Determine for which values of m the function variant Φ(x) = x^m is a solution to the given equation. a. 3x^2 (d^2y/dx^2) + 11x(dy/dx) - 3y = 0 b. x^2 (d^2y/dx^2) - x(dy/dx) - 5y = 0

Answer:

-60

Step-by-step explanation:

The objective is to state whether or not the following limit exists

                                [tex]\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2}[/tex].

First, we simplify the expression in the numerator of the fraction.

[tex]3(2x-1)^2 -75 = 3(4x^2 - 4x +1) -75 = 12x^2 - 12x + 3 - 75 = 12x^2 - 12x -72[/tex]

Now, we obtain

                         [tex]12(x^2-x-6) = 12(x+2)(x-3)[/tex]

and the fraction is transformed into

                       [tex]\frac{3(2x-1)^2 - 75}{x+2} = \frac{12(x+2)(x-3)}{x+2} = 12 (x-3)[/tex]

Therefore, the following limit is

       [tex]\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2} = \lim_{x \to -2} 12(x-3) = 12 \lim_{x \to -2} (x-3)[/tex]

You can plug in [tex]-2[/tex] in the equation, hence

                        [tex]12 \lim_{x \to -2} (x-3) = 12 (-2-3) = -60[/tex]

The limit does not exist.

To find the limit of the given function as x approaches -2, we can simply substitute -2 into the function and simplify.

Start by replacing x with -2 in the function:

lim as x → -2 (3(2x - 1)2 - 75) / (x + 2)

Substitute -2 for x:

(3(2(-2) - 1)2 - 75) / (-2 + 2)

Simplify:

(3(-4 - 1)2 - 75) / 0

Continue simplifying:

(3(-5)2 - 75) / 0

(3(25) - 75) / 0

(75 - 75) / 0

0 / 0

Since we end up with 0/0, the limit is undefined, or it does not exist.

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Answer:

Population:  The computer chips coming off of an assembly line.

Sample: The selected 36 chips coming off an assembly line.

Step-by-step explanation:

In the undergoing study the population is always consists of set of all possible units and sample is a part or subset of population. The population of interest in the study conducted by engineer to check the whether the computer chips coming off of an assembly line are within the specification consists of all the computer chips coming off of an assembly line. The sample in the undergoing study are the selected 36 computer chips coming off of an assembly line.

In this scenario, the population is all the computer chips coming off the assembly line and the sample is the 36 chips the engineer randomly selects and tests.

In this study, the population refers to all the computer chips coming off of the assembly line. It's important to note that the population in statistical terms does not have to represent an actual, concrete group of physical objects or individuals. Instead, it represents the total set of observations that can be made. In this case, all the chips that come off the assembly line - both those chosen for testing and those not chosen - constitute the population.

On the other hand, the sample in this study would be the 36 random chips that the engineer tests. A sample is a subset of the population that is selected for study. The characteristics of this sample are then used to infer information about the overall population.

So, in the context of this study, the objective is to infer from the testing of 36 chips (sample) whether all chips coming off the assembly line (population) are within specifications.

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The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. This calculation gives the percentage of the theoretical yield that is actually obtained in the reaction.

The percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100. This calculation gives the percentage of the theoretical yield that is actually obtained in the reaction. The formula for percent yield is:

Percent Yield = (Actual Yield / Theoretical Yield) x 100

Actual and theoretical yields can be expressed as masses or molar amounts as long as they are in the same units. The percent yield allows us to quantify the efficiency of a reaction and determine how much product was obtained compared to the maximum potential.

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Therefore, the particular solution is:

[tex]\[ y = \frac{1}{2}e^{-e^{-2y}} + \ln(4) - \frac{1}{2}e^{-\frac{1}{4}} \][/tex]

To find the particular solution of the given differential equation, we need to integrate both sides with respect to x. However, since the equation is not separable, we can use the method of integrating factors.

First, let's rewrite the equation in the form:

[tex]\[\frac{dy}{dx} - e^{-2y}(x-4) = 0\][/tex]

To find the integrating factor, we consider the term multiplying dy/dx, which is [tex]\(-e^{-2y}\).[/tex]

The integrating factor, denoted by [tex]\( \mu \)[/tex], is given by the exponential of the integral of [tex]\(-e^{-2y}\):[/tex]

[tex]\[ \mu = e^{\int -e^{-2y} dx} \][/tex]

[tex]\[ = e^{\frac{1}{2}e^{-2y}} \][/tex]

Multiplying both sides of the differential equation by the integrating factor [tex]\( \mu \)[/tex], we get:

[tex]\[ e^{\frac{1}{2}e^{-2y}}\frac{dy}{dx} - (x-4)e^{-\frac{1}{2}e^{-2y}} = 0 \][/tex]

This can be written as the derivative of a product:

[tex]\[ \frac{d}{dx}\left( e^{\frac{1}{2}e^{-2y}}y \right) = (x-4)e^{-\frac{1}{2}e^{-2y}} \][/tex]

Now, integrating both sides with respect to x, we get:

[tex]\[ e^{\frac{1}{2}e^{-2y}}y = \int (x-4)e^{-\frac{1}{2}e^{-2y}} dx + C \][/tex]

[tex]\[ e^{\frac{1}{2}e^{-2y}}y = \int (x-4)e^{-\frac{1}{2}e^{-2y}} dx + C \][/tex]

[tex]\[ e^{\frac{1}{2}e^{-2y}}y = \int (x-4)e^{-\frac{1}{2}e^{-2y}} dx + C \][/tex]

At this point, it seems difficult to directly integrate the right-hand side. So, let's substitute [tex]\( u = e^{-\frac{1}{2}e^{-2y}} \), then \( du = -\frac{1}{2}e^{-2y}e^{-\frac{1}{2}e^{-2y}} dy \).[/tex]

After making this substitution, the equation becomes:

[tex]\[ y = \int (x-4) du + C \][/tex]

[tex]\[ y = \frac{1}{2}u^2 + C \][/tex]

[tex]\[ y = \frac{1}{2}e^{-e^{-2y}} + C \][/tex]

To solve for  C , we use the initial condition [tex]\( y(4) = \ln(4) \):[/tex]

[tex]\[ \ln(4) = \frac{1}{2}e^{-e^{-2\ln(4)}} + C \][/tex]

[tex]\[ \ln(4) = \frac{1}{2}e^{-\frac{1}{4}} + C \][/tex]

[tex]\[ C = \ln(4) - \frac{1}{2}e^{-\frac{1}{4}} \][/tex]

Therefore, the particular solution is:

[tex]\[ y = \frac{1}{2}e^{-e^{-2y}} + \ln(4) - \frac{1}{2}e^{-\frac{1}{4}} \][/tex]

The Correct question is:

Find the particular solution of the differential equation

dydx=(x−4)e^(−2y) satisfying the initial condition y(4)=ln(4).

Answer: y=

The particular solution of the differential equation [tex]\(\frac{dy}{dx} = (x - 5)e^{-2y}\)[/tex] satisfying the initial condition [tex]\(y(5) = \ln(5)\)[/tex] is given by the implicit equation [tex]\(e^{2y} - xe^{2y} + 2y = 2\ln(5) + 5\)[/tex].

To find the particular solution, we start by separating the variables in the differential equation:

[tex]\[\frac{dy}{dx} = (x - 5)e^{-2y}\][/tex]

[tex]\[e^{2y} dy = (x - 5) dx\][/tex]

Now, we integrate both sides:

[tex]\[\int e^{2y} dy = \int (x - 5) dx\][/tex]

[tex]\[\frac{1}{2}e^{2y} = \frac{1}{2}x^2 - 5x + C\][/tex]

To find the constant of integration [tex]\(C\)[/tex], we use the initial condition [tex]\(y(5) = \ln(5)\)[/tex]:

[tex]\[\frac{1}{2}e^{2\ln(5)} = \frac{1}{2}(5)^2 - 5(5) + C\][/tex]

[tex]\[\frac{1}{2}e^{\ln(25)} = \frac{1}{2}(25) - 25 + C\][/tex]

[tex]\[\frac{1}{2}(25) = \frac{1}{2}(25) - 25 + C\][/tex]

[tex]\[C = 25\][/tex]

Substituting [tex]\(C\)[/tex] back into the equation, we get:

[tex]\[\frac{1}{2}e^{2y} = \frac{1}{2}x^2 - 5x + 25\][/tex]

Multiplying through by 2 to clear the fraction:

[tex]\[e^{2y} = x^2 - 10x + 50\][/tex]

Now, we add [tex]\(2y\)[/tex] to both sides to isolate [tex]\(e^{2y}\)[/tex]:

[tex]\[e^{2y} + 2y = x^2 - 10x + 50 + 2y\][/tex]

Since [tex]\(e^{2y} - xe^{2y} + 2y = e^{2y} + 2y - xe^{2y}\)[/tex], we can rewrite the equation as:

[tex]\[e^{2y} - xe^{2y} + 2y = 50 - 10x + 2y\][/tex]

Using the initial condition [tex]\(y(5) = \ln(5)\)[/tex] again, we have:

[tex]\[e^{2\ln(5)} - 5e^{2\ln(5)} + 2\ln(5) = 50 - 10(5) + 2\ln(5)\][/tex]

[tex]\[25 - 5(25) + 2\ln(5) = 50 - 50 + 2\ln(5)\][/tex]

[tex]\[25 - 125 + 2\ln(5) = 2\ln(5)\][/tex]

[tex]\[-100 + 2\ln(5) = 2\ln(5)\][/tex]

This confirms that the constant [tex]\(C\)[/tex] is correct. Therefore, the particular solution of the differential equation satisfying the initial condition is:

[tex]\[e^{2y} - xe^{2y} + 2y = 2\ln(5) + 5\][/tex]

Answer:

The coefficient is 3 and the constant is 4 in

relation to the equation mx + c where m is the coefficient of x and c is the constant.

Answer: (a). prob = 0.1543

(b). Prob = 2⁵⁰/ \left[\begin{array}{ccc}100\\50\end{array}\right].

Step-by-step explanation:

The information in the question tells us that the United States contains two senators from each of the 50 states, this means there is a total of a 100 senators in the senate.

(a). this question tells us to find the probability that it will contain one of the the two senators from a certain specified state.

Explanation:

Consider selecting a committee of 8 senators at random,

the total number of ways to select 8 senators from the Senate of 100 senators is \left[\begin{array}{ccc}100\\8\end{array}\right].

Next is to consider an event Q that at least one of the senators from the specified state is added to the senate,

Let us call the two senators F and G.

where F is the number of possible ways to select 7 members from 98 senetors; \left[\begin{array}{ccc}98\\7\end{array}\right].

Also, G is the number of possible ways to select 6 members from 98 senators; \left[\begin{array}{ccc}98\\6\end{array}\right].

The probability that committee of 8 senators will contain at least one of the two senators from a certain specified state is given thus;

Prob (Q) = \left[\begin{array}{ccc}2\\1\end{array}\right]\left[\begin{array}{ccc}98\\7\end{array}\right]  / \left[\begin{array}{ccc}100\\8\end{array}\right]   + \left[\begin{array}{ccc}2\\2\end{array}\right] \left[\begin{array}{ccc}98\\6\end{array}\right] / \left[\begin{array}{ccc}100\\8\end{array}\right]

Prob (Q) = 0.1487 + 0.0057 = 0.1543

Prob (Q) = 0.1543

(b).  the questions tells us to find the probability it will contain one senator each from each of the states.

consider that a group of 50 senators is selected in random, the total number of ways to select 50 senators from the senate of 100 senators is

\left[\begin{array}{ccc}100\\50\end{array}\right].

Now let us consider that an event Q, that a total of 50 groups contains exactly one senator from each state. for any state, there are two ways for exactly one senator to be in a group of 50.

∴ the probability becomes;

Pr (Q) = 2⁵⁰/ \left[\begin{array}{ccc}100\\50\end{array}\right].

cheers, i hope this helps.

This response provides a methodology for calculating probabilities related to the selection of senators from the U.S. Senate. It explains that the probability that a randomly selected eight-person committee includes a senator from a particular state is simply one minus the ratio of groups of eight not from that state to the total groups of eight. Furthermore, the probability of randomly selecting a fifty-person group that includes one senator from each state is 1.

The subject of this question is the calculation of probability in the context of selections from the United States Senate. Probability describes the likelihood of an event occurring and is often represented as a fraction or decimal, where 1 (or 100%) represents certainty, and 0 (or 0%) indicates impossibility.

(a) To find the probability that a committee of eight senators includes at least one senator from a particular state, first consider the total number of possible committees: C(100,8), or combinations of 100 things taken 8 at a time. The easiest way to ensure at least one senator from a particular state is to envision the opposite scenario: the committee consists of senators from any state but that one, calculated as C(98,8). The desired probability is then 1 minus this ratio, or 1 - C(98,8) / C(100,8).

(b) The probability that a group of 50 senators selected at random will contain one senator from each state is more straightforward, as there is only one way to draw one senator from each of the 50 states out of a pool of 100 senators. So, the probability for such an event to occur is C(100,50) / C(100,50) = 1.

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Answer:

b. False.

See explanation below.

Step-by-step explanation:

b. False.

When we have a significant result that means [tex] P_v < \alpha[/tex] where [tex] P_v[/tex] represent the p value for the test and [tex]\alpha[/tex] the significance level assumed at the begin of the hypothesis test.

For this case we have a null hypothesis [tex]H0[/tex] and an alternative hypothesis [tex]H_1[/tex] for a parameter of interest let's say [tex]\theta[/tex], and using the test we conclude thar we reject the null hypothesis, so on this case we need to have that [tex] p_v <\alpha[/tex], so then that means that we have a significant difference.

And when we have this situation we can't say that the difference between the sample statistic and the hypothesized parameter is just due to chance, since we are obtaining singificant results that are showing difference between the two values on statistical terms

Answer:

mi/hr • min does not equal mi. You must convert 30 minutes to hours.

Step-by-step explanation:

I got it correct on TTM

The area of the region is bounded by the curve [tex]\rm y=e^2x[/tex] the x-axis the y axis, and the line x=2 is equal to [tex]\rm \dfrac{e^4}{2}-\dfrac{1}{2}[/tex].

Given that,

The area of the region bounded by the curve [tex]\rm y=e^2x[/tex],

We have to determine,

The x-axis the y axis and the line x=2 is equal to.

According to the question,

The area of the region bounded by the curve

[tex]\rm y=e^2x[/tex]

The area of the region bounded by the curve is determined by integrating the curve at x = 0 to x = 2.

Integrating the curve on both sides,

[tex]\rm Area=\int\limits^2_0 { e^{2x}} \, dx\\\\Area=[ \dfrac{e^{2x}}{2}]^2_0\\\\Area= [ \dfrac{e^{2(2)}}{2}- \dfrac{e^{2(0)}}{2}]\\\\Area = \dfrac{e^4}{2}-\dfrac{e^0}{2}\\\\Area = \dfrac{e^4}{2}-\dfrac{1}{2}[/tex]

Hence, The area of the region is bounded by the curve [tex]\rm y=e^2x[/tex] the x-axis the y axis, and the line x=2 is equal to [tex]\rm \dfrac{e^4}{2}-\dfrac{1}{2}[/tex].

To know more about Area under the curve click the link given below.

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Answer: A differentiable function [tex]f(x)[/tex] has a local minimum at the point [tex]x_0[/tex] if two conditions are met: the value of its first derivative is equal to zero at that point and the value of its second derivative is negative at that point.

Step-by-step explanation: The procedure for finding the local minima of the function [tex]f(x)[/tex] is the following.

Step 1. Find the first derivative of the function [tex]f(x)[/tex], denoted by [tex]f'(x)[/tex] according to the rules of derivation.

Step 2. Find all [tex]x[/tex] such that [tex]f'(x)=0.[/tex] Denote these solutons by [tex]x_1, x_2\ldots[/tex].

Step 3. Find the second derivative of the function [tex]f(x)[/tex], denoted by [tex]f''(x)[/tex]. Evaluate this derivative at each point found in step 2. Only If, say [tex]f''(x_1)>0[/tex] then [tex]x_1[/tex] is the local minimum and the same goes for all other values of [tex]x[/tex] you found in step 2.

For what value(s) of x does f(x) have a local minimum?

Using the example below to explain

f(x) = x2 − 6x + 5.  

Answer:

The point x on the function f(x) is a local minimum if and only if the following conditions are satisfied

1. f'(x) = 0 (at that point df(x)/dx must be equal to zero)

2. f"(x)>0 (the second derivative of the function must be greater than zero, it must be positive)

Using the example below to explain

f(x) = x2 − 6x + 5.  

Since f'(x)= 0 and f"(x) greater than 0 (positive), then we can now confirm that the function f(x) has a local minimum at x = 3

Step-by-step explanation:

The point x on the function f(x) is a local minimum if and only if the following conditions are satisfied

1. f'(x) = 0 (at that point df(x)/dx must be equal to zero)

2. f"(x)>0 (the second derivative of the function must be greater than zero, it must be positive)

For the example above:

f(x) = x2 − 6x + 5

f'(x) = 2x - 6

Condition 1:

f'(x) = 0

So,

f'(x) = 2x - 6 = 0

Solving for x

2x - 6 = 0

2x = 6

x = 3

Therefore, at x = 3, f(x) has a critical point.

We need to determine whether it is a local minimum, local maximum or saddle point.

Condition 2:

f"(x) > 0

f"(x) = f'(f'(x)) = d/dx (2x - 6) = 2

So,

f"(x) = 2 >0

Note: in some cases we would need to substitute x into f"(x) to determine the value.

Since f'(x)= 0 and f"(x) greater than 0 (positive), then we can now confirm that the function f(x) has a local minimum at x = 3

Answer:

(5.83, 2.11, 2)

Step-by-step explanation:

To convert from rectangular coordinates to cylindrical coordinates we use

[tex]x=rcos(u)[/tex]

[tex]y=rsin(u)[/tex]

[tex]r=\sqrt{x^2+y^2}[/tex]

Therefore (-3,5,2):

[tex]r=\sqrt{(-3)^2+5^2}=5.83[/tex]

[tex]cosu=x/r=-3/5.83=-0.51[/tex]

[tex]u=2.11 radians[/tex]

So the coordinates are (5.83, 2.11, 2)

Answer: 0.025

Step-by-step explanation: we reject null hypothesis if p<0.05

Answer:

I also gave aadded the complete question for the expressions of the function f(x)

a) Exact value of the net signed area = 1/2 or 0.5

b) Exact average value of f = 53/30 = 1.7667

Step-by-step explanation:

The step by step explanation and calculation is attached below.

Answer:

we CANNOT DIVIDE 3 with 6.

Step-by-step explanation:

Here,as given in the question:

Starting value = 6

Constant ratio = 1/3

Now, exponential function is obtained by the product of starting value and the constant ratio repeatedly.

⇒ f(x) = (Starting value) x (ratio)... x times

[tex]\implies f(x) = 6 (\frac{1}{3} )^x = 6 (\frac{1}{3} ) (\frac{1}{3} ) (\frac{1}{3} ) .... x[/tex]  

Now, we CANNOT DIVIDE 3 with 6 as it is in the power of x.

Hence, [tex]\implies f(x) = 6 (\frac{1}{3} )^x[/tex] and [tex]f(x) \neq 2^x[/tex]

Answer:

The student made an error by applying the exponent to the product of a and b instead of just b. The final answer should be ​f(x)=6(1/3)^x

Step-by-step explanation:

U can't multiply 6 by 1/3

Answer:

Step-by-step explanation:

Be the points Pa=(-1,-6,-8) ; Pb=(0,-4,-5) we have calculate the middle point or center

[tex]c=(\frac{x1+x2}{2}, \frac{y1+y2}{2}, \frac{z1+z2}{2})=(\frac{-1}{2}, -5,\frac{-13}{2})[/tex]

Now we must to find d=r (view graph)

[tex]r=\sqrt{(Cx-x2)^{2}+(Cy-y2)^{2}+(Cz-z2)^{2}}\\ r=\sqrt{(\frac{-1}{2} )^{2}+(-5+4)^{2}+(\frac{-13}{2}+5 )^{2}}\\r=\sqrt{\frac{1}{4} +1+\frac{9}{4}}=\sqrt{\frac{14}{4}}=r^{2}=\frac{14}{4}[/tex]

We find the canonical sphere equation

[tex](x-h)^{2}+(y-k)^{2}+(z-l)^{2}=r^{2}\\(x+\frac{1}{2})^{2}+(y+5)^{2}+(z+\frac{13}{2} )^{2}=\frac{14}{4}\\x^{2}+x+y^{2}+10y+z^{2}+13z+64=0[/tex]

Note: The Pa=(-1,-6,-8) can also be used  in c

Answer:

Step-by-step explanation:

The values of x are given as

0, 1, 2, 3

The corresponding values of y are given as

0,5 10,15

Let k represent constant of proportionality

Therefore,

When x = 0, y = 0

When x = 1, y = 5

y/x = k

k = 5/1 = 5

When x = 2, y = 10

y/x = k

k = 10/2 = 5

When x = 3, y = 15

y/x = k

k = 15/3 = 5

Therefore, the constant of proportionality is 5

The equation to represent the table is

y = 5x

Answer:

Step-by-step explanation:

1. Un = 3^ n

U1 = 3, U2 = 9, U3 = 27, U4 = 81, U5 = 243

2. Un = (-2/5)^n

U1 = -2/5, U2 = 4/25, U3 = 8/125, U4 = 16/625, U5 = 32/3125

3. Un = sin npi/2

U1 = 1, U2 = 0, U3 = -1, U4 = 0, U5 = -1

4. Un = 3n/n+4

U1 = 3/5, U2 = 1, U3 = 9/7, U4 = 3/2, U5 = 5/3

5. Un =  (-1)^n+1(2/n)

U1 = 1, U2 = 2, U3 = -1/3, U4 = 3/2, U5 = -3/5

6. Un = 2 + 2/n - 1/n^2

U1 = 3, U2 = 11/4, U3 = 23/9, U4 = 39/16, U5 = 59/25

Answer:Amy used 4 41-cent stamps and 8 6-cent stamps.

Step-by-step explanation:

Let x represent the number of 41-cent stamps that Amy used. Let y represent the number of 6-cent stamps that Amy used.

41 cents = 41/100 = $0.41

6 cents = 6/100 = $0.06

She uses 41-cent stamps and 6-cent stamps to pay $2.12 in postage. It means that

0.41x + 0.06y = 2.12

Multiplying through by 100, it becomes

41x + 6y = 212

6y = 212 -41x

We would test for corresponding values of x and y that satisfies the equation and they must be whole numbers.

If x = 3,

6y = 212 - 41 × 3 = 89

y = 89/6 = 14.8333

If x = 4,

6y = 212 - 41 × 4 = 48

y = 48/6 = 8

To solve this problem, we establish two linear equations based on the given information. After logical assumption, we deduce Amy used five 41-cent stamps (X) and seven 6-cent stamps (Y).

This is a problem of linear equations. We need to figure out the number of 41-cent stamps (let's denote them as X) and 6-cent stamps (we'll call them Y) Amy used to total the postage price of $2.12 We can write the following two equations based on the given in the question:

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Answer:

24

Step-by-step explanation:

A 2024-T4 aluminum tube with an outside diameter of 3.6 in. will be used to support a 24-kip load. If the axial stress in the member must be limited to 6.4 ksi, determine the wall thickness required for the tube.

Answer:[tex]\frac{n!}{r!(n-r)!}[/tex]

ways

Step-by-step explanation:

Given that a shelf contains n separate compartments. There are r indistinguishable marbles

The marbles are identical so they can be placed in any order.

Let us consider the places available for placing these r marbles

No of compartments available =n

Marbles to be placed = r

Since marbles are identical and order does not matter

number of ways the r marbles can be placed in the n compartments

= nCr

=[tex]\frac{n!}{r!(n-r)!}[/tex]

Answer:mean= 2905.83, median=3398, mode=5456

Step-by-step explanation:

The mean:

Mean=summation of six machine produced/ n

Mean=(2567+5456+3769+2245+3398)/6

Mean=17435/6

Mean=2905.8333333

b. The median

Firstly we have to rearranged the machine product in order:

2245, 2567, 3398, 3769, 5456

So 3398 is at the middle, so median is 3398

c. The mode

The machine produces the highest number (frequency) is mode. So the mode is 5456

Answer:

(a) Proved

(b) discussed

(c) There are infinite number of solutions.

Step-by-step explanation:

It will be easier just to give a solution that satisfies the differential equation, but that will not suffice.

These are first order Nonlinear Differential Equations whose solutions are not as straightforward as they might seem. Two questions must be asked:

1. Does the solution to the differential equation exist?

2. If it exists, is it unique?

I will explain the general case, and then explain how they correlate with your work.

EXISTENCE

Suppose F(t, x) is a continuous function. Then the initial value problem

x'= F(t, x), x(t_0) = a

has a solution x = f(t) that is, at least, defined for some δ > 0.

This guarantees the existence of solution to the initial value problem, at

least for infinitesimal times (t). In some cases, this is the most that can be said, although in many cases the maximal interval α < t < β of the existence of solution might be much larger, possibly infinite, −∞ < t < ∞, resulting in a general solution.

The interval of existence of a solution strongly depends upon both the equation and the particular initial values. For instance, even though its right hand side is defined everywhere, the solutions to the scalar initial value problem x' = x^⅓ only exist up until time 1/(x_0) (1/0 in this case, which is infinity), and so, the larger the initial value, the shorter the time of existence.

UNIQUENESS

having talked about the importance of existence of solution, we need to ask ourselves, does the initial value problem

have more than one solution? If it does, changes will happen everytime, and we cannot use the differential equation to predict the future state of the system. The continuity of the right hand side of the differential equation will ensure the existence of a solution, but it is not sufficient to guarantee uniqueness of the solution to the initial value problem. The difficulty can be appreciated by looking at the first differential equation you gave.

x' = x^⅓ , x(0) = 0

From the explanation above, since the right hand side is a continuous function, there exists a solution, at least for t close to 0. This equation can be easily solved by the method of integration:

dx/dt = x^⅓

dx/(x^⅓) = dt

Int{x^(-⅓)dx} = dt

(x^⅔)/(⅔) = t + c

(3/2)x^⅔ = t + c

x = (⅔t + c1)^(3/2)

Applying the initial condition x(0) = 0

implies that c1 = 0, and hence,

x = ⅔t^(3/2) is a solution to the initial value problem.

But again, since the right hand side of the differential equation vanishes at x = 0, the constant function x(t) ≡ 0 is an equilibrium solution to the differential equation. Moreover, the equilibrium solution has the same initial value x(0) = 0. Therefore, we have two different solutions to the initial value problem, which invalidates its uniqueness. In fact, there is an infinite number of solutions to the initial value problem. For any positive a, the function

x(t) = 0 for 0 ≤ t ≤ a,

= (⅔t − a)^(3/2) for 2t ≥ 3a,

is differentiable at every point.

This explains the situation of questions (a) and (b).

For question (c) x' = x/t² for x(0) = 0.1

This is quite different

Solving by integration, we have

dx/x = t^(-2) dt

ln x = -1/t + c

x = kexp(-1/t)

Applying the initial condition, we realise that as n approaches 0, the lim n approaches negative infinity.

Which also means there are infinitely many solutions.

I hope this helps

There are infinitely many solutions for the differential equation x' = x^1/3 satisfying x(0) = 0. For x' = x/t, there is a unique solution for any initial condition x(0) = x0. For x' = x/t^2, there are infinitely many solutions for different values of A.

(a) To prove that there are infinitely many different solutions of the differential equation x' = x1/3 satisfying x(0) = 0, we can consider the function x = 0 and the function x = t3/2. Both functions satisfy the differential equation and the initial condition. Since they are different functions, this proves that there are infinitely many solutions.

(b) For x' = x/t with x(0) = x0, it can be shown that the solution is given by x = t * ln(t) + x0. Hence, there is a unique solution for any initial condition x(0) = x0.

(c) For x' = x/t2 with x(0) = 0, the solution is given by x = Ae1/t, where A is an arbitrary constant. This implies that there are infinitely many solutions for different values of A.

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Answer:

a) 6,455 L

b) V(t) = (18/3) * t^(1.5)

Step-by-step explanation:

Part a

Integrating the given relation from t = 0 to t= 105 mins

[tex]V = \int {9 * \sqrt{t} } \, dt\\V = {\frac{9*2}{3}*t^(1.5) = \frac{18}{3}*t^(1.5)\\\\V = \frac{18}{3}*105^(1.5)\\\\\\V = 6,455L[/tex]

Part b

[tex]V(t) = \frac{18}{3} * t^(1.5) + C\\Since V = 0 @ t = 0 ; hence, C = 0\\\\V(t) = \frac{18}{3} * t^(1.5)[/tex]

a) Amount of water that flows into the cistern in 1.75 hours is 6450 liters. b) Function for the amount of water in the tank at any time [tex]\( t \geq 0 \)[/tex] is [tex]\( Q(t) = 6 t^{3/2} \)[/tex].

We'll follow the steps for each part and include the necessary calculations and graphs.

Part a: Amount of water that flows into the cistern in 1.75 hours

1. Convert 1.75 hours to minutes:

[tex]\[ 1.75 \text{ hours} = 1.75 \times 60 \text{ minutes} = 105 \text{ minutes} \][/tex]

2. Integrate the rate function [tex]\( Q'(t) = 9\sqrt{t} \)[/tex] from [tex]\( t = 0 \)[/tex] to [tex]\( t = 105 \)[/tex]:

[tex]\[ \int_{0}^{105} 9\sqrt{t} \, dt \][/tex]

Integration:

[tex]\[\int 9\sqrt{t} \, dt = \int 9t^{1/2} \, dt = 9 \int t^{1/2} \, dt = 9 \left( \frac{t^{3/2}}{3/2} \right) = 9 \left( \frac{2}{3} t^{3/2} \right) = 6 t^{3/2}\][/tex]

Evaluating this from [tex]\( t = 0 \)[/tex] to [tex]\( t = 105 \)[/tex]:

[tex]\[\left[ 6 t^{3/2} \right]_0^{105} = 6 \left( 105^{3/2} \right)\][/tex]

Calculating [tex]\( 105^{3/2} \)[/tex]:

[tex]\[105^{3/2} = (105)^{1.5} = 105 \times \sqrt{105} \approx 105 \times 10.24695 \approx 1075\][/tex]

Therefore:

[tex]\[6 \times 1075 = 6450 \text{ liters}\][/tex]

So, the amount of water that flows into the cistern in 1.75 hours is 6450 liters.

Part b: Function for the amount of water in the tank at any time [tex]\( t \)[/tex]

We already have:

[tex]\[ Q'(t) = 9\sqrt{t} \][/tex]

Integrating to find [tex]\( Q(t) \)[/tex]:

[tex]\[ Q(t) = \int 9\sqrt{t} \, dt = 6 t^{3/2} + C \][/tex]

Given that the tank is empty when [tex]\( t = 0 \)[/tex], we have [tex]\( Q(0) = 0 \)[/tex]:

[tex]\[ 0 = 6 \times 0^{3/2} + C \][/tex]

[tex]\[ C = 0 \][/tex]

Thus, the function that gives the amount of water in the tank at any time [tex]\( t \geq 0 \)[/tex] is:

[tex]\[ Q(t) = 6 t^{3/2} \][/tex]

Answer:

[tex]x=\sqrt{21}[/tex]

[tex]y=\sqrt{21}[/tex]

Step-by-step explanation:

Let x  and y are the two positive real numbers

product of two numbers is 21

[tex]x y=21[/tex]

[tex]y=\frac{21}{x}[/tex]

The sum of the two numbers is f(x) =x+y

Replace y with 21/x

[tex]f(x) =x+\frac{21}{x}[/tex]

we need to find smallest possible sum , so we take derivative using power rule

[tex]f'(x)= 1-\frac{21}{x^2}[/tex]

when sum is minimum then the derivative is equal to 0

[tex]0= 1-\frac{21}{x^2}[/tex]

[tex]0=\frac{x^2-21}{x^2}[/tex]

multiply both sides by x^2

[tex]x^2-21=0[/tex]

[tex]x^2=21[/tex]

Take square root on both sides

[tex]x=\sqrt{21}[/tex]

[tex]y=\frac{21}{x}[/tex]

[tex]y=\frac{21}{\sqrt{21}}[/tex]

[tex]y=\sqrt{21}[/tex]

The smallest possible sum of two positive real numbers whose product equals 21 is attained when both numbers are equal to the square root of 21.

To solve this problem, we can use algebra and calculus. To minimize the sum of the two numbers, call them x and y, subject to the constraint that their product equals 21 (x*y=21), we can use the formula for a minimum of a function. We first change the problem into the equation x + y = sum and then transform it into a function with a single variable: f(y) = y + 21/y. Solving the derivative of y with respect to this function and setting the result equal to zero, we find that the minimum sum is attained when y = sqrt(21), and consequently, x = sqrt(21) as well. Therefore, the two positive real numbers whose product is 21 have the smallest possible sum are both sqrt(21).

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Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

[tex]\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}[/tex]

Solve for S(t):

[tex]\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}[/tex]

[tex]e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}[/tex]

The left side is the derivative of a product:

[tex]\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}[/tex]

Integrate both sides:

[tex]e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt[/tex]

[tex]e^{t/800}S(t)=64e^{t/800}+C[/tex]

[tex]S(t)=64+Ce^{-t/800}[/tex]

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

[tex]0=64+C\impleis C=-64[/tex]

and so (b) the amount of sugar in the tank at time t is

[tex]S(t)=64\left(1-e^{-t/800}\right)[/tex]

As [tex]t\to\infty[/tex], the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

The tank initially contains 64 kg of sugar. The amount of sugar after t minutes is given by the equation: Amount of sugar = Initial amount of sugar + (Rate of sugar entering - Rate of solution leaving) × t. As t becomes large, the value of y(t) approaches the concentration of sugar in the solution entering the tank (0.04 kg/L).

(a) How much sugar is in the tank at the beginning?

To find the amount of sugar in the tank at the beginning, we need to calculate the total mass of sugar in the tank.

Mass of sugar = Volume of solution × Concentration of sugar = 1600 L × 0.04 kg/L = 64 kg

Therefore, there is 64 kg of sugar in the tank at the beginning.

(b) Find the amount of sugar after t minutes.

To find the amount of sugar after t minutes, we need to know the rate of sugar entering the tank and the rate of solution leaving the tank.

The rate of sugar entering the tank is given as 0.04 kg/L.

The rate of solution entering and leaving the tank is given as 2 L/min.

Therefore, the amount of sugar after t minutes is given by the equation: Amount of sugar = Initial amount of sugar + (Rate of sugar entering - Rate of solution leaving) × t = 64 kg + (0.04 kg/L - 2 L/min) × t

(c) As t becomes large, what value is y(t) approaching?

As t becomes large, the value of y(t) is approaching a constant value, which is the concentration of sugar in the solution entering the tank.

In this case, the concentration of sugar in the solution entering the tank is 0.04 kg/L.

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Answer:

B. Yes, it is equivalent

Step-by-step explanation:

A = (-3, 2, 3)

B = (-3, 5, 2)

/AB/ = (-3-(-3), 2-5, 3-2)

= (0, -3, 1)

P = (2, -3, 2)

Q = (2, 0, 1)

/PQ/ = (2-2, -3-0, 2-1)

= (0, -3, 1)

So, /AB/ is equivalent to /PQ/

Final answer:

The convection heat transfer coefficient of air at the upper surface of the plate is found to be 133.33 W/m·K by using the Fourier's law of heat conduction in conjunction with Newton's law of cooling and rearranging the expression to solve for the convection heat transfer coefficient.

Explanation:

To determine the convection heat transfer coefficient of air at the upper surface of the plate, we apply Fourier's law of heat conduction for a steady-state situation and Newton's law of cooling:

Fourier's law gives us:

Q = -kA(dT/dx)

Where:

Q is the rate of heat transfer (Watts),

k is the thermal conductivity of the material (W/m·K),

A is the surface area (m²),

(dT/dx) is the temperature gradient through the material (K/m).

We then use Newton's law of cooling:

Q = hA(T_surface - T_fluid)

Where:

h is the convection heat transfer coefficient (W/m·K),

T_surface is the surface temperature (K),

T_fluid is the fluid temperature (K).

We know the temperature difference across the plate (dT) is 10°C (60°C - 50°C) and assuming the heat transfer rate (Q) is the same throughout the plate due to steady-state conditions, we can rearrange and solve for h:

h = (k/d) (dT/(T_surface - T_fluid))

Given the values:

k = 80 W/m·K,

d = 15 cm = 0.15 m,

50°C (T_surface) - 10°C (T_fluid) = 40 K.

Substituting these into the equation gives us:

h = (80/0.15) (10/40) = 133.33 W/m·K

The convection heat transfer coefficient of air at the upper surface is [tex]\( 133.33 \) W/m^\²\cdotK.[/tex]

[tex]\[ q = h_1 (T_{s1} - T_{\infty}) = \frac{k}{L} (T_{s1} - T_{s2}) = h_2 (T_{s2} - T_{\infty}) \][/tex]

where:

- [tex]\( q \)[/tex] is the heat transfer rate per unit area[tex](W/m^\²)[/tex],

- [tex]\( h_1 \)[/tex] and [tex]\( h_2 \)[/tex] are the convection heat transfer coefficients at the upper and lower surfaces, respectively [tex](W/m^\²K)[/tex],

- [tex]\( T_{s1} \)[/tex] and [tex]\( T_{s2} \)[/tex] are the temperatures at the upper and lower surfaces of the plate, respectively (°C),

-[tex]\( T_{\infty} \)[/tex] is the ambient air temperature (°C),

- [tex]\( k \)[/tex] is the thermal conductivity of the plate [tex](W/m\cdot K)[/tex],

- [tex]\( L \)[/tex] is the thickness of the plate (m).

Given:

- [tex]\( T_{s1} = 50 \)\°C[/tex] (temperature at the upper surface),

- [tex]\( T_{s2} = 60 \)\°C[/tex] (temperature at the lower surface),

- [tex]\( T_{\infty} = 10 \)\°C[/tex] (air temperature),

- [tex]\( k = 80 \) W/m\cdotK[/tex] (thermal conductivity of the plate),

- [tex]\( L = 0.15 \)[/tex] m (thickness of the plate).

Since we want to find [tex]\( h_1 \)[/tex], we can equate the expressions for [tex]\( q \)[/tex] that involve [tex]\( h_1 \)[/tex] and [tex]\( k \)[/tex]:

[tex]\[ h_1 (T_{s1} - T_{\infty}) = \frac{k}{L} (T_{s1} - T_{s2}) \][/tex]

Now, solve for [tex]\( h_1 \)[/tex]:

[tex]\[ h_1 = \frac{k}{L} \cdot \frac{(T_{s1} - T_{s2})}{(T_{s1} - T_{\infty})} \][/tex]

Substitute the given values:

[tex]\[ h_1 = \frac{80 \text{ W/m\cdotK}}{0.15 \text{ m}} \cdot \frac{(50 \text{ \°C} - 60 \text{ \°C})}{(50 \text{ \°C} - 10 \text{ \°C})} \][/tex]

[tex]\[ h_1 = \frac{80}{0.15} \cdot \frac{-10}{40} \][/tex]

[tex]\[ h_1 = 533.33 \cdot \frac{-1}{4} \][/tex]

[tex]\[ h_1 = -133.33 \text{ W/m}^2\text{\cdotK} \][/tex]

The negative sign indicates that the heat transfer is from the plate to the air, which is expected. However, the convection heat transfer coefficient is typically reported as a positive value, so we take the absolute value:

[tex]h_1 = -133.33 {(\frac{W}{m}})^2} { \cdot K}[/tex]

Answer:

Definitely

Step-by-step explanation:

In order to completely surround a circle, you need six circles to do that while here in the question it is mentioned that currently there are only five circles surrounding the circle, hence there is enough room to easily fit one more circle.

I just hope that you are satisfied with the answer, Best of Luck.

Each star has a 20% difference.

A four star rating would be above the bottom 60% ( 1, 2 and 3 stars) but be below the top 20%. (5 stars).

A four-star mutual fund is considered to be a good choice within its investment class, indicating that it has performed well relative to its peers but is not quite in the top 20% of performers like a five-star fund.

A four-star mutual fund is typically considered to be above average in its investment class. Here's the interpretation:

One-star mutual fund: This fund is in the bottom 20% of its investment class, which means it has performed poorly compared to most other funds in the same category.

Two-star mutual fund: This fund is also below average but may have performed slightly better than one-star funds.

Three-star mutual fund: A three-star fund is considered to be a neutral or average performer within its investment class. It neither significantly outperforms nor underperforms its peers.

Four-star mutual fund: A four-star fund is above average within its investment class. It has likely delivered solid returns and may have consistently outperformed the majority of other funds in its category.

Five-star mutual fund: This is the top 20% of funds in its investment class, indicating that it is among the best-performing funds in its category. A five-star fund is often associated with excellent performance and consistent returns.

for such more question on average

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Answer:

a) There is a 24.14% probability that both marbles are red.

b) There is a 5.56% probability of first selecting a blue marble, then a green marble.

Step-by-step explanation:

a) Two marbles are selected with replacement. Find the probability that both marbles are red.

Initially, there are 30 marbles, of which 15 are red. So there is a 15/30 probability that the first marble selected is red.

After a red marble is selected, there are 29 marbles, of which 14 are red. So there is a 14/29 probability that the second marble selected is red.

The probability that both marbles are red is:

[tex]P = \frac{15}{30}*\frac{14}{29} = 0.2414[/tex]

There is a 24.14% probability that both marbles are red.

b) Two marbles are selected without replacement. Find the probability of first selecting a blue marble, then a green marble

There are 30 marbles, of which 10 are blue and 5 are green.

So, there is a 10/30 probability of selecting a blue marble and a 5/30 probability of selecting a red marble.

The probability of selecting a blue marble and then a green marble is:

[tex]P = \frac{10}{30}*\frac{5}{30} = 0.0556[/tex]

There is a 5.56% probability of first selecting a blue marble, then a green marble.