The cash operating expenses of the regional phone companies during the first half of 1994 were distributed about a mean of $29.93 per access line per month, with a standard deviation of $2.65. Company A's operating expenses were $27.00 per access line per month. Assuming a normal distribution of operating expenses, estimate the percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean. (Round your answer to two decimal places.)

Answer:

Step-by-step explanation:

Given than n = 25 , mean = 52800 sd = 4600

1) P(X<50000) , so please keep the z tables ready

we must first convert this into a z score so that we can look for probability values in the z table

using the formula

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

checkng the value in the z table

P ( Z>−0.608 )=P ( Z<0.608 )=0.7291

b)

again using the same formula and converting to z score

we need to calculate P(X<50000)

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

P ( Z<−0.608 )=1−P ( Z<0.608 )=1−0.7291=0.2709

proportion is 27% approx

c)

When your data points are measurements on a numerical scale you have variables data , here yield stress is numeric in nature , hence its a sampling by variable plan

Answer:

(b) a one-way between-subjects ANOVA

That's the correct option since we have one factor (class grade) and we have more than two groups.

Step-by-step explanation:

(a) a two-independent sample t test

We can't apply a two independnet t test since we are comparing more than two groups (Freshman, sophomore, Junior and senior). And for this case when we have more than two groups, the most powerful method is the one way ANOVA between subjects.

(b) a one-way between-subjects ANOVA

That's the correct option since we have one factor (class grade) and we have more than two groups.

One way Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]

And we have this property

[tex]SST=SS_{between}+SS_{within}[/tex]

(c) a two-way between-subjects ANOVA

We can't apply a two way ANOVA since we have just one factor (or variable of interest) the class grades measured with a score. So then is not appropiate use this method for this case.

(d) both a two-independent sample t test and a one-way between-subjects ANOVA

False since we can't apply the two way ANOVA, that's not correct.

Answer:

D. All of the above

Step-by-step explanation:

Linear regression models relate to some assumptions about the distribution of error terms. If they are violated violently, the model is not suitable for drawing conclusions. Therefore, it is important to consider the suitability of the model for information before further analysis can be performed based on this model.

The fit of the model is related to the remaining behavior complying with the basic assumptions for error values in the model. When a regression model is constructed from a series of data, it should be shown that the model responds to the standard statistical assumptions of the linear model because of conducting inference. Residual analysis is an effective tool to investigate hypotheses. This method is used to test the following statistical assumptions for a simple linear regression model:

-Regression function is linear in parameters,

-Error conditions have constant variance,

-Error conditions are normally distributed,

-Error conditions are independent.

If no statistical hypothesis of the model is fulfilled, the model is not suitable for data. The fourth hypothesis (independence of error conditions) relates to the regulation of time series data. It is now used some simple graphical methods to analyze analysis, the feasibility of a model, and some formal statistical tests. In addition, when a model fails to meet these assumptions, some data changes may be made to ensure that the assumptions are reasonable for the modified model.

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  

Answer:

B. II only

Step-by-step explanation:

For this case the correct options would be:

B. II only

We analyze one by one the statements:

I.This study provides good evidence that walking is effective in controlling cholesterol

That's FALSE because there are many other factors that can influence the cholesterol and we just have two possible conditions (women who walk at least 10 miles and women who do not exercise at all), and we have just two averages obtained from two samples that we don't know if is a paired values or independent values. And is not appropiate to conclude this with the lack of info on this case.

II. This is an observational study, not an experiment

Correct we don't have a design for this experiment or factors selected in order to test the hypothesis of interest. So for this case we can conclude that we have an observational study for this case.

III. Although the study was conducted only on women, we can confidently generalize the results to men in the same age group.

False, we can't generalize results from women to men since they are different groups of people and with different characteristics.

Answer:

Step-by-step explanation:

1) The diagram is a polygon with unequal sides. The number of sides and angles is 5. This means that it is an irregular Pentagon. The formula for the sum of interior angles of a polygon is expressed as

(n - 2)180

Where n is the number if sides of the polygon. Since the number of sides of the given polygon is 5, the sum if the interior angles would be

(5-2)×180 = 540 degrees. Therefore,

10x - 3 + 5x + 2 + 7x - 11 + 13x - 31 + 8x - 19 = 540

43x - 62 = 540

43x = 540 + 62 = 602

x = 602/43 = 14

Angle S = 13 - 31 = 13×14 - 31 = 182 - 31

Angle S = 151 degrees.

4) The diagram is a rectangle. Opposite sides are equal. Triangle JNM is an isosceles triangle. It means that its base angles, Angle NMJ and angle NJM are equal. Therefore

3x + 38 = 7x - 2

7x - 3x = 38 + 2

4x = 40

x = 40/4

x = 10

Angle NMJ = 7x - 2 = 7×10 - 2 = 68 degrees.

Angle JML = 90 degrees ( the four angles in a rectangle are right angles). Therefore,

Angle NML = 90 - 68 = 22 degrees

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]

To learn more on dissolution processes, we kindly invite to check this verified question: https://brainly.com/question/25752764

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]    

Answer:

[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]

[tex]x_{1} = -10[/tex]

[tex]x_{2} = -3.95[/tex]

Step-by-step explanation:

Generally, the Newton-Raphson method can be used to find the solutions to polynomial equations of different orders. The formula for the solution is:

[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]

We are given that:

f(x) = [tex]x^{2} + 21[/tex]; [tex]x_{0} = -21[/tex]

[tex]f^{'} (x)[/tex] = df(x)/dx = 2x

Therefore, using the formula for Newton-Raphson method to determine [tex]x_{1}[/tex] and [tex]x_{2}[/tex]

[tex]x_{1} = x_{0} - \frac{f(x_{0} )}{f^{'}(x_{0})}[/tex]

[tex]f(x_{0}) = x_{0} ^{2} + 21 = (-21)^{2} + 21 = 462[/tex]

[tex]f^{'}(x_{0}) = 2*(-21) = -42[/tex]

Therefore:

[tex]x_{1} = -21 - \frac{462}{-42} = -21 + 11 = -10[/tex]

Similarly,

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

[tex]f(x_{1}) = (-10)^{2} + 21 = 100+21 = 121[/tex]

[tex]f^{'}(x_{1}) = 2*(-10) = -20[/tex]

Therefore:

[tex]x_{2} = -10 - \frac{121}{20} = -10+6.05 = -3.95[/tex]

Answer:0.05

Step-by-step explanation:

There are 5 different bank accounts

Probability that A will give maximum benefit i.e. Jake save most in account A

[tex]P_1=\frac{1}{5}[/tex]

so picking 1 bank we are left with 4 banks . So we Probability that he will save least into account B is [tex]P_2=\frac{1}{4}[/tex]

Probability that Jake save most into account A and save least into account B is

[tex]=P_1\times P_2[/tex]

[tex]=\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}[/tex]

Answer:

B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]

The correct answer is:

B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.

As regards using the normal model, the correct answer is D. No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.

In sampling distributions, the normal model can be used if np ≥ 10 and n (1 - p) ≥ 10.

In this case, those with weak bones are:

= 8.5% x 82

= 6.97 people which is less than 10

= 1% x 593

= 5.93 people

We do not have 10 or more people for the sample sizes so the normal model will not be a good fit.

Find out more on the normal model at https://brainly.com/question/15399601.

Answer: It's neither arithmetic or a geometric sequence

Explanation: For an arithmetic sequence, to get the common difference, we subtract the first term from the second, third term from second which should give the same value. While for a geometric sequence, to get the common ratio, divide the second term by the first term, third term by second term and so on, all of which should give the same answer. But in the above sequence, it doesn't follow this pattern

The sequence 45, 59, 65, 70, 85 is neither arithmetic nor geometric.

To determine if a sequence is arithmetic, one must check if the difference between consecutive terms is constant. For this sequence:

- The difference between the second and first terms is 59 - 45 = 14.

- The difference between the third and second terms is 65 - 59 = 6.

- The difference between the fourth and third terms is 70 - 65 = 5.

- The difference between the fifth and fourth terms is 85 - 70 = 15.

Since these differences are not the same, the sequence is not arithmetic.

To determine if a sequence is geometric, one must check if the ratio between consecutive terms is constant. For this sequence:

- The ratio between the second and first terms is 59/45.

- The ratio between the third and second terms is 65/59.

- The ratio between the fourth and third terms is 70/65.

- The ratio between the fifth and fourth terms is 85/70.

Simplifying these ratios:

- 59/45 = 1.3111

- 65/59 = 1.1017

- 70/65 = 1.0769

- 85/70 = 1.2143

Since these ratios are not the same, the sequence is not geometric.

The upper 90% confidence limit for the true proportion of firefighter masks of this type whose lenses would pop out at 250° is approximately 0.2949.

To construct a 90% upper confidence limit for the true proportion of firefighter's masks whose lenses would pop out at 250°, firstly, we need to calculate the sample proportion (p) which is the ratio of the number of masks that had lenses popping out (12) to the total number of masks tested (60). Thus, the sample proportion is 12/60 = 0.2.

Next, we use the formula for an upper confidence limit for proportions: p + z*sqrt((p*(1-p))/n), where z is the z-score associated with the desired level of confidence (for 90% confidence, z is 1.645), p is the sample proportion, and n is the sample size.

So the upper 90% confidence limit would be calculated as: 0.2 + 1.645*sqrt((0.2*0.8)/60) = 0.2 + 1.645*0.05774 = 0.2949322. Rounding to four decimal places, we get an upper confidence limit of 0.2949.

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Answer:

a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]

b) [tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/tex]  

Step-by-step explanation:

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".  

Solution

Let put some notation  

x=value for right arm , y = value for left arm

x: 102, 101,94,79,79

y: 175,169,182,146,144

The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:

d: 73, 68, 88, 67, 65

Part a

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]

The third step would be calculate the standard deviation for the differences, and we got:

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]

Part b

The next step is calculate the degrees of freedom given by:

[tex]df=n-1=5-1=4[/tex]

Now we need to calculate the critical value on the t distribution with 4 degrees of freedom. The value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], so we need a quantile that accumulates on each tail of the t distribution 0.05 of the area.

We can use the following excel code to find it:"=T.INV(0.05;4)" or "=T.INV(1-0.05;4)". And we got [tex]t_{\alpha/2}=\pm 2.13[/tex]

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

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

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

[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.331[/tex]  

[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.069[/tex]  

So on this case the 90% confidence interval would be given by (63.331;81.069).

[tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/tex]  

The carrying capacity K for the logistic model is 45. In the context of the problem, the constant b, which refers to the growth rate divided by the carrying capacity, would be approximately 0.00444. The logistic model representing P vs. m under these conditions would be P(m) = 45/ (1+ (45/15-1) * e^(-0.00444m)).

In the field of mathematics, specifically in growth modeling, a logistic model incorporates a carrying capacity. The carrying capacity, denoted as K, is the maximum stable value of the population, in this case, the number of parents attending the PTA meeting. The carrying capacity is expected to be 45 in this instance.

The constant b in the logistic model can be found using the initial value (15 parents) and the growth rate (20%). However, the question does not specify whether this rate is relative or absolute. Assuming relative growth, the initial growth rate r is 0.2 and the constant b = r/K would be 0.00444. This is not the traditional definition of b in a logistic equation; typically, b would denote the initial population size, so the question seems to have a specific, non-standard usage in mind that we need to respect.

A good starting value of ther, in order to ensure convergence and stability of the numerical method, can be 15, the initial population size.

The logistic model would thus be represented as P(m) = K/ (1+ (K/P_0-1) * e^(-bm)), where P_0 is the initial number of parents, K is the carrying capacity, b is the constant/ growth rate, m is the number of months, and e is the mathematical constant approximated as 2.718.

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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:

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:

[tex]t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738[/tex]    

[tex]p_v =P(t_{(62)}>2.738)=0.0040[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=20.49[/tex] represent the mean time for the sample  

[tex]s=1.42[/tex] represent the sample standard deviation for the sample  

[tex]n=63[/tex] sample size  

[tex]\mu_o =20[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean time is actually higher than 20 min, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 20[/tex]  

Alternative hypothesis:[tex]\mu > 20[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=63-1=62[/tex]  

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

[tex]p_v =P(t_{(62)}>2.738)=0.0040[/tex]  

And we can use the following excel code to find it:

"=1-T.DIST(2.738,62,TRUE)"

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .  

To perform the hypothesis test, set up null and alternative hypotheses, calculate the test statistic, and compare it to the critical value from the t-distribution table.

To perform a hypothesis test, we need to set up the null and alternative hypotheses. The null hypothesis, H0, states that the mean delivery time is 20 minutes or less. The alternative hypothesis, Ha, states that the mean delivery time is more than 20 minutes. We can perform a one-sample t-test using the sample mean, standard deviation, sample size, and the desired level of significance. Calculate the test statistic and compare it to the critical value from the t-distribution table. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis.

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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:

a) Null hypothesis: [tex]\mu =0[/tex]

Alternative hypothesis: [tex]\mu \neq 0[/tex]

b) t =-7.44

c) [tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]

d) Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.

We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.

Step-by-step explanation:

Assuming this output:

[tex]t_{difference}=-0.395[/tex]

t(observed value) =-7.44

t(Critical value )= 1.984

DF = 98

p value (two tailed) < 0.0001

[tex]\alpha =0.05[/tex]

a) What are the null and alternative hypothesis

Null hypothesis: [tex]\mu =0[/tex]

Alternative hypothesis: [tex]\mu \neq 0[/tex]

The reason is because the output says a bilateral test so for this case w eselect this option.

b) Identify the statistic

The correct formula for the statistic is given by:

[tex]t=\frac{\bar X_1 -\bar X_2 -0}{\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}}[/tex]

Based on the output the calculated value its t =-7.44

c) Identify the p value

We have the degrees of freedom given 98

Based on the alternative hypothesis the p value is given by:

[tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]

d) State the final conclusion that addresses the original claim

Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.

We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.

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.

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

Answer: [tex]\$3108[/tex]

Step-by-step explanation:

Mr. Kim, his wife and two children are 4 persons. Now, he needs to buy 3 plane tickets for each one, which means he has to buy 12 plane tickets:

[tex](3 tickets)(4)=12 tickets[/tex]

On the other hand, we know each ticket costs [tex]\$259[/tex] per person. So, if we have 12 tickets to buy, we will have to multiply [tex]\$259[/tex] by 12:

[tex](12)(\$259)=\$3108[/tex] This is the amount Mr. Kim will spend in the plane tickets

Answer

The answer and procedures of the exercise are attached in the following archives.

Explanation  

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

The pairs (C,r) where the expected profit from a warranty is zero are calculated by balancing the warranty cost to the store for stoves failing within the warranty period with the revenue from selling the warranties. Reasonable values should consider customer appeal, market competition, and business risk.

The question involves calculating the expected profit from a warranty which depends on an exponential random variable representing the lifetime of a stove. The lifetime of the stove, modeled as an exponential random variable, is typically used for modeling the lifespan of objects like mechanical or electronic devices whose failure rate is constant over time. This property is also known as the memoryless property. The distribution is defined by the parameter lambda (λ). In this case, λ equals 1/10, which suggests the average lifespan of the stove is 10 years.

The cost to replace the stove is $800, and the price charged for the warranty is denoted as C. The question asks for pairs of (C,r) where the expected profit is zero. This occurs when the cost of the warranty C is equal to the cost of replacing the stove, $800, over the time r in years for which the warranty is valid. It is also important to remember that not every stove will require a replacement, only those that fail within the warranty period r. The probability that a stove fails within the r years is computed using the exponential distribution's cumulative probability function as P(X

The expected profit is zero when the total warranty cost compensates for all the stoves replaced within their warranty periods i.e., C*P(X>r) = $800*P(X

Reasonable choices for C and r would depend on many factors such as the company's risk tolerance, competition in the warranty market, and customers' willingness to pay for extended warranties. A higher warranty price C increases profit but may discourage customers from buying the warranty. A longer warranty period r increases customer appeal but also the company's costs if more stoves fail and need to be replaced.

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A. 15917 ft   B.∠D=88.35°  C.0.0532 rad

Step-by-step explanation:

A. Given  that angle ∠ACB = 88.60° and the distance from C to B is 389 ft then triangle ABC is right ,90° at B. Applying the formula for tangent of an angle which is;

Tan of an angle = opposite side length/adjacent side length

Tan Ф = O/A =AB/389 ft

Tan 88.60°= AB/389

AB=389*tan 88.60° = 389×40.92 =15916.87 ft ⇒15917 ft

B.

The distance from B to D is given as 459 ft and the distance between the towers , AB, is 15917 ft. To get angle ∠D apply the formula for tangent of an angle where ;

Tan ∠D=O/A =15917/459 =34.6775599129

∠D =tan⁻(34.6775599129)

∠D=88.35°

C. To get angle ∠A subtract the sum of angle ∠C and ∠D from 180°. Apply the sum of angles in a triangle theorem

 ∠A =180° - (88.60°+88.35°)

   ∠A = 180°-(176.95°)=3.05°

    Changing degrees to radians you multiply the degree value with 0.0174533

3.05°=3.05*0.0174533=0.05323254 rad

To 4 decimal places

=0.0532 rad

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Answer:

Step-by-step explanation:

Answer:

Upper Riemann Sum is 9/16

Step-by-step explanation:

Final answer:

To compute the upper Riemann sum for the function f(x) = x^2 over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1], we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval. The upper Riemann sum is obtained by summing up all these values.

Explanation:

To compute the upper Riemann sum, we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval.

Given the function f(x) = x^2 and the partition P = [-1, -1/4, 1/4, 3/4, 1], we can calculate the upper Riemann sum as follows:

Calculate the width of each subinterval: Δx = (1 - (-1)) / 4 = 1/2.

Find the value of the function at each partition point:

Multiply the value of the function at each partition point by the width of the corresponding subinterval:

Sum up all the values obtained: 1/2 + 1/32 + 1/32 + 9/32 + 1/2 = 17/16

Therefore, the upper Riemann sum for the given function over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1] is 17/16.

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:

[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: A) .1587

Step-by-step explanation:

Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.

i.e. [tex]\mu=12.30[/tex] and [tex]\sigma=0.20[/tex]

Let x denotes the amount of soda in any can.

Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.

Then, the probability that a randomly selected can will need to go through the mentioned process =  probability that a randomly selected can has more than 12.50 ounces of soda poured into it =

[tex]P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587[/tex]

Hence, the required probability= A) 0.1587

Final answer:

The probability that a randomly selected can will need a special cleaning process because it holds more than 12.50 ounces of soda is calculated using the Z-score. The correct answer is A) 0.1587.

Explanation:

The question asks us to calculate the probability that a randomly selected can will require a special cleaning process if it holds more than 12.50 ounces of soda, given a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. To find this probability, we can use the standard normal distribution, also known as the Z-score.

First, we calculate the Z-score for 12.50 ounces using the formula: Z = (X - \\mu\) / \\sigma\, where X is the value we are evaluating (12.50 ounces), \mu is the mean (12.30 ounces), and \sigma is the standard deviation (0.20 ounce). Plugging in the values gives us Z = (12.50 - 12.30) / 0.20 = 1. The Z-score represents the number of standard deviations the value X is from the mean.

Next, to find the probability that a can will overflow (have more than 12.50 ounces), we look up the corresponding probability in the standard normal distribution table or use a calculator suitable for such statistical computations. The probability corresponding to a Z-score of 1 is approximately 0.8413. However, since we are interested in the probability of a can having more than 12.50 ounces, we need the area to the right of the Z-score, which is 1 - 0.8413 = 0.1587. Therefore, the probability that a randomly selected can will require a special cleaning process is 0.1587.

The correct answer to the question is A) 0.1587.