Find the instantaneous rate of change for the function at the given value. f (x )equals x squared plus 3 x at xequalsnegative 3 The instantaneous rate of change at xequalsnegative 3 is nothing.

Answer:

If Tristan rolled the dice first the probability that Tristan wins is 0.474.

Step-by-step explanation:

The probability of an event E is computed using the formula:

[tex]P(E)=\frac{Favorable\ otucomes}{Total\ outcomes}[/tex]

Given:

Tristan and Iseult play a game where they roll a pair of dice alternatively until Tristan wins by rolling a sum 9 or Iseult wins by rolling a sum of 6.

The sample space of rolling a pair of dice consists of a total of 36 outcomes.

The favorable outcomes for Tristan winning is:

S (Tristan) = {(3, 6), (4, 5), (5, 4) and (6, 3)} = 4 outcomes

The favorable outcomes for Iseult winning is:

S (Iseult) = {(1, 5), (2, 4), (3, 3), (4, 2) and (5, 1)} = 5 outcomes

Compute the probability that Tristan wins as follows:

[tex]P(E)=\frac{Favorable\ otucomes}{Total\ outcomes}\\ P(Tristan\ wins)=\frac{4}{36}\\P(T)=\frac{1}{9} \\\approx0.1111[/tex]

Compute the probability that Iseult wins as follows:

[tex]P(E)=\frac{Favorable\ otucomes}{Total\ outcomes}\\ P(Iseult\ wins)=\frac{4}{36}\\P(I)=\frac{1}{9} \\\approx0.1111[/tex]

If Tristan plays first, then the probability that Tristan wins is:

= P(T) + P(T')P(I')P(T) + P(T')P(I')P(T')P(I')P(T)+...

=P(T) + [(1-P(T))(1-P(I))P(T)]+[(1-P(T))(1-P(I))(1-P(T))(1-P(I))P(T)]+...

[tex]=0.1111+(0.8889\times0.8611\times0.1111)+(0.8889\times0.8611\times0.8889\times0.8611\times0.1111))+...\\=0.1111[1+(0.8889\times0.8611)+(0.8889\times0.8611)^{2}+...]\\[/tex]This is an infinite geometric series.

The first term is, a = 0.1111 and the common ratio is, r = (0.8889×0.8611).

The sum of infinite geometric series is:

[tex]S_{\infty}=\frac{a}{1-r}\\ =\frac{0.1111}{1-(0.8889\times0.8611}\\ =0.47364\\\approx0.474[/tex]

Thus, the probability that Tristan wins if he rolled the die first is 0.474.

Answer:

Step-by-step explanation:

Based on the given information the required probabilities are as follows:

(a) Probability that all of the next three vehicles pass: 0.343

(b) Probability that at least one of the next three vehicles fails: 0.657

Given that,

70% of all vehicles examined at a certain emissions inspection station pass the inspection.

Successive vehicles pass or fail independently of one another.

(a) To find the probability that all three vehicles pass,

Multiply the individual probabilities.

Given that 70% of vehicles pass,

The probability that a single vehicle passes is 0.7.

So, the probability that all three vehicles pass is 0.7³ = 0.343.

(b) To find the probability that at least one of the next three vehicles fails, We can find the complement probability and subtract it from 1.

The complement probability is the probability that all three vehicles pass, which we calculated in the previous part.

So, the probability that at least one vehicle fails is 1 - 0.343 = 0.657.

(c) To find the probability that exactly one of the next three vehicles passes,

Use the binomial probability formula:

[tex]P(X=k) = ^nC_k p^k (1-p)^{(n-k)}[/tex],

Where n is the number of trials,

k is the number of successes,

p is the probability of success, and

C(n,k) is the combination of n and k.

In this case,

n = 3,

k = 1,

p = 0.7

Plugging these values into the formula,

We get [tex]P(X=1) = ^3C_1\times 0.7 \times (1-0.7)^2[/tex]

P(X=1) = 3x0.7x0.3²

P(X=1) = 0.189

So, the probabilities are:

(a) P(all of the next three vehicles inspected pass) = 0.343

(b) P(at least one of the next three inspected fails) = 0.657

(c) P(exactly one of the next three inspected passes) = 0.189

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To calculate the probabilities, we use the binomial probability formula. The probability of all three vehicles passing is 0.343, the probability of at least one vehicle failing is 0.657, and the probability of exactly one vehicle passing is 0.189.

To calculate the probabilities, we can use the binomial probability formula. Let's solve each part step by step:

(a) P(all of the next three vehicles inspected pass):

The probability of each vehicle passing is 70%, so the probability of all three passing is 0.7 x 0.7 x 0.7 = 0.343.

(b) P(at least one of the next three inspected fails):

The probability of a vehicle failing is 30%, so the probability of all three passing is 1 - (0.7 x 0.7 x 0.7) = 0.657.



(c) P(exactly one of the next three inspected passes):

There are three possible scenarios where exactly one vehicle passes: (Pass, Fail, Fail), (Fail, Pass, Fail), (Fail, Fail, Pass). The probability of each scenario is 0.7 x 0.3 x 0.3 = 0.063. Since there are three scenarios, the total probability is 0.063 x 3 = 0.189.

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Answer:

b. can be calculated by subtracting the fitted values from the actual values.

Step-by-step explanation:

OLS residuals -  it stands for ordinary least square. it is used to determine the missing value in the regression analysis. OLS works on one purpose that is to minimize the difference between the observed response and predict response.

The basic difference between Residual sum of square(RSS) and OLS is that RSS is used to predict how good is model while OLS  is considered as the method which is used to construct model>

Answer:

Kamala had the higher Z-score, so she had the best GPA when compared to other students at his school.

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:

Three students, graded on different curves. I will find whoever has the higher Z-score, and this is the one which had the best GPA.

Thuy 2.7 3.2 0.8

So the student GPA is 2.7, the Average GPA at the school was 3.2 and the standard deviation was 0.8.

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

[tex]Z = \frac{2.7 - 3.2}{0.8}[/tex]

[tex]Z = -0.625[/tex]

Vichet 87 75 20

So the student GPA is 87, the Average GPA at the school was 75 and the standard deviation was 20.

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

[tex]Z = \frac{87 - 75}{20}[/tex]

[tex]Z = 0.6[/tex]

Kamala 8.6 8 0.4

So the student GPA is 8.6, the Average GPA at the school was 8 and the standard deviation was 0.4.

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

[tex]Z = \frac{8.6 - 8}{0.4}[/tex]

[tex]Z = 1.5[/tex]

Kamala had the higher Z-score, so she had the best GPA when compared to other students at his school.

Answer:

(a) 1 in 365 or 0.2740%

(b) 0.8227%

Step-by-step explanation:

(a) For any given birthday date of the first person, there is a 1 in 365 chance that the second person shares the same birthday, therefore the probability that the first two people share a birthday is:

[tex]P = \frac{1}{365}=0.2740\%[/tex]

(b) There are four possibilities that at least two people share a birthday, first and second, first and third, second and third, all three share a birthday. Therefore, the probability that at least two people share a birthday is:

[tex]P =3* \frac{1}{365}+ (\frac{1}{365})^2\\ P=0.8227\%[/tex]

a) The probability that the first two people share a birthday is approximately 0.0027. b) The probability that at least two people share a birthday is approximately 0.9901.

let's solve each part step by step:

To calculate this probability, we can consider the scenario where the first person is born on any day of the year (365 possibilities) and the second person must share the same birthday. So, the probability that the second person shares the same birthday as the first is 1/365.

Therefore, the probability that the first two people share a birthday is [tex]\( \frac{1}{365} \).[/tex]

To find this probability, we can use the complement rule: the probability of the event happening is 1 minus the probability of the event not happening.

The probability that no two people share a birthday can be found by considering the birthday of each person and ensuring that they all have different birthdays. For the first person, any of the 365 days is possible. For the second person, there are 364 days remaining. For the third person, there are 363 days remaining. So, the probability that no two people share a birthday is:

[tex]\[ \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \][/tex]

To find the probability that at least two people share a birthday, we subtract this probability from 1:

[tex]\[ 1 - \left( \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \right) \][/tex]

[tex]\[ = 1 - \frac{365 \times 364 \times 363}{365^3} \][/tex]

[tex]\[ \approx 1 - \frac{479,664,580}{48,627,125} \][/tex]

[tex]\[ \approx 1 - 0.009882 \][/tex]

[tex]\[ \approx 0.990118 \][/tex]

So, the probability that at least two people share a birthday is approximately ( 0.990118 ).

There are 640 possible three-digit phone prefixes that do not have 0 or 1 in the first or second digits.

The number of three-digit phone prefixes are possible that have no 0 or 1 in the first or second digits.

Since each digit in a phone number can be a number from 0-9, but the first two digits cannot be 0 or 1, we have 8 choices (2-9) for the first digit, 8 choices (2-9) for the second digit, and 10 choices (0-9) for the third digit because the third digit has no such restriction.

Therefore, the total number of possible phone prefixes is calculated by multiplying the number of choices for each digit,

8 (choices for the first digit) × 8 (choices for the second digit) × 10 (choices for the third digit) = 640 possible three-digit phone prefixes.

Answer:

a) [tex]E(A)=\frac{1+6}{2}=3.5 ppm[/tex]

b) [tex] P(2.875 <X < 3.5) = F(3.5) -F(2.875) = \frac{3.5-1}{5}- \frac{2.875-1}{5}= \frac{1}{8}= 0.125[/tex]

c) [tex] P(X<4.125) = F(4.125) = \frac{4.125-1}{5}= 0.625[/tex]

Step-by-step explanation:

If we work with the limits defined from 5 to 10 then part b and c from this question not makes sense. If we work with the limits 1 and 6 all the parts for the question makes sense because if we work with 5 and 10 the only thing that we can find is the expected value [tex] E(A) = \frac{5+10}{2}= 7.5[/tex]

Assuming the following correct question : "A publication released the results of a study of the evolution of a certain mineral in the​ Earth's crust. Researchers estimate that the trace amount of this mineral x in reservoirs follows a uniform distribution ranging between 1 and 6 parts per million"

Solution to the problem

Let A the random variable that represent " amount of the mineral x ". And we know that the distribution of A is given by:

[tex]A\sim Uniform(1 ,6)[/tex]

Part a

For this uniform distribution the expected value is given by [tex]E(X) =\frac{a+b}{2}[/tex] where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

[tex]E(A)=\frac{1+6}{2}=3.5 ppm[/tex]

Part b

For this case we can use the cumulative distribution function for the uniform distribution given by:

[tex] F(X=x)= \frac{x-a}{b-a} = \frac{x}{6-1} =\frac{x-1}{5} , 1 \leq X \leq 6[/tex]

And we want this probability:[tex] P(2.875 <X < 3.5) = F(3.5) -F(2.875) = \frac{3.5-1}{5}- \frac{2.875-1}{5}= \frac{1}{8}= 0.125[/tex]Part c

For this case we want this probability:

[tex] P(X<4.125) = F(4.125) = \frac{4.125-1}{5}= 0.625[/tex]

Answer:

no, the 550 each year is best offer

Step-by-step explanation:

Answer:

I’ll just stay with the 550 a year

Step-by-step explanation:

Answer:

A. 1. P(A∪B)=0.85

2. P(A∩C)=0.045

3. P(A/B)=0.3

4. P(B∪C)=0.70

B. Event B and Event C are dependent

Step-by-step explanation:

A. As events are mutually exclusive, so,

P(A∪B)=P(A)+P(B)

P(A∩B)=P(A)*P(B)

1. P(A∪B)=?

P(A∪B)=P(A)+P(B)=0.3+0.55=0.85

P(A∪B)=0.85

2. P(A∩C)

P(A∩C)=P(A)*P(C)=0.30*0.15=0.045

P(A∩C)=0.045

3. P(A/B)

P(A/B)=P(A∩B)/P(B)

P(A∩B)=P(A)*P(B)=0.30*0.55=0.165

P(A/B)=P(A∩B)/P(B)=0.165/0.55=0.3

P(A/B)=0.3

4.  P(B∪C)

P(B∪C)=P(B)+P(C)=0.55+0.15=0.70

P(B∪C)=0.70

B.

The event B and C are mutually exclusive and events B and event C are dependent i.e. P(B and C)≠P(B)P(C)

The events are mutually exclusive i.e. P(B and C)=0

whereas  P(B)*P(C)=0.55*0.15=0.0825

Mutually exclusive events are independent only if either one of two or both events has zero probability of occurring.

Thus, event B and C are dependent

A) 1:P(A∪B)=0.85

2: P(A∩C)=0.045

3: P(A/B)=0.3

4: P(B∪C)=0.70

B) Events B and C are dependent events.

Since all three events are mutually exclusive:

So, P(A∪B)=P(A)+P(B)

P(A∪B) = 0.30+0.55

P(A∪B) = 0.85

P(A∩B)=P(A)P(B)

P(A∩B) = 0.30*0.55 = 0.165

P(A/B)=P(A∩B)/P(B)

P(A/B) = 0.165/0.55 = 0.3

Similarly, P(A∩C) =0.045

P(BUC) = 0.70

Events B and C are dependent events because they will be independent only if there is zero possibility of their occurrence.

Therefore, A) 1:P(A∪B)=0.85

2: P(A∩C)=0.045

3: P(A/B)=0.3

4: P(B∪C)=0.70

B) Events B and C are dependent events.

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Answer:

They are in contact with 20 people at least once in a year.

Step-by-step explanation:

We are given the basic summary statistics:

Mean = 20.2

Mode = 10

Median = 10

Standard deviation = 28.7

1st quartile = 5

3rd quartile = 25

Minimum = 0

Maximum = 300.

Out of all these statistics, we know that the median is the middle value or mid-value. Thus, by formula, we have that:

median = n/2. Thus,

==> 10 = n/2

==> n = 2*10 = 20

NB: From the distribution of the summary statistics, we can clearly see that there is evidence of outlier in the series. Thus, median is the most appropriate statistic (because is not affected by outlier) to give a true picture of the data series or sets.

Answer:

a) Let X the random variable that represent the blood pressure for people of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(128,23)[/tex]  

Where [tex]\mu=128[/tex] and [tex]\sigma=23[/tex]

b) [tex]P(X\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 0.304)=1-P(Z<0.304)[/tex]

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

[tex]P(Z\geq 0.304)=1-P(Z<0.304)= 1-0.619=0.381 [/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".  

Part a

Let X the random variable that represent the blood pressure for people of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(128,23)[/tex]  

Where [tex]\mu=128[/tex] and [tex]\sigma=23[/tex]

Part b

We are interested on this probability

[tex]P(X\geq 135)[/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\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 0.304)=1-P(Z<0.304)[/tex]

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

[tex]P(Z\geq 0.304)=1-P(Z<0.304)= 1-0.619=0.381 [/tex]

In this context, the random variable is the blood pressure of people in China. The probability that a person in China has a blood pressure of 135 mmHg or more is about 38.21%, calculated using the Z-score and standard Z-table.

The subject of this question is the study of normal distribution and probability in statistics, a branch of mathematics.

a.) The random variable in this context is the blood pressure of people in China.

b.) To find the probability that a person in China has a blood pressure of 135 mmHg or more, we need to convert this to a Z-score. The Z-score is calculated by subtracting the mean from the individual score and then dividing by the standard deviation. Therefore, Z = (135 - 128) / 23 = 0.30.

Using a standard Z-table, the probability corresponding to Z=0.30 is about 0.6179. However, because we want the probability of a person having a blood pressure that's 135 mmHg or more (greater than the mean), we must subtract this value from 1. Thus, the probability is about 1 - 0.6179 = 0.3821, or 38.21%.

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Since we're given the femour's length, we'll have to use the first function.

If we substitute [tex]f=49[/tex] in the expression we have

[tex]g(f)=2.56f+47.24 \implies g(49)=2.56\cdot 49+47.24=125.44+47.24=172.68[/tex]

The height of the woman with a femur length of 49 cm is 172.68 cm.

A function is a relationship between inputs where each input is related to exactly one output.

Example:

f(x) = 2x + 1

f(1) = 2 + 1 = 3

f(2) = 2 x 2 + 1 = 4 + 1 = 5

The outputs of the functions are 3 and 5

The inputs of the function are 1 and 2.

We have,

G(f) = 2.56f + 47.24

f = 49 cm

G(49) = 2.56 x 49 + 47.24

G(49) = 125.44 + 47.24

G(49) = 172.68  cm

Thus,

The height of the woman is 172.68 cm.

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Answer:

Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20

Step-by-step explanation:

assuming that each cherry is equally probable to be chosen , since each cherry is independent from the others and sampling is done without replacement , the random variable X= number of cherries that are rotten from the picked ones follows a hyper geometrical distribution , where

P(X=k)= C(M,k) * C(N-M, n-k) / C(N,n)

where

N= population size = 25

n= number of picks

M = total number of rotten cherries =2

k = number of rotten cherries picked =2

C( ) = combination

then

1/20=C(2,2)*C(25-2,n-2)/C(25,n) = 1 * (23!/(n-2)!*(25-n)! / (25!/(n!*(25-n)!

1/20 = n!/(n-2)!  * 1/(24*25)

24*25/20 = n*(n-1)

n²-n-30 =0

n= (1 +√(1+4*1*30))/2 = 12/2= 6

n=6

then Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20

To find the number of cherries Susan should pick, use the combination formula and probability calculation. After setting up an equation with probability equal to 1/20, solve for 'x' using trial and error methods.

To answer the question, we need to use the combination formula. This formula in the field of statistics is used to find the number of possible combinations that can be obtained by taking 'r' elements from a set of 'n' elements.

The formula is: C(n, r) = n! / [(n - r)! * r!]

Given 25 cherries, 2 of which are rotten, Susan wants to choose some cherries such that the probability of getting exactly 2 rotten cherries is 1/20. Let's assume she needs to pick 'x' cherries.

Now we can write the probability equation: Probability = [C(2, 2) * C(23, x - 2)] / C(25, x) = 1/20

Unfortunately, we can't explicitly solve this equation because it would require checking different values of 'x'. However, it can be solved manually or through trial and error using software or a calculator. After checking different values, you can find the 'x' that satisfies the equation.

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Answer:

In the profile here, candidate A wins plurality, B wins anti plurality, C wins Borda count, and D wins vote-for-two.

Step-by-step explanation:

             2  3  2  4   3

             A  A  B  C  D

             D  C  D  D  C

             B  B  C  B  B

             C  D  A  A  A                        

An election profile can be created where each of 4 candidates can win a unique voting method: For plurality where the most top ranked votes win, for Borda count where points are given based on positions, for a positional voting method where points awarded to lower place candidates changes, thereby awarding the win to different candidates.

In the context of voting theory and social choice theory, we can create a profile for an election with 4 candidates (let's call them A, B, C and D) where each candidate can win based on different positional voting methods.

Consider this profile for the 20 voters:

   4 voters prefer D > A > B > C      

For plurality method (where the candidate ranked first by most voters wins), A would win with 6 votes. For Borda count method (where points are given based on ranks), Candidate B would gain the most points and win. For positional method where points are assigned 3 for 1st place, 2 for 2nd place, 1 for last two places, C would win. For positional method where points are assigned differently - 4 points for 1st position, 2 for 2nd, 1 for 3rd and 0 points for fourth, D wins. The unique winner for each method thus satisfies the given condition.

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Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know 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 = 11, \sigma = 3[/tex]

a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that [tex]n = 25, s = \frac{3}{\sqrt{25}} = 0.6[/tex]

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

[tex]Z = \frac{11.2 - 11}{0.6}[/tex]

[tex]Z = 0.33[/tex]

[tex]Z = 0.33[/tex] has a pvalue of 0.6293.

X = 10.8

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

[tex]Z = \frac{10.8 - 11}{0.6}[/tex]

[tex]Z = -0.33[/tex]

[tex]Z = -0.33[/tex] has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.5 and 11 ​minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

[tex]Z = \frac{11 - 11}{0.6}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.5.

X = 10.5

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

[tex]Z = \frac{10.5 - 11}{0.6}[/tex]

[tex]Z = -0.83[/tex]

[tex]Z = -0.83[/tex] has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that [tex]n = 100, s = \frac{3}{\sqrt{100}} = 0.3[/tex]

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

[tex]Z = \frac{11.2 - 11}{0.3}[/tex]

[tex]Z = 0.67[/tex]

[tex]Z = 0.67[/tex] has a pvalue of 0.7486.

X = 10.8

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

[tex]Z = \frac{10.8 - 11}{0.3}[/tex]

[tex]Z = -0.67[/tex]

[tex]Z = -0.67[/tex] has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

To find the probability that the sample mean is between eight minutes and 8.5 minutes, calculate the z-scores for both values and find the area under the standard normal distribution curve between these z-scores.

To find the probability that the sample mean is between eight minutes and 8.5 minutes, we need to calculate the z-scores for both values and then find the area under the standard normal distribution curve between these two z-scores.

The formula to calculate the z-score is: z = (x - mu) / (sigma / sqrt(n))

where x is the sample mean, mu is the population mean, sigma is the population standard deviation, and n is the sample size.

Using the given information, we can calculate the z-scores as follows:

z1 = (8 - 11) / (3 / sqrt(25))

z2 = (8.5 - 11) / (3 / sqrt(25))

Next, we use a standard normal distribution table or a calculator to find the area between these two z-scores, which represents the probability that the sample mean is between eight minutes and 8.5 minutes.

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Answer:

c₁ = 1/2 cos⁻¹ (2/π) = 0.44

c₂ = -1/2 cos⁻¹ (2/π) = -0.44

Step-by-step explanation:

the average value of f(x)=2 cos(2x) on ( − π/ 4 , π/ 4 ) is

av f(x) =∫[2*cos(2x)] dx /(∫dx) between limits of integration − π/ 4 and π/ 4

thus

av f(x) =∫[cos(2x)] dx /(∫dx) = [sin(2 * π/ 4 ) - sin(2 *(- π/ 4 )] /[ π/ 4 -  (-π/ 4)]

= 2*sin (π/2) /(π/2) = 4/π

then the average value of f(x) is 4/π . Thus the values of c such that f(c)= av f(x) are

 4/π = 2 cos(2c)  

2/π = cos(2c)

c = 1/2 cos⁻¹ (2/π) = 0.44

c= 0.44

since the cosine function is symmetrical  with respect to the y axis then also c= -0.44 satisfy the equation

thus

c₁ = 1/2 cos⁻¹ (2/π) = 0.44

c₂ = -1/2 cos⁻¹ (2/π) = -0.44

The two values are,

[tex]c=-\frac{1}{2} cos^{-1}(\frac{2}{\pi}) or\\ c=\frac{1}{2} cos^{-1}(\frac{2}{\pi}) [/tex]

Given that,

[tex]f(x)=2cos(2x)[/tex]

[tex]f_{avg}=\frac{1}{\frac{\pi}{2} }\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}2cos(2x)dx\\ =\frac{8}{2\pi} sin\frac{\pi}{2} \\ =\frac{4}{\pi} [/tex]

[tex]f(c)=2cos(2c)=\frac{4}{\pi} \\ cos2c=\frac{2}{\pi} \\ c=-\frac{1}{2} cos^{-1}(\frac{2}{\pi})or\\ c=\frac{1}{2} cos^{-1}(\frac{2}{\pi})or\\[/tex]

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The probability that sales will be less than 4,300 oranges is 21.19%

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

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score, \mu=mean, \sigma=standard\ deviation\\\\Given\ that:\\\mu=4700,\sigma=500\\\\For\ x=4300:\\\\z=\frac{4300-4700}{500} =-0.8[/tex]

From the normal distribution table:

P(x < 4300) = P(z < -0.8) = 0.2119 = 21.19%

The probability that sales will be less than 4,300 oranges is 21.19%

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The probability that sales will be less than 4,300 oranges is approximately 21.23%.

To find the probability that sales will be less than 4,300 oranges, we need to standardize the value using the z-score formula.

The z-score formula is given by z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. In this case, x = 4,300, μ = 4,700, and σ = 500.

Substituting these values into the formula, we get z = (4,300 - 4,700) / 500 = -0.8. We can then use a z-table or a calculator to find the probability associated with a z-score of -0.8, which is approximately 0.2123 or 21.23%.

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Answer:

dilation

Step-by-step explanation:

When the pulmonary valve does not work properly, it can interfere with blood flow from the heart to the lungs, as well as force the heart to work harder to carry the blood that is needed to the rest of the body. Some children have heart conditions present at the time of birth and may require repair or replacement of the pulmonary valve, this option has a lower risk of infection, preserves the strength and functioning of the valve, and eliminates the need to take medication.

Final answer:

The root operation for treating pulmonary valve stenosis in newborns using balloon valvuloplasty is the valvuloplasty itself, which is a non-surgical procedure to widen the stenosed heart valve.

Explanation:

The root operation for treating newborn pulmonary valve stenosis with percutaneous balloon pulmonary valvuloplasty is the procedure known as valvuloplasty.

This procedure involves the insertion of a specialized catheter with a balloon at its tip into a blood vessel, usually via the leg, and navigating it to the valve.

The balloon is then inflated to widen the stenosed valve and allow better blood flow. Subsequently, the balloon is deflated and removed, completing the valvuloplasty.

[tex]\boxed{A=4yd^2}[/tex]

I'll assume the dimensions are:

[tex]base \ (b)=1 \frac{1}{3}yards \\ \\ height \ (h)=6yards[/tex]

First of all, let's convert mixed fraction into improper fraction:

[tex]Add \ whole \ part \ and \ fractional \ part: \\ \\ 1\frac{1}{3}=1+\frac{1}{3} \\ \\ \\ Simplifying: \\ \\ 1\frac{1}{3}=\frac{3+1}{3} =\frac{4}{3}[/tex]

The formula for the area of a triangle is:

[tex]A=\frac{1}{2}b\times h \\ \\ A=\frac{1}{2}(\frac{4}{3})(6) \\ \\ \boxed{A=4yd^2}[/tex]

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Answer:

a) Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]

Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]

b) [tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{850 -0}{\frac{1123}{\sqrt{42}}}=4.905[/tex]

The next step is calculate the degrees of freedom given by:

[tex]df=n-1=42-1=41[/tex]

Now we can calculate the p value, since we have a two tailed test the p value is given by:

[tex]p_v =2*P(t_{(41)}>4.905) =7.6x10^{-6}[/tex]

So the p value is lower than any significance level given, so then we can conclude that we can to reject the null hypothesis that the difference between the two mean is equal to 0.

c) The confidence interval is given by:

[tex] \bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

For this case we have a confidence of 1-0.05 = 0.95 so we need 0.05 of the area of the t distribution with 41 df on the tails. So we need 0.025 of the area on each tail, and the critical value would be:

[tex] t_{crit}= 2.02[/tex]

And if we find the interval we got:

[tex] 850- 2.02* \frac{1123}{\sqrt{42}}=499.96[/tex]

[tex] 850+ 2.02* \frac{1123}{\sqrt{42}}=1200.03[/tex]

We are confident 95% that the difference between the two means is between 499.96 and 1200.03

So we have enough evidence to conclude that one mean is higher than the other one, without conduct another hypothesis test because the confidence interval for the difference of means not contain the value of 0. And for this case the groceries would have a higher mean

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Part a

Let put some notation  

x=test value for 1 , y = test value for 2

The system of hypothesis for this case are:

Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]

Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]

The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}[/tex]

This value is given [tex] \bar d = 850[/tex]

The third step would be calculate the standard deviation for the differences.

This value is given also [tex] s_d = 1123[/tex]

Part b

The 4 step is calculate the statistic given by :

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{850 -0}{\frac{1123}{\sqrt{42}}}=4.905[/tex]

The next step is calculate the degrees of freedom given by:

[tex]df=n-1=42-1=41[/tex]

Now we can calculate the p value, since we have a two tailed test the p value is given by:

[tex]p_v =2*P(t_{(41)}>4.905) =7.6x10^{-6}[/tex]

So the p value is lower than any significance level given, so then we can conclude that we can to reject the null hypothesis that the difference between the two mean is equal to 0.

Part c

The confidence interval is given by:

[tex] \bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

For this case we have a confidence of 1-0.05 = 0.95 so we need 0.05 of the area of the t distribution with 41 df on the tails. So we need 0.025 of the area on each tail, and the critical value would be:

[tex] t_{crit}= 2.02[/tex]

And if we find the interval we got:

[tex] 850- 2.02* \frac{1123}{\sqrt{42}}=499.96[/tex]

[tex] 850+ 2.02* \frac{1123}{\sqrt{42}}=1200.03[/tex]

We are confident 95% that the difference between the two means is between 499.96 and 1200.03

So we have enough evidence to conclude that one mean is higher than the other one, without conduct another hypothesis test because the confidence interval for the difference of means not contain the value of 0. And for this case the groceries would have a higher mean

Answer:

H_o : u_d = 0 , H_1 : u_d ≠ 0

H_o rejected , p < 0.01

[ 499.969 < d < 1200.301 ] , d = 850

Step-by-step explanation:

Given:

- Difference in mean d = 850

- Standard deviation s = 1123

- The sample size n = 42

- Significance level a = 0.05

Solution:

- Set up and Hypothesis for the difference in means test as follows:

            H_o : Difference in mean u_d= 0

            H_1 : Difference in mean u_d ≠ 0

- The t test statistics for hypothesis of matched samples is calculated by the following formula:

           t = d / s*sqrt(n)

Hence,

           t = 850 / 1123*sqrt(42)

           t = 4.9053

  Thus, the test statistics t = 4.9053.

- The p-value is the probability of obtaining the value of the test statistics or a value greater.

 Using Table 2, of appendix B determine p with DOF = n - 1 = 42 - 1 = 41 , We get:

            p < 2*0.05 ----> 0.01

 Thus,  p < 0.05  ....... Hence, H_o is rejected

- Set up and Hypothesis for the difference in means test as follows:

            H_o : Difference in mean u_d =< 0

            H_1 : Difference in mean u_d > 0

- The t test statistics for hypothesis of matched samples is calculated by te following formula:

           t = d / s*sqrt(n)

Hence,

           t = 850 / 1123*sqrt(42)

           t = 4.9053

  Thus, the test statistics t = 4.9053.

 Using Table 2, of appendix B determine p with DOF = n - 1 = 42 - 1 = 41 , We get:

            p < 0.005

 Thus,  p < 0.05  ....... Hence, H_o is rejected

Hence, the point estimate is d = $850

- The interval estimate of the difference between two population means is calculated by the following formula:

             d +/- t_a/2*s / sqrt(n)

Where CI = 1 - a = 0.95 , a = 0.05 , a/2 = 0.025

Using Table 2, of appendix B determine p with DOF = n - 1 = 42 - 1 = 41 , We get:

              t_a/2 = t_0.025 = 2.020

Therefore,

              d - t_a/2*s / sqrt(n)

              850 - 2.020*1123 / sqrt(20)

             = 499.969

And,

              d + t_a/2*s / sqrt(n)

              850 + 2.020*1123 / sqrt(20)

             = 1200.031

- The 95% CI of the difference between two population means is:

              [ 499.969 < d < 1200.301 ]

The answer is 9 1/5 simplified.

Too lazy to make up an explanation or change it to a mixed number even tho it is easy...

Answer:

Step-by-step explanation:

5³/10 +3 9/10

Convert to improper fraction

53/10 +39/10

L. C. M =10

=92/10

=9²/10

=9¹/5

Answer:

249 per 100,000

Step-by-step explanation:

The projected incidence rate of heart disease per 100,000 in the United States in 2020 is determined by the number of estimated new cases of heart disease multiplied by 100,000 and divided by the estimated population of the United States in 2020:

[tex]R=\frac{850,000*100,000}{341,672,244}=248.78[/tex]

Rounding up to the next whole unit, the projected incidence is 249 per 100,000.

Final answer:

To calculate the incidence rate of heart disease per 100,000 in the U.S. in 2020, divide the number of new cases (850,000) by the U.S. population (341,672,244), then multiply by 100,000, resulting in an incidence rate of approximately 248.7.

Explanation:

To calculate the projected incidence rate of heart disease per 100,000 in the United States in 2020, follow these steps:

First, determine the total number of new heart disease cases. Researchers estimated 850,000 new cases of heart disease will be recorded.

Second, determine the population of the United States at the midpoint of 2020, which is 341,672,244.

Finally, calculate the incidence rate using the formula: \((\frac{Number\ of\ new\ cases}{Population}) \times 100,000=Incidence\ Rate\ per\ 100,000\). Therefore, \((\frac{850,000}{341,672,244}) \times 100,000\) results in an incidence rate of approximately 248.7 new cases per 100,000 people.

Understanding this incidence rate can help in assessing the burden of heart disease on the U.S. population and guiding health policy decisions.

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) 0.88

b) 0.02

c) 0.01

d) 0.99

Step-by-step explanation:

Step one: State the given parameters

            [tex]P(D_{1} ) = 0.12[/tex]                                   [tex]P(D_{2} ) = 0.07[/tex]

           [tex]P(D_{3} ) = 0.05[/tex]                                    [tex]P (D_{1} U D_{2} ) = 0.13[/tex]

          [tex]P(D_{1}n D_{2}n D_{3}) = 0.01[/tex]                        [tex]P(D_{1} U D_{3}) = 0.14[/tex]

Step 2 : Obtain the probability that a unit does not have a type 1 defect

         [tex]P(\frac{}{D_{1} })[/tex] =  [tex]1 -P(D_{1} )[/tex]

                    = [tex]1 - 0.12[/tex]

                    = 0.88  

Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?

          The probability of the unit having both type 2 and type 3 defect is denoted as [tex]P(D_{2} n D_{3} )[/tex]

   This is calculated as

                    [tex]P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\ = 0.07 + 0,05 - 0.13[/tex]

                    =   0.02

Therefore P(D_{2} n D_{3} ) = 0.02

Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect

                  Let [tex]P(\frac{}{D_{1}} n D_{2} n D_{3} )[/tex] denote the  probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.

This can be calculated as follows :

                      [tex]P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})[/tex]

                                               =   0.02 - 0.01

                                               =  0.01

Step 4 : Obtain the probability that a unit has at most two defects

               P(at most 2 defects)  = 1 - P(all three defects)

                                                  = [tex]1- P(D_{1} n D_{2}nD_{3})[/tex]

                                                  =  1 - 0.01

                                                  = 0.99

Answer: the system of linear equations to represent this scenario are

8x + 4y = 40

3x + 10y = 32

Step-by-step explanation:

Let x represent the cost of one hamburger.

Let y represent the cost of one hotdog.

You spend $40 for 8 hamburgers and 4 hotdogs at a ballgame. This means that

8x + 4y = 40 - - - - - - - - - - - - 1

The next game you spend $32 for 3 hamburgers and 10 hotdogs. This means that

3x + 10y = 32- - - - - - - - - - - - 2

Multiplying equation 1 by 3 and equation 2 by 8, it becomes

24x + 12y = 120

24x + 80y = 256

- 68y = - 136

y = - 136 /- 68

y = 2

Substituting y = 2 into equation 2, it becomes

3x + 10y = 32

3x + 10 × 2 = 32

3x = 32 - 20 = 12

x = 12/3 = 4

Answer:

D. 0.821

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The combinations formula is important to solve this question:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

Desired outcomes

The order is not important. For example, Elisa, Laura and Roze is the same outcome as Roze, Elisa and Laura. This is why we use the combinations formula.

At least 3 girls.

3 girls

3 girls from a set of 5 and 2 boys from a set of 3. So

[tex]C_{5,3}*C_{3,2} = 30[/tex]

4 girls

4 girls from a set of 5 and 1 boy from a set of 3. So

[tex]C_{5,4}*C_{3,1} = 15[/tex]

5 girls

5 girls from a set of 5

[tex]C_{5,5} = 1[/tex]

[tex]D = 30+15+1 = 46[/tex]

Total outcomes

5 from a set of 8. So

[tex]T = C_{8,5} = 56[/tex]

Probability

[tex]P = \frac{D}{T} = \frac{46}{56} = 0.821[/tex]

So the correct answer is:

D. 0.821

Answer:

I got answer choice D

Hope this helps :)

Step-by-step explanation:

Answer:

S={R,A,N}

Step-by-step explanation:

There are three letters in a word RAN i.e. R, A and N. The sample space consists of all possible outcomes of an experiment. So, the sample for selecting a letter from word RAN will be

Sample Space=S={R,A,N}

As, there are three possible letters that can be selected n(S)=3

Thus, the required sample in the set notation is

S={R,A,N}.

Final answer:

The sample space for selecting a letter from the word RAN is S = {R,A,N}, which includes all the individual letters of the word as separate and distinct outcomes.

Explanation:

The sample space for the experiment of randomly selecting one of the letters in the word RAN is a set of all possible outcomes of this experiment. Using set notation, we can describe this sample space as follows:

S = {R,A,N}

Each letter represents a unique outcome in the sample space, which in this case consists of the three letters that make up the word. Since the word RAN has no repeated letters, each letter is a distinct outcome. To list this sample space in set notation we simply enclose the elements within braces and separate them with commas, without spaces.

Answer:

The instantaneous rate of change of h(t) = 2t² + 2 at the point t = −1 is -4.

Step-by-step explanation:

If two distinct points [tex]P(x_1,y_1)[/tex] and [tex]Q(x_2,y_2)[/tex] lie on the curve [tex]y=f(x)[/tex], the slope of the secant line connecting the two points is

[tex]m_{sec}=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x_2)-f(x_1)}{x_2-x_1}[/tex]

If we let the point [tex]x_2[/tex] approach [tex]x_1[/tex], then Q will approach P along the graph f(x). The slope of the secant line through points P and Q will gradually approach the slope of the tangent line through P as

[tex]m_{tan}= \lim_{x_2 \to x_1}\frac{f(x_2)-f(x_1)}{x_2-x_1}[/tex]

And this is the instantaneous rate of change of the function f(x) at the point [tex]x_1[/tex].

From the information given, we know that the point P [tex](-1,2(-1)^2+2)=(-1,4)[/tex] lies on the curve [tex]h(t) = 2t^2 + 2[/tex].

If Q is the point [tex](t, 2t^2 + 2)[/tex] we can find the slope of the secant line PQ for the following values of t. Because we choose values that are getting closer and closer to −1.

[tex]\begin{array}{c}-0.9&-0.99&-0.999&-0.9999\\-1.1&-1.01&-1.001&-1.0001\\\end{array}\right[/tex]

Let the point P be [tex](x_2=-1, y_2=4)[/tex] and the point Q be [tex](x_1=t, y_1=2t^2+2)[/tex]. So,

[tex]m=\frac{4-(2t^2+2)}{-1-t}\\\\m=-\frac{2\left(t+1\right)\left(t-1\right)}{-1-t}\\\\m=2\left(t-1\right)[/tex]

Next, substitute the value of x in the formula of the slope

[tex]m=2(-0.9-1)=-3.8[/tex]

Do this for the other values of x.

Below, there is a table that shows the values of the slope.

From the table, as t approaches -1 from the left side (-0.9 to -0.9999), the slopes are approaching to -4 and as t approaches -1 from the right side (-1.1 to -1.0001), the slopes are approaching to -4. The value of the slope at P(-1,4) is then m = -4.

Final answer:

To estimate the instantaneous rate of change of h(t) at t = -1, we calculate the slopes of secant lines near that point and observe the pattern to approximate the slope of the tangent line, which represents the instantaneous velocity.

Explanation:

To estimate the instantaneous rate of change of the function h(t) = 2t² + 2 at t = −1, we need to calculate the slope of the tangent line at that point. We can approximate this slope using secant lines connecting points increasingly closer to t = −1.

Let's select two points close to t = −1, say t1 = −1.1 and t2 = −0.9, and compute the slope of the secant line:

Slope of secant line = (h(t2) − h(t1)) / (t2 − t1) = (3.62 − 4.42) / (-0.9 + 1.1) = −0.8 / 0.2 = −4.

To get a more accurate approximation, we'd choose points even closer to t = −1 and observe the pattern. We can assume that the slope of the tangent line, which represents the instantaneous velocity, is approximately equal to the slopes of the secant lines as they converge to a single value.

Answer:

f(t)= 11 in - 1.3 in/h *t

Step-by-step explanation:

defining the length of the candle as L , then since the candle burns at a constant rate , then

-dL/dt = 1.3 in/h = a

therefore

-∫dL = a∫dt

-L(t)=a*t + C , C=constant

at t=0 , the length of the candle is L₀= 11 in ,thus

-L₀=a*0 + C →  C= -L₀

replacing the value of C

-L(t)=a*t - L₀

L(t) = L₀ - a*t = 11 in - 1.3 in/h *t

then

f(t)= 11 in - 1.3 in/h *t

Answer:

Option C) 23.6°

Step-by-step explanation:

we know that  

In this problem the triangle ABC is a right triangle

see the attached figure to better understand the problem

[tex]sin(A)=\frac{BC}{AB}[/tex] ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

[tex]sin(A)=\frac{6}{15}[/tex]

using a calculator

[tex]A=sin^{-1}(\frac{6}{15})=23.6^o[/tex]