Solve the equation M=7r2h/19 for r in terms of M and h. Assume r, M and h are all positive.

Answer:

c. Inductive and Strong

Step-by-step explanation:

In inductive reasoning, provided data is analyzed in order to reach a conclusion. In this case, the argument provides data regarding Jane and Nancy's awards and their love for mathematics and then draws a conclusion regarding Nancy's performance in a particular class, this is an example of inductive reasoning.

As for the strength of the argument, it is plausible to infer that Jane and Nancy have similar mathematics skills since they both love calculus and excel academically. Therefore, if Jane does well in the calculus class, it is a strong argument to say that Nancy does as well.

The answer is :

c. Inductive and Strong

Final answer:

The argument is Deductive and Invalid as it does not guarantee Nancy's success in the calculus class based on Jane's performance.

Explanation:

The argument presented is Deductive and Invalid. This is because the conclusion that Nancy will do well in the calculus class based on Jane's performance does not necessarily follow logically. Just because Jane excelled does not guarantee the same result for Nancy.

In deductive reasoning, the conclusion must follow necessarily from the premises, which is not the case here. An example of a valid deductive argument would be 'All humans are mortal, Socrates is a human, therefore Socrates is mortal.'

Because the argument in question does not provide a guarantee of Nancy's success based on Jane's performance, it is categorized as Invalid in the realm of deductive reasoning.

Final answer:

The sample space S for a. is a set of all possible denominations of cards; the sample space S for b. is a set of all possible combinations of two cards with color recorded; event H is defined as the first card being hearts in c.; event DC is defined as the first card being diamonds and second card being red in d.; events C and B are disjoint in e.

Explanation:

a. One card is selected at random and the denomination is recorded. What is the sample space S for the set of possible outcomes?

The sample space S for this scenario is the set of all possible denominations of cards in a standard deck, which includes the numbers 1 through 10, as well as the face cards (J, Q, K) and the Ace (A).

b. Two cards are selected at random and the color is recorded. What is the sample space S for the set of possible outcomes?

The sample space S for this scenario is the set of all possible combinations of two cards from a standard deck, with the colors (red or black) of each card being recorded. Since there are 26 red cards and 26 black cards, the number of possible outcomes is 26 * 26 = 676.

c. Two cards are selected at random and the denomination is recorded. The event H is defined as the event that the first card is hearts. What defines event H?

Event H is defined as the event where the first card selected is a heart. In terms of the sample space, it can be defined as the subset of sample space S that includes all outcomes where the first card is a heart.

d. Two cards are selected at random and the denomination is recorded. The event D is defined as the event that the first card is diamonds and the second card is red. What defines event DC?

Event DC is defined as the event where the first card selected is a diamond and the second card selected is red. In terms of the sample space, it can be defined as the subset of sample space S that includes all outcomes where the first card is a diamond and the second card is red.

e. Two cards are selected at random. Event C is defined as the event that the first card is clubs, event R as the event that the first card is red, and event B as the event that the second card is black. Which events are disjoint?

Events C and B are disjoint because the first card cannot be both a club and black at the same time. However, events R and B are not disjoint because there are red clubs in the deck, which would satisfy both events.

The sample spaces for the following are as follows:  a) Hearts, Diamonds Clubs, Spades. b) Red Red,  Red Black, Black Red, Black Black.

c) Hearts - Hearts, Hearts - Diamonds, Hearts - Clubs, Hearts - Spades.

d) Diamond - Black, Black - Black, Black - Red, Hearts - Any.

e) The events C and R are disjoint.

A standard deck of cards has 52 cards, with 26 red and 26 black cards across four suits: hearts, diamonds, clubs, and spades, each containing 13 cards.

a. One card is selected at random and the denomination is recorded.

The sample space for the denomination of a single card is:

Hearts

Diamonds

Clubs

Spades

b. Two cards are selected at random and the color is recorded.

The sample space for the color of two cards selected can be:

Red, Red

Red, Black

Black, Red

Black, Black

c. Two cards are selected at random and the denomination is recorded. T

Event H is defined as drawing a heart for the first card. The possible outcomes for the second card can be any of the four denominations: hearts, diamonds, clubs, or spades. So, the event H sample space is:

Hearts - Hearts

Hearts - Diamonds

Hearts - Clubs

Hearts - Spades

d. Two cards are selected at random and the denomination is recorded.

Event D is defined as drawing a diamond for the first card and a red card for the second card. Since a red card can only be a diamond or heart, event DC (complement of event D) includes any draw combinations not fitting this pattern:

Diamond - Black

Black - Black

Black - Red

Hearts - Any

e. Two cards are selected at random.

Events C (first card is clubs) and R (first card is red) are disjoint since a card cannot simultaneously be clubs (black) and red. However, either of these events can be paired with event B (second card is black). Therefore, events C and R are disjoint.

The sample spaces for the following are as follows:  a) Hearts, Diamonds Clubs, Spades. b) Red Red,  Red Black, Black Red, Black Black.

c) Hearts - Hearts, Hearts - Diamonds, Hearts - Clubs, Hearts - Spades.

d) Diamond - Black, Black - Black, Black - Red, Hearts - Any.

e) The events C and R are disjoint.

Answer:

a) 0.659

b) 0.341

c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.

Step-by-step explanation:

Considering the left first

Play goes through the left 30% of the time

- The right guard takes a balanced stance 90% of the time when play goes through the left, that is, 0.9 × 0.3 = 27% overall.

- He takes a shift stance 10% of the time that play goes through the left = 0.1 × 0.3 = 3% overall.

Play goes through the right, 70% of the time.

- He takes a balanced stance 20% of the time that play goes through the right, that is, 0.7 × 0.2 = 14% overall

- He takes a shift stance 80% of the time that play goes through the right, that is, 0.7 × 0.8 = 56% overall.

Then the right guard takes a balanced stance 27% + 14% of all the time = 41% overall.

a) Probability that play goes through the left with the right guard in a balanced stance = 27/41 = 0.659

b) Probability that play goes through the right with the right guard in a balanced stance = 14/41 = 0.341

c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.

Using conditional probability, we can find the probabilities of the play going left or right. The probability of the play going left is 0.3, and the probability of the play going right is 0.7. If the linebacker sees a balanced stance, the probability of the play going left is approximately 0.658, so the linebacker should prepare to defend against a play going left.

To find the probabilities of the play going left or right, we need to use conditional probability. Let's denote L as the event of the play going left, and R as the event of the play going right. We are given P(L) = 0.3 and P(R) = 0.7. We also have the following conditional probabilities: P(B|L) = 0.9 (balanced stance when play goes left) P(S|L) = 0.1 (shift stance when play goes left) P(B|R) = 0.2 (balanced stance when play goes right) P(S|R) = 0.8 (shift stance when play goes right)

a. To find the probability that the play will go left, we can use the Law of Total Probability. P(L) = P(L∩B) + P(L∩S) = P(L)P(B|L) + P(L)P(S|L) = 0.3 X 0.9 + 0.3 * 0.1 = 0.27 + 0.03 = 0.3.

b. Similarly, to find the probability that the play will go right, P(R) = P(R∩B) + P(R∩S) = P(R)P(B|R) + P(R)P(S|R) = 0.7 X 0.2 + 0.7 X 0.8 = 0.14 + 0.56 = 0.7.

c. If the linebacker sees a balanced stance, it means that P(B) = P(B|L)P(L) + P(B|R)P(R) = 0.9 X 0.3 + 0.2 X 0.7 = 0.27 + 0.14 = 0.41. Now, we can compare the conditional probability of the play going left given a balanced stance, P(L|B) = P(B|L)P(L)/P(B) = (0.9 X 0.3)/0.41 = 0.27/0.41 ≈ 0.658. Therefore, the linebacker should prepare to defend against a play going to the left.

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Answer:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]  

And rounded up we have that n=601

Step-by-step explanation:

Previous concepts

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

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

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

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that [tex]ME =\pm 0.04[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

The estimated proportion is [tex] \hat p =0.5[/tex]. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]  

And rounded up we have that n=601

In this question, we cannot provide most of the answer, as it requires the use of Powerpoint and the personal experiences of each participant. However, we are able to talk about some of the personal conflicts and work conflicts that students might have faced in their lives. Some examples that you could use in your text are:

Answer:

$1 USD = $1.20 Canadian dollars

Step-by-step explanation:

Use proportions of Canadian dollars over US dollars so you get

300 CAD/250 USD = x CAD / 1 USD

solve for x to get 1.2

So $1 USD is equal to $1.20 Canadian dollars.

Answer:

$1.20

Step-by-step explanation:check work by any of these ways.

300/1.20=250,

300/250=1.20,

Or 1.20*250=300

Answer:

We conclude that 4095 different teams can be selected.

Step-by-step explanation:

We know that a company has 24 employees: 10 women and 14 men.

A team of 4 employees must be selected. The team must have two women and two men.

So from 10 women we choose 2 women:

[tex]C_2^{10}=\frac{10!}{2!(10-2)!}=45\\[/tex]

So from 14 men we choose 2 men:

[tex]C_2^{14}=\frac{14!}{2!(14-2)!}=91\\[/tex]

We get: 45 · 91 = 4095.

We conclude that 4095 different teams can be selected.

Final answer:

The total number of different teams that can be selected from the department to represent at the companywide meeting is 4095.

Explanation:

To select a team of 4 employees with 2 women and 2 men from a department with 10 women and 14 men, we can use the concept of combinations. We need to choose 2 women out of 10 and 2 men out of 14.

The number of ways to choose 2 women from 10 is given by 10C2 = 10! / (2! * (10-2)!) = 45.

The number of ways to choose 2 men from 14 is given by 14C2 = 14! / (2! * (14-2)!) = 91.

Therefore, the total number of different teams that can be selected is 45 * 91 = 4095.

The arc length of AB = 8.37 meters

Solution:

Degree of AB (θ) = 60°

Radius of the circle = 8 m

Let us find the arc length of AB.

Arc length formula:

[tex]$\text{Arc length}=2 \pi r\left(\frac{\theta}{360^\circ}\right)[/tex]

                 [tex]$=2 \times 3.14 \times 8 \left(\frac{60^\circ}{360^\circ}\right)[/tex]

                 [tex]$=2 \times 3.14 \times 8 \left(\frac{1}{6}\right)[/tex]

Arc length = 8.37 m

Hence the arc length of AB is 8.37 meters.

The function f(x) = x3 - 9x2 - 21x + 8 increases in the intervals (-∞, -1) and (7, ∞), and decreases in the interval (-1, 7). It achieves a local minimum at -190 and a local maximum at -23. The inflection point is at x = 3, with the function being concave up for x < 3 and concave down for x > 3.

Given the function f(x) = x3 - 9x2 - 21x + 8, we need to find intervals where the function is increasing or decreasing, and determine its local minima/maxima, inflection points, and where it is concave up or down.

To find the intervals, we need to first take the derivative of f(x): f'(x) = 3x2 - 18x - 21 which becomes 3(x - 7)(x + 1). The critical points are then 7 and -1.

The function increases in the interval (-∞, -1) and decreases in the interval (-1, 7). The function again increases in the interval (7, ∞).

Local minima and maxima can be found by evaluating the function at the critical points. f(-1) = -23 and f(7) = -190, so the local maximum value of f is -23 and local minimum is -190.

To find the inflection point, we compute the second derivative f''(x) = 6x - 18, which simplifies to 6(x - 3), giving us an inflection point at x = 3.

The function is concave up for x < 3 and concave down for x > 3.

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a) f(x) is increasing for x in (- ∞, - 1) cup(7, infty)

f(x) is decreasing for x \in (- 1, 7)

b)Local minimum=-237 Local maximum=19

c)Point of inflection: (x, y) = (3, - 109)

d)Interval for concave upward curve: x \in (3, ∞)

e)Interval for concave upward curve: x \in (- ∞, 3)

The given function is [tex]f(x) = x ^ 3 - 9x ^ 2 - 21x + 8[/tex]

[tex]f' * (x) = 3x ^ 2 - 18x - 21[/tex]

f^ prime prime (x) = 6x - 18

a)f(x) is increasing for

f' * (x) > 0

[tex]3x ^ 2 - 18x - 21 > 0[/tex]

[tex]x ^ 2 - 6x - 7 > 0[/tex]

[tex]x ^ 2 - 7x + x - 7 > 0[/tex]

x(x - 7) + 1(x - 7) > 0

(x - 7)(x + 1) > 0

x > 7 and x > - 1or x < 7 and x < - 1

f(x) is decreasing for

f' * (x) < 0

3x ^ 2 - 18x - 21 < 0

x ^ 2 - 6x - 7 < 0

x ^ 2 - 7x + x - 7 < 0

x(x - 7) + 1(x - 7) < 0

(x - 7)(x + 1) < 0

x > 7 and x < - 1 or x < 7 and x > - 1

x=(-1,7)

b)for local extrema

f' * (x) = 0

3x ^ 2 - 18x - 21 = 0

x ^ 2 - 6x - 7 < 0

x ^ 2 - 7x + x - 7 = 0

x(x - 7) + 1(x - 7) = 0

(x - 7)(x + 1) = 0

x = 7 and x = - 1

At x=-1:

f(- 1) = (- 1) ^ 3 - 9 * (- 1) ^ 2 - 21(- 1) + 8 = 19

At x=7:

f(7) = (7) ^ 3 - 9 * (7) ^ 2 - 21(7) + 8 = - 237

Local minimum=-237

Local maximum=19

c) For point of inflection:

f^ prime prime (x) = 0

Rightarrow 6x - 18 = 0

Rightarrow x = 3

y = f(3) = (3) ^ 3 - 9 * (3) ^ 2 - 21(3) + 8 = - 109

(x, y) = (3, - 109)

d)Interval for concave upward curve:

f^ prime prime (x) > 0

6x - 18 > 0

x \in (3, ∞)

e)Interval for concave upward curve:

f^ prime prime (x) < 0

6x - 18 < 0

X€(−0,3)

for such more question on Interval

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Answer:

[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]  

[tex]p_v =P(z>2.67)=0.0038[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=18 represent the adults who rate Pepsi as being "concerned with my health"

[tex]\hat p=\frac{18}{100}=0.18[/tex] estimated proportion of adults who rate Pepsi as being "concerned with my health"

[tex]p_o=0.10[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.1.:  

Null hypothesis:[tex]p \leq 0.1[/tex]  

Alternative hypothesis:[tex]p > 0.1[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.67)=0.0038[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1

The student is asked to perform a hypothesis test at the  alpha = 0.05 level to determine if the percentage of participants rating Pepsi as 'concerned with my health' has statistically increased from 10% to 18% in a follow-up survey.

The question involves conducting a hypothesis test to determine if the percentage of participants rating Pepsi as being "concerned with my health" has increased after a new initiative. With an initial percentage of 10%, a follow-up survey showed that 18 out of 100 (or 18%) now rate Pepsi in this manner. To determine if this is a statistically significant increase, at the alpha = 0.05 significance level, a hypothesis test can be performed using the binomial or normal approximation to the binomial distribution.

Null hypothesis (H0): p = 0.10 (The percentage of those who believe Pepsi is concerned with their health has not changed)
Alternative hypothesis (Ha): p > 0.10 (The percentage has increased)

If the calculated p-value is less than the significance level (0.05), we can reject the null hypothesis in favor of the alternative hypothesis, thus concluding that the percentage has indeed increased.

The initial reactant concentration is 0.150 M. Using the first order kinetics equation, we converted the time to seconds and plugged in the given values for initial concentration, time and rate constant. The final concentration of the reactant after 19.0 minutes is 4.92 x 10^-5 M.

The initial concentration of the reactant is 0.150 M and the rate constant, k = 8.70×10−3 s−1. We use the formula for first order kinetics, [A] = [A]0 * e^-kt, where [A] is the final concentration, [A]0 is the initial concentration, k is the rate constant and t is the time.

First, convert the time from minutes to seconds (since the rate constant is in s^-1). Therefore, t = 19 min * 60 s/min = 1140 s.

Then, substitute the given values into the equation: [A] = 0.150 M * e^-((8.70 x 10^-3 s^-1) * (1140 s)) = 0.150 M * e^-9.918 = 4.92 x 10^-5 M.

So, the concentration of the reactant after 19.0 minutes will be 4.92 x 10^-5 M.

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Part A: Concentration ≈ 0.186 MM after 8.00 minutes.

Part B: Initial concentration ≈ 5.90 MM for zero-order reaction.

Part A:

Using the first-order integrated rate law equation:

[tex]\[ [A] = [A]_0 \times e^{-kt} \][/tex]

[tex]\[ [A] = 0.650 \times e^{-(3.70 \times 10^{-3} \times 8 \times 60)} \][/tex]

[tex]\[ [A] \approx 0.186 \, MM \][/tex]

Part B:

Using the zero-order integrated rate law equation:

[tex]\[ [A] = -kt + [A]_0 \][/tex]

[tex]\[ [A]_0 = -kt + [A] \][/tex]

[tex]\[ [A]_0 = -(3.00 \times 10^{-4} \times 80) + 3.50 \times 10^{-2} \][/tex]

[tex]\[ [A]_0 = 3.50 \times 10^{-2} + 2.40 \times 10^{-2} \][/tex]

[tex]\[ [A]_0 \approx 5.90 \, MM \][/tex]

The correct question is:

Part A

The rate constant for a certain reaction is [tex]$k k k=3.70 \times 10^{-3} \mathrm{~s}-1 \mathrm{~s}-1$[/tex]. If the initial reactant concentration was [tex]$0.650 \mathrm{MM}$[/tex], what will the concentration be after 8.00 minutes?

Part B

A zero-order reaction has a constant rate of [tex]$3.00 \times 10^{-4} \mathrm{M} / \mathrm{sM} / \mathrm{s}$[/tex]. If after 80.0 seconds the concentration has dropped to [tex]$3.50 \times 10^{-2} \mathrm{MM}$[/tex], what was the initial concentration?

Final answer:

The standard deviation of the sample of numbers is approximately 10.68, after calculating the mean, finding the deviations, squaring them, averaging the squared deviations to get the variance, and then taking the square root of the variance.

Explanation:

The student is asking how to calculate the standard deviation of a sample of numbers. The sample given consists of the counts of bacteria found in ocean water, which are 41, 62, 45, 48, 69. To calculate the standard deviation, follow these steps:

Performing the calculations, you get:

To approximate the probability, we can use the normal approximation to the binomial distribution. We need to calculate the mean and standard deviation, and then use the normal distribution to find the cumulative probability for each number of errors. Finally, we add up the probabilities to get the probability of at least 2 errors.

To approximate the probability, we can use the normal approximation to the binomial distribution. We know that there are 50 pages with errors out of 1000, so the probability of success is p = 50/1000 = 0.05. The probability of failure is q = 1 - p = 1 - 0.05 = 0.95.

The sample size is 100 pages. To find the probability that at least 2 of the pages in error are in the sample, we need to find the probability of getting 2 errors, 3 errors, 4 errors, and so on, up to 100 errors.

We can use the normal approximation to the binomial distribution by calculating the mean and standard deviation. The mean is given by np = 100 * 0.05 = 5, and the standard deviation is given by sqrt(npq) = sqrt(100 * 0.05 * 0.95) ≈ 2.179.

Next, we can use the normal distribution to find the probability. We need to calculate the z-scores for each number of errors, and then find the cumulative probability from the z-score table or using a calculator. Finally, we add up the probabilities for each number of errors to get the probability of at least 2 errors.

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Given:

m∠T = 61°

m∠S = 5x - 15

m ext∠TUV = 8x - 8

To find:

The measure of exterior angle TUV

Solution:

STU is a triangle.

By exterior angle of triangle property:

The measure of an exterior angle is equal to the sum of the opposite interior angles.

m ext∠TUV = m∠T + m∠S

8x - 8 = 61 + 5x - 15

8x - 8 = 46 + 5x

Add 8 on both sides.

8x - 8 + 8 = 46 + 5x + 8

8x = 54 + 5x

Subtract 5x from both sides.

8x - 5x = 54 + 5x - 5x

3x = 54

Divide by 3 on both sides.

[tex]$\frac{3x}{3} =\frac{54}{3}[/tex]

x = 18

Substitute x = 18 in m ext∠TUV.

m ext∠TUV = 8(18) - 8

                   = 144 - 8

                   = 136

Therefore measure of exterior angle ∠TUV is 136°.

Answer:

a) y = -¼

Step-by-step explanation:

(-2x^6+5x+8)/(8x^6+6x+5)

Just divide the leading coefficients

-2/8 = -¼

Horizontal asymptote:

y = -¼

Answer:a) y = -¼

Step-by-step explanation:

(-2x^6+5x+8)/(8x^6+6x+5)

Just divide the leading coefficients

Answer:

A hypothesis is a proposed statement explaining the relationship between two or more variable which can be verifiable through experiment. In this case a testable hypothesis would be.

When yeast and sugar are mixed together during formation, it will cause the sugar to undergo a high rate of glycolysis and CO[tex]^{2}[/tex] will be present in the fermentation tube.

Answer:

The answer to your question is Circle's area = 113.04 in²  Slice' area = 14.13 in²

Step-by-step explanation:

Data

diameter = 12 in

8 slices

Area = ?

Area of a slice = ?

Process

1.- Calculate the area of the pizza

radius = 12/2

radius = 6 in

Area = πr²

-Substitution

Area = (3.14)(6)²

-Simplification

Area = 113.04 in²

2.- Calculate the area of the slice

We can calculate this area by two methods

a) Divide the area of the number of slices

       Area of the slice = 113.04 / 8

                                   = 14.13 in²

b) Using the area of a sector

-Find the angle of each slice

               360 / 8 = 45°

-Convert 45° to rad

                180° ------------------- πrad

                  45° ------------------- x

                  x = 45πrad/180

                  x = 1/4πrad

- Calculate the area of the sector

Area = 1/4(3.14)(6)²/2

Area = 113.04/8

Area = 14.13 in²                        

Answer:

Step-by-step explanation:

Assuming the pizza is circular, we would apply the formula for determining the area of a circle. It is expressed as

Area = πr²

Where

r represents the radius of the circle.

π is a constant whose value is 3.14

From the information given,

Diameter = 12 inches

Radius = 12/2 = 6 inches

Area = 3.14 × 6² = 113.04 inches²

The formula for determining the area of a sector is expressed as

Area of sector = θ/360 × πr²

Where θ represents the central angle.

The complete circle or pizza is 360°. Since it is divided into 8 equal parts, then the central angle formed by each sector is

360/8 = 45°

Therefore,

Area of sector = 45/360 × 3.14 × 6²

= 14.13 inches²

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

y = P(1 + r/n)^nt

Where

y = the value of the investment at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount invested

From the information given,

P = $4700

r = 4.75% = 4.75/100 = 0.0475

n = 1 because it was compounded once in a year.

Therefore, the exponential function showing the relationship between y and t is

y = 4700(1 + 0.0475/1)^1 × t

y = 4700(1.0475)^t

The differential equations for x and y are:

x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))

y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))

Here's how to solve the problem:

(a) Write the system as a differential equation for x:

Eliminate y: Substitute the second equation into the first equation to eliminate y.

(D+5)x + D2(t2 + 3t) = cost

Rearrange and define operator: Define P(D) = (D+5) + D^2 (D+9).

P(D) x = cost - D^2 (t2 + 3t)

Write as differential equation:

x = (cost - D^2 (t2 + 3t)) / P(D)

(b) Write the system as a differential equation for y:

Eliminate x: Substitute the first equation into the second equation to eliminate x.

D2((cost - D2y) / (D+5)) + (D+9)y = t2 + 3t

Rearrange and define operator: Define Q(D) = D2 / (D+5) + (D+9).

Q(D) y = t2 + 3t - D2 cost

Write as differential equation:

y = (t2 + 3t - D2 cost) / Q(D)

Therefore, the differential equations for x and y are:

x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))

y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))

(a)[tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)\)[/tex]

(b)[tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\)[/tex]   by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations .

The given system of equations can be transformed into differential equations using operator (D), representing differentiation with respect to a variable, often time.

(a) To derive the differential equation for (x), we isolate (x) in terms of the differential operator \(D\). Rearranging the first equation gives us [tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\).[/tex]Substituting the expression for (y) from the second equation into this yields[tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2}{D+5}\left(\frac{t^2 + 3t}{D+9}\right)\).[/tex] Simplifying further leads to [tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)[/tex]\).

(b) Similarly, isolating \(y\) from the given equations leads to [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}x\).[/tex]Substituting the expression for (x) from the first equation into this yields [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}\left(\frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\right)\).[/tex]Further simplification results in [tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\).[/tex]

Therefore, by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations as specified in the final answer.

Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

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 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4[/tex]

What is the probability that the sample average will be less than 182

This is the pvalue of Z when X = 182. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{182 - 187}{4}[/tex]

[tex]Z = -1.2[/tex]

[tex]Z = -1.2[/tex] has a pvalue of 0.1151

11.51% probability that the sample average will be less than 182

Let x* denote Peter's score. Then

P(X > x*) = 0.025

P((X - 896)/174 > (x* - 896)/174) = 0.025

P(Z > z*) = 1 - P(Z < z*) = 0.025

P(Z < z*) = 0.975

where Z follows the standard normal distribution (mean 0 and std dev 1).

Using the inverse CDF, we find

P(Z < z*) = 0.975  ==>  z* = 1.96

Then solve for x*:

(x* - 896)/174 = 1.96  ==>  x* = 1237.04

so Peter's score is roughly 1237.

Answer:

Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member

Step-by-step explanation:

Let suppose the attached histogram

a) Total members whose weight 110 = 5

b) Total members whose weight 120 = 4

c) Total members whose weight 130 = 2

d) Total members whose weight 140 = 0

e) Total members whose weight 150 = 5

f) Total members whose weight 160 = 0

g) Total members whose weight 170 = 4

Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member

A histogram is a statistical graphical representation of data with the help of bars with different heights.

Answer:

The type of sampling is stratified.

Step-by-step explanation:

Samples types are classified as:

Random: Basically, put all the options into a hat and drawn some of them.

Systematic: Every kth element is taken. For example, you want to survey something on the street, you interview every 5th person, for example.

Cluster: Divides population into groups, called clusters, and each element in the cluster is surveyed.

Stratified: Also divides the population into groups. However, then only some elements of the group are surveyed.

In this problem, we have that:

The drivers are divided into two groups according to their ages.

There are thousands and thousand of drivers both under and over 30 years of age, and 500 for each group(some elements of the group) are selected.

So the type of sampling is stratified.

The type of sampling used in the example is stratified sampling. The researcher divided the total population of drivers into two groups (under 30 and over 30), and selected 500 drivers from each group.

In this example, the type of sampling used by the market researcher is called stratified sampling. In stratified sampling, the population is divided into different subgroups, or 'strata', and a sample is taken from each stratum. In this case, the researcher has divided the population of drivers into two strata based on their age: drivers under 30 and drivers over 30. 500 drivers are then selected from each of these strata.

Stratified sampling is a common method used in market research because it allows for a more accurate representation of the population, as each subgroup is adequately represented in the sample.

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Answer:

73.30% probability that the sample mean score will be within 10 of the population mean

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

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], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 510, \sigma = 90, n = 100, s = \frac{90}{\sqrt{100}} = 9[/tex]

What is the probability that the sample mean score will be (a) within 10 of the population mean

This is the pvalue of Z when X = 510 + 10 = 520 subtracted by the pvalue of Z when X = 510 - 10 = 500.

X = 520

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

By the Central Limit Theorem

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

[tex]Z = \frac{520 - 510}{9}[/tex]

[tex]Z = 1.11[/tex]

[tex]Z = 1.11[/tex] has a pvalue of 0.8665

X = 500

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

[tex]Z = \frac{500 - 510}{9}[/tex]

[tex]Z = -1.11[/tex]

[tex]Z = -1.11[/tex] has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that the sample mean score will be within 10 of the population mean

Answer:

[tex]z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01[/tex]  

[tex]p_v =P(z>1.01)=0.156[/tex]  

So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

Step-by-step explanation:

Data given and notation

n=195 represent the random sample taken

X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy

[tex]\hat p=\frac{65}{195}=0.333[/tex] estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy

[tex]p_o=0.3[/tex] is the value that we want to test

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

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.3.:  

Null hypothesis:[tex]p\leq 0.3[/tex]  

Alternative hypothesis:[tex]p > 0.3[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>1.01)=0.156[/tex]  

So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

To find the p-value for testing the hypothesis that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, we can use the binomial probability formula.

To find the p-value for testing the hypothesis that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, we can use the binomial probability formula.

The formula is:

P(X ≥ k) = 1 - P(X < k)

where X is the number of women in the erythromycin group who complain of nausea, and k is the number of complaints of nausea in the control group. In this case, X = 65 and k = 0, because the question specifies that there are no complaints of nausea in a typical pregnant woman. By plugging these values into the formula, we can calculate the p-value.

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Answer:

a)

[tex]k = \dfrac{1}{21}[/tex]

b) 0.476

c) 0.667    

Step-by-step explanation:

We are given the following in the question:

Y = the number of forms required of the next applicant.

Y: 1, 2, 3, 4, 5, 6

The probability is given by:

[tex]P(y) = ky[/tex]

a) Property of discrete probability distribution:

[tex]\displaystyle\sum P(y_i) = 1\\\\\Rightarrow k(1+2+3+4+5+6) = 1\\\\\Rightarrow k(21) = 1\\\\\Rightarrow k = \dfrac{1}{21}[/tex]

b) at most four forms are required

[tex]P(y \leq 4) = \displaystyle\sum^{y=4}_{y=1}P(y_i)\\\\P(y \leq 4) = \dfrac{1}{21}(1+2+3+4) = \dfrac{10}{21} = 0.476[/tex]

c) probability that between two and five forms (inclusive) are required

[tex]P(2\leq y \leq 5) = \displaystyle\sum^{y=5}_{y=2}P(y_i)\\\\P(2\leq y \leq 5) = \dfrac{1}{21}(2+3+4+5) = \dfrac{14}{21} = 0.667[/tex]

The constant of proportionality (k) equals 1/21. The probability that at most four forms are required is 10/21. The probability that between two and five forms (inclusive) are required is 14/21 or 2/3 when simplified.

The student has provided information regarding a probability distribution which indicates that the probability (p(y)) is proportional to the number of forms required (y), with y ranging from 1 to 6. Since we know probabilities must sum to 1, this allows us to find the constant of proportionality (k).

Part (a): Finding the Value of k

We can use the formula sum of all probabilities equals 1: 1 = k * (1 + 2 + 3 + 4 + 5 + 6). The sum of the numbers from 1 to 6 is 21, therefore 1 = 21k. Solving for k gives us k = 1/21.

Part (b): Probability of At Most Four Forms Required

To find the probability of at most four forms being required, we need to sum the probabilities for 1, 2, 3 and 4 forms: P(x ≤ 4) = k + 2k + 3k + 4k. Substituting the value of k, we get P(x ≤ 4) = (1/21)(1 + 2 + 3 + 4) = 10/21.

Part (c): Probability Between Two and Five Forms Required (Inclusive)

We calculate this by adding the probabilities for 2, 3, 4, and 5 forms: P(2 ≤ x ≤ 5) = 2k + 3k + 4k + 5k. With k = 1/21, this probability is P(2 ≤ x ≤ 5) = (1/21)(2 + 3 + 4 + 5) = 14/21 or 2/3 when simplified.

Answer:

405150

Step-by-step explanation:

3 people are to be selected from 75 people. This could be done using the combination function, [tex]\binom{75}{3}[/tex]. But thus function only gives the number of ways of choosing any 3 out of the 75 customers. For any single selection of any 3 customers, there are 3! ways of sharing the three prizes among them. Hence the total number of ways of sharing the prizes is

[tex]\binom{75}{3}\times3! = 405150[/tex]

In fact, this is the permutation function given by

[tex]{}^{75}P_3 = \dfrac{75!}{(75-3)! = 405150}[/tex]

Answer:

Step-by-step explanation:

Hello!

X: number of gypsy moths in a randomly selected trap.

This variable is strongly right-skewed. with a standard deviation of 1.4 moths/trap.

The mean number is 1.2 moths/trap, but several have more.

a.

The population is the number of moths found in traps places by the agriculture departments.

The population mean μ= 1.2 moths per trap

The population standard deviation δ= 1.4 moths per trap

b.

There was a random sample of 60 traps,

The sample mean obtained is X[bar]= 1

And the sample standard deviation is S= 2.4

c.

As the text says, this variable is strongly right-skewed, if it is so, then you would expect that the data obtained from the population will also be right-skewed.

d. and e.

Because you have a sample size of 60, you can apply the Central Limit Theorem and approximate the distribution of the sampling mean to normal:

X[bar]≈N(μ;σ²/n)

The mean of the distribution is μ= 1.2

And the standard deviation is σ/√n= 1.4/50= 0.028

f. and g.

Normally the distribution of the sample mean has the same shape of the distribution of the original study variable. If the sample size is large enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient you can apply the theorem and approximate the distribution of the sample mean to normal.

You have a sample size of n=10 so it is most likely that the sample mean will have a right-skewed distribution as the study variable. The sample size is too small to use the Central Limit Theorem, that is why you cannot use the Z table to calculate the asked probability.

I hope it helps!

Answer:

Avis Company is a car rental company that is located three miles from the Los Angeles airport (LAX). Avis is dispatching a bus from its offices to the airport every 2 minutes. The average traveling time (round-trip) is 20 minutes.

a. How many Avis buses are traveling to and from the airport?

b. The branch manager wants to improve the service and suggests dispatching buses every 0.5 minutes. She argues that this will reduce the average traveling time from the airport to Avis offices to 2.5 minutes. Is she correct? If your answer is negative, then what will the average traveling time be?

The answers to the question are

a. 10 buses

b. Yes, increasing the fleet by a factor of 4 will reduce the apparent average travel time from the airport to Avid offices from 10 minutes to 2.5 minutes

Step-by-step explanation:

Number of buses dispatched every 20 minutes= 20÷2=10

Therefore 10 Avid buses are travelling to and from the Los Angeles airport

b. By dispatching every 0.5 minutes we have 20÷0.5 = 40 buses.

Since with 10 buses it takes 10 minutes to make the journey from the airport to Avid offices, then it would take 40 buses 10÷4 = 2.5 minutes time to deliver what it took 10 buses, 10 minutes to deliver.

The Avis Company has 10 buses traveling to and from LAX airport with an average round trip time of 20 minutes and a dispatch frequency of 2 minutes. Increasing the dispatch frequency to every 0.5 minutes does not reduce the average travel time; it remains at 20 minutes as it is dependent on the route's distance and traffic conditions, not the dispatch frequency.

Part A: Number of Avis Buses

To calculate the number of buses traveling to and from the airport, we use the average traveling time for a round trip and the dispatch frequency. With a round trip time of 20 minutes and buses leaving every 2 minutes, we divide the round trip time by the dispatch interval:

Total Buses = Round Trip Time / Dispatch Frequency

Total Buses = 20 minutes / 2 minutes

Total Buses = 10

Therefore, Avis has 10 buses traveling to and from the airport.

Part B: Average Traveling Time with Increased Dispatch Frequency

Increasing the dispatch frequency to every 0.5 minutes does not affect the average traveling time for a round trip. The travel time is determined by the distance between the Avis office and the airport and the traffic conditions, not merely the frequency of bus dispatches. If Avis dispatches buses every 0.5 minutes, there will be more buses in operation, but the average round trip time remains the same at 20 minutes, assuming no changes in traffic conditions or route speeds.

Answer:

[tex]a. \ P(X\leq 8)=0.932\\b. \ P(X=8)=0.065\\c. \ P(9\leq X)=0.068\\d. \ P(5\leq X \leq 8)=0.491\\[/tex]

e. P(5<X<8)=0.251

Step-by-step explanation:

Given that boiler follows a Poisson distribution,$\sim$

[tex]X $\sim$Poi(\mu=5)[/tex]

Poisson Distribution formula is given by the expression:-

[tex]p(x,\mu)=\frac{e^{-\mu}\mu^x}{x!}[/tex]

Our probabilities will be calculated as below:

a.

[tex]P(X=x)=\frac{5^xe^{-5}}{x!}\\P(X\leq 8)=P(X=0)+P(X=1)+...+P(X=8)\\=\frac{5^0e^{-5}}{1!}+...+\frac{5^8e^{-5}}{8!}\\=0.006738+...+0.065278\\=0.961[/tex]

b.

[tex]P(X=8)=\frac{5^8e^{-5}}{8!}\\=0.065[/tex]

c.

[tex]P(9\leq X)=1-P(X=0)-P(X=1)-...-P(X=8)\\=0.068[/tex]

d.

[tex]P(5\leq X\leq 8)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\=0.491[/tex]

e.P(5<X<8)

[tex]P(X=6)+P(X=7)\\=0.251[/tex]