Every morning I take either bus number 5 or bus number 8 to work. Every morning the waiting time for the number 5 is exponential with mean 10 minutes, while the waiting time for the number 8 is exponential with mean 20 minutes. Assume all waiting times are independent of each other. Let S be the total amount of bus-waiting (in minutes) that I have done duringn mornings, and let T, be the number of times I have taken the number 5 bus during n mornings. a) Find the limit i, PS, s 7nl (b) Find the limit lim P(T 0.6n). 1-00 Hint. Recall Examples 6.33 and 6.34.

Answers

Answer 1

Answer

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

Step-by-step 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.  

Every Morning I Take Either Bus Number 5 Or Bus Number 8 To Work. Every Morning The Waiting Time For

Related Questions

Given an actual demand this period of 61, a forecast value for this period of 58, and an alpha of 0.3, what is the experimential smoothing forecast for the next period?

Answers

Answer:

Experimental smoothing forecast for the next period will be 58.9

Step-by-step explanation:

We have given actual demand is [tex]A_{t-1}=61[/tex]

Initial forecast [tex]f_{t-1}=61[/tex] and [tex]\alpha =0.3[/tex]

We have to find the experimental smoothing forecast [tex]f_t[/tex]

Experimental forecast is given by

[tex]f_t=(1-\alpha )f_{t-1}+\alpha A_{t-1}=(1-0.3)\times 58+0.3\times 61=58.9[/tex]

So experimental smoothing forecast for the next period will be 58.9

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

(a) Use a linear approximation to estimate g(2.9) and g(3.1). g(2.9) ≈ g(3.1) ≈
(b) Are your estimates in part (a) too large or too small? Explain.

Answers

The value of the function g(x) at x = 2.9 and x = 3.1 will be -6.57 and -3.37, respectively. And the estimation is too small in part (a).

What is integration?

Integration is a way of finding the total by adding or summing the components. It's a reversal of differentiation, in which we break down functions into pieces. This approach is used to calculate the total on a large scale.

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Integrate the function, then we have

∫g'(x) = ∫(x² + 7) dx

g(x) = x³/3 + 7x + c

g(3) = 3³ / 3 + 7 (3) + c

- 5 = 9 + 21 + c

c = - 30 - 5

c = -35

Then the function is given as,

g(x) = x³/3 + 7x - 35

At x = 2.9, we have

g(2.9) = (2.9)³/3 + 7(2.9) - 35

g(2.9) = -6.57

At x = 3.1, we have

g(3.1) = (3.1)³/3 + 7(3.1) - 35

g(3.1) = -3.37

The value of the function g(x) at x = 2.9 and x = 3.1 will be -6.57 and -3.37, respectively. And the estimation is too small in part (a).

More about the integration link is given below.

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Does involving a statistician to help with statistical methods improve the chance that a medical research paper will be published? A study of papers submitted to two medical journals found that 135 of 190 papers that lacked statistical assistance were rejected without even being reviewed in detail. In contrast, 293 of the 514 papers with statistical help were sent back without review. Find the value of the z-test statistic. Give your answer to two decimal places.

Answers

Answer:

z-test statistic is 3.38

Step-by-step explanation:

Null and alternative hypotheses are:

[tex]H_{0}[/tex]: the proportion of papers that lacked statistical assistance  sent back without review is the same as the proportion of papers with statistical help sent back without review

[tex]H_{a}[/tex]: the proportion of papers that lacked statistical assistance sent back without review is different than the proportion of papers with statistical help sent back without review

Test statistic can be found using the equation:

[tex]z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

p1 is the sample proportion of papers that lacked statistical assistance  sent back without review ( ([tex]\frac{135}{190}=0.71[/tex])p2 is the sample proportion of papers with statistical help sent back without review  ([tex]\frac{293}{514}=0.57[/tex])p is the pool proportion of p1 and p2 ([tex]\frac{135+293}{190+514}=0.61[/tex])n1 is the sample size of papers that lacked statistical assistance (190)n2 is the sample size of papers with statistical help (514)

Then [tex]z=\frac{0.71-0.57}{\sqrt{{0.61*0.39*(\frac{1}{190} +\frac{1}{514}) }}}[/tex] ≈ 3.38

Patterson Electronics supplies microcomputer circuitry to a company that incorporates microprocessors into refrigerators and other home appliances. One of the components has an annual demand of 250 units, and this is constant throughout the year. Carrying cost is estimated to be $1 per unit per year, and the ordering cost is $30 per order.a. To minimize cost, how many units should be ordered each time an order is placed?


b. How many orders per year are needed with the optimal policy?


c. What is the average inventory if costs are minimized?


d. Suppose that the ordering cost is not $30, and B. Brady has been ordering 250 units each time an order is placed. For this order policy (of Q = 250 to be optimal, determine what the ordering cost would have to be.

Answers

Answer:

a) Q = 122 units/order

b) Number of orders = 2.05 orders/year

c) Average inventory = 61 units

d) Ordering costs = 125 $/order

Step-by-step explanation:

The economic quantity order (EOQ) formula allow us to minimize the ordering cost, in function of the demand, ordering cost and holding cost.

The EOQ formula is:

[tex]EOQ=\sqrt{\frac{2DS}{H} }[/tex]

where:

D: demand in units/year

S: Order costs, per order

H: holding or carrying cost, per unit a year

a) In this case:

D: 250 u/year

S: 30 $/order

H: 1 $/year-unit

[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*30}{1} }=\sqrt{15000}=122.47\approx122[/tex]

b) If we have a demand of 250 units/year and we place orders of 122 units, the amount of orders/year is:

[tex]\#orders=\frac{D}{EOQ}=\frac{250\,units/year}{122\,units/order}=2.05\, \frac{orders}{year}[/tex]

c) We assume that there is no safety stock, so everytime the stock hits 0 units, a new order enter the inventory.

In this case, the average inventory can be estimated as the average between the inventory when a new order enters the inventory (122 u.) and the inventory right before a order enters (0 u.)

[tex]\#av.inventory=\frac{122+0}{2}=61[/tex]

The average inventory is 61 units.

d) If 250 units is the optimal quantity for an order, it means it is equal to the EOQ. We can calculate the new ordering costs as:

[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*S}{1} }=250\\\\2*250*S=250^2\\\\S=250/2=125\,\$/order[/tex]

A) To minimize cost, the number of units that should be ordered each time an order is placed is; 122 units

B) The number of orders per year needed with the optimal policy is; 2 orders per year.

C)  The average inventory if costs are minimized is; 61 units

D) For the order policy of Q = 250 to be optimal, the ordering cost would have to be; $125 per order

The formula for economic quantity order (EOQ) is given as;

EOQ = √(2DS/H)

Where;

D is demand rate S is set up costs H is holding cost

We are given;

Annual demand; D = 250 units/year

Holding Cost; H = $1 per unit per year

Set up costs; S = $30 per order

A) EOQ here is;

EOQ = √(2 × 250 × 30/1)

EOQ = 122.47

But EOQ has to be a whole number and so we approximate to the nearest whole number to get;

EOQ = 122

B) With the optimal policy, the number of orders per year is gotten from the formula;

n = D/EOQ

Plugging in the relevant values gives;

n = 250/122

n = 2.049

But number of orders has to be a whole number. Thus, we approximate to the nearest whole number to get;

n = 2 orders per year

C) If costs are minimized, the average inventory is defined as the average between of the inventory when a new order enters and the inventory just before a new order enters.

Before a new order enters the inventory is 0 if we assume that there is no safe stock. Thus;

average inventory = (122 + 0)/2 = 61 units

D) We are told that ordering cost is not $30 but Optimal order EOQ is 250 and so;

EOQ = √(2DS/H)

⇒ 250 = √(2 × 250 × S/1)

Square both sides to get;

250² = 500S

S = 250²/500

S = $125 per order

Read more about economic quantity order (EOQ) at; https://brainly.com/question/16395657

A crossover trial is a type of experiment used to compare two drugs, Subjects take one drug for a period of time, thenswitch to the other. The responses of the subjects are then compared using matched pair methods In an experiment tocompare two pain relievers, seven subjects took one pain reliever_'___———Sublet!1 1 3 4 5 6 7____'_._Drug A 6 3 4 S 7 l 4Drug B 5 1 5 5 5 2 2_____'_'__Can you conclude that the mean response differs between the two dmgs?

Answers

Answer:

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]

[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/tex]

So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between Drug A and Drug B is equal to 0.

Step-by-step explanation:

Data given

              1   2   3    4    5   6   7

Drug A   6   3   4   5    7    1    4

Drug B   5    1   5   5    5   2    2

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value for A , y = test value for B

The system of hypothesis for this case are:

Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]

Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]

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

d: -1, -2, 1, 0, -2, 1, -2

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714[/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} =1.38[/tex]

The 4 step is calculate the statistic given by :

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]

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

[tex]df=n-1=7-1=6[/tex]

Now we can calculate the p value, since we have a two tailed test the p value is given by:

[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/tex]

So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between Drug A and Drug B is equal to 0.

Help me please and thank you

Answers

Answer:

Based on the model, the length of the wall is [tex]\frac{9}{8}[/tex] ft, the width of the wall is [tex]\frac{1}{2}[/tex] ft,  and the height of the wall is [tex]\frac{11}{8}[/tex] ft. The volume of the portion of security wall that  Tim has constructed so far is [tex]\frac{99}{128}[/tex]  cu ft.

Step-by-step explanation:

Given:

The figure constructed shows a rectangular prism made up of small wooden cubes of length [tex]\frac{1}{8}\ ft[/tex].

Width of the prism = [tex]\frac{1}{2}\ ft[/tex].

To find length , height and volume of the figure.

Solution:

From the figure we can conclude that :

Length side of the prism counts 9 cubes.

Thus, length of the prism will be given as :

⇒ [tex]\textrm{Length of each cube}\times \textrm{Number of cubes}[/tex]

⇒ [tex]\frac{1}{8}\ ft\times 9[/tex]

⇒ [tex]\frac{9}{8}\ ft[/tex]  (Answer)

Height side of the prism counts 11 cubes.

Thus, height of the prism will be given as :

⇒ [tex]\textrm{Length of each cube}\times \textrm{Number of cubes}[/tex]

⇒ [tex]\frac{1}{8}\ ft\times 11[/tex]

⇒ [tex]\frac{11}{8}\ ft[/tex] (Answer)

Volume of the prism can be given as :

⇒ [tex]Length\times width\times height[/tex]

⇒ [tex]\frac{9}{8}\ ft\times \frac{1}{2}\ ft\times \frac{11}{8}\ ft  [/tex]

⇒ [tex]\frac{99}{128}\ ft^3[/tex] (Answer)

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of 150 sec and a standard deviation of 25 sec. The fastest 10% are to be given advanced training. What task times qualify individuals for such training? (Round the answer to one decimal place.)

________ seconds or less

Answers

Final answer:

The cutoff time to be in the fastest 10% of job applicants is approximately 118.0 seconds or less, calculated using a z-score of the 10th percentile in a normal distribution.

Explanation:

The student is asking about finding the cutoff time for the fastest 10% of job applicants performing a certain task, given that the time taken is normally distributed with a mean of 150 seconds and a standard deviation of 25 seconds. To solve this problem, we will use the z-score corresponding to the fastest 10% (the 10th percentile) in a normal distribution.

First, we look up the z-score for the 10th percentile in the z-table, which is approximately -1.28. Then we use the z-score formula:

z = (X - μ) / σ

Plugging in the known values:

-1.28 = (X - 150) / 25

Now, we solve for X:

X = -1.28 × 25 + 150

X = -32 + 150

X = 118 seconds

So, job applicants must complete the task in approximately 118.0 seconds or less to qualify for the advanced training.

4. A food company is concerned that its 16-ounce can of sliced pears is being overfilled. The quality-control department took a sample of 35 cans and found that the sample mean weight was 16.04 ounces, with a sample standard deviation of 0.08 ounces. Test at a 5% level of significance to see if the mean weight is greater than 16 ounces. What is your conclusion?

Answers

Answer:

The mean weight is not greater than 16 ounces

Step-by-step explanation:

Null hypothesis: Mean weight is greater than 16 ounces

Alternate hypothesis: Mean weight is not greater than 16 ounces.

Sample mean= 16.04 ounces

Assumed mean= 16 ounces

Significance level = 0.05

Sample standard deviation = 0.08 ounces

Number of samples= 35

To test the claim about the mean, Z = (sample mean - assumed mean) ÷ (standard deviation÷ √number of samples)

Z = (16.04 - 16) ÷ (0.08 ÷ √35)

Z = 0.04 ÷ (0.08 ÷ 5.92)

Z= 0.04 ÷ 0.014 = 2.86

Z = 2.86

It is a one-tailed test, the critical region is an area of 0.05 (the significance level). The critical value of the critical region is 1.64

Since the computed Z 2.86 is greater than the critical value 1.64, the null hypothesis is rejected.

Conclusion

The mean weight is not greater than 16 ounces

A newsletter for investors recently reported that the average stock price for a blue chip stock over the past 12 months was​ $72. No standard deviation was given. Is the standard deviation more likely to be​ $6, $26, or​ $60? Explain.

Answers

Answer:

$6

Explanation:

The standard deviation gives an idea of the dispertion of values in those 12 month. A hight value of standard deviation means that the prices changed in big increments ( for example one month is $20, other month is $60 and other $85). If that  would be the case, is important to investors to know it and the deviation should be reported.

So, given that it isnt reported, we can say that the price vary but only a few dollars, resulting in a small standard deviation such as $6.

Final answer:

A standard deviation of $6 for a blue chip stock is more likely than $26 or $60, as blue chip stocks are typically stable with less price volatility.

Explanation:

Regarding the standard deviation of the blue chip stock price, it is unlikely that the standard deviation would be very high considering the nature of blue chip stocks. Blue chip stocks are known for their stability and are typically less volatile than other types of stocks. Therefore, a standard deviation of $60 would suggest a very high level of volatility, which is uncharacteristic of blue chip stocks. A standard deviation of $6 is more plausible because it indicates minor fluctuations around the average price of $72, which is more in line with the expected behavior of blue chip stocks. A standard deviation of $26 is possible but less likely than $6, given that such a value still represents a fairly high level of volatility for blue chip stocks, which are commonly considered as safe, long-term investments with relatively steady prices.

I really need help with this!

Answers

Answer:

Step-by-step explanation:

1) in a rhombus, the diagonals bisect each other at the midpoint, forming 4 right angles.

MK = NK + NM

Since MK = 24 and NK = NM, then,

NK = 24/2 = 12

JL = NJ + NL

Since JL = 20 and NJ = NMlL, then,

NL = 20/2 = 10

Looking at triangle LMN, it is a right angle triangle. Applying Pythagoras theorem,

ML^2 = 10^2 + 12^2 = 100 + 144 = 244

ML = √244 = 15.62

Since all the sides of the rhombus are equal, then

MJ = ML = 15.62

Angle KNL = angle JML = 90 degrees. This is so because the diagonals are perpendicular)

Angle KJL = 90 - angle MJL

Angle KJL = 90 - 50 = 40 degrees.

Since the opposite angles of the rhombus are equal,

Angle MLK = angle angle MJK = 50 + 40 = 90 degrees

2) since all the sides of a rhombus are equal,

5x + 16 = 9x - 32

9x - 5x = 32 + 16

4x = 48

x = 48/4 = 12

Since PQ = NR = 5x + 16, then

PQ = 5×12 + 16 = 60 + 16

PQ = 76

3) Triangle XYZ is an isosceles triangle. This means that its base angles, XZY and ZXY are equal. The sum of angles in a triangle is 180 degrees. Therefore,

Angle XZY + angle ZXY = 180 - 136

Angle XZY + angle ZXY = 44

Angle XZY = 44/2 = 22

Therefore,

10x - 8 = 22

10x = 22 + 8 = 30

x = 30/10 = 3

Suppose that prices of a gallon of milk at various stores in one town have a mean of $3.73$⁢3.73 with a standard deviation of $0.10$⁢0.10. Using Chebyshev's Theorem, what is the minimum percentage of stores that sell a gallon of milk for between $3.43$⁢3.43 and $4.03$⁢4.03? Round your answer to one decimal place.

Answers

Answer: At-least 88.89%

Step-by-step explanation:

As per given , we have

Population mean : [tex]\mu=\$3.73[/tex]

Standard deviation :  [tex]\sigma=\$0.10[/tex]

Now , $3.43= $⁢3.73- 3(0.10) = [tex]\mu-3\sigma[/tex]

$⁢4.03 = $⁢3.73+3(0.10) = [tex]\mu+3\sigma[/tex]

i.e. $3.43 is 3 standard deviations below mean and $⁢4.03 is 3 standard deviations above mean .

To find :  the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03.

i.e. to find minimum percentage of stores that sell a gallon of milk  lies within 3 standard deviations from mean.

According to Chebyshev, At-least [tex](1-\dfrac{1}{k^2})[/tex] of the values  lies with in [tex]k\sigma[/tex] from mean.

For k= 3

At-least [tex](1-\dfrac{1}{3^2})[/tex] of the values lies within [tex]3\sigma[/tex] from mean.

[tex]1-\dfrac{1}{3^2}=1-\dfrac{1}{9}=\dfrac{8}{9}[/tex]

In percent = [tex]\dfrac{8}{9}\times100\%\approx88.89\%[/tex]

Hence, the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03 =  At-least  88.89%

Final answer:

Using Chebyshev's Theorem, the minimum percentage of stores selling a gallon of milk between $3.43 and $4.03 is at least 88.9% when the mean price is $3.73 and the standard deviation is $0.10.

Explanation:

Using Chebyshev's Theorem, we can determine the minimum percentage of stores selling a gallon of milk within a certain range of prices given the mean and standard deviation. The theorem states that for any number k, where k > 1, at least (1 - 1/k²) of the data values will fall within k standard deviations of the mean. In this case, the range of prices is from $3.43 to $4.03, which is $0.30 away from the mean of $3.73 on either side.

To find k, we divide the distance from the mean by the standard deviation: k = 0.30 / 0.10 = 3. Thus, at least (1 - 1/3²) or (1 - 1/9) of the stores sell a gallon of milk within this range. Calculating this, we get at least (1 - 1/9) = 8/9 or approximately 88.9% of stores.

Therefore, according to Chebyshev's Theorem, the minimum percentage of stores that sell a gallon of milk for between $3.43 and $4.03 is at least 88.9%.

Put the equation in slope intercept form y-10=-2(x-3)

Answers

Answer:

y= -2x + 16

Step-by-step explanation:

multiple the -2 with everything in parentheses then get -2x and 6 add 10 to both sides then then bring everything down to get y= -2x + 16

What type of variable is the number of auto accidents reported in a given month?

Select one:
A. Interval
B. Ratio
C. Continuous
D. Discrete

Answers

Answer:

D. Discrete

Step-by-step explanation:

The number of reported car accidents is a countable variable, that can only be natural values.

You can have 0, 1, 2, 3, 4, ..., 100, ..., 1000 reported car accidents in a given month.

You cannnot have 4.5 reported car accidents in a given month, for example.

So the correct answer is:

D. Discrete

Final answer:

The type of variable that represents the number of auto accidents reported in a given month is a discrete variable, as it can only take certain values (0, 1, 2, etc.), and never a fraction or decimal.

Explanation:

The type of variable that best describes the number of auto accidents reported in a given month is Discrete.

In statistics, variables are classified into different types based on their properties. A discrete variable is one that can only take certain values. This is the case for the number of auto accidents in a month, which could be 0, 1, 2, 3, etc., but never a fraction or decimal value like 2.5.

In contrast, a continuous variable could take any value within a certain range, including fractions and decimals. Meanwhile, interval and ratio are types of data measurements that don't apply in this context.

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find a and b from the picture​

Answers

Answer:a is 40 degrees

b is 140 degrees

Step-by-step explanation:

The given polygon has 5 irregular sides. This means that it is an irregular pentagon. The sum of the exterior angles of a polygon is 360 degrees

The exterior angles of the given pentagon are a, 75, 65, 60 and 120. Therefore

a + 75 + 65 + 60 + 120 = 360

a + 320 = 360

Subtracting 320 from both sides of the equation, it becomes

a = 360 - 320 = 40 degrees

The sum of angles on a straight line is 180/degrees. Therefore,

a + b = 180

b = 180 - a = 180 - 40 = 140 degrees

Find an equation of the sphere with center s23, 2, 5d and radius 4. What is the intersection of this sphere with the yz-plane?

Answers

Answer:

Equation: [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]

Intersection: [tex](y-2)^2+(z-5)^2=7[/tex]

Step-by-step explanation:        

We are asked to write an equation of the sphere with center center [tex](-3,2,5)[/tex] and radius 4.    

We know that equation of a sphere with radius 'r' and center at [tex](h,k,l)[/tex] is in form:

[tex](x-h)^2+(y-k)^2+(z-l)^2=r^2[/tex]

Since center of the given sphere is at point [tex](-3,2,5)[/tex], so we will substitute [tex]h=-3[/tex], [tex]k=2[/tex], [tex]l=5[/tex] and [tex]r=4[/tex] in above equation as:

[tex](x-(-3))^2+(y-2)^2+(z-5)^2=4^2[/tex]

[tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]

Therefore, our required equation would be [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex].

To find the intersection of our sphere with the yz-plane, we will substitute [tex]x=0[/tex] in our equation as:

[tex](0+3)^2+(y-2)^2+(z-5)^2=16[/tex]

[tex]9+(y-2)^2+(z-5)^2=16[/tex]

[tex]9-9+(y-2)^2+(z-5)^2=16-9[/tex]

[tex](y-2)^2+(z-5)^2=7[/tex]

Therefore, the intersection of the given sphere with the yz-plane would be [tex](y-2)^2+(z-5)^2=7[/tex].

A city has 1,091,953 residents. A recent census showed that 364,656 of these residents regularly use the city's public transportation system. A survey is being conducted in which 1,047 of the city's 1,091,953 residents will be randomly selected. This question relates to the number of people in the survey that is made up of people that do use the city's public transport. The number of people in the survey that do use public transport approximately follows a normal distribution. Calculate the standard deviation of this distribution. Give your answer rounded to 2 decimal places.

Answers

Answer:

The standard deviation of the distribution is 492.847

Step-by-step explanation:

Consider the provided information.

A city has 1,091,953 residents. A recent census showed that 364,656 of these residents regularly use the city's public transportation system.

Therefore the value of n is 1,091,953

The probability of success is: [tex]\frac{364,656 }{1,091,953}=0.33395[/tex]

Calculate the standard deviation using the formula: [tex]\sigma=\sqrt{np(1-p)}[/tex]

Substitute the respective values.

[tex]\sigma=\sqrt{1091953(0.334)(1-0.334)}[/tex]

[tex]\sigma=\sqrt{242898.3931}[/tex]

[tex]\sigma=492.847[/tex]

Hence, the standard deviation of the distribution is 492.847

Use a parameterization of the lower portion cut from the sphere x squared plus y squared plus z squared equals 4x2+y2+z2=4 by the conez equals StartFraction 1 Over StartRoot 3 EndRoot EndFraction StartRoot x squared plus y squared EndRootz=13x2+y2 to express the area of the surface as a double integral. Then evaluate the integral.

Answers

Substituting [tex]z[/tex] from the cone's equation,

[tex]z=\dfrac13\sqrt{x^2+y^2}[/tex]

into the equation of the sphere,

[tex]x^2+y^2+z^2=4[/tex]

gives the intersection of the two surfaces,

[tex]x^2+y^2+\left(\dfrac13\sqrt{x^2+y^2}\right)^2=4\implies x^2+y^2=\dfrac{18}5[/tex]

which is a circle of radius [tex]\sqrt{\frac{18}5}[/tex] centered at [tex]\left(0,0,\frac13\sqrt{\frac{18}5}\right)[/tex].

We parameterize this part of the sphere outside the cone (call it [tex]S[/tex]) by

[tex]\vec s(u,v)=\langle2\cos u\sin v,2\sin u\sin v,2\cos v\rangle[/tex]

with [tex]0\le u\le2\pi[/tex] and [tex]\cos^{-1}\frac1{\sqrt{10}}\le v\le\pi[/tex].

Take the normal vector to [tex]S[/tex] to be

[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle4\cos u\sin^2v,4\sin u\sin^2v,4\cos v\sin v\rangle[/tex]

Then the area of [tex]S[/tex] is

[tex]\displaystyle\iint_S\mathrm dA=\int_0^{2\pi}\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|\,\mathrm dv\,\mathrm du[/tex]

[tex]=\displaystyle2\pi\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi4\sin v\,\mathrm dv=\boxed{\frac{40+4\sqrt{10}}5\pi}[/tex]

Sue and Kathy were doing their algebra homework. They were asked to write the equation of
the line that passes through the points (-3,4) and (6,1). Sue wrote y - 4 = -(x + 3) and Kathy
wrote y = -x + 3. Justify why both students are correct.

Answers

Answer:both students are incorrect

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)

Slope = (y2 - y1)/(x2 - x1)

y2 = 1

y1 = 4

x2 = 6

x1 = - 3

Slope = (1 - 4)/(6 - -3) = -3/9 = -1/3

To determine the intercept, we would substitute m = -1/3, x = 6 and y = 1 into y = mx + c. It becomes

1 = -1/3 × 6 + c = -2 + c

c = 1 + 2 = 3

The equation becomes

y = -x/3 + 3

If the equation was written in the slope intercept form which is expressed as

y - y1 = m(x - x1)

It becomes

y - 4 = -1/3(x - - 3)

y - 4 = -1/3(x + 3)

Both students are incorrect

The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 600 to 2400.

Suppose that PrepIt! is a company that offers classes to help students prepare for the SAT exam. In their ad, PrepIt! claims to produce "statistically significant" increases in SAT scores. This claim comes from a study in which 427 PrepIt! students took the SAT before and after PrepIt! classes. These students are compared to 2,733 students who took the SAT twice, without any type of formal preparation between tries.

We also conduct a hypothesis test with this data and find that students who retake the SAT without PrepIt! also do significantly better (p-value < 0.0001). So now we want to determine if PrepIt! students improve more than students who retake the SAT without going through the PrepIt! program. In a hypothesis test, the difference in sample mean improvement ("PrepIt! gain" minus "control gain") gives a p-value of 0.004. A 90% confidence interval based on this sample difference is 3.0 to 13.0.

What can we conclude?

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

Answers

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

[tex]n_1 [/tex] represent the sample after the preparation

[tex]n_2 [/tex] represent the sample without preparation  

[tex]\bar x_1 =678[/tex] represent the mean sample after preparation

[tex]\bar x_2 =1837[/tex] represent the mean sample without preparation

[tex]s_1 =197[/tex] represent the sample deviation after preparation

[tex]s_2 =328[/tex] represent the sample deviation without preparation

[tex]\alpha=0.1[/tex] represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex]

The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

[tex]SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}[/tex]  

After replace in the formula for the confidence interval we got this:

[tex]3.0 < \mu_1 -\mu_2 <13.0 [/tex]

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

A researcher wants to estimate the mean weekly family expenditure on clothing purchases and maintenance. First, she needs to determine the number of families that must be sampled, in order to estimate the mean weekly expenditure on clothing to within $15 at alpha = 0.01. From prior similar studies she is confident that the standard deviation of weekly family expenditures on clothing purchases and maintenance is $125. The required sample size to be taken is___________.

A. 21
B. 22
C. 461
D. 460
E. 460.46

Answers

Answer: C. 461

Step-by-step explanation:

We know that the formula to find the sample size is given by :-

[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]

, where [tex]\sigma[/tex] = Population standard deviation from prior study.

E = margin of error.

z* = Critical value.

As per given , we have

[tex]\sigma=125[/tex]

E= 15

Significance level : [tex]\alpha=0.01[/tex]

Critical value (Two tailed)=[tex]z^*=z_{\alpha/2}=z_{0.005}=2.576[/tex]

Now , Required sample size = [tex]n=(\dfrac{(2.576)\cdot 125}{15})^2[/tex]

[tex]n=(21.4666666667)^2\\\\ n=460.817777779\approx461[/tex] [Round to next integer]

Hence, the required sample size to be taken is 461.

Correct answer = C. 461

Determine the point estimate of the population mean and margin of error for the confidence interval.Lower bound is 22, upper bound is 28The point estimate of the population mean is ___The margin of error for the confidence interval is ___

Answers

Answer: The point estimate of the population mean is 25.

The margin of error for the confidence interval is 3 .

Step-by-step explanation:

The confidence interval for population mean is given by :-

[tex](\overline{x}-E , \overline{x}+E)[/tex] , here [tex]\overline{x}[/tex] is the point estimate of the population mean and E is the margin of error .

As per given , we have

Lower bound of CI = [tex]\overline{x}-E =22[/tex]   (1)

Upper bound of CI =  [tex]\overline{x}+E =28[/tex]  (2)

Add (1) and (2) , we get

[tex]2\overline{x}=50\\\Rightarrow\ \overline{x}=25[/tex]

Subtract (1) from (2) , we get

[tex]2E=6\\\Rightarrow\ E=3[/tex]

Hence, the point estimate of the population mean is 25.

The margin of error for the confidence interval is 3 .

Answer:

1. The point estimate for population mean is 25.

2)

[tex]\text{Margin of error} = \pm 3[/tex]

Step-by-step explanation:

We are given the following information in the question:

Confidence interval: (22,28)

Confidence interval is calculated as:

[tex]\text{Sample mean }\pm \text{ Margin of error}[/tex]

Thus, we can write the equations:

[tex]\bar{x} - \text{Margin of error} = 22\\\bar{x} + \text{Margin of error} = 28[/tex]

1) The point estimate of the population mean

To calculate the point estimate of the population mean we solve the two equations, to find the sample mean

Adding the two equations we get:

[tex]2\bar{x} = 22+ 28 = 50\\\bar{x} = 25[/tex]

Thus, the point estimate for population mean is 25.

2) The margin of error for the confidence interval

Putting the values from the equation, we get:

[tex]\text{Margin of error} = 28 - 25 = 3[/tex]

Thus, the margin of error f the given confidence interval is

[tex]\text{Margin of error} = \pm 3[/tex]

Shane receives an hourly wage of $30.40 an hour as an emergency room nurse. How much does he make if he works 30 hours during the day at the normal work rate?​

Answers

Multiply his rate by number of hours:

30.40 x 30 = $912

Final answer:

Shane would make $912 if he works 30 hours at his normal work rate.

Explanation:

To calculate how much Shane makes if he works 30 hours at an hourly wage of $30.40, we can multiply his hourly wage by the number of hours he works:

$30.40/hour x 30 hours = $912

Therefore, Shane would make $912 if he works 30 hours at his normal work rate.

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Sydney is playing a game on a map with a marked grid. To start, she places a game piece at (1, 4). She moves 3 units east and 2 units north and places a second game piece at (4, 6). She continues playing game pieces so that each piece is 3 units east and 2 units north in relation to the previous piece. What is the location of the third game piece she places?

Answers

Answer:

(7, 8)

Step-by-step explanation:

If Sidney places each piece 3 units east and 2 units north of the last one, we can figure out which coordinate comes first by subtracting the position of the second piece (S) by the position of the first piece (F):

[tex]S - F = (4,6) - (1,4) = (3,2)[/tex]

We can conclude that coordinates are in the (East, North) format.

Therefore, the location of the third piece (T) is:

[tex]T = S(E,N) +(3,2)= (4,6)+(3,2)\\T = (7,8)[/tex]

All areas of the country use a BAC of 0.100.10 ​g/dL as the legal intoxication level. Is it possible that the mean BAC of all drivers involved in fatal accidents who are found to have positive BAC values is less than the legal intoxication​ level? a. No, it is not possible.b. Yes, and it is highly probable.c. Yes, but it is not likely.

Answers

Answer:

The answer is b): Yes, and it is highly probable.

Step-by-step Explanation:

Yes, it is highly probable that the mean BAC of all drivers involved in fatal accidents and found to have positive BAC values, is less than the legal intoxication level because blood alcohol concentration (BAC) level is not the most/only factor that determines fatal accidents. Other types of human errors are the main causes of fatal accidents; they include: side distractions, avoiding the use of helmets and seat belts, over-speeding, beating traffic red lights, overtaking in an inappropriate manner, using the wrong lane, etc. Most times, these errors are not caused by BAC levels.

The Chapin Social Insight Test evaluates how accurately the subject appraises other people. In the reference population used to develop the test, scores are approximately normally distributed with mean 25 and population standard deviation five. The range of possible scores is between 0 to 41. Determine the standardized value (z-value) for the score of 28.

Answers

Answer: 0.6

Step-by-step explanation:

Formula to get the standardized value :

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]  (1)

, where x= random variable , [tex]\mu[/tex] = Population mean and [tex]\sigma[/tex] = population standard deviation.

As per given , we have

[tex]\mu=25[/tex]

[tex]\sigma=5[/tex]

To find standardized value for the score of x=28

[tex]z=\dfrac{28-25}{5}=0.6[/tex]   (substitute values in (1))

Hence, the standardized value for the score of 28 is 0.6 .

The amount of water in a bottle is approximately normally distributed with a mean of 2.55 liters with a standard deviation of 0.035 liter. Complete parts (a) through (d) below. a. What is the probability that an individual bottle contains less than 2.52 liters? Round to three decimal places as needed.) b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded) c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.) d. Explain the difference in the results of (a) and (c)

Answers

Answer:

a) [tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]

b) [tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]

c) [tex]P(\bar X <2.52) = P(Z<-4.286)=0.000[/tex]

d) For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.

If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]

Let X the random variable that represent the amount of water in a bottle of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(2.55,0.035)[/tex]

a. What is the probability that an individual bottle contains less than 2.52 liters?

We are interested on this probability

[tex]P(X<2.52)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<2.52)=P(\frac{X-\mu}{\sigma}<\frac{2.52-\mu}{\sigma})[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196[/tex]

b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded)

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

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

On this case  [tex]\bar X \sim N(2.55,\frac{0.035}{\sqrt{4}})[/tex]

The z score on this case is given by this formula:

[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{4}}}=-1.714[/tex]

For this case we can use a table or excel to find the probability required:

[tex]P(\bar X <2.52) = P(Z<-1.714)=0.043[/tex]

c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.)

The z score on this case is given by this formula:

[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

[tex]z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{25}}}=-4.286[/tex]

For this case we can use a table or excel to find the probability required:

[tex]P(\bar X <2.52) = P(Z<-4.286)=0.0000091[/tex]

d. Explain the difference in the results of (a) and (c)

For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.

If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.  

Final answer:

The problem involves using z-scores, normal distribution, central limit theorem and law of large numbers for probability calculations related to the amount of water in sampled bottles. The increase in sample size from 1 to 25 should show a higher probability for the sample's mean to be close to the population mean.

Explanation:

In this problem, we will apply the concept of the normal distribution and the central limit theorem to calculate probabilities and compare results. First, we will calculate the z-scores for (a), (b), and (c). The z-score formula is Z = (X - μ) / σ. In (a), X is 2.52 liters, μ (mean) is 2.55 liters, and σ (standard deviation) is 0.035 liters.

In (b) and (c), the σ of a sample mean is σ/√n, where n is the sample size (4 in (b) and 25 in (c)).

Next, we look up the z-score in the z-score table (or use a normal distribution calculator) to get the probabilities.

To answer part (d), probabilistically, as sample size increases, the sample mean tends to get closer to the population mean according to the law of large numbers. Hence, the probabilities in (c) should be higher than (a).

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what is the function equation in function notation?
enter your answer in the box show your work. explain how you found numbers for the slope and y intercept ​

Answers

Answer:

[tex] y= 2x +1[/tex]

By direct comparison we can see that m =2 (slope) and b =1 (intercept)

Step-by-step explanation:

Assuming the following function:

[tex] y-2x =1[/tex]

We want to find the following general equation for a linear model:

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

On this case we just need to add 2x on both sides of the original equation and we got:

[tex] y= 2x +1[/tex]

By direct comparison we can see that m =2 and b =1

[tex] mx + b = 2x +1[/tex]

[tex]mx=2x , m =2[/tex]

[tex] b =1[/tex]

the value of m on this case represnt the slope and b the intercept.

The slope is defined by the following formula:

[tex] m =\frac{\Delta y}{\Delta x}[/tex]

And is the interpretation is the rate of change of y respect to x, can be positive or negative. Or the increase/decrease of y when x increase 1 unit.

And the value b=1 represent the y intercept, that means if x=0 then y =1

In which of the following situations would the use of sampling be most appropriate?Multiple Choice
a. The need for precise information is less important.
b. The number of items comprising the population is smaller.
c. The likelihood of selecting a representative sample is relatively low.
d. The use of sampling would be appropriate in all of these situations.

Answers

Final answer:

Sampling is most appropriate

option c

Explanation:

In the given multiple choice situations, the use of sampling would most probably be appropriate in the situation where 'The need for precise information is less important'. Sampling is a procedure of statistical analysis where a predetermined number of observations are taken from a larger population. It is typically used when it is impracticable or expensive to study the entire population. It's an ideal method when having precise and exact information is not as crucial, however, the results will be sufficient to make accurate predictions or conclusions about a larger group.

On the other hand, sampling may not be substantially beneficial for smaller populations or where the likelihood of selecting a representative sample is low because the results might be skewed or non-representative of the larger group. In such conditions, examining the entire population or utilizing another form of data collection would be more advantageous.

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A study was conducted to examine whether the proportion of females was the same for five groups (Groups A, B, C, D, and E). How many degrees of freedom would the χ2 test statistic have when testing the hypothesis that the proportions in each group are all equal?a. 4 b. 1 c. 0.20 d. 5

Answers

Answer: a. 4

Step-by-step explanation:

The number of degrees of freedom represents the number of data-values in the evaluation of a test-statistic that are independent to vary.

The degree of freedom for chi-square test = n-1 , n= Sample size.

Given : A study was conducted to examine whether the proportion of females was the same for five groups (Groups A, B, C, D, and E).

i.e. n= 5

Then, the number of degrees of freedom the χ2 test statistic have when testing the hypothesis that the proportions in each group are all equal = 5-1=4

Hence, the correct answer is OPTION a. 4 .

In a study of annual salaries of employees, random samples were selected from two companies to test if there is a difference in average salaries. For Company "X", the sample was size 65, the sample mean was $47,000 and the population standard deviation is assumed to be $11,000. For Company "Y", the sample size was 55, the sample mean was $44,000 and the population standard deviation is assumed to be $10,000. Test for a difference in average salaries at a 5% level of significance. What is your conclusion?

Answers

Answer:

Step-by-step explanation:

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