Use the given information to find the length of a circular arc. Round to two decimal places.the arc of a circle of radius 11 inches subtended by the central angle of pie/4
Answer is in inches(in)

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:

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:

c. Yes. The outcomes can be classified into two categories: the trials are fixed, and the events are independent.

Since we can calculate the following probabilities:

p= probability that live entertainment had increased the amount of money they spent

q =probability that live entertainment had not increased the amount of money they spent

n = 85

And independence is satisfied.

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

In order to apply the binomial distirbution we need to satisfy some conditions given here:

1) Independence between the trials of the experiment

2)  The number of observations n is fixed for this case n=85

3) Each observation represents one of two outcomes "success" or "failure" and the probability of success is defined.

So then based on this and on the information given we can conclude that the best option for this case is:

c. Yes. The outcomes can be classified into two categories: the trials are fixed, and the events are independent.

Since we can calculate the following probabilities:

p= probability that live entertainment had increased the amount of money they spent

q =probability that live entertainment had not increased the amount of money they spent

n = 85

And independence is satisfied.

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].

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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:

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:

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/

Answer:

Given that

A = A ∩ S -- (1)

S = B ∪ B -- (2)

To prove:

A = (A ∩ B) ∪ (A ∩ B)

(A ∩ B) ∪ (A ∩ B)

= [(A∪A) ∩ (A∪B)] ∩ [(B∪A) ∩ (B∪B)]

=[A ∩ (A∪B)] ∩ [(A∪B) ∩ S]

=A ∩ (A∪B)

=A

Hence proved.

2) If B ⊂ A then A = B ∪ (A ∩ B)

R.H.S = B ∪ (A ∩ B)

= (B ∪ A) ∩ (B∪B) --(3)

As B is subset of A so

(B ∪ A) = A

From (2)

(B ∪ B) = S

(3) becomes

=A ∩ S

from (1)

A ∩ S = A

Hence proved

Answer: (a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.

(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.

Step-by-step explanation:

(a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.

Explanation: If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals.

(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.

Explanation: The 99% of the confidence intervals includes the population proportion value, it means, the remaining (100% – 99%) 1% of the intervals does not includes the population proportion.

If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals and 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.

Answer:

(a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.

(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.

Step-by-step explanation:

(a). 99 percent of the sample proportions results in a 99% confidence interval that includes the population proportion.

Explanation: If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals.

(b). 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.

Explanation: The 99% of the confidence intervals includes the population proportion value, it means, the remaining (100% - 99%) 1% of the intervals does not includes the population proportion.

If multiple samples were drawn from the same population and a 99% CI calculated for each sample, we would expect the population proportion to be found within 99% of these confidence intervals and 1 percent of the sample proportions results in a 99% confidence interval that does not include the population proportion.

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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:

[tex][p-|p|*10^{-3} \, , \, p+|p|* 10^-3][/tex]

Step-by-step explanation

The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is

r = |p*-p|/|p|

we want r to be at most 10⁻³, thus

|p*-p|/|p| ≤ 10⁻³

|p*-p| ≤ |p|* 10⁻³

therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval  [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]

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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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]

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.

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Answer:

Option A)  As the Exam 1 score increases by 1 point, the student's Exam 3 grade will increase, on average, by 0.4845 points.

Step-by-step explanation:

We are given the following in the equation:

[tex]y(x) = 50.57+0.4845x[/tex]

where, above equation is a  linear regression equation to predict the Exam 3 score of students in STA 2023 based on their Exam 1 score.

Here,

y is the dependent variable that is score of exam 3.

x is the independent variable that is the score of exam 1.

Comparing the given equation to a linear equation, we have,

[tex]y = mx + c[/tex]

Slope, m = 0.4845

Intercept, c = 50.57

We define the slope as rate of change.

If there is a increase in x by 1 unit, then,

[tex]y(x) = 50.57+0.4845x\\y(x+1) = 50.57+0.4845(x+1)\\y(x+1)-y(x) = 50.57+0.4845(x+1)-50.57-0.4845x\\y(x+1)-y(x) = 0.4845(x+1-x)\\y(x+1)-y(x) = 0.4845[/tex]

Thus, we can interpret the slope of the line as

Option A) As the Exam 1 score increases by 1 point, the student's Exam 3 grade will increase, on average, by 0.4845 points.

Answer: A) 0,1,1,2,2,4,5,8,9

Step-by-step explanation:

We know that in a stem leaf plot ,

The stem represents the tens value of the term and leaves represent the ones values of the data.

Given data of the amount of time in minutes students spend studying for a quiz:

10, 11, 11, 12, 12, 14, 15, 18, 19, 20, 22, 24, 39, 40, 41, 44, 46, 50, 52, 52, 53, 55, 70.

Here , the least tens value is 1.                 (10, 11, 11, 12, 12, 14, 15, 18, 19)

So the first stem would have value.

Then the leaf of the first stem if you were splitting the stems contains all the ones-values corresponding to tens value as 1 (10, 11, 11, 12, 12, 14, 15, 18, 19).

= 0 , 1, 1,2, 2, 4 , 5 , 8 , 9

Hence, the correct answer is A) 0,1,1,2,2,4,5,8,9

In a stem-and-leaf plot for the given data, the leaves for the first stem (1, or numbers in the 10s) would be 0,1,1,2,2,4,5,8,9, hence the answer is Option A.

In a stem-and-leaf plot, the data is organized by place value. The stem represents the tens digit, and the leaf represents the ones digit. Considering the given data set which ranges from 10 to 70, the first stem represents '1', indicating a range of 10s.

For the values in the 10s: 10, 11, 11, 12, 12, 14, 15, 18, 19, the corresponding leaf units would be 0, 1, 1, 2, 2, 4, 5, 8, 9. So, the correct answer is Option A: 0,1,1,2,2,4,5,8,9 which are the units of the numbers in the 10s place.

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Answer:

5x-9

Step-by-step explanation:

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:

Alternate Hypothesis determines which type of test is used.                  

Step-by-step explanation:

We have find the hypothesis that helps us o determine the nature of test.

[tex]\text{Two tailed test}\\H_{0}: \mu = \mu_0\\H_A: \mu \neq\mu_0\\\\\text{Left tailed test}\\H_{0}: \mu = \mu_0\\H_A: \mu < \mu_0\\\\\text{Right tailed test}\\H_{0}: \mu = \mu_0\\H_A: \mu > \mu_0[/tex]

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:

1.1,000,000,

2, 1 minute 40 secs

3.10^-6 secs

Step-by-step explanation:

sorting algorithm takes 1 second to sort n =1000 items.

1) How many operations will be performed if the sorting algorithm is O(n2) (approximately)?

2) How long will it take to sort 10,000 items if the sorting algorithm is O(n2)?

3) How much time will one operation take if the sorting algorithm is O(n2)?

algorithm takes time proportional to n^2,

1. then 1,000^2=1,000,000,

2. if it takes 1 secs to generate 1000 items

yhen n^2=1000^2=1000000  and 10,000^2=100,000,000.

Dividing  by  100. Therefore, the sorting algorithm would take

1 minute and 40 seconds to sort 10,000 items.

3. How much time will one operation take if the sorting algorithm is O(n2)?

1/1000^2

10^-6 secs to sort 1 operations

A O(n^2) sorting algorithm will perform about 1,000,000 operations in 1 second for 1000 items. It will take around 100 seconds for 10,000 items. The time taken per operation is roughly 1 microsecond.

This question is about Big O Notation, a concept used in Computer Science for analyzing an algorithm's running time by characterizing the number of operations it will perform as a function of the input size (n).

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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:

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.

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:

1. Five classes in the frequency distribution

2.The mid point of the third class is 24.5

3.The total number of frequency is 100

4.The relative frequency of the third class is (20/100=0.2)

5. We use n-1 because we are dealing with a sample data

6. The mean is 24.5

7. The value of the standard deviation is 11

Step-by-step explanation:

See attached picture for the solutions

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:

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:

a) Discrete Variable

b) Discrete Variable

c) Discrete Variable

d) Continuous Variable

Step-by-step explanation:

We have to identify the given variable as discrete r continuous.

Discrete Variables:

Continuous Variables:

a) The number of countries ever visited

Since number of countries will always be expressed in whole numbers and not decimals. Also, they will always be counted and not measured. Thus, it is a discrete variable.

b) The number of sons

Since number of sons will always be expressed in whole numbers and not decimals. Also, they will always be counted and not measured. Thus, it is a discrete variable.

c) Shoe size

Shoe size are expressed in whole number. The underlying measure is length of feet which is a continuous variable but shoe size are always given in whole number. Thus, they cannot take any value within an interval. Thus, it is a discrete variable.

d) Body temperature

Body temperature can be expressed in decimal. A Body temperature of 42.5 makes sense. Thus, they can take any value within an interval. Also, it is measured not counted. Thus, it is a discrete variable.

Answer:3,2

Step-by-step explanation:32/10

Phil paid $3.20 per pound for the jellybeans.

To find out how much Phil paid per pound of jellybeans, we need to divide the total cost by the total weight of the jellybeans. Phil paid $32 for 10 pounds of jellybeans, so the calculation is as follows:

Cost per pound = Total cost / Total weight

Cost per pound = $32 / 10 pounds

Cost per pound = $3.20

Therefore, Phil paid $3.20 per pound for the jellybeans.