The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of a bridge is 1270 ft long and 157 ft high. The parabola y=0.00039x^2 gives a good fit to the shape of the cables, where IxI less than of equal to 635, and x and y are measured in feet. Approximate the length of the cables that stretch between the tops of the two towers.

Answer:

Answer = I. Chi-squared test of independence

Step-by-step explanation:

The test of independence uses the contingency table format. The analysis of a two-way contingency table helps to answer the question whether two variables are related or independent of each other. Thus, the chi-square statistic measures how much the observed frequencies differ from the expected frequencies when variables are independent. The first procedure is to state the null and alternative hypothesis  

H0: No relationship exists between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

H1: There is a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

To be able to decide whether to reject the null hypothesis or not, we need to compare the calculated test statistic with its chi-square table or critical value using a given level of significance and degree of freedom and then decide using the decision rule (Reject H0 if the calculated chi-square statistic is greater than the critical value otherwise, accept H0)

Final answer:

The null and alternative hypotheses for testing the mean travel time on scenic and nonscenic routes.

Explanation:

a. In this case, the hypotheses to be tested are:

Null hypothesis (H0): The mean travel time for the nonscenic route is not shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 ≤ 10.

Alternative hypothesis (Ha): The mean travel time for the nonscenic route is shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 > 10.

b. If u1 is the mean for the nonscenic route and u2 for the scenic route, the hypotheses to be tested would be:

Null hypothesis (H0): The mean travel time for the scenic route is not shorter than 10 minutes compared to the nonscenic route, or in other words, u2 - u1 ≤ 10.

Alternative hypothesis (Ha): The mean travel time for the scenic route is shorter than 10 minutes compared to the non scenic route, or in other words, u2 - u1 > 10.

The correct options are : - (a) The correct hypothesis is 2. [tex]\(\mu_1 - \mu_2 < -10\)[/tex] and (b) The correct hypothesis is 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]

Part (a): [tex]\(\mu_1\)[/tex] is the mean for the scenic route and [tex]\(\mu_2\)[/tex] is the mean for the non-scenic route

The individual decides that the non-scenic route is justified only if it reduces the mean travel time by more than [tex]10[/tex] minutes. In this context, we need to test whether the mean travel time for the scenic route [tex]\(\mu_1\)[/tex] minus the mean travel time for the non-scenic route [tex](\(\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]

Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = -10\)[/tex]

Alternative Hypothesis[tex](\(H_a\)) : \(\mu_1 - \mu_2 < -10\)[/tex]

So, the correct option for (a) is: 2.[tex]\(\mu_1 - \mu_2 < -10\)[/tex]

Part (b): [tex]\(\mu_1\)[/tex] is the mean for the non-scenic route and [tex]\(\mu_2\)[/tex] is the mean for the scenic route

In this case, we need to test whether the mean travel time for the non-scenic route [tex](\(\mu_1\))[/tex] minus the mean travel time for the scenic route [tex](\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]

Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = 10\)[/tex]

Alternative Hypothesis [tex](\(H_a\)) : \(\mu_1 - \mu_2 < 10\)[/tex]

So, the correct option for (b) is: 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]

The complete Question is

An individual can take either a scenic route to work or a non-scenic route. She decides that use of the non-scenic route can be justified only if it reduces the mean travel time by more than 10 minutes.

(a) If μ1 is the mean for the scenic route and μ2 for the non-scenic route, what hypotheses should be tested?

1. μ1 − μ2 = −10

2. μ1 − μ2 < −10μ1 − μ2 = −10

3. μ1 − μ2 > −10 μ1 − μ2 = −10

4. μ1 − μ2 ≠ −10μ1 − μ2 = 10

5. μ1 − μ2 < 10μ1 − μ2 = 10

6. μ1 − μ2 > 10

(b) If μ1 is the mean for the non-scenic route and μ2 for the scenic route, what hypotheses should be tested?

1. μ1 − μ2 = −10

2. μ1 − μ2 < −10μ1 − μ2 = −10

3. μ1 − μ2 > −10 μ1 − μ2 = −10

4. μ1 − μ2 ≠ −10μ1 − μ2 = 10

5. μ1 − μ2 < 10μ1 − μ2 = 10

6. μ1 − μ2 > 10

To use z-procedures for a confidence interval for a population proportion, three conditions must be met. The sample size must be large (at least 30 or 50), the population must be at least 10 or 20 times as large as the sample, and both npˆ and n(1 - pˆ) must be at least 10, ensuring the sampling distribution approximates to normal.

The process of constructing a confidence interval for a population proportion through z-procedures demands certain conditions to be met. These conditions directly shape the accuracy of the findings. There are three key requirements:

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The correct condition for using z-procedures to construct a confidence interval for a population proportion is:

C. Both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] must be at least 10 (some texts say greater than 5).

To use z-procedures to construct a confidence interval for a population proportion, we typically rely on the normal approximation to the binomial distribution. For this approximation to be valid, both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] should be sufficiently large.

- [tex]\( n \)[/tex] represents the sample size.

- [tex]\( \hat{p} \)[/tex] represents the sample proportion.

Both conditions ensure that the sampling distribution of the sample proportion is approximately normal.

1. [tex]\( np \hat{p} \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected successes in the sample.

2.[tex]\( n(1 - \hat{p}) \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected failures in the sample.

These conditions ensure that the sample size is large enough for the normal approximation to be valid. They are more precise than A and B, which are common rules of thumb but not absolute requirements for using z-procedures.

Answer:

the position equation of the projectile are

[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]

[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]

in x and y direction

Step-by-step explanation:

let the mass of the projectile be m, initial velocity be u

the wind applies a force of 3 newton in east direction.

therefore acceleration due to the force in east direction =[tex]\frac{force}{mass}[/tex]

= [tex]\frac{3}{m} =\frac{3}{1}[/tex]

acceleration due to gravity is in south direction = g

let east be x direction and north be y direction.

therefore acceleration in x direction = 3[tex]\frac{m}{s^{2} }[/tex] and in y direction = -g[tex]\frac{m}{s^{2} }[/tex]

writing equation of motion in x and y direction:

[tex]x = u_{x} t + \frac{1}{2} at^{2}[/tex]

[tex]y = u_{y} t + \frac{1}{2} at^{2}[/tex]

[tex]u_{x}[/tex]= ucos45 = [tex]\frac{300}{\sqrt{2} }[/tex]

[tex]u_{y}[/tex]= usin45= [tex]\frac{300}{\sqrt{2} }[/tex]

therefore

[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]

[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]

here 2 is added as the projectile already 2 meter above the ground

Answer:

You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.

Step-by-step explanation:

You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.

Here Andy is basing himself on one anecdotal evidence he heard. So a sample size of 1. If he were to base his opinion from a large sample size, it is highly likely that 99% of the people were saved by the seatbelt and 1% got hurt. Since Andy only heard the anecdotal evidence, he may be basing himself on a very unlikely event.

Andy's refusal to wear a seat belt based on a single incident is an example of reliance on anecdotal evidence, a logical fallacy. This means he is making a general rule based on a singular event, which is misleading as it does not take into account the reality that seat belts vastly improve safety in car crashes. It's important to base decisions on sound evidence rather than isolated incidents.

Andy is relying on anecdotal evidence, which is a logical fallacy. A logical fallacy is a flaw in reasoning that makes an argument invalid. In this case, Andy has heard a story about a particular event, and is applying it as a universal rule.

However, just because a seatbelt caused harm in a single incident does not mean that seat belts are generally harmful or risky. According to various studies, your chance of being killed or seriously injured in a car crash is much higher if you are not wearing a seatbelt.

Therefore, it is illogical to avoid wearing seat belts based on a single unfortunate incident. It's important to remember that anecdotal evidence can be misleading, as it's prone to bias and doesn't take into account the full range of possibilities and outcomes.

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a) There is a mass of 25 kilograms of salt in the tank after 3.5 hours.

b) The concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute.

Salt concentration (c(t)), in kilograms per liter, is usually modelled by first order non-homogeneous linear differential equation, whose form is derived from principle of mass conservation and described below:

[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V} \cdot c(t) = \frac{\dot V\cdot c_{o}}{V}[/tex]   (1)

Where:

The solution of this differential equation is:

[tex]c(t) = \left(c_{i}-c_{o}\right)\cdot e^{-\frac{\dot V}{V}\cdot t }+c_{o}[/tex]   (2)

Where:

[tex]c_{i} = \frac{m_{s}}{V}[/tex]   (3)

Where [tex]m_{s}[/tex] is the initial salt mass in the tank, in kilograms and t is the time, in minutes.

a) The amount of salt is found by multiplying the volume occupied by the water and the salt concentration in the tank. If we know that [tex]m_{s} = 50\,kg[/tex], [tex]V = 2000\,L[/tex], [tex]\dot V = 7\,\frac{L}{min}[/tex], [tex]c_{o} = 0.0125\,\frac{kg}{L}[/tex] and [tex]t = 3.5\,h[/tex], then the amount of salt in the tank:

[tex]c_{i} = \frac{50\,kg}{2000\,L}[/tex]

[tex]c_{i} = 0.025\,\frac{kg}{L}[/tex]

[tex]c(12600) = (0.025-0.0125)\cdot e^{-\frac{7}{2000}\cdot (12600)}+0.0125[/tex]

[tex]c(12600) = 0.0125[/tex]

And the salt mass in the tank is:

[tex]m = \left(0.0125\,\frac{kg}{L} \right)\cdot \left(2000\,L\right)[/tex]

[tex]m = 25\,kg[/tex]

There is a mass of 25 kilograms of salt in the tank after 3.5 hours. [tex]\blacksquare[/tex]

b) For [tex]t \to +\infty[/tex], [tex]c(t) \to c_{o}[/tex]. Therefore, the concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute. [tex]\blacksquare[/tex]

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a) The amount of salt in the tank after 3.5hr is 25Kg

b) The concentration of the salt as time approaches infinity is 0.125Kg/min.

Let y represents the total amount of salt at time(t)

dy/dt = rate of solution in - rate of solution out of the tank

But, concentration = amount of salt at time t/ volume of water at time t

= y/ 2000

dy/dt = 0.0875 - 7y/2000dy/dt

dy/dt = (175 - 7y/2000)dy/dt

separating variables

(1/7y-175)dy = - 1/200dtIn7y - 175

Integration both sides

y = 175 + e^(–1/2000t) + k/7

when the beginning y = 50, t = 0, k = 175

substituting the values

At t = 3.5 hrs

y = 25kg

When t approaches to infinity

Concentration is 0.125kg/min

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Answer:I am not very sure but I think [tex]\frac{1}{51.5}[/tex]

Step-by-step explanation:

The probability of picking the fake coin given it lands heads [tex]\(n\)[/tex] times is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]

To find the probability that you have picked the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row, we can use Bayes' theorem.

Let [tex]\(F\)[/tex] be the event of picking the fake coin, and [tex]\(H\)[/tex] be the event of getting heads [tex]\(n\)[/tex] times in a row.

We are asked to find [tex]\(P(F|H)\)[/tex], the probability of picking the fake coin given that [tex]\(n\)[/tex] heads are observed in a row.

By Bayes' theorem:

[tex]\[ P(F|H) = \frac{P(H|F) \times P(F)}{P(H)} \][/tex]

We know:

[tex]- \(P(H|F) = 1\)[/tex] (since the fake coin has two heads).

[tex]- \(P(F) = \frac{1}{100}\)[/tex] (since there is only one fake coin out of 100).

[tex]- \(P(H)\)[/tex] is the probability of getting [tex]\(n\)[/tex] heads in a row, which is the same for both the fake and normal coins. This is [tex]\(0.5^n\).[/tex]

Now, substituting the values:

[tex]\[ P(F|H) = \frac{1 \times \frac{1}{100}}{0.5^n} \][/tex]

[tex]\[ P(F|H) = \frac{1}{100 \times 0.5^n} \][/tex]

[tex]\[ P(F|H) = \frac{1}{100 \times 2^n} \][/tex]

So, the probability of picking the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]

Answer:

B. 1635

Step-by-step explanation:

We have been given that the seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row.

We can see that the seating order is in form of arithmetic sequence, whose first term is 11 and common difference is 3.

Since there are 30 rows altogether, so we need to find the sum of 30 first terms of the sequence using sum formula.

[tex]S_n=\frac{n}{2}[2a+(n-1)d][/tex], where,

[tex]S_n=[/tex] Sum of n terms,

n = Number of terms,

a = First term,

d = Common difference.

Upon substituting our given value is above formula, we will get:

[tex]S_n=\frac{30}{2}[2(11)+(30-1)3][/tex]

[tex]S_n=15[22+(29)3][/tex]

[tex]S_n=15[22+87][/tex]

[tex]S_n=15[109][/tex]

[tex]S_n=1635[/tex]

Therefore, there are 1635 seats in all and option B is the correct choice.

Answer: the total number of seats is 1635

Step-by-step explanation:

For each additional row after the first row, there are 3 more seats than there are in the previous row. This means that the seats in each row is increasing in arithmetic progression with a common difference of 3. The formula for determining the sum of n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the sequence.

a represents the first term,

d represents the common difference.

From the information given,

a = 11

n = 30

d = 3

Therefore,

S30 = 30/2[2×11 + (30 - 1)3]

S30 = 15[22 + 29×3]

S30 = 15 × 109= 1635

Answer:

Average value of temperature will be [tex]8.335^{\circ}C[/tex]

Step-by-step explanation:

We have given the temperature in degree Celsius [tex]T(x,y)=19-4x^2-y^2[/tex]

It is given that x varies between 0 to 2

So [tex]0\leq x\leq 2[/tex]

And y varies between 0 and 4

So [tex]0\leq y\leq 4[/tex]

So area between the region A = 4×2 = 8[tex]cm^2[/tex]

Now average temperature is given by

[tex]Average\ value=\frac{1}{A}\int \int T(x,y)dA[/tex]

[tex]Average\ value=\frac{1}{8}\int \int (19-4x^2-y^2)dxdy[/tex]

[tex]=\frac{1}{8}\int \int (19x-4\frac{x^3}{3}-y^2x)dy[/tex]

As limit of x is 0 to 2

[tex]=\frac{1}{8}\int  (19\times 2-4\times \frac{2^3}{3}-y^2\times 2)-(19\times 0-4\times \frac{0^3}{3}-y^2\times 0))dy=\frac{1}{8}\int(38-\frac{32}{3}-2y^2)dy[/tex]

=[tex]=\frac{1}{8}(38y-\frac{32}{3}y-\frac{2y^3}{3})[/tex]

As limit of y is 0 to 4

So [tex]Average\ value=\frac{1}{8}(38\times 4-\frac{16}{3}\times 4-\frac{2\times 4^3}{3})-0=8.335^{\circ}C[/tex]

Answer:

Therefore, Michael concludes option C)

C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]

Step-by-step explanation:

Given:

1. DG = 3 and the area of square DEFG is 9.

2. AG = 4 and the area of square GHIA is 16.

3. DA = 5 and the area of square ABCD is 25.

So we have,

[tex](DG)^{2}=3^{2}=9\\ \\(AG)^{2}=4^{2}=16\\\\(DA)^{2}=5^{2}=25\\[/tex]

Now Add DG² and AG² we get

[tex](DG)^{2}+(AG)^{2}=9+16=25=(DA)^{2}[/tex]

Which is also called as  Pythagoras theorem i.e

[tex](\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}[/tex]

Therefore, Michael concludes option C)

C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]

Answer:

Attached as image png.

Step-by-step explanation:

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. An article gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. 62 51 53 58 42 54 56 61 59 64 51 53 64 62 50 68 54 55 57 50 55 50 56 55 46 56 54 55 53 47 48 56 58 48 63 58 57 56 54 59 54 52 50 55 60 51 56 59 Construct a boxplot of the data.

The boxplot is a method to represent a series of numerical data through their quartiles. In this way, the box diagram shows at a glance the median and quartiles of the data.

Answer:

The 95% confidence interval with the normla method would be given by (0.657;0.979)

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)

Step-by-step explanation:

1) Notation and definitions

[tex]X=18[/tex] number of blocks that were sufficiently strong

[tex]n=22[/tex] random sample taken

[tex]\hat p=\frac{18}{22}=0.818[/tex] estimated proportion of blocks that were sufficiently strong

[tex]p[/tex] true population proportion of blocks that were sufficiently strong

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

2) Confidence interval (Normal method)

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.657[/tex]

[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.979[/tex]

The 95% confidence interval with the normla method would be given by (0.657;0.979)

3) Confidence interval (Small sample method)

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n} \frac{N-n}{N-1}}[/tex]

But we need to know the total size for the population, and on this case we don't know but let's assumed that N=2000 for example .

If we replace the values obtained we got:

[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.784[/tex]

[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.852[/tex]

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)

Answer:

a) [tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]

b) [tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

[tex]Rate change = In-Out[/tex]

And we can name the rate of change as [tex]\frac{ds}{dt}=rate change[/tex]

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is [tex]\frac{s}{50gal}[/tex]

And now we can find the expression for the amount of salt after time t like this:

[tex]\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}[/tex]

And we have this differential equation:

[tex]\frac{dS}{dt} +\frac{1}{10} s = 40[/tex]

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

[tex]\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}[/tex]

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

[tex]e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt[/tex]

[tex]e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C[/tex]

Now we can divide both sides by [tex] e^{\frac{1}{25} t} [/tex] and we got:

[tex]s(t) =400 + C e^{-\frac{1}{10} t}[/tex]

Now we can apply the initial condition in order to solve for the constant C like this:

[tex]15 = 400+C[/tex]

[tex]C=-385[/tex]

And then our function would be given by:

[tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

[tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]

a) The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex].

b) There are 4400 ounces of salt in the tank after 25 minutes.

In this question we must model the salt concentration in the tank ([tex]c(t)[/tex]), in ounces per gallon, as a function of time ([tex]t[/tex]), in minutes, in which salt is dissolved into the tank due to a constant inflow rate ([tex]\dot V[/tex]), in gallons. Likewise, an equal outflow rate exists with resulting concentration.

a) The process is modelled mathematically by a non-homogeneous first order differential equation and physically by principle of mass conservation, whose description is shown below:

[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V_{T}}\cdot c(t) = \frac{\dot V}{V_{T}}\cdot c_{in}[/tex]   (1)

Where:

The solution of this differential equation is:

[tex]c(t) = c_{in} + \left(\frac{m_{T}}{V_{T}}-c_{in} \right)\cdot e^{-\frac{\dot V}{V_{T}}\cdot t }[/tex]   (2)

Where [tex]m_{T}[/tex] is the initial salt mass of the tank, in ounces.

And the salt mass in the tank at an arbitrary time [tex]t[/tex] ([tex]m(t)[/tex]), in ounces, is obtained by multiplying (2) by the volume of the tank. That is to say:

[tex]m(t) = c(t)\cdot V_{T}[/tex]   (3)

By replacing [tex]c(t)[/tex] in (3) by (2), we have the following expression:

[tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]   (4)

The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]. [tex]\blacksquare[/tex]

b) If we know that [tex]V_{T} = 50\,gal[/tex], [tex]\dot V = 55\,\frac{gal}{min}[/tex], [tex]c_{in} = 88\,\frac{oz}{gal}[/tex], [tex]m_{T} = 15\,oz[/tex] and [tex]t = 25\,min[/tex], then the quantity of salt is:

[tex]m(25) = (50)\cdot (88)+[15-(50)\cdot (88)]\cdot e^{-\left(\frac{55}{50} \right)\cdot (25)}[/tex]

[tex]m(25) = 4400\,oz[/tex]

There are 4400 ounces of salt in the tank after 25 minutes. [tex]\blacksquare[/tex]

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Answer:

The fourth one

Step-by-step explanation:

Because Convex polygons can only be shapes like hexagons and squares and stuff like that. They cant be curved or zig zagged.

Answer:

The fourth one

Step-by-step explanation:

Convex polygons can be hexagons and stuff like that.

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

[tex]Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}[/tex] [tex]= 60.67[/tex]

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: [tex]s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}[/tex]

Put value in formula of Standard Deviation,

[tex]s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}[/tex] = 40.75

Step-3: Then, we have to find the critical value by chi-square.

[tex]X_{1-\alpha/2}^{2}=3.82[/tex]

[tex]X_{1-\alpha/2}^{2}=21.92[/tex]

Then, find the confidence interval which is 95%.

[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9[/tex]

[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9[/tex]

The 95% confidence interval for the population standard deviation, based on the given sample data, is estimated to be approximately between 2.9 mi/h and 6.9 mi/h. Here option B is correct.

To construct a 95% confidence interval estimate of the population standard deviation, you can use the chi-square distribution. The formula for the confidence interval is given by:

[tex]\[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \][/tex]

where:

- n is the sample size,

- s is the sample standard deviation,

- [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] are the critical values from the chi-square distribution for the lower and upper bounds of the confidence interval, respectively, and

- α is the significance level (0.05 for a 95% confidence interval).

Given that the sample data is: 62 61 61 57 61 54 59 58 59 69 60 67, the sample size n is 12, and the sample standard deviation s is approximately 4.75.

Now, you need to find the critical values [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex]. For a 95% confidence interval and df = n-1 = 11, you can consult a chi-square distribution table or use a statistical software/tool to find these critical values.

Assuming [tex]\( \chi^2_{\alpha/2} \)[/tex] is 19.675 and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] is 2.201, plug these values into the formula:

[tex]\[ \left( \sqrt{\frac{(12-1) \times 4.75^2}{19.675}}, \sqrt{\frac{(12-1) \times 4.75^2}{2.201}} \right) \][/tex]

Calculating this expression gives approximately (2.9, 6.9). Here option B is correct.

To learn more about standard deviation

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Answer:

1. Plane False

2. Sphere False

3. Ellipsoid  False

4. Circular cylinder  True

Step-by-step explanation:

For this case we have the following curve [tex]C(t) = (cos t , sin t , t[/tex]

And we can express like this the terms for the curve or each component:

[tex] x= cos t, y= sin t , z =t[/tex]

1. Plane False

The general equation for a plane is given by:

a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.

For this case we don't satisfy this since have sinusoidal functions and this equation is never satisfied.

2. Sphere False

The general equation for a sphere is given by:

(x - a)² + (y - b)² + (z - c)² = r²

And for this case if we see our parametric equation again that is not satisfied since we have two cosenoidal functions. And another function z=t

3. Ellipsoid  False

The general equation for an ellipsoid is given by:

x^2/a2 + y^2/b2 + z^2/c2 = 1

And for this case again that's not satisfied since we have

[tex]\frac{cos^2 t}{a^2} + \frac{sin^2 t}{b^2}+\frac{t^2}{c^2} \neq 1[/tex]

4. Circular cylinder  True

The general equation for a circular cylinder is given by:

[tex]x^2 +y^2 = r^2[/tex]

And if we replace the equations that we have we got:

[tex] cos^2 t + sin^2 t = 1[/tex] from the fundamental trigonometry property.

So then we see that our function satisfy the condition and is the most appropiate option.

Answer:

Half half for both since all three are fair coins.

Step-by-step explanation:

Final answer:

The cumulative probability of getting a certain number of heads in 4 coin flips can be calculated with the pbinom() function in R, which uses the binomial distribution, suitable for independent Bernoulli trials with two outcomes and a constant success probability.

Explanation:

Binomial Probability Distribution in R

To calculate the cumulative probability of getting a certain number of heads in multiple coin flips using binomial distribution, the pbinom() function in R can be used. We are looking at a scenario where a coin is flipped 4 times, and we want to find the cumulative probability of getting 0, 1, 2, 3, or 4 heads.

The binomial distribution is appropriate here because coin flipping is a Bernoulli trial: there are two possible outcomes (heads or tails), the probability of success (head in this case) is constant, and each flip is independent of others.

The cumulative probability can be calculated as follows:

Probability of no more than 2 heads: pbinom(2, 4, 0.5)

Probability of no more than 4 heads (which is certain): pbinom(4, 4, 0.5) equals to 1

You will get results that match options A through E provided in the question.

Answer:

[tex]5.52*10^{26}[/tex]

Step-by-step explanation:

For the columns with 5 slots, there are 15 distinct options to fill in. The number of ways to fill them in is

15 * 14 * 13 * 12 * 11 = 360360 ways (order matters)

For the column with 4 slots and 15 options. The number of ways to fill them in is

15 * 14 * 13 * 12 = 32760  ways (order matters)

Since a bingo card has 4 columns of 5 slots and 1 column with 4 slots, the total number of combination there is:

[tex]360360^4*32760 \approx 5.52*10^{26}[/tex] (order matters)

Answer:

[tex]t=\pm 2.95[/tex]

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The t distribution or Student’s t-distribution is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

Data given

Confidence =0.99 or 99%

[tex]\alpha=1-0.99=0.01[/tex] represent the significance level

n =16 represent the sample size

We don't know the population deviation [tex]\sigma[/tex]

Solution for the problem

For this case since we don't know the population deviation and our sample size is <30 we can't use the normal distribution. We neeed to use on this case the t distribution, first we need to calculate the degrees of freedom given by:

[tex]df=n-1=16-1=15[/tex]

We know that [tex]\alpha=0.01[/tex] so then [tex]\alpha/2=0.005[/tex] and we can find on the t distribution with 15 degrees of freedom a value that accumulates 0.005 of the area on the left tail. We can use the following excel code to find it:

"=T.INV(0.005;15)" and we got [tex]t_{\alpha/2}=-2.95[/tex] on this case since the distribution is symmetric we know that the other critical value is [tex]t_{\alpha/2}=2.95[/tex]

Answer:5.25

Step-by-step explanation:

Final answer:

To determine the speed limit for a banked road curve, use the formula: speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle)). For a curve with a radius of 280 m and a bank angle of 7.0°, the speed limit should be approximately 23.8 m/s.

Explanation:

When determining the speed limit for a banked road curve, we need to consider the radius of the curve and the bank angle. In this case, the radius of the curve is 280 m and the bank angle is 7.0°.

To calculate the speed limit, we can use the formula:

speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle))

Substituting the given values into the formula, we get:

speed limit = square root of (280 * 9.8 * tan(7.0))

Using a calculator, we find that the speed limit should be approximately 23.8 m/s.

Answer:

[tex]9\pi[/tex]

Step-by-step explanation:

given are two parabolas with vertex as (3,0) and (0.0)

[tex]y^2 =2(x-3)\\y^2 =x[/tex]

These two intersect at x=6

Volume of II  curve rotated about x axis - volume of I curve rotated about x axis = Volume of solid of revolution

For the second curve limits for x are from 0 to 6 and for I curve it is from 3 to 6

V2 =\pi [tex]\int\limits^6_0 {y^2} \, dx \\=\pi\int\limits^6_0 {x} \, dx\\= \pi\frac{x^2}{2} \\=18\pi[/tex]

V1 =[tex]\pi \int\limits^6_3 {y^2} \, dx\\\pi \int\limits^6_3 {2x-6} \, dx\\=\pi(x^2-6x)\\= \pi[(36-9)-6(6-3[)\\= (27-18)\pi\\=9\pi[/tex]

Volume of solid of revolutin = V2-V1 = [tex]9\pi[/tex]

Answer:

y= -3x +10

Step-by-step explanation:

you already have your slop so plug it in and your y intercept is 10

The maximum possible error in the area is approximately 0.18 square inches.

Calculate the nominal area:

Using the given measurements, the nominal area is (1/2) * 3 * 4 * sin(π/4) ≈ 3 square inches.

Find the partial derivatives of the area formula:

A = (1/2)ab sin(C), where a and b are the sides and C is the included angle.

∂A/∂a = (1/2)b sin(C)

∂A/∂b = (1/2)a sin(C)

∂A/∂C = (1/2)ab cos(C)

Estimate the maximum possible errors for each variable:

Δa ≈ 1/18 inches

Δb ≈ 1/18 inches

ΔC ≈ 0.05 radians

Approximate the maximum error using differentials:

ΔA ≈ |∂A/∂a|Δa + |∂A/∂b|Δb + |∂A/∂C|ΔC

ΔA ≈ (1/2)(4)(sin(π/4))(1/18) + (1/2)(3)(sin(π/4))(1/18) + (1/2)(3)(4)(cos(π/4))(0.05)

ΔA ≈ 0.18 square inches

Round the answer:

The maximum possible error in the area is approximately 0.18 square inches.

Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

The equation of the given line is

y=2x+2

Comparing with the slope intercept form, slope = 2

If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (2,-1) is 2

To determine the intercept, we would substitute m = 2, x = 2 and y = -1 into y = mx + c. It becomes

- 1 = 2×2 + c = 4 + c

c = - 1 - 4 = - 5

The equation becomes

y = 2x - 5

Answer:

[tex]10.108 < \mu < 13.892[/tex]    

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

We have the following distribution for the random variable:

[tex]X \sim N(\mu , \sigma=0.45)[/tex]

And by the central theorem we know that the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

2) Confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=12[/tex]

The sample deviation calculated [tex]s=2.268[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=8-1=7[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that [tex]t_{\alpha/2}=2.36[/tex]

Now we have everything in order to replace into formula (1):

[tex]12-2.36\frac{2.268}{\sqrt{8}}=10.108[/tex]    

[tex]12+2.36\frac{2.268}{\sqrt{8}}=13.892[/tex]

So on this case the 95% confidence interval would be given by (10.108;13.892)

[tex]10.108 < \mu < 13.892[/tex]    

The correct conclusion is to not reject the null hypothesis.

If the p-value for a hypothesis test is 0.07 and the chosen level of significance is α = 0.05, then the correct conclusion is to not reject the null hypothesis.

When conducting a hypothesis test, the p-value represents the probability of obtaining the observed data or more extreme values assuming the null hypothesis is true. If the p-value is greater than the significance level (α), which is the threshold for rejecting the null hypothesis, we fail to reject the null hypothesis.

In this case, the p-value of 0.07 is greater than the significance level of 0.05. Therefore, we cannot reject the null hypothesis.

Answer: 325 boxes of candy will be produced on each working day in October

Step-by-step explanation:

The initial number of boxes of candy produced is 100. In each month following this, the shop produced 25 more boxes of candy per working day in addition to the previous month. This means that the number of boxes produced each month is increasing in arithmetic progression. The formula for the nth term of an arithmetic sequence is expressed as

Tn = a + (n-1)d

Where

a is the first term of the sequence

n is the number of terms

d is the common difference.

From the given information,

a = 100

d = 25

n = number of months from January to October = 10

Tn = 100 + (10 - 1)25

Tn = 100 + 9×25

Tn = 100 + 225 = 325

Answer:

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The individual with the higher IQ is the one with the higher z-score. So

Test A has a mean of 100 and a standard deviation of 13. An individual who scores 123 on Test A.

So [tex]\mu = 100, \sigma = 13, X = 123[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{123 - 100}{13}[/tex]

[tex]Z = 1.77[/tex]

Test B has a mean of 100 and a standard deviation of 18. An individual who scores 121 on Test B.

So [tex]\mu = 100, \sigma = 18, X = 121[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{121 - 100}{18}[/tex]

[tex]Z = 1.17[/tex]

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.

Answer:

Answer explained below

Step-by-step explanation:

Let a be the working hours of plant A.

Let b be the working hours of plant H.

Let c be the working hours of plant M.

We have to minimize , Z = 60a + 92b + 140c

subject to constraint , 40a + 20b + 40c >=350

                20a + 40b + 40c >=240

where a>=0 , b>=0 ,c>=0

So ,by solving this , a = 7.67 , b= 2.17 , c = 0

So ,working hours of plant A = 8 hrs

working hours of plant H = 3 hrs

working hours of plant M = 0 hrs

Answer:

A = 8hrs  

H = 3hrs  

M = 0hrs  

Step-by-step explanation:

Let a be the working hours of plant A.  

Let b be the working hours of plant H.  

Let c be the working hours of plant M.  

∴ Hours of plants A,H,M are a,b,c respectively.  

We have to minimize the cost of production, Z = 60a + 92b + 140c  

Regular ice cream: 40a + 20b + 40c = 350  

Deluxe ice cream: 20a + 40b +40c =240  

Where a>=0, b>=0, c>=0  

So, by solving this, a= 7.67, b= 2.77,c= 0  

So working hours of plant A = 8 hrs  

Working hours of plant H = 3 hrs  

Working hours of plant M = 0 hrs