A retail store owner offers a discount on product A and predicts that, the customers would purchase products B and C in addition to product A. Identify the technique used to make such a prediction. a. Data query b. Simulation c. Data mining d. Data dashboards

Answer:

114.3

Step-by-step explanation:

If a 70kg person ingests 2L of water per day containing 8 mg/L of  dinitrotoluene, the concentration of  dinitrotoluene on that person's body is:

[tex]C=2\frac{L}{day}*8\frac{mg}{L} *\frac{1}{70\ kg}\\C=0.22857\frac{mg}{kg-day}[/tex]

The hazard ratio is defined by dividing the intake dosage (C) by the reference dose (RfD)

[tex]H=\frac{0.22857}{2*10^{-3}}\\H=114.3[/tex]

The hazard ratio is most nearly 114.3.

Answer:

a. 4)

b. 4)

Step-by-step explanation:

Hello!

An observational study is one where the investigator has no control or intervenes on it. He just defines the variable of interest and merely collects and documents the information.

An experimental study or experiment is one where the investigator intervenes by defining the variable of interest and artificially manipulates the study factor. It also one of its characteristics the randomization of cases or subjects in groups (two or more, depending on what is the hypothesis of study).

(a) The U.S. Census Bureau tracks population age. In 1900, the percentage of the population that was 19 years old or younger was 44.4%. In 1930, the percentage was 38.8%; in 1970, the percentage was 37.9%; and in 2000, the percentage in the age group was down to 28.5% (The First Measured Century, T. Caplow, L. Hicks, B. J. Wattenberg).

In this item, the researcher merely obtained the records of the population that were 19 or younger over several years and compared the obtained percentages. This is a clear example of an observational study, the researcher did not manipulate any factor or variable, he just looked up the information and documented it.

(b) After receiving the same lessons, a class of 100 students was randomly divided into two groups of 50 each. One group was given a multiple-choice exam covering the material in the lessons. The other group was given an essay exam. The average test scores for the two groups were then compared.

In this item, the students were randomly assigned to one of two groups and each group was given a test, group one: multiple choice and group two: essay. This is an example of an experimental study, after the class, the researcher controlled the variable "type of test" giving one type to each group and the subjects were randomly selected fro the groups assuring that they will not receive biased results. And at the end of the experiment, the response variables "test scores" where compared.

I hope it helps!

Final answer:

The studies presented about U.S. Census Bureau population tracking and exam type comparison in a classroom represent an observational study and an experiment, respectively. The former is so because no treatment was applied, while the latter involved purposefully assigning different tests to observe outcomes.

Explanation:

When classifying the given studies as either an observational study or an experiment, it's important to understand the key characteristics of each. An observational study involves monitoring subjects without any intervention, while an experiment involves the deliberate imposition of a treatment to observe potential changes.

(a) The U.S. Census Bureau tracking population age is an example of an observational study because the data on the percentage of the population that was 19 years old or younger in various years was collected without any manipulation or treatment being applied to the subjects.

(b) Dividing a class into two groups and administering different types of exams constitutes an experiment. This is because a treatment (the type of exam) was deliberately imposed on the students to observe the difference in test scores, which we interpret as the outcome of interest.

Answer:

(a) Ratio of A to B = 17566 : 5827

(b) Ratio of B to A = 5827 : 17566

(c) A is 301.46% percent of B.

Step-by-step explanation:

We are given A = 52,698 number of deaths due to a deadly disease in the United States in 2005   and   B = 17,481 number of deaths due to the same disease in the United States in 2009.

(a) Ratio of A to B = [tex]\frac{A}{B}[/tex] = [tex]\frac{52,698}{17,481}[/tex] = 17566 : 5827

(b) Ratio of B to A = [tex]\frac{B}{A}[/tex] = [tex]\frac{17,481}{52,698}[/tex] = 5827 : 17566

(c) Let A is x% of B so the equation formed will be;

    A = x% of B

     52,698 = x% of 17481

  Therefore x = [tex]\frac{52,698}{17,481}*100[/tex] = 301.46%

Hence, A is 301.46% of B.

The ratio of A to B is 3:1, the ratio of B to A is 1:3, and A is approximately 301.45 percent of B.

a. Find the ratio of A to B:

We can find the ratio of A to B by dividing A by B, which gives us 52,698/17,481. Evaluating this division gives a ratio of approximately 3:1, meaning that for every 3 deaths in 2005, there was 1 death in 2009.

b. Find the ratio of B to A:

To find the ratio of B to A, we divide B by A, which gives us 17,481/52,698. Simplifying this division gives a ratio of approximately 1:3, which is the inverse of the previous ratio.

c. Complete the sentence: A is ____ percent of B:

To find the percentage of A relative to B, we divide A by B, then multiply by 100. Evaluating this division gives us (52,698/17,481) * 100, which is approximately 301.45%. Therefore, we can complete the sentence by saying 'A is approximately 301.45 percent of B'.

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Answer: 1

Step-by-step explanation:

If a random variable x is uniformly distributed in [a,b] the

Mean = [tex]\dfrac{a+b}{2}[/tex]

Standard deviation : [tex]\sqrt{\dfrac{(b-a)^2}{12}}[/tex]

Let x =  Times required for a cable company to fix cable problems

As per given.

x is  uniformly distributed between 40 minutes and 65 minutes.

Then , mean = [tex]\dfrac{65+40}{2}=52.5[/tex] minutes

Standard deviation : [tex]\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22[/tex]minutes

Consider , P (mean- 2(Standard deviation) < X < mean+2(Standard deviation) )

= P(52.5-2(7.22)< X <  52.5+2(7.22))

=P(38.06 <X < 66.94 ).

But x lies between 40 minutes and 65 minutes.

Also,   [40 minutes,  65 minutes]⊂ [38.06 minutes , 66.94 minutes]

Therefore ,P(38.06 <X < 66.94 ) =1

∴ The probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean is 1.

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean,

Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

z = (x - μ)/σ

where x is the raw score, μ is the mean and σ is the standard deviation.

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean.

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Answer:

y = - x + 5

Step-by-step explanation:

L(5.0), M(0,5)

y = mx + b

m = (5 - 0) / (0 - 5) = 5 / -5 = - 1

b = y - mx = 5 - ((-1) x 0) = 5

y = - x + 5

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

[tex]v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}[/tex]

Where:

[tex]x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval[/tex]

a. Let's find h(3) and h(4) using the data provided by the problem:

[tex]h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f[/tex]

The average velocity over the interval [3, 4] is :

[tex]v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s[/tex]

b. Let's find h(3.5) using the data provided by the problem:

[tex]h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f[/tex]

The average velocity over the interval [3, 3.5] is :

[tex]v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s[/tex]

c. Let's find h(3.1) using the data provided by the problem:

[tex]h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f[/tex]

The average velocity over the interval [3, 3.1] is :

[tex]v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s[/tex]

d. Let's find h(3.01) using the data provided by the problem:

[tex]h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f[/tex]

The average velocity over the interval [3, 3.01] is :

[tex]v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s[/tex]

e. Let's find h(3.001) using the data provided by the problem:

[tex]h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f[/tex]

[tex]v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s[/tex]

The average velocity over a given time [3,4] is 38.19 m/sec, the average velocity over a given time [3,3.5] is 38.605 m/sec, the average velocity over a given time [3,3.1] is 38.937 m/sec and this can be determined by using the given data.

Given :

Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by [tex]\rm h(t) = 44t-0.83t^2[/tex].

a) [3,4]

At time t = 3 and 4, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(4) = 44(4)-0.83(4)^2 = 176-13.28=162.72[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{162.72-124.53}{4-3}=38.19[/tex]

b) [3,3.5]

At time t = 3 and 3.5, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.5) = 44(3.5)-0.83(3.5)^2 = 154-10.1675=143.8325[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{143.8325-124.53}{3.5-3}=38.605[/tex]

c) [3,3.1]

At time t = 3 and 3.1, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(4) = 44(3.1)-0.83(3.1)^2 = 136.4-7.9763=128.4237[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{128.4237-124.53}{3.1-3}=38.937[/tex]

d) [3,3.01]

At time t = 3 and 3.01, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.01) = 44(3.01)-0.83(3.01)^2 = 132.44-7.519883=124.920117[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{124.920117-124.53}{3.01-3}=39.0117[/tex]

e) [3,3.001]

At time t = 3 and 3.001, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.001) = 44(3.001)-0.83(3.001)^2 = 132.044-7.47498083=124.5690192[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{124.5690192-124.53}{3.001-3}=39.01917[/tex]

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Answer:

The critical value would be: [tex]\chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                      F          B         S      Bs     Total

home wins    39      156      25      83      303

Visitor wins   31       98        19      75       223

Total              70      254     44      158      526

We need to conduct a chi square test in order to check the following hypothesis:

H0: The number of home team and visiting team losses is independent of the sport.

H1: The number of home team and visiting team losses is dependent of the sport.

The level of significance assumed for this case is [tex]\alpha=0.01[/tex]

The statistic to check the hypothesis is given by:

[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{70*303}{526}=40.32[/tex]

[tex]E_{2} =\frac{254*303}{526}=146.32[/tex]

[tex]E_{3} =\frac{44*303}{526}=25.35[/tex]

[tex]E_{4} =\frac{158*303}{526}=91.02[/tex]

[tex]E_{5} =\frac{70*223}{526}=29.68[/tex]

[tex]E_{6} =\frac{254*223}{526}=107.68[/tex]

[tex]E_{7} =\frac{44*223}{526}=18.65[/tex]

[tex]E_{8} =\frac{158*223}{526}=66.98[/tex]

And the expected values are given by:

                        F           B            S          Bs     Total

home wins    40.32   146.32    25.35    91.02    303

Visitor wins   29.68   107.68    18.65    66.98    223

Total                70         254         44        158      526

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(39-40.32)^2}{40.32}+\frac{(156-146.32)^2}{146.32}+\frac{(25-25.35)^2}{25.35}+\frac{(83-91)^2}{91}+\frac{(31-29.68)^2}{29.68}+\frac{(98-107.68)^2}{107.68}+\frac{(19-18.65)^2}{18.65}+\frac{(75-66.98)^2}{66.98} =3.29[/tex]Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(4-1)(2-1)=3[/tex]

The critical value would be: [tex]chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Answer:

[tex]T(p) = \frac{1}{4}p[/tex]

The new price is $6.75

Step-by-step explanation:

The new price of the toy after the discount (T) is given by the original price (p) subtracted by the discounted amount (3/4 of p):

[tex]T(p) = p-\frac{3}{4}p \\T(p) = (1-\frac{3}{4})p\\T(p) = \frac{1}{4}p[/tex]

If the original price was $26.99, the new price is:

[tex]T= \frac{1}{4}*\$26.99\\T=\$6.75[/tex]

Answer: 24 feet

Step-by-step explanation:

By using Pythagoras rule:

Let x be the high up the house does the ladder reached.

X^2 + 5^2= 25^2

X^2 = 25^2 - 5^2

x^2 = 625 - 25

x^2 = 600

Square both side

x = sqrt(600)

x= 24.495

x = 24 feet

Answer:

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Answer:

Yes, one of the solutions of this differential equation is a constant solution equation.

Step-by-step explanation:

9xy' + 5y = 10.

If y' = 0, 9xy' = 0

5y = 10

y = 2 = c.

So, for all real values of x such that -∞ < x < ∞, 9xy' will be 0 and one of the solutions of the differential equation will be y = 2.

Hope this helps!

Answer:

Option C and D are false

Step-by-step explanation:

All the mentioned option are correct in the given scenario except option C and D.

The reason is that dose is categorized as nil, low and high so, dose is categorical variable. Also, number of tumors is quantitative variable because it can be meaningfully interpreted in numerical form. The number if tumors is discrete quantitative variable.

Now consider all options

A) Gender is categorical ; dose is ordinal

This option is true because gender can be categorized into male and female and also dose is ordinal because it has order i.e. nil,low and high.

B) Gender is discrete; weight is continuous

This option is false because gender can be a discrete variable and weight is continuous variable because it is measurable. So, the statement is true.

Option C and D are already discussed an option E is discussed in option B.

The false statement is C) Number of tumors is categorical. The number of tumors is a discrete variable that involves countable values, not a categorical variable.

In this experiment, we have different types of variables. The statement C) Number of tumors is categorical is false. The number of tumors is actually a discrete variable as it involves countable values. The other statements are correct: A) Gender is categorical; dose is ordinal, as gender is divided into male and female, which is a categorical classification, and the dose is ranked as nil, low, high which makes it an ordinal variable. Statement B) Gender is discrete; weight is continuous is also correct because gender is a discrete variable (only two possible values, male or female), and weight is a continuous variable as it can take any value within a certain range. Statement D) Dose is discrete is correct, as the dose can only take certain values (nil, low, high) it is considered a discrete variable. Lastly, E) Weight is continuous is accurate as Weight can take any value within a range and it involves measurement.

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Answer:

[tex] P(X>9) = 0.3593[/tex]

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=25, p=0.3089)[/tex]  

The probability mass function for the Binomial distribution is given as:  

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

For this case we want this probability:

[tex] P(X >9)[/tex]

And we can use the complement rule like this:

[tex] P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)] [/tex]And we can find the individual probabilities like this:

[tex] P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974[/tex]  

[tex] P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011[/tex]  

[tex] P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584[/tex]  

[tex] P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02[/tex]  

[tex] P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049[/tex]  

[tex] P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092[/tex]  

[tex]P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138[/tex]

[tex] P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167[/tex]

[tex] P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168[/tex]

And in order to do the operations we can use the following excel code:

"=1-BINOM.DIST(8,25,0.3089,TRUE)"  

And we got:

[tex] P(X>9) = 0.3593[/tex]

Answer:

Step-by-step explanation:

let A represent the amount of the 10% fungicide solution

let B represent the amount of the 50% fungicide solution

B milliliters = 50% of B = (50/100)A = 0.5B

Total milliliters = 26% of 200 milliliters = 0.26 * 200 = 52

A milliliters + B milliliters = 200 milliliters

0.1A + 0.5B = 52

using substitution method to solve A and B

0.1(200-B) + 0.5B = 52

20 - 0.1B + 0.5B = 52

20 + 0.4B = 52

0.4B = 52 -20 = 32

0.4B = 32

B = 32/0.4 = 80

since B = 80

A = 200 -B = 200 - 80 = 120

Answer: he must use 120 milliliters of the 10% solution and 80 milliliters of the 50% solution.

Step-by-step explanation:

Let x represent the amount of 10% fungicide solution that Ngoc must use.

Let y represent the amount of 50% fungicide solution that Ngoc must use.

The total volume of the fungicide solution that he wants to create is 200 milliliters. It means that

x + y = 200

Ngoc needs to mix a 10% fungicide solution with a 50% fungicide solution to create 200 millileters of a 26% solution. This means that

(10/100 × x) + (50/100 × y) = (26/100 × 200)

0.1x + 0.5y = 52 - - - - - - - - - -1

Substituting x = 200 - y into equation 1, it becomes

0.1(200 - y) + 0.5y = 52

20 - 0.1y + 0.5y = 52

- 0.1y + 0.5y = 52 - 20

0.4y = 32

y = 32/0.4 = 80 milliliters

x = 200 - y = 200 - 80

x = 120 milliliters

Answer:

variance = (27*24656384-20182^2)/(27*26) =368104.3

standard devaition SD= sqrt(368104.3) =606.716

maximum possible amount that could be awarded under the two-standard-deviation rule = mean +2*SD

= (20182/27)+(2*606.716)

= 1960.913

=$1960913

Answer:

(a) Yes, f(x) is a valid probability mass function.

(b) The probability that you will walk at least two dogs this week = 0.74.

(c)  The expected number of dogs you will walk this week = 3 dogs.

(d) The expected value of X2 = 11.29

(e) Var[x] = E(X2) – (E[X])2 = 3.2811

Step-by-step explanation:

We are given with the probability mass function (pmf),f(x), o'is defined as follows:

Firstly let X = Number of dogs you walk this week

    X                 P(X = x)

    0                      0.14

    1                       0.12

    2                      0.15

    3                      0.23

    4                      0.18

    5                      0.09

    6                      0.08

    7                       0.01

(a) Now f(x) to be a valid probability mass function, two conditions should be    met :

So, First condition is already met as all values are positive and for second condition = 0.14 + 0.12+ 0.15+ 0.23+ 0.18+ 0.09+ 0.08+ 0.01 = 1

Hence both the conditions are satisfied so f(x) is a valid probability mass function.

(b) Probability that we will walk at least two dogs this week = P(X>=2)

      = 1 - P(X = 0) - P(X = 1) = 1 -  0.14 - 0.12 = 0.74

(c) To Compute the expected number of dogs you will walk this week we will use expectation formula which says:

           E(X) = [tex]\frac{\sum X\times P(X=x)}{\sum P(X=x)}[/tex] = [tex]\frac{0*0.14 + 1*0.12 + 2*0.15 + 3*0.23 + 4*0.18 + 5*0.09 + 6*0.08 + 7*0.01}{1}[/tex]

                                           = 2.83 or 3 after rounding off.

Therefore the expected number of dogs you will walk this week are 3 dogs.

(d) The expected value of X2 [E(X2)] =   [tex]\frac{\sum X^{2} \times P(X=x)}{\sum P(X=x)}[/tex]

      = [tex]\frac{0^{2} *0.14 + 1^{2} *0.12 + 2^{2} *0.15 + 3^{2} *0.23 + 4^{2} *0.18 + 5^{2} *0.09 + 6^{2} *0.08 + 7^{2} *0.01}{1}[/tex] = 11.29

(e) Var[x] = E(X2) – (E[X])2

     We have E(X2) = 11.29 and E(X) = 2.83

      Var[x] = [tex]11.29 - (2.83)^{2}[/tex] = 3.2811.

The random variable X represents the number of dogs walked in a week for a dog walking business. By verifying that the sum of the probabilities equals 1, we've confirmed the pmf is valid. The probability of walking at least two dogs is 0.74, and calculations for the expected number and variance involve using the given probabilities and applying the respective formulas.

Understanding Probability Mass Functions

In the context of a dog walking business, let us define the random variable X as the number of dogs walked in a week. The values that X can take on are 0, 1, 2, 3, 4, 5, 6, and 7. The probabilities of these events are given by P(X = x), where x are the respective values X can take.

To verify if the given probability mass function (pmf) is valid, the sum of all probabilities P(X = x) should be equal to 1. We calculate the sum as follows:

Adding them gives 1 which confirms it is a valid pmf.

To find the probability of walking at least two dogs this week, which is P(X ≥ 2), we need to add up the probabilities for all events where X is 2 or more:

P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)
= 0.15 + 0.23 + 0.18 + 0.09 + 0.08 + 0.01
= 0.74

To compute the expected number of dogs walked in a week, or E[X], we use the formula:

E[X] = Σ x * P(X = x)

For E[X2], we calculate the expectation of X-squared by multiplying each value of x by itself and then by its probability, before summing these products.

Finally, the variance Var[X] of the random variable X is given by E(X2) - (E[X])2. This involves first calculating E[X2] as explained, then squaring the previously found E[X] value, and subtracting that square from E[X2].

Answer:

The answer to your question is

Step-by-step explanation:

Inequality 1

                             4(3x - 4) < 32

                             12x - 16 < 32

                             12x < 32 + 16

                              12x < 48

                                 x < 48/12

                                 x < 4

Inequality 2

                           2x + 1 ≤ 8x + 25

                           2x - 8x ≤ 25 - 1

                               - 6x ≤ 24

                                   x ≥ 24/-6

                                   x ≥ - 4

- See the graph below

- Interval [-4, 4)

To solve the inequality system, we divide both sides of each inequality by the respective coefficient to isolate the variable. The solutions are x < 4 and -4 ≤ x. The graph of the solution is a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

To solve the inequality 4(3x - 4) < 32, we divide both sides of the inequality by 4 to isolate the variable. This gives us 3x - 4 < 8. Adding 4 to both sides of the inequality gives us 3x < 12.

Finally, dividing both sides of the inequality by 3 gives us x < 4.

For the inequality 2x + 1 ≤ 8x + 25, we subtract 2x from both sides of the inequality to isolate the variable. This gives us 1 ≤ 6x + 25. Subtracting 25 from both sides of the inequality gives us -24 ≤ 6x.

Finally, dividing both sides of the inequality by 6 gives us -4 ≤ x.

The solutions to the inequality system are x < 4 and -4 ≤ x. The graph of the solution would be a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

The interval notation for the solution is (-∞, 4) and [-4, ∞).

Answer:

We conclude,  Mohammed's opportunity cost knitting one sweater is 48 cookies.

Step-by-step explanation:

We have that  Mohammed can knit 5 sweaters or bake 240 cookies.

In one week Aisha can knit 15 sweaters or bake 480 cookies.  

We calculate how much is Mohammed's opportunity cost knitting one sweater. We get  

\frac{240}{5}= 48.

We conclude,  Mohammed's opportunity cost knitting one sweater is 48 cookies.

Answer:

There is a 55.97% that a male can find a full time job in his chosen profession.

Step-by-step explanation:

We have these following probabilities:

A 60% probability that a college graduates cannot find a full time job in their chosen profession.

A 40% probability that a college graduates can find a full time job in their chosen profession.

57% of the college graduates who cannot find a job are female

43% of the college graduates who cannot find a job are male

18% of the college graduates who can find a job are female

82% of the college who can find a job are male.

Given a male college graduate, find the probability he can find a full time job in his chosen profession?

The total males are 43% of 60%(Those who cannot find a job) and 82% of 40%(Those who can find a job). So the percentage of males is [tex]P(M) = 0.43*0.60 + 0.82*0.40 = 0.586[/tex]

Those who are males and find a job in their chosen profession are 82% of 40%. So [tex]P(M \cap J) = 0.82*0.40 = 0.328[/tex]

[tex]P = \frac{P(M \cap J)}{P(M)} = \frac{0.328}{0.586} = 0.5597[/tex]

There is a 55.97% that a male can find a full time job in his chosen profession.

Answer:

a) Lower inner fence = 77.6168 = 77.62 to 2 d.p

Lower outer fence = 48.9044 = 48.90 to 2 d.p

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159

Step-by-step explanation:

Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Q₁ = 25th percentile = lower quartile

IQR = Inter quartile Range = Q₃ - Q₁

Q₃ = 75th percentile = upper quartile

To calculate Q₁ for a normal distribution with only mean and standard deviation known,

We need the standardized score whose probability is 0.25 P(z) = 0.25

From the normal distribution table

z = (± 0.674)

z = (x - xbar)/σ

x = the value in the data we're interested in,

xbar = mean = 115.9

σ = standard deviation = 14.2

Lower quartile corresponds to (z = - 0.674)

- 0.674 = (x - 115.9)/14.2

Q₁ = X = 106.3292

The upper quartile, Q₃ corresponds to z = (+0.674)

Q₃ = 125.4708

IQR = 125.4708 - 106.3292 = 19.1416

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168

Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)

We standardize 74 by obtaining its z-score

z = (x - xbar)/σ

z = (74 - 115.9)/14.2 = - 2.95

P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)

Final answer:

The lower fence for the IQ data, which determines whether an IQ score is an outlier, is calculated as the mean minus two times the standard deviation. In this case, the lower fence is 87.5, which makes an IQ score of 74 an outlier as it falls significantly below this threshold.

Explanation:

To determine if an IQ score is an outlier, we often use the interquartile range (IQR) and calculate the fences. However, since the data is approximately normally distributed and we have the mean and standard deviation, we can also consider an IQ score to be an outlier if it falls more than two standard deviations from the mean. In this question, we do not have the IQR, so we'll use standard deviations to calculate the outlier threshold.

The mean IQ score is 115.9 and the standard deviation is 14.2. An outlier is typically defined as a value that is more than two or three standard deviations away from the mean. These thresholds are sometimes called the outer fences in statistical outlier detection. Using two standard deviations, we can calculate the lower limit as follows:

Lower Limit = Mean - 2 × Standard Deviation
Lower Limit = 115.9 - 2(14.2)
Lower Limit = 115.9 - 28.4
Lower Limit = 87.5

Therefore, an IQ score of 74 is considerably lower than the lower limit of 87.5, suggesting that it could indeed be considered an outlier.

Answer:

The correct option is (a).

Step-by-step explanation:

The hypothesis of the study can be defined as:

H₀: The divorce rate is 50% for couples who had a child within the first two years of marriage, i.e. p = 0.50

Hₐ: The divorce rate is different from 50% for couples who had a child within the first two years of marriage, i.e. p ≠ 0.50

The 90% confidence interval is: 0.402 ± 0.056 = (0.346, 0.458) ≈ (0.35, 0.46)

The confidence level is 90%, the significance level (α) is:

[tex]\alpha =1-\frac{Confidence\ level}{100}\\=1-\frac{90}{100}\\ =0.10\ or\ 10\%[/tex]

Decision Rule:

If the null hypothesis value is not contained in the 90% confidence interval then the null hypothesis will be rejected and vice-versa.

Interpretation of the Confidence interval:

Thus all the options are correct.

The 90% confidence interval for the proportion of couples who had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056. Based on this interval, we can conclude that the divorce rate is between 35% and 46%.

Based on the given information, the sociologist selected a random sample of 200 couples who had a child within the first two years of marriage. Out of these couples, 80 were found to be divorced within five years. The 90% confidence interval for the proportion of all couples that had a child within the first two years of marriage and are divorced within five years is given as 0.402 ± 0.056.

This means that we can be 90% confident that the true proportion of couples who had a child within the first two years of marriage and are divorced within five years lies between 0.402 - 0.056 and 0.402 + 0.056.

Therefore, the correct statement based on this interval is that the divorce rate is between 35% and 46%.

Answer:a) observational study

b) Randomization

c)Observational study

Step-by-step explanation:

a) The participants in (a) are monitored closely and data collected are reviewed.

b) Participants are randomly assigned to work and study.

a) The research method described is an observational study. b) The research method described is a randomized experiment. c) The research method described is an observational study.

a) The research method described is an observational study. The students are simply keeping a log of the hours they spend studying in a group and reporting it after the course is completed. The researcher is not manipulating any variables or assigning students to different groups.

b) The research method described is a randomized experiment. The students are randomly assigned to study groups and the teacher tells each group how often to meet. This allows for the manipulation of the independent variable (number of hours studying in a group) and the measurement of its effect on the dependent variable (final course grade).

c) The research method described is an observational study. The students voluntarily join groups based on how often the groups will meet. The researcher is not manipulating any variables or assigning students to different groups. The students' choice of groups is a naturally occurring phenomenon that is being observed.

You can't deduce the length of a side from the perimeters of a rectangle.

Say that [tex]s[/tex] and [tex]S[/tex] are, respectively, the short and long side of the rectangle.

So, we know that

[tex]2s+2S=130 \iff 2(s+S)=130 \iff s+S=65[/tex]

But we can't solve exactly for [tex]s[/tex] nor for [tex]S[/tex], unless more information is given.

The length of one of the shorter sides of the window cannot be explicitly determined without additional information. However, considering the window has a rectangular shape, the length of the shorter side should be less than half of the total perimeter, in this case, less than 65 inches.

To find the length of one of the shorter sides of the window, we must first understand that the perimeter of a rectangle is calculated by the formula: 2*length + 2*width = Perimeter.

However, the problem doesn't specify the dimensions of the box, but it does tell us that one pair of sides (the length and the width) are not equal. This suggests that the window is a rectangle. If we assume that the length of the window is longer than the width (length > width), then the shorter sides of the rectangle (the widths) will be of equal length.

Without more details, we can't calculate the exact measurement of the length of one of the shorter sides, but we can say that it should be less than half of the total perimeter, which is less than 65 inches.

https://brainly.com/question/29595517

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Answer:

A. Control group  

Step-by-step explanation:

A control group is a group in an experiment or study that does not receive experimental procedure during such that it is then used as a benchmark to measure how the other tested subjects do.

An experimental group is a group in an experiment or study that receives an experimental procedure. The values of gotten from the test are recorded and the effect of independent variables on the dependent variables are determined.

Control experiment can be used to determine whether or not the customers are friendly. Experimental group will be the customers whom the waiter is extra friendly toward. The control group will be the alternate customer whom the waiter is not extra friendly toward.          

Answer:

Step-by-step explanation:

a) The standard deviation is usually preferred over the range because it is calculated from all of the data and will not be impacted as much as the range when they are outliers,and the standard deviation uses all of the data.

b)The IQR sometimes referred to the standard deviation when there is an outlier because the IQR is less sensitive to this features than standard deviation.

that is the IQR is not affected by an outlier,while the standard deviation is affected by an outlier.

c)The advantage of the standard deviation over the IQR is the standard deviation takes into account the values of all observation,while the IQR uses only some of the data.

Answer:

The probability that the good exam belongs to student X is 0.8571.

Step-by-step explanation:

It is provided that the probability that X did well in the exam is, P (X) = 0.90 and the probability that X did well in the exam is, P (Y) = 0.40,

Compute the probability that exactly one student does well in the exam as follows:

[tex]P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56[/tex]

Then the probability that X is the one who did well in the exam is:

[tex]P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571[/tex]

Thus, the probability that the good exam belongs to student X is 0.8571.

Answer:

1 hr 52 minutes

Step-by-step explanation:

As per Newton law of cooling we have

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

where T0 is the initial temperature of the body

Ts = temperature of surrounding

t = time lapsed

k = constant

Using this we find that T0 = 98.6 : Ts= 65

Let x hours be lapsed before the body was found.

Then we have

[tex]T(x) = 65 +(98.6-65)e^{-kx} = 85\\e^{-kx}=\frac{20}{33.8} =0.5917[/tex]

Next after 1 hour temperature was 80

[tex]T(x+1) = 65+33.6(e^{-k(x+1)}=80\\e^{-k(x+1) =0.4464[/tex]

Dividing we get

[tex]e^k = 1.325408\\k = 0.2817[/tex]

Substitute this in

[tex]e^{-kx} =0.5917\\x=\frac{ln 0.5917}{-k} \\=1.863[/tex]

approximately 1 hour 52 minutes have lapsed.

Answer:

Type I error: The correct option is (C).

Type II error: The correct option is (D).

Step-by-step explanation:

The type-I-error is the probability of rejecting the null hypothesis when the null hypothesis is true.

The type-II-error is the probability of filing to reject the null hypothesis when in fact it is false.

The hypothesis in this problem can be defined as follows:

Null hypothesis (H₀): The percentage of adults who have a job is equal to 88%.

Alternate Hypothesis (Hₐ): The percentage of adults who have a job is different from 88%.

The type-I-error in this case will be committed when we conclude that the percentage of adults who have a job is different from 88% when in fact it is equal to 88%.

The type-II-error in this case will be committed when we conclude that the percentage of adults who have a job is equal to 88% when in fact it is different than 88%.

Answer:

Step-by-step explanation:

given is a vector as (5,9,3)

a = (5,9,3)

To find out direction cosines

First let us calculate modulus of vector a

[tex]||a|| =\sqrt{5^2+9^2+3^2} \\=\sqrt{25+81+9} \\=\sqrt{115}[/tex]

Direction ratios are (5,9,3)

Magnitude of vector a = [tex]\sqrt{115}[/tex]

So direction cosines would be

[tex](\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

Angles would be

[tex](\alpha, \beta, \gamma) = arccos ((\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

=cos inverse  (0.4662, 0.8393, 0.2798)

= (62.21, 32.93,32,94)

Answer:

15.65

Step-by-step explanation:

Suppose we have two points:

[tex]A = (x_{1}, y_{1})[/tex]

[tex]B = (x_{2}, y_{2})[/tex]

The distance between these points is:

[tex]D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}[/tex]

So, for points (-5, -10) and (2, 4)

[tex]D = \sqrt{(2 - (-5))^{2} + (4 - (-10))^{2}}[/tex]

[tex]D = \sqrt{7^{2} + 14^{2}[/tex]

[tex]D = 15.65[/tex]

So the correct answer is:

15.65