Let H be an upper Hessenberg matrix. Show that the flop count for computing the QR decomposition of H is O(n2), assuming that the factor Q is not assembled but left as a product of rotators.

Answer:

-1

Step-by-step explanation:

We evaluate [tex]\sin(2\pi+3\pi/2)[/tex]

In [tex]2\pi+3\pi/2[/tex], [tex]2\pi[/tex] is a complete revolution and is the same as 0. So we have

[tex]\sin3\pi/2 = \sin(\pi+\pi/2)[/tex]

One [tex]\pi[/tex] is a half revolution, putting the point at (-1, 0). [tex]\pi/2[/tex] is a quarter of a revolution. A quarter circle from (-1, 0) anticlockwise is (0, -1).

The sine is the y-coordinate of a point along a unit circle.

Hence, [tex]\sin(2\pi+3\pi/2)=-1[/tex]

Using the cyclical nature of the unit circle and the sine function, the mathematical expression sin (2π + 3π/2) can be simplified to sin (3π/2). travelling 3π/2 around the unit circle takes us to the point where sin is -1.

The question requires the evaluation of the mathematical expression sin (2π + 3π/2). This can be solved by utilizing the cyclical nature of the unit circle and the sine function. Since the distance around the unit circle is 2π, adding or subtracting multiples of 2π from the angle doesn't change the result of the sin function.

So, we can simplify the function sin (2π + 3π/2) to sin (3π/2). Because every π/2 around the unit circle the sine function repeats, sin (π/2) is 1, sin (2π/2) or sin (π) is 0, sin (3π/2) is -1, and sin (4π/2) or sin (2π) is 0. So sin (3π/2) equals -1.

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Answer:

The correct optiion is C

Step-by-step explanation:

Beer =$1 and amount sold is x

so $1×x= $x which is the profit on beer

Wine=$2 and amount sold is y

so, $2×y= $2y which is the profit on wine

so an algebraic formulation of the profit function will be,

the sum off both the profit of the beer and wine which is

x+2y= max(x+2y)

Answer:

note:

please find the attachment

Answer:

(-2, 6)

Step-by-step explanation:

f(x) = x² + 4x + 10

f(x) = x² + 4x + 4 + 6

f(x) = (x + 2)² + 6

The vertex is at (-2, 6).

Answer:

(a) Probability = 0.29947

Step-by-step explanation:

The probability of the hand containing exactly one ace would be:

Number of ways this can happen = 4C1 * 48C4      (using combinations)

Number of ways this can happen = 4 * 194580

Number of ways this can happen = 778,320

Total number possible hands = 2,598,960 (as stated in question)

Total probability of exactly one ace = Number of ways to get an ace / total number of ways

Total probability = 778320 / 2598960 = 0.29947

Thus, the probability of the hand containing exactly one ace will be 0.2994

Another way to solve this:

Probability of one ace and 5 other cards = [tex]\frac{4}{52}*\frac{48}{51}*\frac{47}{50}*\frac{46}{49}*\frac{45}{48}[/tex] = 0.059894

Number of ways to arrange 1 ace and 4 other cards = 5

Total probability = 0.0598 * 5 = 0.29947

The probability of getting exactly one ace in a five-card poker hand is 0.2556 (rounded to four decimal places).

To find the probability of getting exactly one ace in a five-card poker hand, we need to determine the number of favorable outcomes and divide it by the total number of possible outcomes. In a standard 52-card deck, there are 4 aces, and we need to choose 1 ace out of the 4. The remaining 4 cards in the hand can be chosen from the remaining 48 non-ace cards in the deck. Hence, the number of favorable outcomes is C(4,1) * C(48,4). The probability can be calculated as:

P(exactly one ace) = (C(4,1) * C(48,4)) / C(52,5)

Substituting the values and evaluating the expression, we get:

P(exactly one ace) = (4 * 171,230) / 2,598,960 = 0.2556 (rounded to four decimal places)

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Answer:

A. Data

Step-by-step explanation: Data is a term used to describe facts, information or statistics that are collected together in order for it to be used as a reference or for analysis.

An effective data collection is one of the most important aspects in research,experiments or statistics as it helps to guarantee a reliable and effective outcome.

Data collection should be done in such a way that it helps to solve the problem which is being studied or handled.

Answer:

                                          [tex]C = 1/18[/tex]

Step-by-step explanation:

Remember that for a probability density function

                                     [tex]\int_{-\infty}^{\infty } f(x) dx = 1[/tex]

Since [tex]f(x) = 0[/tex]   outside [tex][0,2][/tex]   we would have that

                                       [tex]\int_{0}^{2} C(10-x) dx = 1[/tex]

Therefore  

                                              [tex]18C = 1 \\C = 1/18[/tex]

Step-by-step explanation:

Hope it helps you in your learning process.

Answer:

53

Step-by-step explanation:

47+6=53

Answer:

53

Step-by-step explanation:

Answer:

Step-by-step explanation:

The total number of lines, n(U) = 18

Let the number of lins with verb be n(V) = 11

Let the number of lines with adjectives be n(A) = 13

n(V n A) = 8

Find the number of lines that have a verb but no adjective, that is, n(V n A')

Mathematically, according to sets theory,

n(V) = n(V n A) + n(V n A')

So,

n(V n A') = n(V) - n(V n A) = 11 - 8 = 3.

Hence, only 3 lines have a verb but no adjectives.

Answer:

[tex]t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5[/tex]    

[tex]p_v =P(t_{(24)}>5.5)=0.00000589[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

Step-by-step explanation:

Data given and notation  

[tex]\bar X=201[/tex] represent the mean

[tex]s=10[/tex] represent the sample standard deviation for the sample  

[tex]n=25[/tex] sample size  

[tex]\mu_o =190[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 190, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 190[/tex]  

Alternative hypothesis:[tex]\mu > 190[/tex]  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=25-1=24[/tex]  

Since is a one side test the p value would be:  

[tex]p_v =P(t_{(24)}>5.5)=0.00000589[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

12x-5y=-20

y=x+4

12x-5(x+4)=-20

12x-5x-20=-20

7x=0,

So, we get: x=0 and y=4

Answer:

We conclude that the true mean life of a biomedical device is greater than 5200 hours.

Step-by-step explanation:

We are given that the life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed. For this a random sample of 15 devices is selected and found to have an average life of 5323.8 hours and a sample standard deviation of 220.9 hours.

We have to test that the true mean life of a biomedical device is greater than 5200 or not.

Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 5200 {means that the true mean life of a biomedical device is less than or equal to 5200 hours}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 5200 {means that the true mean life of a biomedical device is greater than 5200 hours}

The test statistics that will be used here is;

        T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, Xbar = sample average life = 5323.8 hours

               s = sample standard deviation = 220.9 hours

               n = sample devices = 15

So, test statistics = [tex]\frac{5323.8-5200}{\frac{220.9}{\sqrt{15} } }[/tex] ~ [tex]t_1_4[/tex]

                            = 2.171

Since, we are not given with the significance level, so we assume it to be 5%, now the critical value of t at 14 degree of freedom in t table is given as 1.761. Since our test statistics is more than the critical value of t which means our test statistics will lie in the rejection region. So, we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the true mean life of a biomedical device is greater than 5200 hours.

Answer:

Ted will need [tex]3\frac{3}{4}[/tex] cups of pretzels to make 15 cups of trail mix.

Step-by-step explanation:

Ted is making trail mix for a party. He mixes 1 1/2 cups of nuts, 1/4 cup of raisins, and 1/4 cup of pretzels.

So, 1/4 cup of pretzels to make 1 trail mix

       x cups of pretzels to make 15 trail mix

Using the ratio and proportional

∴ x = (1/4) * 15 = 3.75 = [tex]3\frac{3}{4}[/tex] cups.

Ted will need [tex]3\frac{3}{4}[/tex] cups of pretzels to make 15 cups of trail mix.

Answer:

95% confidence interval for the mean amount of the active chemical in the drug = [ 90.088 , 94.312 ]

Step-by-step explanation:

We are given that a lab is testing the amount of a certain active chemical compound in a particular drug that has been recently developed. It is known that the standard deviation in the amount of the chemical is 6 mg.

A random sample of 31 batches of the new drug is tested and found to have a sample mean concentration of 92.2 mg of the active chemical i.e.;

Population standard deviation, [tex]\sigma[/tex] = 6 mg

Sample mean, [tex]Xbar[/tex] = 92.2 mg

Sample size, n = 31

Now, the pivotal quantity for 95% confidence interval is given by;

          [tex]\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

So, 95% confidence interval for the mean amount of the active chemical in the drug is given by;

P(-1.96 < N(0,1) < 1.96) = 0.95

P(-1.96 < [tex]\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95

P(-1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] < [tex]Xbar - \mu[/tex] < 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

P(Xbar - 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < Xbar + 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [Xbar - 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] , Xbar + 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ]

                                                  = [ 92.2 - 1.96 * [tex]\frac{6}{\sqrt{31} }[/tex] , 92.2 + 1.96 * [tex]\frac{6}{\sqrt{31} }[/tex] ]

                                                  = [ 90.088 , 94.312 ]

To calculate the 95% confidence interval for the mean amount of the active chemical in the drug, substitute the sample mean concentration, standard deviation, and sample size into the formula: 95% Confidence Interval = Sample Mean ± (Z * (Standard Deviation / √Sample Size)).

To calculate the 95% confidence interval for the mean amount of the active chemical in the drug, we can use the formula:

95% Confidence Interval = Sample Mean ± (Z * (Standard Deviation / √Sample Size))

Given that the sample mean concentration is 92.2 mg, the standard deviation is 6 mg, and the sample size is 31 batches, we can substitute these values into the formula:

95% Confidence Interval = 92.2 mg ± (1.96 * (6 mg / √31))

Calculating this expression gives us a 95% confidence interval of approximately 90.855 mg to 93.545 mg.

Answer:

56 different tests

Step-by-step explanation:

Given:

Number of wires available (n) = 8

Number of wires taken at a time for testing (r) = 5

In order to find the number of different tests required for every possible pairing of five wires, we need to find the combination rather than their permutation as order of wires doesn't disturb the testing.

So, finding the combination of 5 pairs of wires from a total of 8 wires is given as:

[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Plug in the given values and solve. This gives,

[tex]^8C_5=\frac{8!}{5!(8-5)!}\\\\^8C_5=\frac{8\times 7\times 6\times 5!}{5!\times 3\times2\times1}\\\\^8C_5=56[/tex]

Therefore, 56 different tests are required for every possible pairing of five ​wires.

A total of 56 different tests are needed to check every possible combination of five wires out of eight in the cable.

The question is asking for the number of different tests required to test every possible pairing of five wires in an eight-wire cable. This is a combination problem, where we're looking for how many different ways we can combine five items from a group of eight. The formula for combinations is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and ! represents factorial. Plugging into this formula, we get C(8, 5) = 8! / [5!(8 - 5)!] = 56. Therefore, a total of 56 different tests are needed to check every possible combination of five wires out of eight.

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Answer:

We need to check the conditions in order to use the normal approximation.

[tex]np \geq 15[/tex]

[tex]n(1-p) \geq 15[/tex]

If we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np[/tex]

[tex]\sigma=\sqrt{np(1-p)}[/tex]

[tex] X \sim N (\mu = np, \sigma=\sqrt{np(1-p)}) [/tex]

So then the correct answer for this case would be:

E. Normal

Step-by-step explanation:

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".

Let X the random variable of interest, on this case we assume that:

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

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

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

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

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

We need to check the conditions in order to use the normal approximation.

[tex]np \geq 15[/tex]

[tex]n(1-p) \geq 15[/tex]

If we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np[/tex]

[tex]\sigma=\sqrt{np(1-p)}[/tex]

[tex] X \sim N (\mu = np, \sigma=\sqrt{np(1-p)}) [/tex]

So then the correct answer for this case would be:

E. Normal

The approximate shape of the sampling distribution of the sample will be normal.

Option E is correct.

It is a statistic that determines the probability of an event based on data from a small group within a large population.

Given that, [tex]np[/tex] is greater than or equal to 15 and [tex]n(1-p)[/tex] is greater than or equal to 15.

     [tex]np\geq 15[/tex]  and [tex]n(1-p)\geq 15[/tex]

So that the new mean and standard deviation will be,

           [tex]mean=\mu=np\\\\Deviation=\sigma=\sqrt{np(1-p)}[/tex]

Thus, the approximate shape of the sampling distribution of the sample will be normal.

Learn more about the sampling distribution here:

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Answer:

We need to check the conditions in order to use the normal approximation.

[tex]np=13*0.4=5.2 \geq 5[/tex]

[tex]n(1-p)=13*(1-0.4)=7.8 \geq 5[/tex]

Assuming that each trial is independent and we have a sample obtained from a random sampling method.

Then we can conclude that we can use the normal approximation since all the conditions are satisfied.

Step-by-step explanation:

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".

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=13, p=0.4)[/tex]

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

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

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

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

We need to check the conditions in order to use the normal approximation.

[tex]np=13*0.4=5.2 \geq 5[/tex]

[tex]n(1-p)=13*(1-0.4)=7.8 \geq 5[/tex]

Assuming that each trial is independent and we have a sample obtained from a random sampling method.

Then we can conclude that we can use the normal approximation since all the conditions are satisfied.

The area of the circular fire pit is 4096 square inches.

Explanation:

Given that the rectangular fire pit is 7 feet by 6 feet.

We need to determine the area of the largest circular fire that can be made in this fire pit.

The diameter of the circular fire is 6 feet

The radius is given by

[tex]r=\frac{6}{2} =3[/tex]

Radius is 3 feet.

The area of the largest circular fire pit can be determined using the formula,

[tex]Area=\pi r^2[/tex]

Substituting the values in the formula, we have,

[tex]Area = (3.14)(3)^2[/tex]

        [tex]=(3.14)(9)[/tex]

[tex]Area= 28.26 \ ft^2[/tex]

We need to convert feet to inches by multiplying by 12, we get,

[tex]Area = 28.26\times (12)^2[/tex]

[tex]Area = 4096.44 \ in^2[/tex]

Rounding off to the nearest square inch, we get,

[tex]Area= 4096 \ in^2[/tex]

Thus, the area of the circular fire pit is 4096 square inches.

Answer: 95 servings of applesauce

Step-by-step explanation:

It takes approximately 4 medium apples to make 3 servings of homemade apple sauce. It means that the number of servings of homemade apple sauce that can be made from 1 medium apple is

3/4 = 0.75 servings

Approximately 126 medium apples were purchased. Therefore, the number of servings of applesauce that they can make with the bushel is

126 × 0.75 = 95 servings of applesauce rounded to the nearest whole number.

Answer:

Correct option: (a)

Step-by-step explanation:

A confidence interval is an interval estimate of the parameter value.

A (1 - α)% confidence interval implies that the confidence interval has a (1 - α)% probability of consisting the true parameter value.

OR

If 100 such confidence intervals are made then (1 - α) of these intervals would consist the true parameter value.

The 92% confidence interval for the mean annual phone charge of all Vopstra customers is:

[tex]\$470\pm \$65=(\$405, \$535)[/tex]

This confidence interval implies that true mean annual phone charge of all Vopstra customers is contained in the interval ($405, $535) with 0.92 probability.

Thus, the correct option is (a).

Answer:

In the Step-by-step explanation.

Step-by-step explanation:

If a new sample is taken out of the same population,

a. The population proportion, p will not change. It is an unknown value that is estimated with the statistics of the samples.

b. The sample proportion, p^ is expected to change, because it is a new sample that has its own statistic value that may or may not be equal to the first sample.

c. The standard deviation of p^ is expected to change, because it depends on the sample and its size.

d. The standard error of p^ will change, because it depends on the sample.

e. The sampling distribution of p^, including its shape, mean, and standard deviation, will not change, because it is estimated with the data of the previous sample and it is supposed to be a property of the population and the sample size. Although the new information can be used to review the sample mean and standard deviation.

Answer:

Reject  There is sufficient evidence to support the claim that true mean is actually higher than the claim amount $1800.

Step-by-step explanation:

Based on the decision rule, the test statistic is lies in the rejection region. So reject the null hypothesis at 5% level of significance.

There is sufficient evidence to support the claim that the true mean is actually higher than the claim amount $1800.

Solution is attached below

Answer:

X is a discrete random variable.

X can take values from 0 to 12:

[tex]X\in[0,1,2,3,4,5,6,7,8,9,10,11,12][/tex]

Step-by-step explanation:

(a) X = the number of unbroken eggs in a randomly chosen standard egg carton

X is a discrete random variable.

The minimum amount of eggs broken is 0 and the maximum amount of eggs broken is 12 (assuming a dozen egg carton).

Then, X can take values from 0 to 12:

[tex]X\in[0,1,2,3,4,5,6,7,8,9,10,11,12][/tex]

About the probability ditribution nothing can be said, because there is no information about it (it can be a binomial, uniform or non-standard distribution).

The set of possible values for the random variable X, representing the number of unbroken eggs in an egg carton, is {0, 1, 2, 3, ..., 12}. X is a discrete random variable because it takes on distinct, countable values with no gaps in between.

The random variable X represents the number of unbroken eggs in a randomly chosen standard egg carton. To describe the set of possible values for this variable, we can consider the number of eggs in a standard carton, which is typically 12. Therefore, the set of possible values for X would be {0, 1, 2, 3, ..., 12}, as it encompasses all possible outcomes, ranging from having no unbroken eggs (0) to having all 12 unbroken eggs.

Now, let's determine whether the variable X is discrete or not. A discrete random variable is one that can take on a countable number of distinct values with gaps in between.

In this case, X is indeed a discrete random variable because it can only take on integer values from 0 to 12, and there are no intermediate values between these whole numbers. Each value represents a distinct and countable outcome based on the number of unbroken eggs, making it a discrete random variable.

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Answer: it is cheaper to buy for sweatshirts at $9.69 each.

Step-by-step explanation:

The regular price for one sweatshirt is $12.95. if a customer bought three shirts at the regular price of $12.95, a fourth sweatshirt was free

It means that the cost of buying 4 shirts is

12.95 × 3 = $38.85

Due to a special, the price of one sweatshirt was $9.69. It means that the cost of buying 4 shirts at this price is

9.69 × 4 = $38.76

Therefore, it is cheaper to buy for sweatshirts at $9.69 each than to buy at $12.95 each and get a free shirt

Answer: it is cheaper to buy for sweatshirts at $9.69 each.

Answer:

The probability that a randomly selected student is female or a physics major is 0.575.

Step-by-step explanation:

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

Desired outcomes:

Students that are female or physics majors.

17 female

10 physics majors, of which 4 are female.

This means that there are 10 total physics majors and 17-4 = 13 non physics majors female. So

[tex]D = 13 + 10 = 23[/tex]

Total outcomes:

The class has 40 students, so [tex]T = 40[/tex]

Probability

[tex]P = \frac{23}{40} = 0.575[/tex]

The probability that a randomly selected student is female or a physics major is 0.575.

Answer:

(a) No

(b) 0.21 or 21%

Step-by-step explanation:

(a) Since the outcome of the first game influences in the probability of winning the second game, the two games are not independent.

(b) The probability of losing both games is given by the product of the probability of losing the first game and the probability of losing the second game given that you have lost the first:

[tex]P = (1-0.7)*(1-0.3)\\P=0.21=21\%[/tex]

The probability you lose both games is 21%

Given Information:

Probability of wining 1st game = p₁ = 0.7

Probability of wining 2nd game given 1st game won = p₂|p₁ = 0.5

Probability of wining 2nd game given 1st game lost = p₂|q₁ = 0.3

Required Information:

(a) Are the two games independent = ?

(b) Probability of losing both games = ?

Answer:

(a) Are the two games independent = No

(b) Probability of losing both games = 0.21

Step-by-step explanation:

(a) Independent Events:

Two events are said to be independent when the success of one event is not affected by the success or failure of another event.

In this case, the probability of 2nd game depends on the success or failure of the 1st game, therefore, the two games are not independent.

(b) Probability of losing both games

The probability of losing the both games is the product of the probabilities of losing each game.

Probability of losing 1st game = 1 - Probability of wining 1st game

Probability of losing 1st game = 1 - 0.7 = 0.30

Probability of losing 2nd game = 1 - Probability of wining 2nd game given 1st game lost

Probability of losing 2nd game = 1 - 0.3 = 0.70

Please note that since we are finding the probability of losing both games that's why we used the condition of 1st game lost

Probability of losing both games = Probability of losing 1st game*Probability of losing 2nd game

Probability of losing both games = 0.30*0.70

Probability of losing both games = 0.21

Answer: 5.5hours

Step-by-step explanation:

Total distance = 226miles

Total time = 7 hours

Let b represent total time spent while walking.

Distance (walking) = 4b

Distance ( riding) = 40(7-b)

Total distance 226 = 4b + 40(7-b)

226 = 4b + 280 - 40b

226 = 280-36b

b = 54/36

b = 1.5hours

Amount of time spent walking = 1.5hours

Amount of time spent riding = 7-1.5 = 5.5hours

Amount of time spent on bicycle = 5.5hours

Answer:

[tex]0.482 - 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.454[/tex]  

[tex]0.482 + 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.510[/tex]  

And the 90% confidence interval would be given (0.454;0.510).  

B. 0.4542 to 0.5105.

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".  

Solution to the problem

The estimated proportion of residents in the community that support the property tax levis is given by:

[tex] \hat p =\frac{x}{n}= \frac{410}{850}= 0.482[/tex]

The confidence interval for a proportion is given by this formula  

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

For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

And replacing into the confidence interval formula we got:  

[tex]0.482 - 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.4542[/tex]  

[tex]0.482 + 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.5105[/tex]  

And the 90% confidence interval would be given (0.4542;0.5105)

B. 0.4542 to 0.5105.

Answer:

90% confidence interval for p is [0.4542 , 0.5105] .

Step-by-step explanation:

We are given that a local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. Of the 850 residents surveyed, 410 supported the property tax levy.

Let p = proportion of residents in the community that support the property tax levy

[tex]\hat p[/tex] = proportion of residents in the community that support the property tax levy in a survey of 850 residents = [tex]\frac{410}{850}[/tex] = [tex]\frac{41}{85}[/tex]

The pivotal quantity that will be used here population proportion p is;

         P.Q. = [tex]\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

So, 90% confidence interval for p is given by;

P(-1.6449 < N(0,1) < 1.6449) = 0.90 {At 10% significance level the z table give

                                                            critical value of 1.6449)

P(-1.6449 < [tex]\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.6449) = 0.90

P( [tex]-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] < [tex]{\hat p - p}[/tex] < [tex]1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ) = 0.90

P( [tex]\hat p -1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] < p < [tex]\hat p +1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ) = 0.90

90% confidence interval for p = [ [tex]\hat p -1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] , [tex]\hat p +1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] ]

                                       = [ [tex]\frac{41}{85} -1.6449 \times {\sqrt{\frac{\frac{41}{85} (1-\frac{41}{85} )}{850} }[/tex] , [tex]\frac{41}{85} +1.6449 \times {\sqrt{\frac{\frac{41}{85} (1-\frac{41}{85} )}{850} }[/tex] ]

                                       = [ 0.4542 , 0.5105 ]

Therefore, 90% confidence interval for p is [0.4542 , 0.5105] .

Answer: The addition of the new point alters the previous standard deviation greatly

Step-by-step explanation:

Let the initial five points be : 2 3 4 5 and 6. In order to calculate the standard deviation for this data, we will need to calculate the mean first.

Mean = summation of scores/number of scores.

The mean is therefore: (2+3+4+5+6)/5 = 20/5 = 4.

We'll also need the sum of the squares of the deviations of the mean from all the scores.

Since mean = 4, deviation of the mean from the score "2" = score(2) - mean (4)

For score 3, it is -1

For 4, it's 0

For 5 it's 1

For 6 it's 2.

The squares for -2, -1, 0, 1, and 2 respectively will be 4, 1 , 0, 1, 4. Summing them up we have 10 i.e (4+1+0+1+4=10).

Calculating the standard deviation, we apply the formula:

√(summation of (x - deviation of mean)^2)/N

Where N means the number of scores.

The standard deviation = √(10/5) = 1.4142

If we add another score or point that is far away from the original points, say 40, what happens to the standard deviation. Let's calculate to find out.

i.e we now have scores: 2, 3, 4, 5, 6 and 40

We calculate by undergoing same steps.

Firstly mean. The new mean = (2+3+4+5+6+40)/6 = 60/6 = 10.

The mean deviations for the scores : 2, 3, 4, 5, 6 and 40 are -8, -7, -6, -5, -4 and 30 respectively. The squares of these deviations are also 64, 49, 36, 25, 16 and 900 respectively as well. Their sum will then be 1090. i.e. (64+49+36+25+16+900 = 1090).

The new standard deviation is then=

√(1090/6)

= √181.67

= 13.478.

It's clear that the addition of a point that's far away from the original points greatly alters the size of the standard deviation as seen /witnessed in this particular instance where the standard deviation rises from 1.412 to 13.478