In the envelope game, there are two players and two envelopes. One of the envelopes is marked ''player 1 " and the other is marked "player 2." At the beginning of the game, each envelope contains one dollar. Player 1 is given the choice between stopping the game and continuing. If he chooses to stop, then each player receives the money in his own envelope and the game ends. If player 1 chooses to continue, then a dollar is removed from his envelope and two dollars are added to player 2's envelope. Then player 2 must choose between stopping the game and continuing. If he stops, then the game ends and each player keeps the money in his own envelope. If player 2 elects to continue, then a dollar is removed from his envelope and two dollars are added to player 1 's envelope. Play continues like this, alternating between the players, until either one of them decides to stop or k rounds of play have elapsed. If neither player chooses to stop by the end of the kth round, then both players obtain zero. Assume players want to maximize the amount of money they earn.

(a) Draw this game's extensive-form tree for k = 5.

(b) Use backward induction to find the subgame perfect equilibrium.

(c) Describe the backward induction outcome of this game for any finite integer k.

Answer:

In general, 50 % of the values in a data set lie at or below the median. 75 % of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 1600 test scores, 0.5*1600 = 800 of them would be at or below the median. If a sample consists of 1600 test scores, 0.75*800 = 1200 of them would be at or above the first quartile (Q1)

Step-by-step explanation:

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.

The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.

The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.

The answer is:

In general, 50 % of the values in a data set lie at or below the median. 75 % of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 1600 test scores, 0.5*1600 = 800 of them would be at or below the median. If a sample consists of 1600 test scores, 0.75*800 = 1200 of them would be at or above the first quartile (Q1)

Using the principle of quartile distribution, the percentage amount of data at or below the median and third quartile are 50% and 75% respectively. Number of scores below the median and at or above the first quartile is 800 and 1200 respectively.

The median of 1600 can be calculated thus :

Above the first quartile :

Therefore, the score above the first quartile is 1200.

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Answer:

1) This random variable should be modelled using a binomial distribution since we have independence between the events and a bernoulli trial each time when the experiment is conducted, a fixd value for the sample size n and for the probability of success.

Let X the random variable of interest, on this case th distribution would be given by:

[tex]X \sim Binom(n=10, p=0.02)[/tex]

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

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x} = (10Cx) (0.02)^x (1-0.02)^{10-x}[/tex]

2) For this case we don't have a sample size provided and we just have an average rate for a given period, so then we can assume that the best distribution for this case is the Poisson distribution.

Let X the random variable that represent the number of claims per car. We know that [tex]X \sim Poisson(\lambda)[/tex]

The probability mass function for the random variable is given by:

[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]

Where [tex]\lambda=0.2[/tex] represent the mean of occurrences in the interval of 2 years provided.

And f(x)=0 for other case.

Step-by-step explanation:

Previous concepts

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

Random variable 1

This random variable should be modelled using a binomial distribution since we have independence between the events and a bernoulli trial each time when the experiment is conducted, a fixd value for the sample size n and for the probability of success.

Let X the random variable "number of failed drawers", on this case th distribution would be given by:

[tex]X \sim Binom(n=10, p=0.02)[/tex]

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

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x} = (10Cx) (0.02)^x (1-0.02)^{10-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

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

Random variable 2

For this case we don't have a sample size provided and we just have an average rate for a given period, so then we can assume that the best distribution for this case is the Poisson distribution.

Let X the random variable that represent the number of claims per car. We know that [tex]X \sim Poisson(\lambda)[/tex]

The probability mass function for the random variable is given by:

[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]

Where [tex]\lambda=0.2[/tex] represent the mean of occurrences in the interval of 2 years provided.

And f(x)=0 for other case.

Answer:

kilo is [tex]10^{3}[/tex]

nano is [tex]10^{-9}[/tex]

micro is [tex]10^{-6}[/tex]

centi is [tex]10^{-2}[/tex]

mega is [tex]10^{6}[/tex]

milli is [tex]10^{-3}[/tex]

Step-by-step explanation:

Examples of power of 10 are:

[tex]100 = 10^{2}[/tex]

[tex]10 = 10^{1}[/tex]

[tex]1 = 10^{0}[/tex]

[tex]0.1 = 10^{-1}[/tex]

[tex]0.01 = 10^{-2}[/tex]

kilo is 1000. So kilo is [tex]10^{3}[/tex]

Nano is 0.000000001. So nano is [tex]10^{-9}[/tex]

micro is 0.000001. So micro is [tex]10^{-6}[/tex]

centi is 0.01. So centi is [tex]10^{-2}[/tex]

Mega is 1000000. So mega is [tex]10^{6}[/tex]

milli is 0.001. So milli is [tex]10^{-3}[/tex]

The following are conversions.

[tex]\rm{kilo}=10^3\\\rm {nano}=10^{-9}\\\rm{micro}=10^{-6}\\\rm{centi}=10^{-2}\\\rm{Mega}=10^6\\\rm{milli}=10^{-3}[/tex]

On the basis of the power of 10 the following categories can be defined that are as follow:

[tex]\rm{kilo}=10^3\\\rm {nano}=10^{-9}\\\rm{micro}=10^{-6}\\\rm{centi}=10^{-2}\\\rm{Mega}=10^6\\\rm{milli}=10^{-3}[/tex]

The values of all are as follows.

kilo is 1000.

Nano is 0.000000001.

micro is 0.000001.

Centi is 0.01.

Mega is 1000000.

Milli is 0.001.

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Answer:

2^c

Step-by-step explanation:

After calculating the interquartile range (IQR) and identifying the thresholds for outliers, it is clear that there are no low outliers but at least one high outlier, as the maximum age exceeds the high outlier threshold. Therefore, only statement (ii) is correct.

The question asks to consider two statements regarding outliers in a five-number summary of the ages of passengers on a cruise ship, which includes a minimum age (Min), first quartile (Q1), median, third quartile (Q3), and maximum age (Max). The statements to consider are:

To determine whether these statements are correct, we use the interquartile range (IQR) method for identifying outliers. The IQR is calculated as Q3 - Q1. An age is considered a low outlier if it is less than Q1 - 1.5 * IQR, and it is a high outlier if it is greater than Q3 + 1.5 * IQR.

Using the given data:

Q1 = 20
Q3 = 38
IQR = Q3 - Q1 = 38 - 20 = 18

Therefore, the cutoff for low outliers is 20 - 1.5 * 18 = 20 - 27 = -7, and for high outliers, it is 38 + 1.5 * 18 = 38 + 27 = 65.

Since the minimum age is 1, which is well above -7, there are no low outliers. However, the maximum age is 80, which is above the high outlier threshold of 65. Hence, there is at least one high outlier.

The correct answer to the question given the provided data is:

Only statement (ii) is correct.

Answer:

i need more information like what is the length width etc.

Step-by-step explanation:

Answer:

a) ordinal

b) ratio

c) nominal

d) interval

e) ratio

f) nominal

Step-by-step explanation:

a) Salesperson performance is ordinal because it is ordered from below average to above average

b) Price of company stock is ratio because it has a defined zero point.

c) Names of new product is nominal because it can be classified into different categories yet cannot be ordered.

d) Temperature in CEO's office is interval because it don't have a defined zero point.  In temperature zero temperature can exists.

e) Again gross income is ratio because it has a defined zero point. Zero gross income will means that there is no gross income.

f) Color of product packaging is again nominal because it can be classified into different categories yet cannot be ordered.

The measurements can be categorized as "ordinal, ratio, nominal, and interval."

(a) Salesperson's performance: below average, average, above average. - Ordinal

(b) Price of company's stock - Ratio

(c) Names of new products - Nominal

(d) Temperature (°F) in CEO's private office - Interval

(e) Gross income for each of the past 5 years - Ratio

(f) Color of product packaging - Nominal

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Answer:

Step-by-step explanation:

Let

                               [tex]f(x,y,z) = z-x^2-y^4-e^{xy}[/tex].

The partial derivatives of this function are

                                    [tex]\frac{\partial f}{\partial x} (x,y,z) = ye^{xy} - 2x \\ \frac{\partial f}{\partial y} (x,y,z) = xe^{xy} - 4y^3 \\\frac{\partial f}{\partial z} (x,y,z) = 1[/tex]

The tangent plane equation through a point [tex]A(x_1,y_1,z_1)[/tex] is given by                     [tex]f'_x (x_1,y_1,z_1)(x-x_1) + f'_y(x_1,y_1,z_1)(y-y_1) + f'_z(x_1,y_1,z_1)(z-z_1) = 0[/tex]

In this case, we have

                                  [tex]x_1 = 1, y_1 = 0, z_1 = 2.[/tex]

The values of the partial derivatives in this point are

                                 [tex]\frac{\partial f}{\partial x} (1,0,2) = 0 \cdot e^{0} - 2\cdot 1 = -2 \\ \frac{\partial f}{\partial y} (1,0,2) = 1 \cdot e^{0} - 0 = 1 \\ \frac{\partial f}{\partial y} (1,0,2) = 1[/tex]

So, the equation is

                        [tex]-2(x-1) + 1 \cdot (y-0) + 1\cdot (z-2) = 0[/tex]

Therefore, the equation for the plane tangent to the surface at the point [tex](1,0,2)[/tex] is given by

                                      [tex]2x-y-z = 0[/tex]

Answer:

a) [tex] E(T) = 0.2 E(Y) -1000= 0.2*20000 -1000=3000[/tex]

b) [tex] Sd(T) = \sqrt{0.2^2 Var(Y)}=\sqrt{0.2^2 8000^2}= 1600[/tex]

c) Assuming 20 million of families and each one with a mean of income of 20000 for each family approximately then total income would be:

[tex] E(T) = 20000000*20000= 40000 millions[/tex]

And if we replace into the formula of T we have:

[tex] T = 0.2*400000x10^6 -1000= 790000 millions[/tex]

Approximately.  

Step-by-step explanation:

For this case we knwo that Y represenet the random variable "Income" and we have the following properties:

[tex] E(Y) = 20000, Sd(Y) = 8000[/tex]

We define a new random variable T "who represent the taxes"

[tex] T = 0.2(Y-5000) = 0.2Y -1000[/tex]

Part a

For this case we need to apply properties of expected value and we have this:

[tex] E(T) = E(0.2 Y -1000)[/tex]

We can distribute the expected value like this:

[tex] E(T) = E(0.2 Y) -E(1000)[/tex]

We can take the 0.2 as a factor since is a constant and the expected value of a constant is the same constant.

[tex] E(T) = 0.2 E(Y) -1000= 0.2*20000 -1000=3000[/tex]

Part b

For this case we need to first find the variance of T we need to remember that if a is a constant and X a random variable [tex] Var(aX) = a^2 Var(X)[/tex]

[tex] Var(T) = Var (0.2Y -1000)[/tex]

[tex] Var(T)= Var(0.2Y) -Var(1000) + 2 Cov(0.2Y, -1000)[/tex]

The covariance between a random variable and a constant is 0 and a constant not have variance so then we have this:

[tex] Var(T) =0.2^2 Var(Y)[/tex]

And the deviation would be:

[tex] Sd(T) = \sqrt{0.2^2 Var(Y)}=\sqrt{0.2^2 8000^2}= 1600[/tex]

Part c

Assuming 20 million of families and each one with a mean of income of 20000 for each family approximately then total income would be:

[tex] E(T) = 20000000*20000= 40000 millions[/tex]

And if we replace into the formula of T we have:

[tex] T = 0.2*400000x10^6 -1000= 790000 millions[/tex]

Approximately.  

Answer:

Option A is the correct answer.

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

From the graph,

y2 = 6

y1 = 2

x2 = 3

x1 = 1

Slope,m = (6 - 2)/(3 - 1) = 4/2 = 2

To determine the intercept, we would substitute x = 3, y = 6 and

m= 2 into y = mx + c. It becomes

6 = 2 × 3 + c = 6 + c

c = 6 - 6 = 0

The equation becomes

y = 2x

Answer:

east

Step-by-step explanation:

Answer:

Step-by-step explanation:

The cost of printing posters from 2 to 100, given the marginal cost function dc/dx = 1/2√x, is calculated by finding the anti-derivative (or integral) of the function from 1 to 100, resulting in a cost of $18.

The given function dc/dx = 1/2√x represents the marginal cost, which is the derivative of the cost function c(x). To find the cost of printing from the 2nd to the 100th poster, we need to find the integral of the marginal cost from 1 to 100, not including the cost of the first poster. This is represented mathematically as ∫ from 1 to 100 of (1/2√x) dx. Here's how to solve this:

This results in 2*(√100 - √1) = 20 - 2 = $18. This is the cost for printing posters 2 to 100.

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Answer:

a)   (y+ Oz)

b) (ab + ac)(abc + cd + ce)

Step-by-step explanation:

a) xy + xOz  (First find the common factor between xy and xOz, and it is x as it appears in both xy and xOz. Factoring out x will result in

 x(y+ Oz)

There are no more common factors thus the answer is x(y + Oz)

Since the question was asking after removing redundant terms it will on be y + Oz

b) (ab0 f + a 0 cf)(a 0 bcf + c 0 f(d + e))  simplfying the expression

(ab0 f + a 0 cf)(aObcf + cOfd + cOfe) common factor Of is factored out as below

Of[ (ab + ac)(abc + cd + ce) ] therefore after removing redundant terms the answer is (ab + ac)(abc + cd + ce)

Final answer:

To solve -30 ÷ (6a) when a = -5, substitute the value of a into the expression. The answer in simplest form is 1.

Explanation:

To solve the expression -30 ÷ (6a), we need to substitute the value of a, which is -5.

Substituting the value of a, -30 ÷ (6a) becomes -30 ÷ (6*(-5)).

Using the order of operations, first, we solve the multiplication inside the parentheses: -30 ÷ (-30).

Dividing a negative number by a negative number results in a positive number, so -30 ÷ (-30) = 1.

Therefore, the answer in simplest form is 1.

Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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:

[tex]n = 10, p = 0.05[/tex]

a) None of the LED light bulbs are defective?

This is P(X = 0).

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

[tex]P(X = 0) = C_{10,0}*(0.05)^{0}*(0.95)^{10} = 0.5987[/tex]

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

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

[tex]P(X = 1) = C_{10,1}*(0.05)^{1}*(0.95)^{9} = 0.3151[/tex]

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

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

[tex]P(X = 2) = C_{10,2}*(0.05)^{2}*(0.95)^{8} = 0.0746[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884[/tex]

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

So

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

[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]

[tex]P(X = 0) = C_{10,0}*(0.95)^{0}*(0.05)^{10}\cong 0[/tex]

[tex]P(X = 1) = C_{10,1}*(0.95)^{1}*(0.05)^{9} \cong 0[/tex]

[tex]P(X = 2) = C_{10,1}*(0.95)^{2}*(0.05)^{8} \cong 0[/tex]

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

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

There is a 100% probability that three or more of the LED light bulbs are not defective.

The question relates to binomial distribution in probability theory. The probabilities calculated include those of none, one, two or less, and three or more LED bulbs being defective out of a random sample of 10.

This question relates to the binomial probability distribution. A binomial distribution is applicable because there are exactly two outcomes in each trial (either the LED bulb is defective or it's not) and the probability of a success remains consistent.

a) In this scenario, 'none of the bulbs being defective' means 10 successes. The formula for probability in a binomial distribution is p(x) = C(n, x) * [p^x] * [(1-p)^(n-x)]. Plugging in the values, we find p(10) = C(10, 10) * [0.95^10] * [0.05^0] = 0.5987 or 59.87%.

b) 'Exactly one of the bulbs being defective' implies 9 successes and 1 failure. Following the same formula, we get p(9) = C(10, 9) * [0.95^9] * [0.05^1] = 0.3151 or 31.51%.

c) 'Two or less bulbs being defective' means 8, 9 or 10 successes. We add the probabilities calculated in (a) and (b) with that of 8 successes to get this probability. Therefore, p(8 or 9 or 10) = p(8) + p(9) + p(10) = 0.95.

d) 'Three or more bulbs are not defective' means anywhere from 3 to 10 successes. As the failure rate is low, it's easier to calculate the case for 0, 1 and 2 successes and subtract it from 1 to find this probability. This gives us p(>=3) = 1 - p(2) - p(1) - p(0) = 0.98.

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Answer:

The value of x increased by 3.25.

Step-by-step explanation:

It is given that the value of x varies from x=2 to x=5.25.

We need to find the change is x.

If value of a variable x varies from a to b, then the change in te value of variable is

Change in x = b - a

For the given problem the variable is x, a=2 and b=5.25.

Change in x = 5.25 - 2

                  = 3.25

If the change is positive, then the value of variable increased and if the change is negative, then the value of variable decreased.

Therefore, the value of x increased by 3.25.

The value of x changed by 3.25 as it varies from x=2 to x=5.25.

The change in the value of x from x=2 to x=5.25 can be calculated by subtracting the initial value from the final value. In this case, the change in x is 5.25 - 2 = 3.25. Thus, the value of x has changed by 3.25.

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Answer:

For each n get its posting then AND them

Step-by-step explanation:

1. Lets presume we have multiple n terms

2. We get the posting from each n term

3. We use the function AND and apply it to each n term

4. We start with the tiniest set and we continue from there

5. We have a pair of n terms

6. We get the posting from 1st n

7. We get the posting from 2nd n

8. We apply n1 AND n2

The a chi-square of the experimental data is 4.32.

Observed data(o)

Purple tall=206

White tall=65

Purple short=83

White short=30

Total=384

Expected (e)

9 Purple tall=216

3 White tall=72

3 Purple short=72

1 White short=24

Chi-square

Purple tall=(206-216)²/216=0.46

White tall=(65-72)²/72=0.68

Purple short=(83-72)²/72=1.68

White short=(30-24)²/24=1.5

X²=0.46+0.68+1.68+1.5

X²=4.32

Inconclusion  a chi-square of the experimental data is 4.32.

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Final answer:

To analyze the dihybrid cross data, we calculate the expected frequencies using the Mendelian 9:3:3:1 ratio, perform a chi-square analysis, and then conclude whether the observed data significantly deviates from the expected phenotypic ratio.

Explanation:

To analyze the given data, we must first establish our expected ratio. For a dihybrid cross involving two independently assorting traits with complete dominance, Mendel's laws predict a 9:3:3:1 phenotypic ratio among the F2 progeny. These numbers represent the ratio of the offspring displaying both dominant traits, one dominant and one recessive trait, one recessive and one dominant trait, and both recessive traits, respectively.

With the given data:

206 purple tall (dominant traits)

65 white tall (one recessive trait)

83 purple short (one recessive trait)

30 white short (recessive traits)

Adding these together, we have a total of 384 plants. To get the expected counts based on the 9:3:3:1 ratio: Expected purple tall = 9/16 * 384 = 216, Expected white tall = 3/16 * 384 = 72, Expected purple short = 3/16 * 384 = 72, and Expected white short = 1/16 * 384 = 24. Now, we can perform a chi-square analysis using these expected counts.

The chi-square formula is: X^2 = ∑ (observed - expected)^2 / expected. Plugging in our values:

X^2 for purple tall = (206 - 216)^2 / 216

X^2 for white tall = (65 - 72)^2 / 72

X^2 for purple short = (83 - 72)^2 / 72

X^2 for white short = (30 - 24)^2 / 24

After calculating these values, they need to be added to get the total chi-square value. We then compare this value to a critical value from a chi-square distribution table (typically at a 0.05 significance level) with the degrees of freedom equal to the number of classes - 1, here df = 3.

Based on this comparison, we determine if the observed ratios significantly deviate from the expected 9:3:3:1 ratio. If the calculated chi-square value is less than or equal to the critical value, our null hypothesis that purple color and tall height are dominant and assort independently holds true. However, if it's greater, we may have to reject the null hypothesis.

Answer:

[tex]P(57<X<59)=P(\frac{57-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{59-\mu}{\sigma})=P(\frac{57-61}{6}<Z<\frac{59-61}{6})=P(-0.67<Z<-0.33)[/tex]

[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)[/tex]

[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)=0.371-0.251=0.119[/tex]

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

[tex]X \sim N(61,6)[/tex]  

Where [tex]\mu=61[/tex] and [tex]\sigma=6[/tex]

We are interested on this probability

[tex]P(57<X<59)[/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(57<X<59)=P(\frac{57-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{59-\mu}{\sigma})=P(\frac{57-61}{6}<Z<\frac{59-61}{6})=P(-0.67<Z<-0.33)[/tex]

And we can find this probability like this:

[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)[/tex]

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

[tex]P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)=0.371-0.251=0.119[/tex]

Answer:

The percentage of inter-rater reliability is 80%.

Step-by-step explanation:

The rate of inter-rater reliability is the division of the number of intervals in which they agreed number of total intervals.

In this problem, we have that:

There are 10 intervals.

The observers agreed on 8 of them.

So the rate of agreement is 8/10 = 0.8.

As a percentage, we multiply the rate by 100, so 0.8*100 = 80%.

The percentage of inter-rater reliability is 80%.

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

[tex]{A}'[/tex] & [tex]{B}'[/tex] is also independent

[tex]{A}'[/tex] = 1-0.8 = 0.2

[tex]{B}'[/tex] = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

[tex]P(\frac{B'}{A})[/tex] = [tex]\frac{P(B'\cap A)}{P(A)}[/tex] = [tex]\frac{0.4\times 0.8}{0.8}[/tex] = 0.4

Part c)

Assume the events are not independent

Given Data

P[tex](\frac{{A}'}{{B}'})[/tex] = 0.4

=[tex]\frac{P({A}'\cap {B}')}{P({B}')}[/tex] = 0.4

[tex]P[/tex][tex]({A}'\cap {B}')[/tex] = 0.4 x P[tex]({B}')[/tex]

= 0.4 x 0.4 = 0.16

[tex]P({A}'\cap {B}')[/tex] = 0.16

i)

The probability that at least one of them is on time

[tex]P(A\cup B)[/tex] = 1- [tex]P({A}'\cap {B}')[/tex]  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P[tex](A\cap B)[/tex] = 1 - [tex]P({A}'\cup {B}')[/tex] = 1 - [tex][P({A}')+P({B}') - P({A}'\cap {B}')][/tex]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

Answer:

To produce 400 gallons of a chemical, it costs 700 dollars.

Step-by-step explanation:

We have that:

The cost, C (in dollars) to produce g gallons of a chemical can be expressed as C=f(g).

The interpretation is:

To produce g gallons of a chemical it costs C dollars.

So

f(400)=700

This means that to produce 400 gallons of a chemical, it costs 700 dollars.

Final answer:

The statement f(400)=700 means that the cost to produce 400 gallons of a chemical is 700 dollars. This cost function indicates the relationship between the number of gallons produced and the total cost, facilitating the calculation of expenses for chemical production.

Explanation:

The statement f(400)=700 means that to produce 400 gallons of a chemical, the cost is 700 dollars. The function f(g) represents the cost C, in dollars, to produce g gallons of the chemical. Therefore, when we input 400 into the function, it outputs the cost associated with producing that quantity, which in this case is 700 dollars.

Understanding the function as a rate of change, the cost to produce an additional gallon can be determined through ordinary algebra. If it were the case that this chemical firm is in a constant cost industry, where the cost per additional unit produced remains the same (e.g., 10 USD/oz), then this would simplify the exercise of calculating the cost for any number of gallons produced.

Conversion factors in chemistry are often used to relate amounts of substances in reactions, similarly in economics, conversion factors can tell us how much of a good can be bought or produced per unit of currency, like dollars per gallon, which is USD/unit.

Answer:

Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.

Step-by-step explanation:

Let number of cups of juice in each packet of orange be 'x'.

Let number of cups of juice in each pack of apples be 'y'.

Given:

On Tuesday;

Number of packets of oranges = 8

Number of packets of apples = 6 packets.

Amount of fruit juice = 34 cups

So we can say that;

Amount of fruit juice is equal to sum of Number of packets of oranges multiplied by number of cups of juice in each packet of orange and Number of packets of apples multiplied by number of cups of juice in each pack of apples.

framing in equation form we get;

[tex]8x+6y=34 \ \ \ \ equation \ 1[/tex]

On Wednesday;

Number of packets of oranges = 14

Number of packets of apples = 6

Amount of fruit juice = 46 cups

So we can say that;

Amount of fruit juice is equal to sum of Number of packets of oranges multiplied by number of cups of juice in each packet of orange and Number of packets of apples multiplied by number of cups of juice in each pack of apples.

framing in equation form we get;

[tex]14x+6y=46 \ \ \ \ equation \ 2[/tex]

Now Subtracting equation 1 from equation we get;

[tex]14x+6y-(8x+6y)=46-34\\\\14x+6y-8x-6y=12\\\\6x=12[/tex]

Now dividing both side by 6 we get;

[tex]\frac{6x}{6}=\frac{12}{6}\\\\x=2\ cups[/tex]

Now we will substitute the value of 'x' in equation 1 we get;

[tex]8x+6y=34\\\\8\times2+6y=34\\\\16+6y=34[/tex]

Subtracting both side by 16 we get;

[tex]16+6y-16=34-16\\\\6y= 18[/tex]

Now dividing both side by 6 we get;

[tex]\frac{6y}{6}=\frac{18}{6}\\\\y=3[/tex]

Hence Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.

Answer:

a) Continuous since the weight can take decimals for example 2.3 pounds.

b) Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters

c) Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous

d) Discrete since that's a binary variable and only take integers.

e) Discrete since the number of errors can't be 4.3 for example, and can't take decimals.

Step-by-step explanation:

Previous concepts

A continuous random variable by definition is a "random variable where the data can take infinitely many values" defined on a interval.

And a discrete random variable is a random variable that only can takes integers and is defined over a domain

Solution to the problem

a. Weight of the book (e.g., 2.3 pounds)

Continuous since the weight can take decimals for example 2.3 pounds.

b. Number of chapters in the book (e.g., 10 chapters)

Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters

c. Width of the book (e.g., 8 inches)

Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous

d. Type of book (0=hardback or 1=paperback)

Discrete since that's a binary variable and only take integers.

e. Number of typographical errors in the book (e.g., 4 errors)

Discrete since the number of errors can't be 4.3 for example, and can't take decimals.

Answer:

[tex] s= \sqrt{80.568}=8.97 [/tex]

[tex] Median =\frac{1+3}{2}=2[/tex]

Step-by-step explanation:

For this case we have the following data:

1.0 -5 3.0 -8 14 -11 12 0 16 -2 12 7

And ordering this data we have:

-11 -8 -5 -2 0 1 3 7 12 12 14 16  

And we are interested in find the standard deviation for the sample data. In order to do this the first step is find the mean given by this formula:

[tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}=\frac{1-5+3-8+14-11+12+0+16-2+12+7}{12}=3.25[/tex]

Now with the sample mean we can calculate the sample variance with the following formula:

[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And if we replace we got:

[tex] s^2 =80.568[/tex]

And for the sample deviation we just need to take the square root of the sample variance and we got:

[tex] s= \sqrt{80.568}=8.97 [/tex]

The median on this case would be given by:

[tex] Median =\frac{1+3}{2}=2[/tex]

Using the positions 5 and 6 for the average since the sample size is an even number.

Final answer:

The standard deviation of the weight changes for the 12 women is approximately 8.1 pounds, after computing the variance and taking the square root.

Explanation:

To calculate the sample standard deviation of the given weight changes, we will follow these steps:

Now let's apply these steps to the given data:

The data points are: 1.0, -5, 3.0, -8, 14, -11, 12, 0, 16, -2, 12, 7

The mean (step 1) is calculated as:

(1.0 - 5 + 3.0 - 8 + 14 - 11 + 12 + 0 + 16 - 2 + 12 + 7) / 12 = (49) / 12 ≈ 4.0833 pounds

Continuing with steps 2 to 5, the squared differences from the mean, their sum, and the variance are computed. Finally, the sample standard deviation is found to be approximately 8.1 pounds.

Therefore, the correct answer is option b, which is 8.1 pounds.

Answer:

Manuel is correct, since adding 10 is the same as the tens digit going up by 1.

Step-by-step explanation:

It is important to know the concepts of units, tenths and cents.

For example

1 = 1 unit

10 = 1*10 + 0 = The tens digit is one the unit digit is 0

21 = 2*10 + 1 = The tens digit is two and the unit digit is 1.

120 = 1*100 + 2*10 + 0 = The cents digit is 1, the tens digit is two and the unit digit is 0.

So

Adding 1 is the same as the unit digit going up by 1.

Adding 10 is the same as the tens digit going up by 1.

Adding 100 is the same as the cents digit going up by 1.

In this problem, we have that:

460,470,480, 490,500,510

Each number is 10 added to the previous, that is, the tens digit going up by 1.

Manuel thinks the tens digit goes up by 1 in these numbers. Do you agree?

Manuel is correct, since adding 10 is the same as the tens digit going up by 1.

The statement was given by Manuel that the tens digit goes up by 1 in these numbers, is true.

The second place from the right of the number before the decimal is the tens place of a number.

If we look at the series of the number that is given to us then we will notice that in series 460,470,480, 490,500,510, the second place of the number is increasing, therefore, from 6 to 7 to 8 and further. In the last value, the tens value becomes one because after 490, when we add 10, the number will become 500.

Hence, the statement was given by Manuel that the tens digit goes up by 1 in these numbers, is true.

Learn more about Tens Place:

https://brainly.com/question/10935564

Answer: The equation of the sphere is:

(x-2)^2 + (y-2)^2 + (z-5/2)^2 = sqrt(245)/2

Step-by-step explanation:

The centre of the sphere is the midpoint of the diameter, which is

1/2

[(-4,7,6) + (8,-3,5)] =(2,2,5/2)

The length of the diameter is

=sqrt |(8,-3,5) - (-4,7,6)|^2

=sqrt (12^2 + (-10)^2 + (-1)^2)

=sqrt (144+100+1)

=sqrt(245)

so the radius of

the sphere is:

1/2(sqrt(245)) = sqrt(245)/2.

The equation of the sphere is:

(x-2)^2 + (y-2)^2 + (z-5/2)^2 = sqrt(245)/2

Answer:

So the parallelogram is in the first and second quadrants.

Step-by-step explanation:

From Exercise we have that the parallelogram have a vetrices:

A(negative 1​,4​) ​B(4​,0​) ​C(12​,3​) ​D(7​,7​). We use a site geogebra.org to plof a graph for the given parallelogram.

From the graphs we can see that the parallelogram is mostly in the first quadrant and smaller in the second quadrant. So the parallelogram is in the first and second quadrants.

Answer:

a) 120

b) 24

c) 18

Step-by-step explanation:

part a

The sample space is defined by the rank of the horses.

Hence, 5 ranks would permute to = 5! = 120 outcomes

part b

Fixing the first position for horse A we are left with four horses and 4 positions, the position are permutated to 4! = 24 outcomes

part c

Fixing the first position for horse A we are left with four horses and 4 positions, and horse B can not finish second hence:

A _ B _ _  the rest can permute hence, 3! = 6 outcomes

A _ _ B _ the rest can permute hence, 3! = 6 outcomes

A _ _ _ B the rest can permute hence, 3! = 6 outcomes

Total outcomes is sum of 3 cases above = 18 outcomes

Answer:

Step-by-step explanation:

To prove sin(a+b)*sin(a-b)=cos^2b-cos^2a

we simplify the left side  sin(a+b)*sin(a-b) first

sin(a+b) = sin a cos b + cos a sin b

sin(a-b) = sin a cos b - cos a sin b

sin(a+b)*sin(a-b) = (sin a cos b + cos a sin b) x (sin a cos b -cos a sin b)

sin a cos b((sin a cos b + cos a sin b) - cos a sin b (sin a cos b + cos a sin b)

open the bracket

sin a cos b(sin a cos b) + sin a cos b(cos a sin b) -cos a sin b (sin a cos b)+ cos a sin b ( cos a sin b)

sin²a cos²b + sin a cos b cos a sin b - cos a sin b sin a cos b + cos²a sin²b

sin²a cos²b  + 0 + cos²a sin²b

sin²a cos²b + cos²a sin²b

(1-cos² a)cos² b - cos² a(1-cos² b)

= cos² b - cos² a cos² b - cos² a +cos² a cos² b

cos² b - cos² a - cos² a cos² b + cos² a cos² b = cos² b - cos² a  + 0

cos² b - cos² a

left hand side equals right hand side