According to a 2010 study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction. That is, these are tweets that are not replied to or retweeted (Sysomos website, January 5, 2015).
Suppose we randomly select 100 tweets.
(a) What is the expected number of these tweets with no reaction?
(b) What are the variance and standard deviation for the number of these tweets with no reaction?

Answer:

6 Full Time Employees

Step-by-step explanation:

It takes 20 minutes to code each record.

Therefore, in 1 hour, number of records per coder = 60/20 = 3 records

Since each coding professional works 7.5 productive hours per day.

Number of Records Per Day for each coder

=(3X7.5) = 22.5 records

Assume a 5-Week Day

Number of Records that must be treated = 600/5 = 120 records a day

Number of FTE Employees needed

= 120 / 22.5 = 5.33

The employer will require at least 6 FTE since Number of Employees is a discrete data.

Final answer:

The coding supervisor will need approximately 5.33 FTEs to code 600 discharges per week, considering the given average time to code each record and daily productive hours. Since a fraction of an FTE isn't practical, the supervisor will need to employ 6 full-time coding professionals.

Explanation:

To determine the number of Full-Time Equivalents (FTEs) needed to code 600 discharges per week, we first need to calculate the total time required to code all records and then divide that time by the productive hours each coding professional works per day.

First, calculate the total weekly coding time: 600 discharges × 20 minutes per discharge = 12,000 minutes per week.

Then convert the total weekly coding time to hours: 12,000 minutes / 60 = 200 hours per week needed for coding.

Considering each coding professional works 7.5 productive hours per day, let's find out the daily capacity for one FTE: 7.5 hours per day × 5 days per week = 37.5 hours per week.

Now, divide the total weekly hours needed by the weekly capacity of one FTE: 200 hours per week / 37.5 hours per FTE per week = approximately 5.33 FTEs.

Therefore, the coding supervisor will need to employ roughly 5.33 FTEs to code 600 discharges per week, given the average coding time and productive hours per day. Since you can't have a fraction of an FTE in reality, the supervisor would need to round up to employ 6 full-time coding professionals.

Final answer:

To calculate how long it will take to double an investment with a 5.5% annual interest rate, you can use the Rule of 72, which gives you approximately 13.09 years for doubling the investment.

Explanation:

To determine how long it will take to double your money with an annual interest rate of 5.5%, you can use the Rule of 72. The Rule of 72 is a simple formula to estimate the number of years required to double the invested money at a given annual fixed interest rate. You divide 72 by the annual interest rate.

In this situation, divide 72 by 5.5 to find out how many years it will take to double the investment:

72 / 5.5 = 13.09 years

Thus, it will take approximately 13.09 years to double your money at an annual interest rate of 5.5%, when the interest is compounded annually. This is a rough approximation and the actual number might vary slightly depending on the compounding method used by the account.

Answer:

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

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex]  the deviation

We know that the z score is given by:

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

And by properties the value that separate the half  area on one side and half is on the other is z=0, since we have this:

[tex] P(Z<0) =0.5[/tex]

[tex] P(Z>0)=0.5[/tex]

So then the correct answer for this case would be z =0.00

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

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

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex]  the deviation

We know that the z score is given by:

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

And by properties the value that separate the half  area on one side and half is on the other is z=0, since we have this:

[tex] P(Z<0) =0.5[/tex]

[tex] P(Z>0)=0.5[/tex]

So then the correct answer for this case would be z =0.00

The z-score that divides the standard normal distribution into two equal halves is 0.00 as the mean of this distribution is also 0. Z-scores are standardized values expressing how many standard deviations a value is above or below the mean and can be determined using a Z-Table of Standard Normal Distribution.

The value of z that divides the standard normal distribution so that half the area is on one side and half is on the other is 0.00. The standard normal distribution is a symmetrical distribution where half of the values are less than the mean and half of the values are greater than the mean. In a standard normal distribution, the mean is 0. Hence, the z-score that will divide the distribution into half is also 0.

A z-score is a standardized value, expressing how many standard deviations a value is above or below the mean. These scores are particularly useful when comparing values from different data sets that have different means and standard deviations, such as exam scores from different classes or schools. They help us understand whether a particular score is common, or exceptionally high or low.

To find this value, you can use a Z-Table of Standard Normal Distribution, which shows the cumulative probability of a random variable from a standard normal distribution (with mean 0 and standard deviation 1), as being less than a certain value.

https://brainly.com/question/30390016

#SPJ12

Answer:

(a) The total number of combinations that can be applied for making potato chips is 36.

(b) The number of combinations that will be used for each type of oil is 12.

(c) Permutations are not an issue because order does not matter.

Step-by-step explanation:

The effects of cooking temperature, cooking time and cooking oil is studied for making potato chips.

The number of different temperatures applied is, n (T) = 3.

The number of different times taken is, n (t) = 4.

The number of different oils used is, n (O) = 3.

If an assignment can be done in n₁ ways and if for this assignment another assignment can be done in n₂ ways then these two assignments can be performed in (n₁ × n₂) ways.

(a)

Compute the total number of combinations that can be applied for making potato chips as follows:

Total number of combinations for making chips = n (T) × n (t) × n (O)

                                                                               [tex]=3\times4\times 3\\=36[/tex]

Thus, the total number of combinations that can be applied for making potato chips is 36.

(b)

To make potato chips 4 different temperatures are used and 3 different oils are used.

Each of the oil type is cooked in 4 different temperatures.

So the number of ways to select each oil type is,

n (T) × n (O) = [tex]4\times3=12[/tex]

Thus, the number of combinations that will be used for each type of oil is 12.

(c)

Permutation is the arrangement of objects in a specified order.

Since in this case ordering of the the three effects, i.e. temperature. time and oil type is not important, permutations are not an issue.

Answer:

a) Total 16 possibilities

MMMM

FFFF

MMMF

MMFM

MFMM

FMMM

FFFM

FFMF

FMFF

MFFF

MMFF

MFMF

MFFM

FFMM

FMMF

FMFM

b) P(MMMM) = 1/16

Final answer:

The equally likely gender combinations for 4 children are listed by considering all possibilities, including MMMM, MMMF, and so on. The probability that all 4 children are male is 0.0625, or 1/16 as a fraction. The probability of having at least one female child is the complement, 0.9375 or 15/16 as a fraction.

Explanation:

Listing the Equally Likely Gender Combinations

To list the equally likely events for the gender of the 4 children in a family where 'M' represents a male child and 'F' represents a female child, consider all possible combinations. These combinations are: MMMM, MMMF, MMFM, MMFF, MFMM, MFMF, MFFM, MFFF, FMMM, FMMF, FMFM, FMFF, FFMM, FFMF, FFMM, FFFF.

Probability of All Male Children

To calculate the probability that all 4 children are male, note that the events are independent, and the probability of each child being male is 0.5. Since the events are independent, multiply the probabilities for each child: 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 or 1/16 as a fraction.

Alternatively, the complement of having all male children is having at least one female child. The probability of at least one female child can be found by subtracting the probability of all male children from 1: 1 - 0.0625 = 0.9375 or 15/16 as a fraction.

Answer:

[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]

[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]

The 94% confidence interval would be given by (0.283;0.361)

Step-by-step explanation:

Notation and definitions

[tex]n=500[/tex] random sample taken

[tex]\hat p=0.322[/tex] estimated proportion of people between 18 to 34 who live with their parents

[tex]p[/tex] true population proportion of people between 18 to 34 who live with their parents  

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

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

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

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 94% of confidence, our significance level would be given by [tex]\alpha=1-0.94=0.06[/tex] and [tex]\alpha/2 =0.03[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.88, z_{1-\alpha/2}=1.88[/tex]

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

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

If we replace the values obtained we got:

[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]

[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]

The 94% confidence interval would be given by (0.283;0.361)

Final answer:

Explanation of expected total times for Carol and the last customer, along with the probability of being last.

Explanation:

(a) Expected total time for Carol to complete her businesses:

Carol will go to the first available teller, who has a service time of 3 minutes (or service rate of 20 per hour).

Hence, the expected total time for Carol to complete her businesses is 3 minutes.

(b) Expected total time until the last customer leaves:

Calculating the expected total time for each customer to complete their businesses gives Anne: 3 minutes, Betty: 6 minutes.

Therefore, the expected total time until the last customer leaves is 6 minutes.

(c) Probability of being the last to leave:

The last to leave will be the customer with the highest expected service time, which is Betty (6 minutes).

The probability for Anne, Betty, and Carol to be the last one to leave is: Anne 0, Betty 1, Carol 0.

(a) Expected total amount of time for Carol to complete her business: the expected total amount of time for Carol to complete her business is 2 minutes.

(b) Expected total time until the last of the three customers leaves:the expected total time until the last of the three customers leaves is 8 minutes.

c) Probability for Anne, Betty, and Carol to be the last one to leave:the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.

Let's solve each part of the problem:

(a) Expected total amount of time for Carol to complete her business:

Since Carol waits for the first available teller, we need to consider the minimum of the service times of the two tellers. Let's denote:

- ( X ) as the service time for the first teller (with mean three minutes).

- ( Y ) as the service time for the second teller (with mean six minutes).

The minimum of two exponentially distributed random variables with means [tex]\( \mu_1 \) and \( \mu_2 \)[/tex] is also exponentially distributed with mean \( \frac{1}[tex]\( \frac{1}[/tex][tex]{\lambda_1 + \lambda_2} \), where \( \lambda_1 \) and \( \lambda_2 \)[/tex] are the rates.

Given:

- The rate for the first teller is [tex]\( \lambda_1 = \frac{1}{3} \)[/tex] per minute.

- The rate for the second teller is [tex]\( \lambda_2 = \frac{1}{6} \)[/tex] per minute.

So, the rate for the minimum service time for Carol is [tex]\( \lambda = \lambda_1 + \lambda_2 = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \)[/tex] per minute.

The expected total amount of time for Carol to complete her business is the reciprocal of the rate, which is [tex]\( \frac{1}{\lambda} = \frac{1}{\frac{1}{2}} = 2 \) minutes.[/tex]

Therefore, the expected total amount of time for Carol to complete her business is 2 minutes.

(b) Expected total time until the last of the three customers leaves:

The total time until the last customer leaves is the maximum of the service times of all three customers. Since Anne and Betty go directly into service, their service times are independent of each other and of Carol's service time.

Let's denote:

- ( Z ) as the service time for Carol (which we found to be 2 minutes).

- ( W ) as the maximum of the service times for Anne and Betty.

Since ( W ) is the maximum of two exponentially distributed random variables with the same rate, it follows that ( W ) is also exponentially distributed with the same rate.

So, the expected total time until the last of the three customers leaves is the sum of the expected service time for Carol and the expected maximum service time for Anne and Betty, which is [tex]\( 2 + 6 = 8 \)[/tex]minutes.

Therefore, the expected total time until the last of the three customers leaves is 8 minutes.

(c) Probability for Anne, Betty, and Carol to be the last one to leave:

Since Anne and Betty go directly into service, their service times are independent of each other and of Carol's service time. Therefore, each of them has an equal chance of being the last one to leave.

So, the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.

Therefore, the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.

Answer:

The inspector's claim has strong statistical evidence.

Step-by-step explanation:

To answer this we have to perform a hypothesis test.

The inspector claimed that the actual proportion of code violations is greater than 0.07, so the null and alternative hypothesis are:

[tex]H_0: \pi\leq0.07\\\\H_a: \pi>0.07[/tex]

We assume a significance level of 0.05.

The sample size is 200 and the proportion of the sample is:

[tex]p=\frac{23}{200}= 0.115[/tex]

The standard deviation is

[tex]\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }= \sqrt{\frac{0.07*0.93}{200}}=0.018[/tex]

The z-value can be calculated as

[tex]z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.115-0.07-0.5/200}{0.018} =\frac{0.0425}{0.018}=2.36[/tex]

The P-value for this z-value is P=0.00914.

This P-value is smaller than the significance level, so the effect is significant and the null hypothesis is rejected.

The inspector's claim has strong statistical evidence.

Answer:

Yes, because the p-value is 0.0062

Step-by-step explanation:

I looked it up on like 5 other websites and they all said this was the answer. I'm way too lazy to do it on my own.

Answer:

The total number of possible complete staffs you could hire is 350.

Step-by-step explanation:

The order that the software engineers are is not important. For example, hiring John and Laura as the software engineers is the same as hiring Laura and John. The same applies for the computer engineers. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

What is the total number of possible complete staffs you could hire?

2 software engineers from a set of 5

3 computer engineers from a set of 7

So

[tex]T = C_{5,2}*C_{7,3} = \frac{5!}{2!3!}*\frac{7!}{3!4!} = 10*35 = 350[/tex]

The total number of possible complete staffs you could hire is 350.

Answer:

Pat's score therfore was the 20th percentiles

Step-by-step explanation:

When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.

In this problem, we have that:

34 scores lower than Pat's.

[tex]\frac{34}{170} = 0.20[/tex]

So Patrick's score was the 20th percentile.

Final answer:

To find the probability that the size of the family is between 2 and 5 inclusive, calculate the proportion of the given sample that falls within that range. In this case, the probability is approximately 0.792.

Explanation:

To find the probability that the size of the family is between 2 and 5 inclusive, we need to calculate the proportion of the given sample that falls within that range. In the provided sample of college math class, the family sizes are:

We can see that there are 19 family sizes falling between 2 and 5, inclusive. Therefore, the probability is:

Probability = Number of favorable outcomes / Total number of outcomes

Probability = 19 / 24 = 0.792

So, the probability that the size of the family is between 2 and 5 inclusive is approximately 0.792.

Part(a):

Then the outcomes can be,

[tex]\{(1,1,1)(1,1,0)(1,0,1)(0,1,1)(1,0,0)(0,1,0)(0,0,1)(0,0,0) \}[/tex]

Part(b):

The correct option is (a).

Part(c):

The outcomes are,

[tex](1,1,1):x=3\\(1,1,0):x=2\\(1,0,1):x=2\\(0,1,1):x=2\\(1,0,0):x=1\\(0,1,0):x=1[/tex]

Experimental probability, also known as Empirical probability, is based on actual experiments and adequate recordings of the happening of events.

Let the ordered pair (a,b,c) denote the outcome with a,b,c taking either 1 if the position is offered

Or 0  if the position is not offered.

Then the outcomes can be,

[tex]\{(1,1,1)(1,1,0)(1,0,1)(0,1,1)(1,0,0)(0,1,0)(0,0,1)(0,0,0) \}[/tex]

Let, [tex]x[/tex] is number probability function offers made.

The variable is discrete taking 0 or 1 or 2 or 3 as values.

So, the correct option is (a)

The outcomes are,

[tex](1,1,1):x=3\\(1,1,0):x=2\\(1,0,1):x=2\\(0,1,1):x=2\\(1,0,0):x=1\\(0,1,0):x=1[/tex]

Learn more about the topic experimental outcomes:

https://brainly.com/question/21493171

There are 8 experimental outcomes for the interviews. The variable representing the number of offers is discrete, taking values between 0 and 3. For the given subsets of experimental outcomes, the value of this variable is the number of 'Y' present.

A) The total number of experimental outcomes is calculated by the rule of product. There are two possible results (Yes or No) for each of the three interviews. So the number of experimental outcomes is 2*2*2, which is 8.

B) The variable x, which is the number of students receiving an offer, is discrete. A variable is discrete if it can only take on a finite or countable number of values. In this case, x can take on only four possible values (0, 1, 2, or 3), depending on the number of students receiving an offer.

C) The value of the random variable X for the subset of experimental outcomes is as follows:
(Y,Y,Y) - X = 3
(Y,N,Y) - X = 2
(N,Y,Y) - X = 2
(N,N,Y) - X = 1
(N,N,N) - X = 0

The above experimental outcomes show the different possible results of the three interviews, with N signifying no job offer and Y signifying a job offer after the interview.

https://brainly.com/question/31538429

#SPJ3

Answer:

90 percent confidence interval = [72.674 ,77.326]

Step-by-step explanation:

We are given that weight of a product is measured in pounds.

A random sample of 50 units is taken from a recent production. The sample yielded y bar = 75 lb, and we know that [tex]\sigma^{2}[/tex] = 100 lb.

The Pivotal quantity for 9% confidence interval is given by;

              [tex]\frac{Ybar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, Y bar = sample mean = 75

                [tex]\sigma[/tex]  = population standard deviation = 10

                 n = sample size = 50

So, 90% confidence interval for population mean,  is given by;

P(-1.6449 < N(0,1) < 1.6449) = 0.90

P(-1.6449 < [tex]\frac{Ybar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.6449) = 0.90

P(-1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]{Ybar - \mu}[/tex] < 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.90

P(Y bar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < Y bar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.90

90% confidence interval for [tex]\mu[/tex] = [ Y bar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] , Y bar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ]

                                              = [ 75 - 1.6449 * [tex]{\frac{10}{\sqrt{50} }[/tex] , 75 + 1.6449 * [tex]{\frac{10}{\sqrt{50} }[/tex] ]

                                              = [ 72.674 , 77.326 ]

Therefore, 90% confidence interval for population mean is [72.674 ,77.326] .

Answer:

a) Probability of no broken bats in a game P(X=0) = 0.3678

b) Probability of at-least broken bats in a game P(X≥2) = 0.2644

Step-by-step explanation:

we will use Poisson distribution

P(X=x) = e^(-λ) λ^ r/r!

a) Probability of no broken bats in a game P(X=0) = e^-1

                                                                                   = 0.3678

b) Probability of at-least two broken bats in a game

                                                                     P(X≥2) = 1-([p(x=0)+P(x=1)]

                                                                               = 1- (e^(-1)+e^(-1))

                                                                                  = 1-(0.3678+0.3678)

                                                                   P(X≥2)= 1- 0.7356

                                                                  P(X≥2)= 0.2644

Answer:

1. 90% 2. 10% 3. 50%

Step-by-step explanation:

Standard Deviation (σ) = 50 days

Average/Mean (μ) = 300days

Probability that it would last more than 300 days = P(Bulb>300 days)

We will assume there are 365 days in a year.

P(Bulb>300 days)  implies that the bulb would

Using the normal equation;

z = standard/normal score = (x-μ)/σ where x is the value to be standardized

P(Bulb>300 days) implies x = 365 days

Therefore z = (365-300)/50 = 1.3

Using the normal graph for z=1.3, probability = 90%

2. P(Bulb<300 days) = 1 - P(Bulb>300 days)\

P(Bulb<300 days) = 1 - 0.9

P(Bulb<300 days)  = 10%

3. P(Bulb=300 days) implies z=0 since x=300

Using the normal graph for z=0, probability =50%

1. The probability that an Acme light bulb will last more than 300 days is 50%. 2. The probability that an Acme light bulb will last less than 300 days is 50%. 3. The probability that an Acme light bulb will last exactly 300 days is zero.

1. To find the probability that an Acme light bulb will last more than 300 days, we need to determine the area under the normal distribution curve to the right of 300 days. We use the z-score formula: z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Substituting the values, we get: z = (300 - 300) / 50 = 0. The area to the right of 300 days is equal to the area to the left of 0. Using a standard normal distribution table, we find that the area to the left of 0 is 0.5. Therefore, the probability that an Acme light bulb will last more than 300 days is 0.5 or 50%.

2. To find the probability that an Acme light bulb will last less than 300 days, we need to determine the area under the normal distribution curve to the left of 300 days. Using the same z-score formula, we get: z = (300 - 300) / 50 = 0. The area to the left of 0 is 0.5. Therefore, the probability that an Acme light bulb will last less than 300 days is 0.5 or 50%.

3. To find the probability that an Acme light bulb will last exactly 300 days, we need to determine the area under the normal distribution curve at 300 days. Since the normal distribution is continuous, the probability of any single value is zero. Therefore, the probability that an Acme light bulb will last exactly 300 days is zero.

Answer:

6435 different communities.

Step-by-step explanation:

This is a combination problem,

We need to know how many ways 8 species out of 15 can be selected for the new small island

15C8 = 15!/(15-8)!(8!) = 15!/(8!)(7!) = 6435 different communities.

Answer:

The standard deviation is 5.83 kg

Step-by-step explanation:

Variance and Standard Deviation

Given a data set of random values, the variance is defined as the average of the squared differences from the mean. We use this formula to calculate the variance:

[tex]\displaystyle \sigma^2=\frac{\sum(x_i-\mu)^2}{n}[/tex]

Where [tex]\mu[/tex] is the mean of the values xi (i=1 to n), n the total number of values.

The standard deviation is known by the symbol [tex]\sigma[/tex] and is the square root of the variance.

We are given the value of the variance:

[tex]\sigma^2=34\ kg^2[/tex]

We now compute the standard deviation

[tex]\sigma=\sqrt{34\ kg^2}=5.83\ kg[/tex]

The standard deviation is 5.83 kg

Answer and Step-by-step explanation:

Obviously addition is closed for R including both infinities

As t+infty = infty +t = infty and t+(-infty) =(-infty)+t =(-infty)

and inverse are -infty for infty and infty for -infty

Hence a group under addition

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

But regarding multiplication

we can say 1/infty = 2/infty =0

Hence infinity*0 is not unique.

Also infinity and -infty do not have multiplicative inverse as there is no t such that t*infty = 1

Hence cannot be a vector space.

Answer:

a.

[tex]\bar p_1=0.05\\\bar p_2=0.067[/tex]

b-Check illustration  below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let [tex]p_1 \& p_2[/tex] denote processes 1 & 2.

For [tex]p_1[/tex]: T1=10,n1=200

For [tex]p_2[/tex]:T2=20,n2=300

Therefore

[tex]\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067[/tex]

b. To test for hypothesis:-

i.

[tex]H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05[/tex]

ii.For a two sample Proportion test

[tex]Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\[/tex]

iii. for [tex]\frac{\alpha}{2}=(-1.96,+1.96)[/tex] (0.5 alpha IS 0.025),

reject [tex]H_o[/tex] if[tex]|Z|>1.96[/tex]

iv. Do not reject [tex]H_o[/tex]. The noncomforting proportions are not significantly different as calculated below:

[tex]z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}[/tex]

z=-0.78

c.[tex](1-\alpha).100\%[/tex] for the p1-p2 is given as:

[tex](\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645 \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\[/tex]

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

A psychologist can use normative data, sequential design, or parent-report questionnaires to estimate the developmental score of a typical 4-year-old child without direct participation from children of that specific age.

To determine the developmental score of a typical 4-year-old child without having 4-year-old children participate in the study, a psychologist may use several statistical techniques grounded in developmental science. One method is to use normative data, which refers to average developmental milestones established through extensive research on children at various ages. By comparing existing data on children older and younger than 4 years, the psychologist can interpolate or approximate the developmental score for the typical 4-year-old based on these norms.

Another approach involves a sequential design, where overlapping cohorts of different age groups are studied over time. This method can help infer the developmental score of 4-year-olds by comparing changes in development across the different cohorts that are adjacent in age. Additionally, parent-report questionnaires can be used to gather data from parents or guardians about typical developmental milestones, which can then be statistically analyzed to estimate the developmental score of children in the age group of interest.

Answer:

a. P(tips from 60 parties > $600)=0.461

b. The total amount that she earns on the best 5% of such weekends is at least $19.43.

Step-by-step explanation:

a. To earn $600 in 60 parties means $10 per party in average.

If we assume a normal distribution of tips, we can calculate the z-value and its probability for this situation:

[tex]z=\frac{x-\mu}{\sigma}=\frac{10.00-9.40}{6.10}=\frac{0.60}{6.10}= 0.098\\\\P(x>10)=P(z>0.098)=0.461[/tex]

There is a probability of 46% that she earns at least $600 over a weekend of work.

b. The best 5% of the weekends corresponds to:

[tex]P(x>x_1)=0.05[/tex]

This probability (5%) corresponds to a z-value of z=1.6449.

In tips, this value represents:

[tex]x=\mu+z*\sigma=9.40+1.6449*6.10=9.40+10.03=19.43[/tex]

The total amount that she earns on the best 5% of such weekends is at least $19.43.

Answer:

6/8

Step-by-step explanation:

The Fraction of mowers that fail the functional performance test is the ratio of the number of mowers that fail to the total number of mowers. Hence the fraction of mowers that fail [tex] \frac{9}{500}[/tex]

The probability of failure can be defined thus :

Therefore, the probability of having x failures in the next 100 mowers is (0.018x)

Learn more :https://brainly.com/question/18405415

Answer:

Since it's a removable discontinuity, we will remove the discontinuity by creating a new function defined by x=3;

So we have;

F(x) = {[x² - 2x - 3]/(x-3); x ≠ 3

         {4, x = 3

Step-by-step explanation:

I have attached my explanation as the system here is not allowing me to save my answer.

To remove the discontinuity at x=3 in the function f(x) = (x^2 - 2x - 3) / (x - 3), simplify the function to f(x) = x + 1, then calculate f(3) = 3 + 1 = 4. Defining f(3) = 4 makes the function continuous at x=3.

To remove the discontinuity of a function such as f(x) = (x^2 - 2x - 3) / (x - 3), you need to find a value for f(3) that would make the function continuous at x = 3. First, let's manipulate the function:

The function f(x) = (x^2 - 2x - 3) / (x - 3) simplifies to f(x) = (x - 3)(x + 1) / (x - 3).

As you can see, the (x - 3) terms cancel out, leaving f(x) = x + 1.

However, this is only valid for x ≠ 3. To find f(3), you can now simply substitute 3 into the simplified function:

f(3) = 3 + 1 = 4.

So, if we define f(3) = 4, then the function becomes continuous at x = 3.

https://brainly.com/question/11001777

#SPJ3

Answer:

a.

[tex]P(X=3)=0.2903\\\\P(X \geq 3)=0.5801\\\\P(X\leq 3)0.7102[/tex]

b.

[tex]P(2\leq x\leq 4 )=0.7451[/tex]

c. mean=2.8

d . standard deviation=1.2961

Step-by-step explanation:

We determine that the accident rates follow a binomial distribution. The rate of success p=0.4 and sample n=7:

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}[/tex]

#the probability of exactly​ three;

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=3)={7\choose 3}0.4^3(0.6)^{4}\\\\=0.2903[/tex]

#At least(more than 2)

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 3)=1-P(X\leq 2)\\\\=1-{7\choose 0}0.4^0(0.6)^{7}-{7\choose 1}0.4^1(0.6)^{6}-{7\choose 2}0.4^2(0.6)^{5}\\\\=1-0.0280-0.1306-0.2613\\\\=0.5801[/tex]

#At most 3;

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X \leq 3)={7\choose 0}0.4^0(0.6)^{7}+{7\choose 1}0.4^1(0.6)^{6}+{7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}\\\\\\\=0.0280+0.1306+0.2613+0.2903\\\\=0.7102[/tex]

b. Between 2 and 4:

Using the binomial expression, this probability is calculated as:

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(2\leq x\leq 4 )={7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}+{7\choose 4}0.4^4(0.6)^{3}\\\\\\\\\=0.2613+0.2903+0.1935\\\\=0.7451[/tex]

Hence,the probability of between 2 and four is 0.7451

c. From a above, we have the values of n=7 and p=0.4.

-We substitute this values in the formula below to calculate the mean:

-The mean of a binomial distribution is calculated as the product of the probability of success by the sample size, mean=np:

[tex]\mu=np, n=7, p=0.4\\\\\mu=7\times 0.4\\\\=2.8[/tex]

Hence, the standard deviation of the sample is 2.8

d. From a above, we have the values of n=7 and p=0.4

--We substitute this values in the formula below to calculate the standard deviation

-The standard deviation a binomial distribution is given as:

[tex]\sigma={\sqrt {np(1-p)}\\\\=\sqrt{7\times 0.4\times 0.6}\\\\=1.2961[/tex]

Hence, the standard deviation of the sample is 1.2961

We can calculate the probabilities of having exactly 3 fatalities, at least 3 fatalities, and at most 3 fatalities using the formula. The mean and standard deviation of the random variable Y can be calculated using specific formulas for a binomial distribution.

To solve this problem, we will use the binomial probability formula. The probability of exactly 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by plugging in the values of n (total number of trials) and p (probability of success in a single trial) into the formula. The probability of at least 3, or more than 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by summing the probabilities of 3, 4, 5, 6, and 7 fatalities. The probability of at most 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by summing the probabilities of 0, 1, 2, and 3 fatalities. The mean of the random variable Y can be calculated by multiplying the number of trials (7) by the probability of success (0.4). The standard deviation of Y can be calculated using the formula for the standard deviation of a binomial distribution.

https://brainly.com/question/33993983

#SPJ3

Answer:

a) The 90% confidence interval is:

[tex]0.0003237 \leq \pi \leq 0.0003239[/tex]

b) Yes. Because the proportion of cancer rate for the cell phone users population is bigger than 0.00324%, which is over the rate of non-cell phone users (0.0251%).

Step-by-step explanation:

a) We will construct the 90% confidence interval based on the information given by the sample taken in this study.

The sample proportion is:

[tex]p=\frac{X}{n}=\frac{136}{420,026} =0.000324[/tex]

The standard deviation is estimated as:

[tex]\sigma=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.000324*0.999676}{420,026}}=\sqrt{7.7\cdot 10^{-10}} = 0.000028[/tex]

As the sample size is big enough, we use the z-score. For a 90% CI, the value of z is z=1.645.

The margin of error of the CI is:

[tex]E=z\sigma/\sqrt{n}=1.645* 0.000028 /\sqrt{420,026}\\\\E= 0.000046 /648= 0.00000007[/tex]

The 90% CI is:

[tex]p-z\sigma/\sqrt{n}\leq \pi\leq p+z\sigma/\sqrt{n}\\\\0.000324- 0.00000007 \leq\pi\leq 0.000324+ 0.00000007\\\\ 0.0003237 \leq \pi \leq 0.0003239[/tex]

Final Answer:

a. The 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, rounded to three decimal places, is approximately: [0.028%, 0.037%]
b. The known rate for those not using cell phones is 0.0251%.

Explanation:

a. To construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, we need to know the number of successes (in this case, the number of people who developed cancer), the total number of trials (the total number of cell phone users), and use the z-score that corresponds to the desired confidence level (90%).
Given:
Number of cell phone users (n) = 420,026
Number of cell phone users who developed cancer (x) = 136
Confidence level (CL) = 90%

First, we calculate the sample proportion (p):
p = x / n = 136 / 420,026 ≈ 0.000324

For a 90% confidence interval, the z-score associated with the two-tailed confidence level is approximately 1.645 (you would find this value in a z-score table or by using statistical software).

Next, we calculate the standard error (SE) for the proportion:
SE = sqrt((p(1 - p)) / n)
SE = sqrt((0.000324(1 - 0.000324)) / 420,026)
SE ≈ sqrt((0.000324 * 0.999676) / 420,026)
SE ≈ sqrt(0.00000032443 / 420,026)
SE ≈ sqrt(7.72476e-10)
SE ≈ 2.779e-5

Now, we calculate the margin of error (ME):
ME = z * SE
ME = 1.645 * 2.779e-5
ME ≈ 4.570635e-5

Finally, we construct the confidence interval:
Lower bound (LB) = p - ME
LB ≈ 0.000324 - 4.570635e-5
LB ≈ 0.000278

Upper bound (UB) = p + ME
UB ≈ 0.000324 + 4.570635e-5
UB ≈ 0.000370

So the 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, rounded to three decimal places, is approximately:
[0.028%, 0.037%]

b. To determine if cell phone users appear to have a rate of cancer that is different from the rate among those not using cell phones, we compare the confidence interval we just calculated with the known rate for non-cell phone users.
The known rate for those not using cell phones is 0.0251%.

Since the entire 90% confidence interval of [0.028%, 0.037%] for cell phone users is above 0.0251%, this suggests that the rate of cancer among cell phone users might be higher than the rate among non-users. However, this does not provide enough evidence to definitively conclude that cell phone use causes a higher rate of cancer since other factors might be at play and causation cannot be determined from this data alone. It is also important to note that the confidence interval is statistically very close to the known rate for non-users, so the difference might not be practically significant. Additionally, further research with more rigorous control conditions would be necessary to establish a cause-and-effect relationship.

Answer:

-8 - 3i

Step-by-step explanation:

To find the Complex Conjugate , Negate the term with i

Hopefully this helps.

Answer:

Research hypothesis

Step-by-step explanation:

A specific, clear, and testable proposition or predictive statement about the possible outcome of a scientific research study based on a particular property of a population, such as presumed differences between groups on a particular variable or relationships between variables is known as a research hypothesis  .

One of the most important steps in planning a scientific quantitative research study is specifying the research hypotheses. A prior expectation about the results of the study in one or more research hypotheses  is usually being stated by a  quantitative researcher before conducting the study, because the design of the research study and the planned research design is often determined by the stated hypotheses.

In the question, it is clearly seen that a prior expectation about the results of the study on the relationship between income and self-esteem was already suggested, before been tested. Thus, it is a research hypothesis

Answer:

(a) Null hypothesis: The new drug reduces the rate of first heart attacks by a percentage same as aspirin.

(b) Alternate hypothesis: The new drug reduces the rate of first heart attacks by a percentage greater than aspirin.

Step-by-step explanation:

(a) A null hypothesis is a statement from the population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

(b) An alternate hypothesis is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

Answer:

The First-In, First-Out (FIFO) inventory costing method assumes that the inventory items ordered first are the first ones sold.

Step-by-step explanation:

The First-In, First-Out inventory costing method assumes that the inventory items ordered first are the first sold. This is ideal for goods that are highly perishable, for example fresh milk. Since no figures or dates are given, we will assume that the month is March 2019 and use any figures to make the example.

Date Item      Quantity of stock Cost Price

01  Opening stock bought on Feb 28  10   100

05  Sale of 5 goods (cost is $10 each)  (5)   50

15 Purchase of stock (20 goods at $20 each) 20   400

25 Sale of 15 goods                     (15)   250  

(5 at $10 each & 10 at $20 each)

31 Closing Stock               10   200

       (20 goods bought on 15th - 10 goods sold on 25th)

The quantity on hand at the end of the month is 10 units.  

Total cost of goods on hand at end of the month = 10 units * $20 = 200.

Total cost of goods purchased during the month = $20 * 20 units = $400

Total cost of goods sold during the month = [($10 *5) + ($10 * 5)+ ($20 * 10)] = $200

Answer:

(a) 95% confidence interval for the population mean body temperature for healthy female is between a lower limit of 98.21 °F and an upper limit of 98.57 °F.

(b) The average is less than 98.6 °F

(c) Yes

Step-by-step explanation:

(a) Confidence interval = mean + or - Margin of Error (E)

mean = 98.39 °F

sd = 0.743 °F

n = 65

degree of freedom = n - 1 = 65 - 1 = 64

confidence level = 95%

t- value corresponding to 64 degrees of freedom and 95% confidence level is 1.9976.

E = t×sd/√n = 1.9976×0.743/√65 = 0.18 °F

Lower limit = mean - E = 98.39 - 0.18 = 98.21 °F

Upper limit = mean + E = 98.39 + 0.18 = 98.57 °F

95% confidence interval is between 98.21 °F and 98.57 °F.

(b) The average is less than 98.6 °F. The lower limit 98.21 °F and the upper limit 98.57 °F are both less than 98.6 °F

(c) It was valid to use the t-distribution approach to find the confidence Interval beci it gives a range of values for the population mean body temperature for healthy female.