Find an equation of the line passing through the pair of points (4 comma 5 )and (7 comma 11 ). Write the equation in the form Ax plus By equals Upper C.

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.

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.

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:

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:

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: 160 liters of the

40% solution and 80 liters of the 70% solution will be used.

Step-by-step explanation:

Let x represent the number of liters of 40% antifreeze solution that should be used.

Let y represent the number of liters of 70% antifreeze solution that should be used.

The volume of the mixture to be mixed is 240 liters. It means that

x + y = 240

The 40% antifreeze solution is to be mixed with a 70% antifreeze

solution to get 240 liters of a 50% solution. This means that

0.4x + 0.7y = 0.5(240)

0.4x + 0.7y = 120 - - - - - - - - - - - -1

Substituting x = 240 - y into equation 1, it becomes

0.4(240 - y) + 0.7y = 120

96 - 0.4y + 0.7y = 120

- 0.4y + 0.7y = 120 - 96

0.3y = 24

y = 24/0.3

y = 80

x = 240 - y = 240 - 80

x = 160

[tex]160 \ litres[/tex] of [tex]40 \%[/tex] antifreeze solution and [tex]80 \ litres[/tex] of [tex]70 \%[/tex] antifreeze solutions will be used.

Given, two solutions namely [tex]40 \%[/tex] antifreeze and [tex]70 \%[/tex] antifreeze solutions.

Let [tex]x[/tex] litres of the [tex]40 \%[/tex] antifreeze solution and [tex]y[/tex] litres of the [tex]70 \%[/tex] antifreeze solutions will be used.

Total volume of the solution,

[tex]x+y=240..........(1)[/tex]

Now, [tex]40\%[/tex] of antifreeze solution is to be mixed with a [tex]70 \%[/tex] of antifreeze

solution to get 240 liters of a [tex]50 \%[/tex] solution,

[tex]0.4x+0.7y=240\times 0.5[/tex]

[tex]0.4x+0.7y=120........(2)[/tex]

From Equation (1)  [tex]y=240-x[/tex], substitute the value of [tex]y[/tex] in Equation (2),we get

[tex]0.4x+0.7(240-x)=120\\0.4x+168-0.7x=120\\-0.3x=120-168\\-0.3x=-48\\x=160[/tex]

Putting the value of [tex]x=160[/tex], we get

[tex]y=240-160[/tex]

[tex]y=80[/tex].

Hence [tex]160 \ litres[/tex] of [tex]40 \%[/tex] antifreeze solution and [tex]80 \ litres[/tex] of [tex]70 \%[/tex] antifreeze solutions will be used.

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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: 60

Step-by-step explanation:

Given : The number of choices for different temperatures = 3

The number of choices for different pressures = 4

The number of choices for different catalyst = 5

Since , the experimenter is studying the effects of temperature, pressure, and different type of catalysts.

Then by using the Fundamental principle of counting , we have

The number of  experimental runs are available = (number of choices for  temperatures ) x (number of choices for pressures) x( number of choices for  catalyst)

= 3 x 4 x 5 = 60

Hence, the number of experimental runs are available = 60

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:

Value of b = 14

Step-by-step explanation:

The detailed calculations with steps is 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: 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:

[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 10th term of the sequence is 3

Step-by-step explanation:

A sequence is a list of numbers or objects in a specific order

The nth term given in the question is not a function of n

i.e aₙ= 3

Since the sequence starts with an index of 1

a₁=3

All other terms in the sequence will also be 3. Meaning that it is an arithmetic progression with a common difference of 0.

The 10th term is given by

a₁₀= 3

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:

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:

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:

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:

a. D'(p) = -10p - 6

b. There is a decrease of 96 units of demand for each dollar increase

Step-by-step explanation:

The demand function is:

[tex]D(p) = - 5p^2-6p+400[/tex]

(a) The derivate of the demand function with respect to price gives us the rate of change of demand:

[tex]\frac{dD(p)}{dp}=D'(p) = -10p-6[/tex]

(b) When p = $9, the rate of change of demand is:

[tex]D'(9) = -10*9-6\\D'(9) = -96\ \frac{units}{\$}[/tex]

This means that, when p = $9,  there is a decrease of 96 units of demand for each dollar increase.

Final answer:

The rate of change of demand with respect to price is given by the derivative D'(p) = -10p - 6. When the price is $9, the rate of change of demand is -96, indicating that for each dollar increase in price, demand decreases by 96 items.

Explanation:

To address the demand model problem, we first need to calculate the rate of change of demand with respect to price. This involves taking the derivative of the demand function D(p) = -5p^2 - 6p + 400 with respect to price p. The derivative, D'(p), is calculated as follows:

Differentiate each term with respect to p:

The derivative of -5p^2 is -10p.

The derivative of -6p is -6.

The derivative of a constant (400) is 0.

Combine these to get the rate of change formula D'(p) = -10p - 6.

To find the rate of change of demand when the price is $9, we substitute p with 9 into the rate of change formula:

D'(9) = -10(9) - 6 = -90 - 6 = -96

The rate of change of demand at a price of $9 is -96 items per dollar. This means that for each one dollar increase in price, the quantity demanded decreases by 96 items.

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.

Final answer:

To identify the score separating risk-averse customers from others in a normally distributed set of questionnaire scores, we find the 10th percentile, which corresponds to a z-score of -1.2816. Using the mean of 49.5 and a standard deviation of 15, we calculate the cutoff score as 30.3.

Explanation:

To find the score that separates the customers who are considered risk averse from those who are not, we must look for the score at the 10th percentile in the normal distribution. Since the scores are approximately normally distributed, we can use the standard z-score table or a statistical calculator to find this value.

The mean (μ) of the distribution is 49.5, and the standard deviation (σ) is 15. We want to find the z-score that corresponds to the cumulative probability of 0.10. Looking at the z-score table or using a calculator, we find that the z-score associated with the bottom 10% of the distribution is approximately -1.2816.

Now we can use the z-score formula:
z = (X - μ) / σ

Where X is the score we are looking for. Substituting the values we have, we get:
-1.2816 = (X - 49.5) / 15

Solving for X:
X = -1.2816 * 15 + 49.5

X ≈ -19.224 + 49.5

X ≈ 30.276

When rounded to one decimal place, we get X = 30.3. Therefore, a score of 30.3 on the questionnaire separates those who are considered risk averse from those who are not.

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:

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:

[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:

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:

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:

The probability of the three blue marbles ending up next to one another (i.e., without any yellow marbles in between them is 1/5 or 0.2.

Step-by-step explanation:

The 6 marbles can be arranged in 6! ways, But, there are 3 identical marbles of similar colour in two separate cases,

So, 6 marbles, consisting of 3 blue and 3 yellow marbles can be arranged in 6!/(3!3!) ways = 20 ways.

But, to arrange the six marbles in such a way that the 3 blue marbles end up next to one another without any yellow marble between them. This can be done by viewing the 3 blue marbles as one. Therefore, there are 4 marbles; 3 identical, blue marbles and 1 special marble.

To arrange that, it is, 4!/(3!1!) = 4

The probability of the three blue marbles ending up next to one another (i.e., without any yellow marbles in between them will be 4/20 = 1/5 or 0.2.

Final answer:

The probability that the three blue marbles will end up next to each other is 1/120. This is calculated by considering the blue marbles as one unit and arranging them with the yellow marbles, leading to 24 total arrangements, but since the blue marbles are indistinguishable, the number of favorable outcomes is the same as the arrangement of the yellow marbles, which is 6. The total number of possible outcomes is 720, calculated by 6 factorial.

Explanation:

The question asks for the probability that three blue marbles will end up next to each other after being dropped and rolling into a crack. To solve this, consider the three blue marbles as a single unit. Since there are also three yellow marbles, we can arrange four units (three blue marbles together as one unit and the three individual yellow marbles) in 4! (4 factorial) ways, which is equal to 24. However, the three blue marbles as a single unit can also be arranged among themselves in 3! (3 factorial) ways, but since they are indistinguishable, we don't consider these arrangements separate. So, there is only one way to arrange the blue block. Therefore the total number of favorable outcomes is the same as the number of ways to arrange the yellow marbles, which is also 3! or 6. To find the probability, we divide the favorable outcomes by the total possible outcomes without restrictions.

The total possible outcomes without any restrictions can be calculated assuming all 6 marbles are distinct, which gives us 6! (6 factorial) arrangements, equal to 720. Applying the probability formula, we have-

Probability = Favorable outcomes / Total outcomes = 6 (the number of ways to arrange the yellow marbles) / 720 possible arrangements = 6/720 = 1/120.

Therefore, the probability that the three blue marbles will end up next to each other is 1/120.

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:

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:

%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%.