours, we plug in t = 3:

P(3) = 6,000(3) + 35,000 = 53,000

Therefore, the estimated amount of dirt in the pile after 3 hours is 53,000 cubic feet.

Part B:

To determine whether our estimate in Part A was an underestimate or an overestimate at t = 3, we need to find [tex]d^2P/dt^2[/tex] and evaluate it at t = 3.

Taking the second derivative of the given differential equation with respect to t, we get:

[tex]d^2P/dt^2 = 1/5\ dP/dt\\\\d^2P/dt^2 = 1/5 (P - 7000)[/tex]

To evaluate this at t = 3, we need to find P(3). Using the equation we found in Part A:

P(3) = 6,000(3) + 35,000 = 53,000

So, we have:

[tex]d^2P/dt^2 = 1/5 (53,000 - 7000) = 8,400[/tex]

Since [tex]d^2P/dt^2[/tex] is positive at t = 3, this means that P is concave up at this point. Therefore, our estimate in Part A (53,000) was an underestimate.

Part C:

To find the general solution to the differential equation dP/dt = 1/5 (P - 7000), we can separate variables and integrate both sides:

dP/(P - 7000) = (1/5) dt

Integrating both sides:

ln|P - 7000| = (1/5) t + C

where C is the constant of integration.

Solving for P, we have:

|P - 7000| = [tex]e^{(t/5 + C)[/tex]

P - 7000 = ±[tex]e^{(t/5 + C)[/tex]

P = 7000 ± [tex]e^{(t/5 + C)[/tex]

where the ± sign indicates that there are two possible solutions depending on the sign of the exponential term.

To find the specific solution that satisfies the initial condition P(0) = 35,000, we can plug in these values:

35,000 = 7000 ± [tex]e^{(0/5 + C)[/tex]

Solving for C, we get:

C = ln(25,000)

Plugging this back into the general solution, we get:

P = 7000 + [tex]e^{(t/5 + ln(25,000))[/tex]

Since [tex]e^{(ln(25,000))} = 25,000[/tex], we can simplify this to:

P = 7000 + 25,000[tex]e^{(t/5)[/tex]

Therefore, the specific solution that satisfies the initial condition P(0) = 35,000 is:

P = 7000 + 25,000[tex]e^{(t/5)[/tex]

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Answer:

A. 800

Step-by-step explanation:

Eight in the number is three places over from the decimal spot. this means the eight is in the hundreds spot. This makes it 800.

A function is considered continuous at a point if its limit exists at that point and is equal to the function's value at that point.

a function is continuous at a point if it has no gaps, jumps, or holes in its graph at that point. Since the integral of f(x) from 2 to 10 is zero, and f(x) is continuous and non-negative on this interval, it follows that f(x) must be identically zero on [2, 10].

Therefore, f(6) = 0

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