Calculate descriptive statistics for the variable (Coin) where each of the thirty-five students in the sample flipped a coin 10 times. Round your answers to three decimal places and write the mean and the standard deviation.

Answer:

The most that a bag can weigh and not need to be repackaged is 15.355 ounces.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 13.50, \sigma = 1.06[/tex]

Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?

Bags in the upper 4% have a pvalue of 1-0.04 = 0.96. So the most that a bag can weight and not need to be repackaged is the value of X when Z has a pvalue of 0.9599. So this is X when [tex]Z = 1.75[/tex]

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

[tex]1.75 = \frac{X - 13.50}{1.06}[/tex]

[tex]X - 13.50 = 1.75*1.06[/tex]

[tex]X = 15.355[/tex]

The most that a bag can weigh and not need to be repackaged is 15.355 ounces.

Answer:

Therefore the required probability is [tex]=\frac{27}{128}[/tex]

Step-by-step explanation:

The probability of success is [tex]\frac{3}{4}[/tex]

The number of trial = 4

X= the items survive out of 4

[tex]P(x=r)=^nC_rq^{n-r}p^r[/tex]        p =the probability of success and q = the probability failure.

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

[tex]\therefore P(X=2)=^4C_2(\frac{1}{4} )^2(\frac{3}{4} )^2[/tex]

                  [tex]=\frac{4!}{2!2!} (\frac{1}{16} )(\frac{9}{16} )[/tex]

                  [tex]=\frac{27}{128}[/tex]

Therefore the required probability is [tex]=\frac{27}{128}[/tex]

Final answer:

The probability that exactly 2 out of 4 components survive a shock test is calculated using the binomial probability formula, which results in approximately 0.2109 when rounded to four decimal places.

Explanation:

The question is asking for the probability that exactly 2 out of 4 components will survive a shock test given that the probability of a single component surviving is 3/4. To solve this, we use the binomial probability formula which is P(X = k) = (n choose k) p^k (1-p)^(n-k), where 'n' is the total number of trials, 'k' is the number of successful trials, and 'p' is the probability of success on a single trial.

Plugging in the given values, we have:

n = 4 (the number of components tested)

k = 2 (the desired number of components to survive)

p = 3/4 (the probability of a component surviving)

Using the formula:

P(X = 2) = (4 choose 2) * (3/4)^2 * (1/4)^(4-2)

P(X = 2) = 6 * (9/16) * (1/16)

P(X = 2) = 6 * (9/256)

P(X = 2) = 54/256

P(X = 2) = 0.2109 when rounded to four decimal places.

Therefore, the probability that exactly 2 of the next 4 components tested survive the shock test is approximately 0.2109.

Answer:

C = 2*Ln (2) - 1

Step-by-step explanation:

Given

x²(dy/dx) = (4x²-x-3)/(x+1)(y+1)

y(1) = 2

We apply separation of variables as follows

(y+1) dy = ((4x²-x-3)/(x+1)(x²)) dx

⇒ ∫(y+1) dy = ∫((4x²-x-3)/(x+1)(x²)) dx

(y²/2) + y + C₁ = 2 ∫(1/(x+1)) dx + ∫((2x-3)/x²) dx

⇒  (y²/2) + y + C₁ = 2 Ln (x+1) 2 Ln (x) + (3/x) + C₂

⇒  C₁ - C₂= Ln (x+1)² + Ln (x)² + (3/x) - (y²/2) - y

⇒  C = Ln ((x+1)²(x)²) + (3/x) - (y²/2) - y

⇒  C = Ln ((x²+x)²) + (3/x) - (y²/2) - y

if y(1) = 2

we get

C = Ln ((1²+1)²) + (3/1) - (2²/2) - 2

⇒    C = 2*Ln (2) + 3 - 4 = 2*Ln (2) - 1

Answer:

then the probability of failure goes between 0.00003 (0.003%) and 0.5212 (52.12%) depending on the system configuration

Step-by-step explanation:

the solution depends on the system configuration, that is , if some component ( lets say A) is run in parallel from other , or is in series

if a component is run in parallel then the system fails only if all the components in parallel fails

but if the system is connected in series , the system will fail only if one of the components the serie fails.

Therefore denoting the events A= fails A , B= fails B , C= fails C , D= fails D , we have:

- lower bound of probability of failure = all components are in parallel

probability of failure P(A∩B∩C∩D)=P(A)*P(B)*P(C)*P(D)= 0.1 * 0.2 * 0.05 * 0.3 = 0.00003 (0.003%)

- upper bound of probability of failure = all components are in parallel

probability of failure P(A∪B∪C∪D)= P(A) + P(B) + P(C) +P(D) - P(A ∩ B) - P(A ∩ C) - P(A ∩ D)- P(B ∩ C) - P(B ∩ D) - P(C ∩ D) + P(A ∩ B ∩ C) + P(A ∩ B ∩ D) + P(A ∩ C ∩ D) + P(B ∩ C ∩ D) - P(A ∩ B ∩ C ∩ D) = (P(A) + P(B) + P(C) +P(D)) - ( P(A)*P(B) + P(A)*P(C) + P(A)*P(D) + P(B)*P(C) + P(B)*P(D) + P(C)*P(D) ) + P(A)*P(B)*P(C)  + P(A)*P(B)*P(D)+  P(A)*P(C)*P(D)+  P(B)*P(C)*P(D) -  P(A)*P(B)*P(C)*P(D)

replacing values

P(A∪B∪C∪D)= 0.5212 (52.12%)

then the probability of failure goes between 0.00003 (0.003%) and 0.5212 (52.12%) depending on the system configuration

Answer:

a) 13.277

Step-by-step explanation:

The chi-square critical value can be assessed using chi-square area table and for this table need value of alpha and degree of freedom.

The value of alpha is given which is 0.01 and degree of freedom can be calculated as

df=k-1

Where k represent the categories which are 5 in the given case.

df=5-1=4.

For alpha=0.01 and df=4, we get the chi-square critical value 13.277.

Answer:

1. At p1 = (100,000 - p2)/1,600 for Ultra Minis

2. At p2 = (150,000 - p1)/1,600 for Big Stack

Step-by-step explanation:

Since we are dealing with demand functions in which there is a negative relationship between price and quantity demanded, the question marks marks in the two demand functions can be assumed to be negative signs. As a result, the equations can be re-stated as follows:

q1(p1, p2) = 100,000 - 800p1 + p2 ................................ (1)

q2(p1, p2) = 150,000 + p1 - 800p2 ............................... (2)

In economics, total revenue (TC) is quantity demanded/sold multiply by price, the TCs for Ultra Mini (TCq1), and the Big Stack (TCq2) can be obtained by multiplying equations (1) and (2) with p1 and p2 as follows:

For q1:

TCq1 = p1*q1(p1, p2) = p1(100,000 - 800p1 + p2)

TCq1 = 100,000p1 - 800p1^2 + p1p2 .............................. (3)

For q2:

TCq2 = p2*q2(p1, p2) = p2(150,000 + p1 - 800p2)

TCq2 = p2150,000 + p1p2 - 800p2^2 .......................... (4)

We will take partial derivatives of each of equations (3) and (4) to obtain the marginal revenue (MR) as follows:

Partial derivative of equation (3) with respect to p1 and equate to zero:

MR = dTCq1/dp1 = 100,000 - 2(800p1) + p2 = 0

                           = 100,000 - 1,600p1 + p2 = 0

By rearranging and solving for p1, we have:

1,600p1 = 100,000 - p2

p1 = (100,000 - p2)/1,600 ....................................... (5)

The p1 in equation (5) is the price that will maximize the total revenue of Ultra Mini.

Partial derivative of equation (4) with respect to p2 and equate to zero:

MR = dTCq2/dp2 = 150,000 + p1 - 2(800p2) = 0

                             = 150,000 - 1,600p2 + p1 = 0

By rearranging and solving for p2, we have:

1,600p2 = 150,000 - p1

p2 = (150,000 - p1)/1,600 ....................................... (6)

The p2 in equation (6) is the price that will maximize the total revenue of Big Stack.

Therefore the prices at which total revenue of the company will be maximized are at p1 = (100,000 - p2)/1,600 for Ultra Minis and at p2 = (150,000 - p1)/1,600 for Big Stack.

To maximize total revenue, we need to find the prices for the Ultra Mini and the Big Stack that will maximize the quantity sold for each speaker model.

To maximize total revenue, we need to find the prices for the Ultra Mini and the Big Stack that will maximize the quantity sold for each speaker model. To do this, we need to find the demand functions for the Ultra Mini and the Big Stack and set their derivatives equal to zero to find the critical points.

Since there are no critical points for either speaker model, we cannot find the prices that will maximize total revenue. The company should consider other factors, such as production costs and market competition, to determine the optimal pricing strategy.

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Answer:

a) 2.2 × 10⁸ beats

b) 2.20 × 10⁸ beats

c) 2.207 × 10⁸ beats

Step-by-step explanation:

Data provided in the question:

Average heart rate of a typical person = 70.0 beats/min

Now,

In the given cases, the significance is on the significant figures after the decimal

Therefore,

the answer is will be provided accordingly

Now,

a) Time = 6.0 years

[since 1 significant figure after decimal. answer will be give in  1 significant figure after decimal ]

time in minutes = 6.0 × 365 × 24 × 60

= 3.1 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.1 × 10⁶

= 2.2 × 10⁸ beats

b)  Time = 6.00 years

[since 2 significant figure after decimal. answer will be give in 2 significant figure after decimal ]

time in minutes = 6.00 × 365 × 24 × 60

= 3.15 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.15 × 10⁶

= 2.20 × 10⁸ beats

c) Time = 6.000 years

[since 3 significant figure after decimal. answer will be give in 3 significant figure after decimal ]

time in minutes = 6.000 × 365 × 24 × 60

= 3.154 × 10⁶ minutes

Total beats = Average heart rate × Time

= 70 × 3.154 × 10⁶

= 2.207 × 10⁸ beats

Final answer:

The total number of samples in the sample space can be found by using the concept of permutations and the rule of product.Therefore total number of samples in the sample space is 352,560.

Explanation:

The total number of samples in the sample space can be found by using the concept of permutations and the rule of product. Since there are 43 objects and we are sampling 4 objects in order, we can use the formula:

nPr = n! / (n - r)!

where n is the total number of objects and r is the number of objects being sampled. Plugging in the values, we get:

43P4 = 43! / (43 - 4)!

Simplifying, we have:

43P4 = 43! / 39!

which can be further simplified to:

43P4 = (43 * 42 * 41 * 40 * 39!) / 39!

The 39! terms cancel out, leaving us with:

43P4 = 43 * 42 * 41 * 40

Evaluating this expression, we find that the total number of samples in the sample space is 352,560.

The total number of samples in the sample space when randomly sampling 4 objects from 43 is calculated using permutations, resulting in 2,906,880 possible samples.

The student has asked about finding the total number of samples in the sample space when randomly sampling 4 objects in order from among 43 objects. This can be calculated using the formula for permutations, given that the order of selection matters and repeats are not allowed. The formula for permutations of n objects taken r at a time is nPr = n! / (n - r)!, where n! denotes the factorial of n, and (n-r)! denotes the factorial of (n-r).

In this case, n = 43 (the total number of objects) and r = 4 (the number of objects being selected). Thus, the calculation will be 43P4 = 43! / (43 - 4)! = 43! / 39!. Carrying out the calculation, 43P4 equals 43 × 42 × 41 × 40, which results in 2,906,880 possible samples.

Answer:

[tex]0.000394[/tex]

Step-by-step explanation:

First we will find the probability of selecting five cards out of pack of cards

Probability of selecting five cards is equal to

[tex]^{52}C_5[/tex]

On expanding we get

[tex]\frac{52!}{47! * 5!} \\[/tex]

[tex]\frac{52 * 51 * 50 * 49 * 48 * 47!}{47 ! * 5*4*3*2*1} \\= 2598960[/tex]

straight high card [tex]9[/tex] means five cards with values lesser than [tex]9[/tex] but adjacent to it are

[tex]9, 8, 7, 6, 5[/tex]

there are four card for each number

Hence, probability of choosing five cards is equal to

[tex]4*4*4*4*4\\= 1024[/tex]

Probability of getting a straight with high card 9 is equal to

[tex]\frac{1024}{2598960}[/tex]

[tex]0.000394[/tex]

Answer:

The shape of the distribution of the sample mean would be bell-shaped(normally distributed).

Step-by-step explanation:

The Central Limit Theorem answers this question.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size, of at least 30, can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that

The population distribution is right-skewed.

Sampling distribution of the sample mean with size 40.

The Central Limit Theorem estabilishes that the sampling distribution of the sample mean will be normally distributed, even if the distribution of the population is not.

So the shape of the distribution of the sample mean would be bell-shaped(normally distributed).

Answer: 26667 miles

Step-by-step explanation:

According to the statement,

In option 1 we have 510 dollars per month plus $0.38/mile.

In option 2 we have $0.65/mile

Let x be the number of miles.

For a whole year, the option 1 is 510*12+ 0.38 x

For a whole year, the option 2 is 0.65 x

Equating both, we get

6120 + 0.38 x = 0.65 x

Solving, we get

x= 6120/ 0.27

x= 22666.67

x= 26667 miles

Answer:

b. false

Step-by-step explanation:

As the car is being towed using a horizontal chain, the tension's direction is horizontal and in opposite with the friction force's direction. The weight of the car, on the other hand, has a vertical direction and downward. Therefore, tension and weight are not related. In order to maintain constant velocity tension needs to be equal to friction force, which may be equal or less than the car weight.

Answer:

3) One-sided,compare two populations

Step-by-step explanation:

A manufacturer wants to investigate the whether there is an increase in the mean gas mileage of mid-sized sedan having new type of tire compared to mean gas mileage of mid-sized sedan not having new type of tire.

So, there are two populations :

population 1 : mid-sized sedan having new type of tire.

population 2 : mid-sized sedan not having new type of tire.

So, two populations are being compared.

Also, A manufacturer is investigating whether there is an increase in mean gas mileage.The alternative hypothesis will be right tailed and so, this corresponds to one sided test.

Final answer:

To determine if the new tires increase fuel efficiency, the manufacturer should use a one-sided hypothesis test to analyze changes in a single population's mean gas mileage.

Explanation:

The automotive manufacturer should perform a one-sided hypothesis test to analyze a change in a single population. This is because they are specifically interested in testing if the mean gas mileage for mid-sized sedans using the new tires is greater than 32 miles per gallon, not whether it is different (either less or greater). A two-sided test would be used if they were interested in any difference from the historic mean, rather than specifically an increase.

Answer:

[tex]\chi^2 =\frac{20-1}{1} 4.84 =91.96[/tex]

[tex]p_v =P(\chi^2 >91.96) \approx 0[/tex]

"=1-CHISQ.DIST(91.96,19,TRUE)"

If we compare the p value and the significance level provided we see that [tex]p_v <<\alpha[/tex] so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And we have evidence to conclude that the sample variance is higher than 1 and indeed that the deviation is higher than 1 mg/dL

Step-by-step explanation:

Assuming this problem:"A lab technician is tested for her consistency by making multiple measurements of the cholesterol level in one blood sample. The target precision is a standard deviation of 1 mg/dL or less. If 20 measurements are taken and the standard deviation is 2.2 mg/dL, is there enough evidence to support the claim that her standard deviation is greater than the target, at a significance level of= .01? "

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

[tex]n=20[/tex] represent the sample size

[tex]\alpha=0.01[/tex] represent the confidence level  

[tex]s^2 =2.2^2 =4.84 [/tex] represent the sample variance obtained

[tex]\sigma^2_0 =1[/tex] represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is violated, so the system of hypothesis would be:

Null Hypothesis: [tex]\sigma^2 \leq 1[/tex]

Alternative hypothesis: [tex]\sigma^2 >1[/tex]

Calculate the statistic  

For this test we can use the following statistic:

[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

[tex]\chi^2 =\frac{20-1}{1} 4.84 =91.96[/tex]

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 20-1=19. And since is a right tailed test the p value would be given by:

[tex]p_v =P(\chi^2 >91.96) \approx 0[/tex]

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(91.96,19,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that [tex]p_v <<\alpha[/tex] so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And we have evidence to conclude that the sample variance is higher than 1 and indeed that the deviation is higher than 1 mg/dL

Answer:

[tex]P(3.9<X<5)=P(\frac{3.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{3.9-4.8}{0.3}<Z<\frac{5-4.8}{0.3})=P(-3<z<0.67)[/tex]

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)[/tex]

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722[/tex]

d. .7462

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(4.8,0.3)[/tex]  

Where [tex]\mu=4.8´[/tex] and [tex]\sigma=0.3[/tex]

We are interested on this probability

[tex]P(3.9<X<5.0)[/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(3.9<X<5)=P(\frac{3.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{3.9-4.8}{0.3}<Z<\frac{5-4.8}{0.3})=P(-3<z<0.67)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)[/tex]

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

[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722[/tex]

And on this case the most accurate answer would be:

d. .7462

Answer: option d is the correct answer.

Step-by-step explanation:

For healthy females, x has a approximately normal distribution. We would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = red blood counts.

µ = mean count

σ = standard deviation

From the information given,

µ = 4.8

σ = 0.3

We want to find the that a healthy female has a red blood count between 3.9 and 5.0. It is expressed as

P(3.9 ≤ x ≤ 5) = P(x ≤ 5) - P(x ≤ 3.9)

For x = 3.9,

z = (3.9 - 4.8)/0.3 = - 3

Looking at the normal distribution table, the probability corresponding to the z score is 0.00135

For x = 5,

z = (5 - 4.8)/0.3 = 0.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.7486

P(3.9 ≤ x ≤ 5) = = 0.7486 - 0.00135

= 0.747

Answer: Interaction

Step-by-step explanation: Interaction in statistics occurs mostly during factorial experiments and analyzes of regression. When two variables interact, it means the

relationship between them and some other variable (dependent), depends on the value of the other interacting variable.Simply put, the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.

Finding interacting variables involve having a factorial design. Here, independent variables are "crossed" with one another so that there are observations at every levels of combination of the independent variables.

Interaction effect refers to the scenario in which the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.

Interaction effect refers to the scenario in which the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.

For example, let's say we are studying the effects of studying time and caffeine consumption on test scores. We find that the interaction effect between studying time and caffeine consumption is significant, indicating that the relationship between studying time and test scores depends on the level of caffeine consumption.

This concept is commonly explored in statistics and regression analysis.

hence, the final answer is Interaction effect .

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Answer: 2/3 hours

Step-by-step explanation: 1 1/3 will equal 4/3 as you have a whole number equaling 3/1 so then you divide by two and get 2/3

Answer: t= 460.52 minutes

Step-by-step explanation:Q'=Q/100

Q'= rate in and out of water

Finding the differential equation

Let Q'(t)= The quantity of dye in the tank for t time

But rate in=0 Q/200 ×2=Q'

Q'/Q=-1/100

Dividing by Q gives

Ln/Q/ + c = -1/100 + c1

Integrating both sides gives

Ln/Q/ = -(1/100)t + c2

But c+c1=C2= A constant

Q=C2e(-t/100)

200e-(t/100)

t= ln200

t=460.52minutes

Final answer:

To find the time it takes for the dye concentration in the tank to reach 1% of its original value, we can use the concept of dilution. By applying the dilution formula, we can calculate that it would take approximately 166.67 hours for the concentration to reduce to 1% of its original value.

Explanation:

To find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value, we can use the concept of dilution. The tank is being rinsed with fresh water flowing in at a rate of 2 L/min, and the well-stirred solution is flowing out at the same rate.

The concentration of dye in the tank after a certain amount of time can be calculated using the formula:

C1V1 = C2V2

Where:

Rearranging the formula to solve for V2:

V2 = (C1V1) / C2 = (1 g/L × 200 L) / (0.01 g/L) = 20000 L

So, it would take 20000 L / 2 L/min = 10000 min = 166.67 hours for the concentration of dye in the tank to reach 1% of its original value.

Answer:

%age of multifactor productivity reduction = 8.33%

Step-by-step explanation:

Total cost for chemicals = $10

Total cost of labor = $40

Total cost of misc = $5

Total initial cost = 10+40+5 = $55

Increase in use of chemical = 50%

Increase in cost of chemical = 10+ (0.5)(10)  =15

[Note (0.5)(10) is 50% of initial chemical cost]

Total increase in cost= $15+ $40 + $5 =%60

[tex]\%age\,\, of\,\, increase\,\, in \,\,cost =\frac{initial\,\, cost}{incresed\,\,cost}\times 100\%\\\\=\frac{55}{60}\times 100\%\\\\\\= 91.67\%[/tex]

%age of multifactor productivity reduction = 100 - 91.67 = 8.33%

Final answer:

The multifactor productivity of the cleaning company has fallen by approximately 9.09% after increasing the use of chemicals by 50%.

Explanation:

The student is asking about how the increase in chemical use affects the multifactor productivity (MFP) of a cleaning company. The original costs of cleaning one house are $10 for chemicals, $40 for labor, and $5 for miscellaneous expenses. With a 50% increase in chemical use, the new cost for chemicals becomes $15.

Firstly, we need to calculate the total initial cost: $10 (chemicals) + $40 (labor) + $5 (misc) = $55. After the increase in chemical use, the new total cost is $15 (chemicals) + $40 (labor) + $5 (misc) = $60. To find the percentage decrease in MFP, we compare the change in input cost to the original input cost.

The productivity decrease is calculated as: (($60 - $55) / $55) * 100 = (5 / 55) * 100 \\approx 9.09%. Thus, the multifactor productivity has fallen by approximately 9.09%.

Answer:

1.45lb of salt is present after 1 minute

Step-by-step explanation:

The detailed steps and derivation from integration is as shown in the attachment.

Answer and Step-by-step explanation:

a) In statistics, an experimental unit is one member of a set of entities being studied. Experimental units are the individuals on whom an experiment is being performed on.

25 pairs of jeans are randomly selected, hence, a single experimental unit is a pair of jeans.

b) Variables are the qualities/topics being investigated. Qualitative variables puts the variables in categories while quantitative variables involve numerical variables.

This question focuses on which state each pair of jeans is being produced, therefore this quality categorizes the jeans according to which state they were produced in. Hence, the variable being measured is a qualitative one.

c) For the pie chart, we need the frequency of the pair's of the jeans according to which state they were produced in.

California, CA, frequency = 9

Arizona, AZ, frequency = 8

Texas, TX, frequency = 8

Total = 25

A pie chart is based on 360°

CA will occupy (9/25) × 360° = 129.6°

AZ will occupy (8/25) × 360° = 115.2°

TX will occupy (8/25) × 360° = 115.2°

The pie chart is drawn with Microsoft Excel and presented in the image attached to this answer.

d) The bar chart is drawn with Microsoft Excel and presented in the image attached to this question. Each bar has equal width, but the height of each bar corresponds to its frequency.

e) The proportion of jeans produced in Texas = 8/25 = 0.32

f) The state that produces the highest number of jeans is the one with the highest frequency. That is California with frequency of 9 out of 25.

(g) The plant that produced more jeans than the others is the state in the

pie chart of part (c) with the largest angle (129.6°) or slice (California) and is the state in the bar graph of part (d) that has the highest bar.

From the data, it is evident that the three states make almost the same number of jeans, but California slightly edges the other two by being the state that produces the most number of jeans. Arizona and Texas produce the similar amounts of Jeans.

Although, this is just a sample out of a whole large quantity of Jeans, the laws of statistics and probability makes this random selection a representative of the whole set of jeans produced.

Hope this helps!

In this case, the experimental unit is the pair of jeans, the variable measured is the production state which is qualitative. The proportional production and most productive state can be determined by counting entries, and pie and bar charts can illustrate production distribution.

a. The experimental unit is the individual pair of jeans.

b. The variable being measured is the state in which the jeans are produced, which is a qualitative variable.

c. and d. To construct the pie and bar charts, you simply need to count the total number of jeans produced in each state and represent this on the charts. This can be useful for visualizing the distribution of production across the states.

e. The proportion of jeans made in Texas can be found by counting the number of 'TX' entries and dividing by the total number of jeans (25).

f. The state that produced the most jeans is the one with the highest count in your data set. You would need to tally the appearances of each state to determine this.

g. The charts from parts c and d can assist you in determining whether there is a noticeable difference in production between the states. If one section of the pie chart or one bar on the bar graph is significantly larger than the others, it would suggest that state produces significantly more jeans. Similarly, if all sections/bars are similar in size, it suggests equal production across the states.

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Answer:

(1) Correct option is (A)

(2) P (H or G) = 0.80

(3) Correct option is (C)

Step-by-step explanation:

Mutually exclusive events are those events that cannot occur at same time.

If events A and B are mutually exclusive then, P (A and B) = 0.

Given: P (G) = 0.50 and P (H) = 0.30

(1)

The statement is: P (H|G) = 0.40.

The conditional probability of event B given event A is:

[tex]P(B|A)=\frac{P(A\cap B)}{P(A)}[/tex]

The probability statement P (H|G) can be written as:

[tex]P(H|G)=\frac{P(H\cap G)}{P(G)}[/tex]

But as H and G are mutually exclusive, P (H ∩ G) = 0.

Hence, P (H|G) = 0.

Thus, the provided statement is false because events H and G are mutually exclusive, which makes P(H ∩ G) = 0.

Option (A) is correct.

(2)

The addition rule of probability states that:

[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]

Compute the value of P (H or G) as follows:

[tex]P(H\cup G)=P(H)+P(G)-P(H\cap G\\=0.30+0.50-0\\=0.80[/tex]

Thus, the value of P (H or G) is 0.80.

(3)

Independent events are those events that are not affected by the occurrence of other events.

Events H and G are mutually exclusive events, i.e. occurrence of one affects the occurrence of other.

Thus, events H and G are dependent events.

Option (C) is correct.

Mutually exclusive events have a probability of 0 when they occur together. To find the probability of G OR H, use P(H) + P(G) - P(H AND G). G and H are independent events because they are mutually exclusive.

G and H are mutually exclusive events, which means they cannot occur at the same time. The probability of two mutually exclusive events happening together, P(H AND G), is always 0.

Therefore, the statement P(H | G) = 0.4 must be false. The correct answer is (A.) The events are mutually exclusive, which makes P(H AND G) = 0; therefore, P(H | G) = 0.

To find P(H OR G), you can use the formula P(H OR G) = P(H) + P(G) - P(H AND G). Plugging in the given values, P(H OR G) = 0.3 + 0.5 - 0 = 0.8.

G and H are mutually exclusive events, which means that the occurrence of one event does not affect the probability of the other event happening.

Therefore, G and H are independent events.

The correct answer is (B.) G and H are independent events because they are mutually exclusive.

Equation 1: 3.75t + 7.50b = 750
Equation 2: b = 2t

We can set up a system of equations to determine the number of tacos and burritos sold by Samantha.

Let's define the variables as follows:
Let "t" represent the number of tacos sold.
Let "b" represent the number of burritos sold.

According to the given information, we know:
1. Each taco is sold for $3.75, and each burrito is sold for $7.50.
2. Samantha made a total of $750 in revenue from all taco and burrito sales.
3. There were twice as many burritos sold as there were tacos sold.

To represent the total revenue, we can set up the equation:
3.75t + 7.50b = 750

To represent the relationship between the number of burritos and tacos sold, we can set up the equation:
b = 2t

Now we have a system of equations:
Equation 1: 3.75t + 7.50b = 750
Equation 2: b = 2t

By solving this system of equations, we can find the values of "t" and "b" which represent the number of tacos and burritos sold by Samantha.

Answer:

[tex]P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)[/tex]

[tex]P(Z>0.8)=1-P(Z\leq 0.8)[/tex]

[tex]P(Z>0.8)=1-0.788=0.212[/tex]

Step-by-step explanation:

Assuming this previous info : "Trucks carry barrels of crude oil from a port in Texas to a distributer in New Mexico on long  trailers. Filling equipment is used to fill the barrels with the oil. When functioning properly,  the actual volume of oil loaded into each barrel by the filling equipment at the port is  approximately normally distributed with a mean of 55 gallons and a standard deviation of 0.5  gallons. If the mean is greater than 55.4 gallons, the filling mechanism is overfilling."

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 amount filling of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(55,0.5)[/tex]  

Where [tex]\mu=55[/tex] and [tex]\sigma=0.5[/tex]

We are interested on this probability

[tex]P(X>55.4)[/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>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z>0.8)=1-P(Z\leq 0.8)[/tex]

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

[tex]P(Z>0.8)=1-0.788=0.212[/tex]

Final answer:

The probability of a randomly selected barrel containing 55.4 gallons or more is calculated using normal distribution and z-scores. Since the value of 55.4 gallons is significantly higher than the mean with a large positive z-score, the probability is very small and close to 0.

Explanation:

The question involves finding the probability that the volume of oil in a randomly selected barrel will be 55.4 gallons or more, given that the normal operating conditions of a tank are 45 gallons with a 3-gallon standard deviation. This is a problem of probability involving normal distribution, where we are looking for the area under the normal curve to the right of 55.4 gallons.

To find this probability, we first calculate the z-score, which is the number of standard deviations away from the mean the value of 55.4 gallons is. The z-score formula is given by Z = (X - μ) / σ, where X is the value of interest (55.4 gallons), μ is the mean (45 gallons), and σ is the standard deviation (3 gallons). Plugging in the numbers, we get Z = (55.4 - 45) / 3 = 3.467.

Next, we use z-score tables or a calculator to find the probability corresponding to the calculated z-score. The probability represents the area to the left of the z-score under the normal curve. Since we are interested in the probability of the volume being more than 55.4 gallons, we need to subtract this value from 1 to find the area to the right of the z-score.

If using a standard normal distribution table, which gives the area to the left, we find the value corresponding to Z = 3.467. If such a precise value is not available in the table (which is likely because 3.467 is quite high for typical z-score tables), we would take the value for the closest z-score provided or use software or a calculator with statistical functions. Most standard normal tables do not extend beyond a z-score of 3.49, which is associated with a probability very close to 1. Therefore, the probability of the barrel having 55.4 gallons or more is very small and could be approximated as P(Z > 3.467) ≈ 0.

Answer:

49.32% probability that the proportion in our sample of green candies will be more than 25%.

Step-by-step explanation:

For each candy, there are only two possible outcomes. Either it is green, or it is not. The probabilities of each candy being green are independent from each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [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]

And p is the probability of X happening.

In this problem we have that:

According to the manufacturer of the candy Skittles, 25% of the candy produced are green. This means that [tex]p = 0.25[/tex]

If we take a random sample of 10 bags of Skittles, what is the probability that the proportion in our sample of green candies will be more than 25%?

10 bags, so [tex]n = 10[/tex]

This is [tex]P(X > 0.25*10) = P(X > 2.5)[/tex]

The number of bags is a discrete number, so [tex]P(X > 2.5) = P(X \geq 3)[/tex].

Either the number of green bags is lower than 3, or it is greater or equal than 3. The sum of these probabilities is decimal 1. Mathematically, this means that:

[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]

We want [tex]P(X \geq 3)[/tex]. So

[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]

In which

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

So

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

[tex]P(X = 0) = C_{10,0}.(0.25)^{0}.(0.75)^{10} = 0.0563[/tex]

[tex]P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1689[/tex]

[tex]P(X = 2) = C_{10,2}.(0.25)^{2}.(0.75)^{8} = 0.2816[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0563 + 0.1689 + 0.2816 = 0.5068[/tex]

Then

[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.5068 = 0.4932[/tex]

49.32% probability that the proportion in our sample of green candies will be more than 25%.

Final answer:

The subject matter revolves around the concept of probability in mathematics, particularly involving calculations using the binomial and Poisson distributions, and statistical methods for analyzing sample data.

Explanation:

The question posed relates to the concept of probability in mathematics, specifically the calculation of the probability of obtaining a certain number of items of a particular color in a random sample from a given population. This field of study typically involves understanding and applying theories of probability distributions, such as the binomial distribution and the Poisson distribution, as well as using statistical methods to estimate population parameters and test hypotheses.

For example, to find the probability of drawing exactly 3 green marbles from a bag containing 50 marbles of which 15 are green, one would need to use the binomial distribution formula P(X = k) = C(n, k) p^k (1-p)^(n-k), where 'C(n, k)' is the combination of 'n' items taken 'k' at a time, 'p' is the probability of success on a single trial, and 'X' is the random variable representing the number of successes in 'n' trials. Similar concepts are applicable to the drawing of candies from a bag and calculating the proportion of colors in a sample to match or compare with theoretical distributions.

Exercises that involve calculating mean and standard deviations for repeated sampling, such as in the exercise with M&M's, are instances of applying statistical methods to analyze data. This process of analysis provides a deeper understanding of the underlying distributions and the behavior of random variables.

Answer:

The value of K is 90461 Pa·s^(0.456) and the value of n is 0.456 (with no units).

Compleated data:

η              ∂vx / ∂y

0,02 750000

0,05 450000

0,1        350000

0,2        200000

0,5        130000

1         100000

2         60000

5         35000

10          28000

20           17000

50          10000

100            8000

Step-by-step explanation:

To solve this problem we can use a Least Squares Approximation with a power function to approximate the data sample.

In this case, we have to do some mathematical work to linearize the function.

For the function selected:

[tex]\eta=K(\partial v_x/\partial y)^{n-1}\\ln(\eta)=ln(K(X)^{m})=ln(K)+ln((\partial v_x/\partial y)^{n-1})\\ln(\eta)=ln(K)+(n-1)ln((\partial v_x/\partial y))[/tex]

Now we can do the next change of variables:

[tex]ln(\eta)=Y\\ln(K)=C\\(n-1)=m\\ln((\partial v_x/\partial y))=X\\[/tex]

Therefore:

[tex]Y=C+mX[/tex]

the matrix resultant of Least Squares Approximation with the data above is:

[tex]Y=\left[\begin{array}{c}-3.91&-3&-2.3\\-1.61&-0.69&0\\0.69&1.61&2.3\\3&3.91&4.61\end{array}\right][/tex] and [tex]\left[\begin{array}{cc}1&13.53\\1&13.02\\1&12.77\\1&12.21\\1&11.78\\1&11.51\\1&11\\1&10.46\\1&10.24\\1&9.74\\1&9.21\\1&8.99\end{array}\right] \cdot \left[\begin{array}{c}C&M\end{array}\right]=A\cdot x[/tex]

We then solve the equation:

[tex]A\cdot x=Y\\A^tA\cdot x=A^tY=b[/tex]

Solving this system of 2x2, we obtain:

C=11.4126741 and m=-0.544

Therefore

[tex]C=11.4126741=ln(K)\rightarrow K=90461\\m=-0.544=(n-1)\rightarrow n=0.456[/tex]

Knowing that the viscosity has as units Pa·s and the shear rate s⁻¹, the units of the constant k is:

[tex]K=90461 Pa\cdot s^{0.456}\\[/tex]

The constant n has no units.

Answer and Step-by-step explanation

η = K ((∂Vx / ∂y)^(n-1))

-Take the natural logarithm of both sides

In η = In {K ((∂Vx / ∂y)^(n-1))}

In η = In K + In ((∂Vx / ∂y)^(n-1))

In η = In K + (n-1) In (∂Vx / ∂y)

In η = (n-1) In (∂Vx / ∂y) + In K

-Compare this relation to the equation of a straight line, y = mx + C

y = In η

m = (n-1)

x = In (∂Vx / ∂y)

C = In K

So, the data missing must be for the Viscosity, η and the shear rate, ∂Vx / ∂y

- First step in the data treatment is to take the natural logarithm of these data sets.

- This leads to a new table of data with In η and In (∂Vx / ∂y).

- Plot this new set of data on a graph with In η on the y-axis and In (∂Vx / ∂y) on the x-axis.

- The slope of this graph, m = (n-1) from the power law relation. Therefore, n = slope + 1

- And the intercept on the y-axis, c = In K, that is, K = (e^c)

So, there goes the answers to the questions, n and K.

n has no units and K has varying units depending on the value of n.

Answer:

Sample Space = {SSS,SSF,SFF,SFS,FSS,FSF,FFS,FFF}

Step-by-step explanation:

Given a sale representative makes presentations about a product in three homes per day.

In each home, Sale is denoted by S and No Sale is denoted by F.

Now the cases for required sample space will be :

So our Required sample space for the experiment

    = {SSS,SSF,SFF,SFS,FSS,FSF,FFS,FFF}.

Answer:

standard deviation of these expected returns = 0.0295 or 2.95%

Step-by-step explanation:

The detailed step is shown in the attachment

To calculate standard deviation of a portfolio's returns, one must compute the average return, find deviations, square them, average those, and take the square root. However, without weights for each stock in the portfolio, the calculation cannot be completed with the provided information.

The student asked what the standard deviation of the expected returns for a portfolio consisting of four stocks is, with returns of minus 9%, 11%, 13%, and 17% respectively over the next four years. To calculate standard deviation, we need to follow several steps:

However, the information given does not include all the necessary data, such as the proportion of each stock within the portfolio. Without individual stock weights, we cannot compute the exact number for the portfolio's standard deviation. Typically, such a calculation would also require historical return data or expected return data for each stock to compute the portfolio variance and standard deviation.

Answer:

20m

Step-by-step explanation:

Say you walk down the street for 10m. Then you take a left and walk for another 10m then go inside a bakery. You walked 10m on one street and 10m on another street to get to the bakery. In total you walked 20m to get to the bakery. We know this because 10 + 10 = 20.

I really do hope that this helps you! Have a blessed day!

Answer:

x = 0.51

Step-by-step explanation:

[tex]4\cdot 8^x = 11.48\\8^x = \frac{11.48}{4}\\ 8^x = 2.87\\\log_8 8^x = \log_8 2.87\\x=\log_8 2.87\\x=0.51[/tex]

Answer:

x=0.51

Step-by-step explanation: