The mean monthly expenditure on gasoline per household in Middletown is determined by selecting a random sample of 100 households. The sample mean is $128, with a sample standard deviation of $38, what is the upper bound of a 90% confidence interval for the mean monthly expenditure on gasoline per household in Middletown? a. $134.66 b. $162.20 c. $129.38 d. $131.42 e. $13287

Answer:

  31.747 GW

Step-by-step explanation:

The total cell capacity (C) manufactured in the time period is the integral of the rate of manufacture. So, we have ...

  [tex]C=\displaystyle\int\limits^3_0 {3.7e^{0.61t}} \, dt=\frac{3.7}{0.61}(e^{0.61\cdot 3}-1)\approx 31.747 \quad\text{GW}[/tex]

About 31.747 GW of generating capacity was manufactured in that time interval.

Answer:

Null hypothesis:[tex]p_{SB} \geq p_{NSB}[/tex]  

Alternative hypothesis:[tex]p_{SB} < \mu_{NSB}[/tex]  

The p value is a very low value and using the significance given [tex]\alpha[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{NSB}=30[/tex] represent the number of people killed not using seat belts

[tex]X_{SB}=20[/tex] represent the number of people killed using seat belts

[tex]n_{NSB}=2902[/tex] sample of people not wearing seat belts

[tex]n_{SB}=7866[/tex] sample of people wearing seat belts

[tex]p_{NSB}=\frac{30}{2902}=0.0103[/tex] represent the proportion of people killed not using seat belts

[tex]p_{SB}=\frac{20}{7866}=0.00254[/tex] represent the proportion of people killed using seat belts

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that seat belts are effective in reducing fatalities (If using seat belts reduce the proportion of deaths we need to see that the proportion of death using seat belts is lower than not using seat belts) , the system of hypothesis would be:  

Null hypothesis:[tex]p_{SB} \geq p_{NSB}[/tex]  

Alternative hypothesis:[tex]p_{SB} < \mu_{NSB}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{SB}-p_{NSB}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{SB}}+\frac{1}{n_{NSB}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{SB}+X_{NSB}}{n_{SB}+n_{NSB}}=\frac{20+30}{7866+2902}=0.00464[/tex]

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.00254-0.0103}{\sqrt{0.00464(1-0.00464)(\frac{1}{7866}+\frac{1}{2902})}}=-5.26[/tex]  

4) Statistical decision

Using the significance level provided [tex]\alpha=0.05[/tex], the next step would be calculate the p value for this test.  

Since is a one side lower test the p value would be:  

[tex]p_v =P(Z<-5.26)=7.2x10^{-8}[/tex]  

So the p value is a very low value and using the significance given [tex]\alpha[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .  

Answer:

0.2773

Step-by-step explanation:

Please see attachment.

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = -  (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] =  20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Answer:

4,200 liters

Step-by-step explanation:

The flow rate is given by:

[tex]r(t) = 200 - 4t[/tex]

Integrating the flow rate expression from t=0 to t=30 minutes, yields the total volume that flows out of the tank during that period:

[tex]\int\limits^{30}_0 {r(t)} = \int\limits^{30}_0 {(200 - 4t} )\, dt \\V=(200t - 2t^2)|_0^{30}\\V= (200*30 -2*30^2)-(200*0 -2*0^2)\\V=4,200\ liters[/tex]

4,200 liters of water flow from the tank during the first 30 minutes.

Answer:

  b. y-y1 = m(x-x1)

Step-by-step explanation:

It's a matter of definition. There are perhaps a dozen useful forms of equations for a line. Each has its own name (and use). Here are some of them.

Adding y1 to the point-slope form puts it in an alternate form that is useful for getting to slope-intercept form faster: y = m(x -x1) +y1. I use this when asked to write the equation of a line with given slope through a point, with the result in slope-intercept form.

The equation of the line, in point-slope form, is given by:

[tex]y - y_1 = m(x - x_1)[/tex]

Option b.

-----------------------------------------

The equation of a line, in point-slope form, is given by:

[tex]y - y_1 = m(x - x_1)[/tex]

In which

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Answer:

80 ft x 40 ft

Step-by-step explanation:

Let 'L' be the length of the longer side and 'W' be the length of the shorter side (or the width).

The equations that compose the linear system are:

[tex]L+2W=160\\L=2W[/tex]

Solving the system:

[tex]L+L=160\\L=80\\W=\frac{L}{2}\\W=40[/tex]

The garden is a rectangle with dimensions 80 ft x 40 ft.

The dimensions of the garden are approximately 53.33 feet in width and 106.66 feet in length according to the given linear equations and conditions.

Given that the fencing of 160 feet encloses a rectangular garden on three sides and one side of the rectangle is parallel with the barn, we can derive the linear equations based on these parameters. Let's denote the width of the rectangle as x and the length of the rectangle as 2x, as stated the length is twice the width.

The perimeter of this garden will equal the amount of fencing, which is 160 feet. Because only three sides of the rectangle are enclosed by the fence, the equation for the perimeter becomes 2x + x = 160. Simplifying this, we get 3x = 160. Solving for x, we divide both sides by 3, hence, x = 53.33 ft approximately. The width of the garden is thus around 53.33 feet and the length (being twice the width) would be 2x = 2*53.33 = 106.66 ft.

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In order to estimate the allowed angle error when measuring the height of a tree, you should first use the given distance and the angle of elevation to calculate the tree's height. Then calculate the potential height range considering the 1.5% error, and figure out the corresponding angles with these heights. The angle error can be estimated by the difference between these two angles.

To solve this problem, one can use the principles of trigonometry. Specifically, the tangent function in trigonometry (tan θ = height / distance) can be used to estimate the height of the tree. Here, the distance from the tree (104 feet) and the angle of elevation (27.5°) are given.

We must first calculate the tree height using tan θ = height / distance. After that, we find that the tree is approximately 58.4 ft tall.

In this case, a 1.5% error in the height measurement would be 0.015 * 58.4 ft = 0.876 feet. Using the equation θ = arctan (height / distance), we can figure out the angle error. Enter the heights 58.4 + 0.876 ft and 58.4 - 0.876 into the equation and use the distance 104104 ft to calculate the corresponding angles. The allowed error for the angle follows as the difference of these two angles.

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To ensure the tree height is within 1.5% accuracy, we calculated that the allowable error in the angle measurement is ±0.22 degrees. This was determined by converting the tolerable height range back to angles using inverse tangent. The primary method used is trigonometry.

To determine the allowed error in the angle measurement for a tree with a height accurate to within 1.5%, we follow these steps:

Given that d = 104 ft and tan(27.5°) ≈ 0.5197, we can calculate: h ≈ 104 × 0.5197 ≈ 54.05 ft

So, 54.05 ft ± 0.015 × 54.05 ft ≈ 54.05 ft ± 0.81 ft

Tree Height thus ranges from approximately 53.24 ft to 54.86 ft.

We convert the bounds of height back to angles using inverse tangent:

The allowed error in the angle measurement is the difference from the original angle:

± |27.5° - 27.28°| ≈ ±0.22° and ± |27.72° - 27.5°| ≈ ±0.22°

Therefore, the allowable error in the angle measurement to ensure a height accuracy within 1.5% is ±0.22 degrees.

Answer: [tex]X\sim N(2689,\560)[/tex]

Step-by-step explanation:

Let X be a random variable that represents the number of calories consumed.

Given : The number of calories consumed by customers at the Chinese buffet is normally distributed with mean 2689 and standard deviation 560.

i.e. Population mean : [tex]\mu=2689[/tex]

Population standard deviation: [tex]\sigma=560[/tex].

We can say X follows normal distribution [tex]\mu=2689[/tex] and  [tex]\sigma=560[/tex]

So , the distribution of X : [tex]X\sim N(\mu,\ \sigma)[/tex]

i.e. [tex]X\sim N(2689,\560)[/tex]

Answer:

36

Step-by-step explanation:

multiply the width & the height

Answer:

  |x| < √7

Step-by-step explanation:

The product of the roots of the given quadratic equation is 2/a, so the other root (the one not given) is 2/(7a). It will be negative, since "a" is negative.

The roots of ax^2 +bx +2 = 0 are the values of x^2 that satisfy ax^4 +bx^2 +2 = 0. That is, roots of the latter equation will be the square root of the roots of the former equation.

We know that two of the zeros of the quartic are ±√7, and the other two are complex, as they are the square roots of a negative number. So, the graph of the quartic opens downward (because a < 0), and has real zeros at x=±√7.

The solution to the inequality must be ...

  -√7 < x < √7

_____

The graph shows an example of the quadratic (green) and quartic (black). The ripple in the quartic changes amplitude with different values of "a", but the locations of the zeros do not change.

Answer:

a) P(X=41)= 0.0086 = 0.86%

b) P(36≤X≤38)== 0.1215 = 12.15%

c) would be unusual that fewer that 20 murders were cleared because P(X≤20) = 0.000081 =0.0081%

Step-by-step explanation:

if the random variable X= number of cleared murders , then X follows a binomial distribution:

P(X=x)= n!/[(n-x)!*x!]*p^x*(1-p)^(n-x)

where P(X=x)= probability of x cleared murders, n= number of murders selected=50 , p= probability for a murder to be cleared (67%)

therefore

a) P(X=41)=50!/[(50-41)!*41!]*0.67^41*0.33^(50-41)= 0.0086

b) P(36≤X≤38)= P(X≤38) - P(X≤36) = 0.9371- 0.8156 = 0.1215

c) P(X≤20) = 0.000081

therefore would be unusual that fewer that 20 murders were cleared

Answer:

a) -0.842

b) $0

c) 11,664 cases

Step-by-step explanation:

a)

The price elasticity of demand E at a given point [tex]\large (p_1,q_1)[/tex] is defined as

[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)[/tex]

and in this case, it would measure the possible response of tissues demand due to small changes in its price when the price is at [tex]\large p_1[/tex]

When the price is set at $32 the demand is

[tex]\large q=(108-32)^2=5,776 [/tex]

cases of tissues, so

[tex]\large (p_1,q_1)=(32,5776)\Rightarrow \frac{p_1}{q_1}=\frac{32}{5776}=0.00554[/tex]

Also, we have

[tex]\large \frac{\text{d}q}{\text{d}p}=-2(108-p)\Rightarrow \frac{\text{d}q}{\text{d}p}(32)=-2(108-32)=-152[/tex]

hence

[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)=0.00554(-152)=-0.842[/tex]

That would mean the demand is going down about 0.842% per 1% increase in price at that price level.

b)

When the price is $108 the demand is 0, so the price should always be less than $108.

On the other hand, the parabola

[tex]\large q=(108-32)^2=5,776 [/tex] is strictly decreasing between 0 and 108, that means the maximum demand would be when the price is 0.

c)

When the price is 0 the demand is  

[tex]\large (108)^2=11,664[/tex] cases

The elasticity for p = 32 is E = -0.84.

The maximum revenue is obtained when p = 32, and the demand for that price is q = 5,184.

It is given by:

[tex]E = \frac{p}{q} *\frac{dq}{dp} (p)[/tex]

if p = $32, then:

q(32) = (108 − 32)^2 = 5,776

and:

dq/dp = -2*(108 - p)

Evaluating that in p = 32 we get:

-2*(108 - 32)  = -152

Then the elasticity is:

E = (32/5,776)*-152 = -0.84.

The revenue is equal to the demand times the price, so we get:

R = p*q = p*(108 - p)^2 = p*(p^2 - 216p + 11,664)

To maximize this we need to find the zeros of the derivation, we have:

R' = 3p^2 - 2*216*p + 11,664

The zeros of that equation are given by:

[tex]p = \frac{432 \pm \sqrt{(-432)^2 - 4*3*11,664} }{2*3} \\\\p = (432 \pm 216)/6[/tex]

Notice that we need to use the smaller of these values, so the demand never becomes zero, then we use:

p = (432 - 216)/6 = 36

(the other root gives p = 108, so that is a minimum).

c) For this price, the number of cases demanded is:

q = (108 - 36)^2 = 5,184

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The true weight, according to a 95% confidence interval, is likely between 3.062 mg and 3.238 mg. According to a 99% confidence interval, it's likely between 3.035 mg and 3.265 mg.

This question is about finding the confidence interval of the true weight based on a series of measurements and given the standard deviation of the instrument's error is 0.1 mg.

The error is normally distributed, thus the average of the measurements is likely to be the true value, and can be calculated by adding the measurements up and divide by the number of measurements, in this case, 5. The result is 3.1502.

(a) To determine a 95 percent confidence interval, we need to use a standard Z-score, which for 95% is ±1.96. The standard error is the standard deviation divided by the square root of the number of measurements, in this case, 0.1/√5 = 0.04472. The 95% confidence interval can be found by multiplying the standard error by the Z-score and subtracting/adding this from/to the mean. In this case, the interval is [3.1502 -1.96(0.04472), 3.1502 + 1.96(0.04472)] = [3.062, 3.238] mg.

(b) For a99 percent confidence interval, the Z-score is 2.576. The calculations are the same as for part (a), resulting in the interval [3.1502 - 2.576(0.04472),  3.1502 + 2.576(0.04472)] = [3.035, 3.265] mg.

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The 95% and 99% confidence intervals for the true weight based on the five readings and the given standard deviation can be calculated using the average weight and the concept of standard error in combination with the respective z-scores for 95% and 99%.

First, calculate the average weight ([tex]\bar{x}[/tex]) from the 5 readings. That is (3.142 + 3.163 + 3.155 + 3.150 + 3.141) / 5 = 3.1502 mg. Our goal is to find the confidence interval around this average.

(a) For a 95 percent confidence interval, we use a z-score of 1.96 (which corresponds to 95% in a standard normal distribution). The standard error (SE) is the standard deviation (s) divided by the square root of the sample size (n), i.e., SE = s / sqrt(n) = 0.1 / sqrt(5). Hence, the 95 percent confidence interval is [tex]\bar{x}[/tex] ± z * SE = 3.1502 ± 1.96 * SE.

(b) For a 99 percent confidence interval, we use a z-score of 2.576. So, the 99 percent confidence interval is [tex]\bar{x}[/tex] ± z * SE = 3.1502 ± 2.576 * SE.

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Answer: [tex]0.12< \sigma<0.42[/tex]

Step-by-step explanation:

Confidence interval for standard deviation is given by :-

[tex]\sqrt{\dfrac{s^2(n-1)}{\chi^2_{\alpha/2}}}< \sigma<\sqrt{\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2}}}[/tex]

Given : Confidence level : [tex]1-\alpha=0.95[/tex]

⇒[tex]\alpha=0.05[/tex]

Sample size : n= 7

Degree of freedom = 6    (df= n-1)

sample standard deviation : s= 0.19

Critical values by using chi-square distribution table :

[tex]\chi^2_{\alpha/2, df}}=\chi^2_{0.025, 6}}=14.4494\\\\\chi^2_{1-\alpha/2, df}}=\chi^2_{0.975, 6}}=1.2373[/tex]

Confidence interval for standard deviation of the weights of the packages prepared by the machine is given by :-

[tex]\sqrt{\dfrac{ 0.19^2(6)}{14.4494}}< \sigma<\sqrt{\dfrac{ 0.19^2(6)}{1.2373}}[/tex]

[tex]\Rightarrow0.12243< \sigma<0.418400[/tex]

[tex]\approx0.12< \sigma<0.42[/tex]

Hence, the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine.  :

[tex]0.12< \sigma<0.42[/tex]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

[tex]T(t)=T_a+(T_0-T_a)e^{-kt}[/tex]

where [tex]T_a[/tex] is the ambient temperature, [tex]T_0[/tex] is the initial temperature, [tex]t[/tex] is the time and [tex]k[/tex] is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say [tex]T_0[/tex]

We know that the ambient temperature is [tex]T_a=65[/tex], so

[tex]T(t)=65+(T_0-65)e^{-kt}[/tex]

We also know that when [tex]t=5 \:min[/tex] the temperature is [tex]T(5)=35[/tex] and when [tex]t=10 \:min[/tex] the temperature is [tex]T(10)=50[/tex] which gives:

[tex]T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}[/tex]

[tex]T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}[/tex]

Rearranging,

[tex]35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}[/tex]

[tex]50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}[/tex]

If we divide these two equations we get

[tex]\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}[/tex]

[tex]\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}[/tex]

Now, that we know the value of [tex]k[/tex] we can use it to find the initial temperature of the beam,

[tex]35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5[/tex]

so the beam started out at 5 °F.

Using Newton's Law of Cooling, the initial temperature of the metal beam is estimated to be 5°F.

Estimating the Initial Temperature of the Beam Using Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature of its surroundings. Mathematically, it can be expressed as:

where:

Given:

Step-by-Step Solution:

Therefore, solving for k: k = ln(1/2) / 5 ≈ -0.1386

Finally, To = 65 - 60 = 5°F

Answer:

E.

simple random sample

Step-by-step explanation:

Answer:

C.

convenience sample

Step-by-step explanation:

Answer:

t= 32.327

Would be a significant difference in the average amount of saturated fat in solid and liquid fats.

Step-by-step explanation:

1) Data given and notation

Stick:[26.2,25.6,25.5,26.1,26.5,26.7]

Liquid:[16.3,16.4,16.3,16.7,16.6,17.7]

[tex]\bar X_{stick}[/tex] represent the mean for the sample Stick

[tex]\bar X_{Liquid}[/tex] represent the mean for the sample Liquid

[tex]s_{stick}[/tex] represent the sample standard deviation for the sample Stick

[tex]s_{Liquid}[/tex] represent the sample standard deviation for the sample Liquid

[tex]n_{stick}=6[/tex] sample size for the group Stick

[tex]n_{Liquid}=6[/tex] sample size for the group Liquid

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null Hypothesis :[tex]\mu_{stick}=\mu_{Liquid}[/tex]

Alternative Hypothesis :[tex]\mu_{stick} \neq \mu_{Liquid}[/tex]

If we analyze the size for the samples both are < 30 so for this case we can apply a t test to compare means, and the statistic formula is:

[tex]t=\frac{\bar X_{stick}-\bar X_{Liquid}}{\sqrt{\frac{s^2_{stick}}{n_{stick}}+\frac{s^2_{Liquid}}{n_{Liquid}}}}[/tex] (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

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)  

3) Calculate the statistic

First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:

[tex]\bar X_{stick}=26.10[/tex] [tex]s_{stick}=0.477[/tex]

[tex]\bar X_{Liquid}=16.67[/tex] [tex]s_{Liquid}=0.532[/tex]

And with this we can replace in formula (1) like this:

[tex]t=\frac{26.10-16.67}{\sqrt{\frac{0.477^2}{6}+\frac{0.532^2}{6}}}}=32.327[/tex]  

4) Statistical decision

For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{stick}+n_{liquid}-2=6+6-2=10[/tex]

Since is a bilateral test the p value would be:

[tex]p_v =2*P(t_{(10)}>32.327)=1.8x10^{-11}[/tex]

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average amount of saturated fat in solid and liquid fats.

Answer:

a) Null hypothesis: [tex]p \geq 0.368[/tex]

Alternative hypothesis: [tex]p<0.368[/tex]

b) A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.

c) A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.

Part a

On this case we want to test if the proportion of students who enroll in her institution have a lower completion rate (0.398), so the system of hypothesis would be:

Null hypothesis: [tex]p \geq 0.368[/tex]

Alternative hypothesis: [tex]p<0.368[/tex]

Part b

A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.

Part c

A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.

The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion. Making a Type I error would mean wrongly rejecting the null hypothesis, while making a Type II error would mean failing to reject the null hypothesis when it is actually false.

(a) The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion of 0.398. The alternative hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is lower than 0.398.

(b) Making a Type I error in this context would mean rejecting the null hypothesis when it is actually true. This would imply that the college has a lower completion rate when in fact it does not.

(c) Making a Type II error in this context would mean failing to reject the null hypothesis when it is actually false. This would imply that the college has the same completion rate as the national proportion when it actually has a lower completion rate.

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Answer:

8

Step-by-step explanation:48 is the total amount of money Julie has it's 8 dollars for admission and 5 dollars per ride subtract the 8 from 48 which will leave you 40 dollars left now divide that from g and the total of ride juile will be able to ride is 8

Answer:

Step-by-step explanation:

(a) The average number of times a customer carries out banking transactions per week is 13 give or take 5 or so.

(b) The 90% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 1.645\times \sigma/\sqrt{n}

CI=13\pm 1.645\times 5/\sqrt{350}

CI=13\pm 0.439645

CI=(12.560355, 13.439645)

The 99% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 2.576\times \sigma/\sqrt{n}

CI=13\pm 2.576\times 5/\sqrt{350}

CI=13\pm 0.688465

CI=(12.311535, 13.688465)

The difference in the margin of error is 0.688465-0.439645=0.24882

The difference in the margin of error will make the confidence interval wider in the second case(99% confidence interval) as compared to the first case(90% confidence interval).

(c) Since, both our confident interval contains the value 10 hence we have sufficient evidence to conclude that the apparent difference in banking habits between the nation (It is given that the national survey suggest that on an average 10 transactions are performed per week by a single person) and the customer of the bank is not significant or is not real and is just due to the random errors/variations in sampling.

(d) A 95% confidence interval gives a range of values for the (A) Population Average which are plausible according to the observed data.

(The confidence interval always gives the range of possible values for the population values)

(e) The sample standard deviation measures how far (A) number of bank operations is from sample average.

The standard deviation for the sample average measures how far (B) Average number of bank operations is from the population average for typical (D) samples.

Answer: 7

Step-by-step explanation: Linear equations can be written in the form y = mx + b where the multiplier, m, represents the slope of the line and the base, b, represents the y-intercept of the line.

The linear equation y = -6x + 7 has a y-intercept equal to 7.

Out of 1200 acres, approximately 504 of them would be expected to yield more than 190 bushels of corn per acre, assuming a normal distribution of yields with a mean of 185.2 bushels and a standard deviation of 23.5 bushels.

Your question pertains to the field of statistics and normal distribution. When given a mean of 185.2 bushels and a standard deviation of 23.5 bushels for corn yield, you're interested in knowing how many acres out of the total surveyed, in this case 1200 acres, would yield more than 190 bushels per acre.

We first need to calculate the z-score of the 190 bushels. The z-score is the number of standard deviations from the mean a data point is. The formula for a z-score is:

Z = (X - μ) / σ

In this instance, X is the value we are comparing to the mean (190), μ represents the mean (185.2), and σ is the standard deviation (23.5). Plugging these values in gives us a z-score of approximately 0.204.

Now, we consult a standard z-table, which gives us the area to the left of the z-score under a standard normal distribution curve. If we look up the z-score of 0.204, we'd find that the proportion of data less than this score is approximately 0.5803 (i.e., 58.03%). So, the proportion that is higher (expected to yield more than 190 bushels) is about 41.97%.

If we apply this percentage to the number of acres in our study (1200), we find that approximately 504 acres are expected to yield more than 190 bushels of corn per acre.

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Answer:

The answer for the given function is f( -12 ) = 240.

Step-by-step explanation:

Given:

[tex]f(x)=-21x-12[/tex]

To Find :

F ( -12 ) = ?

Solution:

Function:

For Example

f ( x ) = x + 3

x = independent variable

f ( x ) = Output for a given input

Here we have

[tex]f(x)=-21x-12\\Put\ x=-21\\\therefore f(-12) = -21\times -12 - 12\\\therefore f(-12) = 252 - 12\\\therefore f(-12) = 240[/tex]

Therefore f(-12) = 240

Answer for f(-12) = 240 for function f(x) = -21x -12

Step-by-step explanation:

Theory: f(x) is the" Function " and defined as a relation between the set of input values and a respective set of output values. set of input values is term as " Domain " and a respective set of output values is term as " Range "

In given question, f(x) = -21x -12

Note: Question didn't mention the domain so that, we can take it as Real numbers.

To find value of f( -12 )

Replacing value of x with (-12) in f(x)

we get,

f(-12) = -21(-12) -12

f(-12) = 240.

Answer:

Step-by-step explanation:

Answer:

Mean = 71.4 years

Standard Deviation = 20.64 years

Step-by-step explanation:

We are given the following data set:

72, 68, 81, 93, 56, 19, 78, 94, 83, 84, 77, 69, 85, 97, 75, 71, 86, 47, 66, 27

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{1428}{20} = 71.4[/tex]

Sum of squares of differences =

0.36 + 11.56 + 92.16 + 466.56 + 237.16 + 2745.76 + 43.56 + 510.76 + 134.56 + 158.76 + 31.36 + 5.76, 184.96 + 655.36 + 12.96 + 0.16 + 213.16 + 595.36 + 29.16 + 1971.36 = 8100.8

[tex]S.D = \sqrt{\displaystyle\frac{8100.8}{19}} = 20.64[/tex]

Mean = 71.4 years

Standard Deviation = 20.64 years

Answer:

P is exactly 3km east from the oil refinery.

Step-by-step explanation:

Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:

[tex]\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}[/tex]

So the cost of laying pipe under water with this distance is

[tex]800000\sqrt{(3 - d)^2 + 4}[/tex]

And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:

[tex]800000\sqrt{(3 - d)^2 + 4} + 400000d[/tex]

We can find the minimum value of this by taking the 1st derivative and set it to 0

[tex]800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0[/tex]

We can move the first term over to the right hand side and divide both sides by 400000

[tex]1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}[/tex]

[tex]\sqrt{(3 - d)^2 + 4} = 6 - 2d[/tex]

From here we can square up both sides

[tex](3 - d)^2 + 4 = (6 - 2d)^2[/tex]

[tex]9 - 6d + d^2 + 4 = 36 - 24d + 4d^2[/tex]

[tex]3d^2-18d+27 = 0[/tex]

[tex]d^2 - 6d + 9 = 0[/tex]

[tex](d - 3)^2 = 0[/tex]

[tex]d -3 = 0[/tex]

d = 3

So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.

Answer:

0.5

Step-by-step explanation:

Since Ashok stays for exactly 15 minutes after his arrival, they will meet if he arrives in the 10 minutes between 9:10 and 9:20 independently of the time Melina arrives.

Since there are 20 minutes from 9:00 and 9:20 and Ashok arrives at a random time, the distribution of his arrival is uniform, so the probability that he arrives between 9:10 and 9:20 is

10/20 = 0.5

Answer:

(a) B. G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

(b) Every function of the form [tex]4x^2+C[/tex] is an antiderivative of 8x

Step-by-step explanation:

A function F is an antiderivative of the function f if

[tex]F'(x)=f(x)[/tex]

for all x in the domain of f.

(a) If [tex]f(x) = 8x[/tex], then [tex]G(x)=4x^2[/tex] is an antiderivative of f because

[tex]G'(x)=8x=f(x)[/tex]

Therefore, G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

Let F be an antiderivative of f. Then, for each constant C, the function F(x) + C is also an antiderivative of f.

(b) Because

[tex]\frac{d}{dx}(4x^2)=8x[/tex]

then [tex]G(x)=4x^2[/tex] is an antiderivative of [tex]f(x) = 8x[/tex]. Therefore, every antiderivative of 8x is of the form [tex]4x^2+C[/tex] for some constant C, and every function of the form [tex]4x^2+C[/tex] is an antiderivative of 8x.

Final answer:

G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x. The general antiderivative of f(x) = 8x is F(x) = 4x^2 + C.

Explanation:

To determine if G(x) is an antiderivative of f(x), we must consider the derivative of G(x), which is G'(x). If G'(x) is the same function as f(x), then by definition G is an antiderivative of f. Starting with G(x) = 4x2, the derivative is G'(x) = 8x, which is exactly the function f(x). Thus, the correct choice is B: G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

Now, to find all antiderivatives of the function f(x) = 8x, we integrate f(x) to get the general antiderivative, which includes a constant of integration, C. The antiderivative of 8x is F(x) = 4x2 + C, representing all antiderivatives of f(x).

Answer:

Step-by-step explanation:

Hello!

You have some p-values and different decisions that can be made using these p-values.

Remember the decision rule for the p-value method.

If p-value > α, you don't reject the null hypothesis.

If p-value ≤ α, you reject the null hypothesis.

With this in mind, I've arranged the decisions with the given p-values.

a. do not reject H₀ at 10% ⇒ p-value: 0.3

b. do reject H₀ at 10% but not at 5% p-value: ⇒ 0.07

c. do reject H₀ at 5% but not at 1% ⇒ p-value: 0.02

d. do reject H₀ at 1% ⇒ p-value: 0.006

I hope it helps!