Find a system of two equations in two variables, x1 and x2, that has the solution set given by the parametric representation x1 = t and x2 = 5t − 6, where t is any real number. (Enter your answer as a comma-separated list of equations.)

The mean of the number of different sweaters worn in a week is 2.76. This is calculated by multiplying each possible outcome (from 1 to 4 different sweaters) by its probability and then adding these up.

The random variable X here is the number of different sweaters you wear in a week. To find the mean (expected value) of a random variable, you need to take each possible outcome, multiply it by its probability, and then add up all these products.

Here are the possibilities:

So, the expected value or mean of X is:

1*(4/1024) + 2*(192/1024) + 3*(432/1024) + 4*(96/1024) + 4*(24/1024) = 2.76

So on an average week, you would expect to wear about 2 to 3 different sweaters.

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Answer:

59.92% probability he will get at least 2 hits in the game.

Step-by-step explanation:

For each at bat, there are only two possible outcomes. Either he gets a hit, or he does not. The probability of a getting a hit in each at bat is independent. So we use the binomial probability distribution to solve this question.

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 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]

And p is the probability of X happening.

In this problem we have that:

[tex]p = 0.212[/tex]

In one game, he gets 9 at bats. What is the probability he will get at least 2 hits in the game?

This is [tex]P(X \geq 2)[/tex] when [tex]n = 9[/tex]

He either gets less than two hits in the game, or he gets at least two hits. The sum of the probabilities of these events is decimal 1. So

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

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

In which

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

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

[tex]P(X = 0) = C_{9,0}.(0.212)^{0}.(0.788)^{9} = 0.1171[/tex]

[tex]P(X = 1) = C_{9,1}.(0.212)^{1}.(0.788)^{8} = 0.2837[/tex]

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1171 + 0.2837 = 0.4008[/tex]

Finally

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4008 = 0.5992[/tex]

59.92% probability he will get at least 2 hits in the game.

Answer:

d. 4.6 years

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 5.8, \sigma = 0.9[/tex]

What warranty should be provided so that the company is replacing at most 10% of their toasters sold?

Only those on the 10th percentile or lower will be replaced.

So the warranty is the value of X when Z has a pvalue of 0.10.

So it is X when [tex]Z = -1.28[/tex]

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

[tex]-1.28 = \frac{X - 5.8}{0.9}[/tex]

[tex]X - 5.8 = -1.28*0.9[/tex]

[tex]X = 4.6[/tex]

So the correct answer is:

d. 4.6 years

Final answer:

To find the warranty length, we need to determine the value that corresponds to the 10th percentile of the normally distributed life span. By solving the equation using the z-score formula, we find that the warranty should be provided for at least 4.6 years.

Explanation:

In order to determine the warranty that should be provided to the toasters, we need to find the value that corresponds to the 10th percentile of the normally distributed life span. To do this, we can use a standard normal distribution table or a calculator. The z-score for the 10th percentile is approximately -1.28. Using the formula z = (x - mean) / standard deviation, we can solve for x to find the warranty length.

-1.28 = (x - 5.8) / 0.9

x - 5.8 = -1.28 × 0.9

x - 5.8 = -1.152

x = 5.8 - 1.152

x = 4.648

Therefore, the warranty should be provided for at least 4.648 years, which is approximately 4.6 years. The correct answer is option d. 4.6 years.

Answer:

Lowest Cost is for Drum set, when compared to others instruments of same type.

Highest cost is for Guitar,when compared to others  instruments of same type.

Order:

Guitar>Piano>Drum Set

Step-by-step explanation:

Consider the normal distribution for all cases.

Formula we are going to use is:

[tex]z=\frac{\bar x-\mu}{\sigma}[/tex]

where:

[tex]\bar x[/tex] is the purchasing cost

[tex]\mu[/tex] is the mean

[tex]\sigma[/tex] is the standard deviation

For Piano:

[tex]z=\frac{3000-4000}{2500}\\ z=-0.4[/tex]

For Guitar:

[tex]z=\frac{550-500}{200}\\ z=0.25[/tex]

For Drum Set:

[tex]z=\frac{600-700}{100}\\ z=-1[/tex]

Lowest Cost is for Drum set, when compared to others instruments of same type.

Highest cost is for Guitar,when compared to others  instruments of same type.

Order:

Guitar>Piano>Drum Set

The drum set, priced at $600, costs the least compared to the average drum set price, being 1.0 standard deviation below the mean. The guitar, priced at $550, costs the most compared to the average guitar price, at 0.25 standard deviation above the mean. These conclusions are drawn using the z-score method to compare the instruments' prices to their respective averages and standard deviations.

To determine which musical instrument costs the least and the most when compared to others of the same type, we calculate how many standard deviations below or above the mean each instrument's cost falls. For the piano, guitar, and drum set, we will use the z-score formula, which is (z = (X - μ) / σ), where X is the observed value, μ is the mean, and σ is the standard deviation.

For the piano, the z-score is ($3,000 - $4,000) / $2,500 = -0.4.

For the guitar, the z-score is ($550 - $500) / $200 = 0.25.

For the drum set, the z-score is ($600 - $700) / $100 = -1.0.

The drum set's cost is the lowest compared to other drums because it is 1.0 standard deviations below the mean. The guitar's cost, being 0.25 standard deviations above the mean, is the highest compared to other guitars. Therefore, Matt's drums cost the least in comparison to his own instrument type, and Becca's guitar costs the most compared to her own instrument type.

sin 40° = 0.6428

Step-by-step explanation:

The complementary ratio of sine is cos and vice-versa.

sin θ = cos (90 - θ)

cos θ = sin (90 - θ)

So cos 50° = sin (90 - 50)°

                   = sin 40°

sin 40° = 0.6428

This indicates that the complement of cos 50° is sin 40° which is equal to 0.6428.

Answer:

A. 13.9 and 14.1 inches

See explanation below.

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

[tex]X \sim N(14,0.1)[/tex]  

Where [tex]\mu=14[/tex] and [tex]\sigma=0.1[/tex]

If we want the middle 68% of the data we need to have on the tails 16% on each one

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.84[/tex]   (a)

[tex]P(X<a)=0.16[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.16 of the area on the left and 0.84 of the area on the right it's z=-0.994. On this case P(Z<-0.994)=0.16 and P(z>-0.994)=0.84

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.16[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.16[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.994<\frac{a-14}{0.1}[/tex]

And if we solve for a we got

[tex]a=14 -0.994*0.1=13.9[/tex]

And since the distribution is symmetrical for the upper limit we can use z = 0.994 and we have:

[tex]z=0.994<\frac{a-14}{0.1}[/tex]

And if we solve for a we got

[tex]a=14 +0.994*0.1=14.1[/tex]

So the correct answer for this case would be:

A. 13.9 and 14.1 inches

Answer:

Step-by-step explanation:

Please kindly check the attached

(p(x))/(q(x))=f(x)+(r(x))/(q(x))

Answer:

Hypothesis Test states that we will accept null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, [tex]\mu_1[/tex] = true mean voltage for type K batteries.

           [tex]\mu_2[/tex] = true mean voltage for type Q batteries.

So, Null Hypothesis, [tex]H_0[/tex] :  [tex]\mu_1 = \mu_2[/tex] {mean voltage for these two types of

                                                        batteries is same}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu_1 \neq \mu_2[/tex] {mean voltage for these two types of

                                                          batteries is different]

The test statistics we use here will be :

                     [tex]\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex]  follows [tex]t_n__1 + n_2 -2[/tex]

where, [tex]X_1bar[/tex] = 8.54         and     [tex]X_2bar[/tex] = 8.69

                [tex]s_1[/tex]  = 0.225       and         [tex]s_2[/tex]     =  0.725

               [tex]n_1[/tex]   =  37           and         [tex]n_2[/tex]     =  58

               [tex]s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2} }{37+58-2} }[/tex]  =  0.585               Here, we use t test statistics because we know nothing about population standard deviations.

     Test statistics =  [tex]\frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58} } }[/tex] follows [tex]t_9_3[/tex]

                             = -1.219

At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that mean voltage for these two types of batteries is same.

Answer:

Step-by-step explanation:

Hello!

You have a bag with 181 peanuts in it.

65 of the shells contain one peanut. ⇒ 16 are cracked and 49 are complete.

111 of the shells contain two peanuts. ⇒ 30 are cracked 81 are complete.

5 of the shells have three peanuts and none of them is cracked.

A. What is the probability the shell is cracked?

To calculate the probability of the shell bein cracked you have to add all possible cases in which it is cracked and divide it by the total of peanuts in the bag:

[tex]P(Cracked)= \frac{(16+30)}{181}= 0.25[/tex]

B. What is the probability the shell is not cracked (Cr) and it contains two peanuts(2p)?

This probability is an intersection P(Cr∩2p), to calculate it you have to divide the total of peanuts that fulfill these characteristics and divide it by the total number of peanuts in the bag.

[tex]P(Crn2p)= \frac{30}{181}= 0.17[/tex]

C. What is the probability the shell is not cracked (Cr') or it contains two peanuts(2p)?

This probability is the union between two events, the event "cracked" and the event "two peanuts", symbolically: P(Cr'∪2p), these two events are not mutually exclusive, this means that they can happen at the same time, you have to apply the following formula to calculate it:

[tex]P(Cr'u2p)= P(Cr') + P(2p) - P(Cr'n2p)[/tex]

Since these events aren't mutually exclusive, some of them fulfill both categories, this means they are counted when you calculate the probability of "not cracked" and again when you calculate the probability of "2 peanuts" therefore you need to subtract their intersection to obtain the correct probability.

[tex]P(Cr'u2p)= \frac{(49+81+5)}{181} + (\frac{111}{181} ) - (\frac{81}{181} )= 0.76+0.61-0.45=0.92[/tex]

D. What is the probability the shell is not cracked given that it contains two peanuts?

This is a conditional probability, this means that both events are dependant and the occurrence of "2p" modifies the probability of the peanut shell to be "Cr'", symbolically: P(Cr'/2p) and you calculate it using the following formula:

[tex]P(Cr'/2p)= \frac{P(Cr'n2p)}{P(2p)}= \frac{0.45}{0.61}= 0.74[/tex]

I hope it helps!

A) The probability the shell is cracked is 25.4%.

B) The probability the shell is not cracked and it contains two peanuts is 44.75%.

C) The probability the shell is not cracked or it contains two peanuts is 91.16%.

D) The probability the shell is not cracked given that it contains two peanuts is 72.97%.

Since a bag of peanuts in their shells contains 181 peanuts. 65 of the shells contain one peanut, 111 of the shells contain two peanuts, and the rest contain three peanuts, and, of the shells with one peanut, 16 of them are cracked, and of the shells with two peanuts, 30 of them are cracked, while none of the shells with three peanuts are cracked, and one peanut shell is randomly selected from the bag, to determine A) what is the probability the shell is cracked; B) what is the probability the shell is not cracked and it contains two peanuts; C) what is the probability the shell is not cracked or it contains two peanuts; and D) what is the probability the shell is not cracked given that it contains two peanuts; the following calculations must be performed:

A)

B)

C)

D)

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The greatest common factor of 4 and 32 would be 4.

The common factor for m^6 and m^5 would be m^5, since 6 is greater than 5.

The GCF becomes 4m^5

Answer:

Option A) Quantitative, because numerical values, found by either measuring or counting, are used to describe the data

Option C) Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.

Step-by-step explanation:

We are given the following in the question:

"The average monthly rainfall in inches for a certain city throughout the year"

Qualitative or​ quantitative:

Thus, the given quantitative data is quantitative because rainfall is always measured and the output is expressed in numerical values.

Option A) Quantitative, because numerical values, found by either measuring or counting, are used to describe the data

Level of measurement:

Since,for the given data true zero exist and a negative values make no sense, thus its is ratio.

Option C) Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.

The data described is quantitative because it is measured in numerical values. The level of measurement for this data is interval because the differences between data points can be meaningfully measured, but the data lacks a true zero point.

The data about the average monthly rainfall in inches for a certain city throughout the year is a Quantitative Data. This is because the data are numerical values which are obtained through measuring. Quantitative data is based on numbers and can be manipulated using statistical methods. The type of measurement used for this data set is Interval. This is because although the differences in the data (rainfall from one month to the next) can be meaningfully measured, the data do not have a 'true zero' point since there isn't a point where there’s absolutely no rainfall.

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Answer:

A) P=2.28%

B) P=28.57%

C) I expect 10 students to be unable to complete the exam in the alloted time.

Step-by-step explanation:

In order to solve this problem, we will need to find the respective z-scores. The z-scores are found by using the following formula:

[tex]z=\frac{x-\mu}{sigma}[/tex]

Where:

z= z-score

x= the value to normalize

[tex]\mu = mean[/tex]

[tex]\sigma[/tex]= standard deviation

The z-score will help us find the area below the normal distribution curve, so in order to solve this problem we need to shade the area we need to find. (See attached picture)

A) First, we find the z-score for 60 minutes, so we get:

[tex]z=\frac{60-80}{10}=-2[/tex]

So now we look for the z-score on our normal distribution table. Be careful with the table you are using since some tables will find areas other than the area between the mean and the desired data. The table I used finds the area between the mean and the value to normalize.

so:

A=0.4772 for a z-score of -2

since we want to find the number of students that take less than 60 minutes, we subtract that decimal number from 0.5, so we get:

0.5-0.4772=0.0228

therefore the probability that a student finishes the exam in less than 60 minutes is:

P=2.28%

B) For this part of the problem, we find the z-score again, but this time for a time of 75 minutes:

[tex]z=\frac{75-80}{10}=-0.5[/tex]

and again we look for this z-score on the table so we get:

A=0.1915 for a z-score of -0.5

Now that we got this area we subtract it from the area we found for the 60 minutes, so we get:

0.4772-0.1915=0.2857

so there is a probability of P=28.57% of chances that the students will finish the test between 60 and 75 minutes.

C) Finally we find the z-score for a time of 90 minutes, so we get:

[tex]z=\frac{90-80}{10}=1[/tex]

We look for this z-score on our table and we get that:

A=0.3413

since we need to find how many students will take longer than 90 minutes to finish the test, we subtract that number we just got from 0.5 so we get:

0.5-0.3413=0.1586

this means there is a 15.86% of probabilities a student will take longer than 90 minutes. Now, since we need to find how many of the 60 students will take longer than the 90 available minutes, then we need to multiply the total amount of students by the percentage we previously found, so we get:

60*0.1586=9.516

so approximately 10 Students will be unavailable to complete the exam in the allotted time.

Answer:

Part A:

Probability is P(z<-2)=1-0.9772=0.0228

Part B:

P(-2<z<-0.5)=0.2857

Part C:

Number of students unable to complete the exam=60-50=10 students

Step-by-step explanation:

Part A:

Mean=μ=80 min

Standard Deviation=σ=10 min

Formula:

[tex]z=\frac{\bar x- \mu}{\sigma}[/tex]

where:

[tex]\bar x=60\ min[/tex]

[tex]z=\frac{60-80}{10}\\ z=-2[/tex]

Probability is P(z<-2)

From the probability distribution tables (Cumulative Standardized normal distribution table)

Probability is P(z<-2)=1-0.9772=0.0228

Part B:

For 75 min:

[tex]z=\frac{75-80}{10}\\ z=-0.5[/tex]

For [tex]\bar x=60\ min[/tex]

[tex]z=\frac{60-80}{10}\\ z=-2[/tex]

From the probability distribution tables (Cumulative Standardized normal distribution table)

P(-2<z<-0.5)=P(z<-0.5)-P(z<-2)

P(-2<z<-0.5)=(1-0.6915)-(1-0.9772)

P(-2<z<-0.5)=0.2857

Part C:

[tex]\bar x=90\ min[/tex]

[tex]z=\frac{90-80}{10}\\ z=1[/tex]

From the probability distribution tables (Cumulative Standardized normal distribution table)

Probability is P(z<1)=0.8413

Number of students=0.8413*60

Number of students to complete the exam=50.478≅50

Number of students unable to complete the exam=60-50=10 students

Answer:

You want to buy some golden apples. Each golden apple cost $2. How much much does golden apples pass cost?

y = total cost

x = amount of apple

Answer: he would need 77 scoops of fertilizer.

Step-by-step explanation:

The measure of Horalco's garden is

10 1/2 ft by 5 1/2 ft. Converting to improper fraction, it becomes 21/2 ft by 11/2 ft. The garden is rectangular in shape and the formula for determining the area of a rectangle is expressed as

Area = length × width.

Therefore, area of the garden is

21/2 × 11/2 = 231/4 ft²

He needs 1 1/3 scoops of fertilizer for each square foot of the garden. Converting to improper fraction, it becomes 4/3 scoops. Therefore, the number of scoops of fertilizer that Horaclo need for the entire garden is

4/3 × 231/4 = 77 scoops of fertilizer.

The distance the in which the water will reach before it shatters will be 1.03ft

Data;

Using integration, the force at any depth y is 62.4y lb/ft^3

[tex]200 = \int\limits^d_0 {62.4y} \, 6dy\\ 200 = \int\limits^d_0 {374.4y} \, dy\\ 200 = \int\limits^6_0 {374.4y} \, dy\\ 200 = [374.4/2 y^2]_0^d\\200 = 187.2(d^2 - 0)\\200 = 187.2d^2\\d^2 = 200/187.2\\d = 1.03ft[/tex]

From the calculations above, the distance in which the water will reach before it shatters is 1.03ft

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Answer: It would take approximately 27 steps to obtain an approximation of the root.

Step-by-step explanation: The bisection method is a numerical procedure to find a root to a equation continuous in an interval [a.b]. It consists in repeatedly halve the interval [a,b], keeping the half for which f(x) changes sign. The repetition will continue until a stopping criteria is reached. To find how many interactions is necessary, we can satisfy the equation:

[tex]\frac{1}{2^{n} }[/tex] · (b - a) ≤ ε, where a and b are the original interval and ε is the stopping criteria.

Substituting the parameters and using log 2 = 0,301, the number of iterations needed is approximately 27.

Answer:

Correct option (A).

Step-by-step explanation:

The probability of an individual catching a flu when he or she has taken vitamin C is, P (F|C) = 0.0342.

The probability of an individual catching a flu when he or she has not taken vitamin C is, P (F|C') = 0.2653.

Th ratio of individuals who caught the flu when they did not take vitamin C to those who took vitamin C is:

[tex]=\frac{P(F|C')}{P(F|C)}\\ =\frac{0.2653}{0.0342} \\=7.7573[/tex]

This implies that:

[tex]P(F|C')=7.7573\times P(F|C)[/tex]

Thus, the the individuals not taking vitamin C are 7.7573. Times more likely to catch the flu than individuals taking vitamin C.

The statement is True.

The statement that individuals not taking vitamin C are 7.7573 times higher at risk of catching flu than those taking it, is true based on the provided numbers in the question.

To determine if the statement is true or false, we need to divide the odds of catching the flu among individuals not taking vitamin C by the odds of those taking it:

To do this, you divide 0.2653 (odds for those not taking vitamin C) by 0.0342 (odds for those taking vitamin C). This calculation results in approximately 7.7573.

Therefore, the statement 'individuals not taking vitamin C are 7.7573 times more likely to catch flu than individuals taking vitamin C is true as the calculation matches the provided number in the statement.

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Answer:

[tex](0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129[/tex]  

[tex](0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787[/tex]  

And the 95% confidence interval would be given (0.00129;0.0787).  

We are confident at 95% that the difference between the two proportions is between [tex]0.00129 \leq p_B -p_A \leq 0.0787[/tex]

And as we can see the confidence interval for the difference on this case not contains the 0.

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

[tex]p_A[/tex] represent the real population proportion for the gift certificate

[tex]\hat p_A =\frac{65}{500}=0.13[/tex] represent the estimated proportion for the gift certificate

[tex]n_A=500[/tex] is the sample size required for the gift certificate

[tex]p_B[/tex] represent the real population proportion for without the gift certificate

[tex]\hat p_B =\frac{45}{500}=0.09[/tex] represent the estimated proportion for without the gift certificate

[tex]n_B=500[/tex] is the sample size required for Brand B

[tex]z[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

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

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]  

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129[/tex]  

[tex](0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787[/tex]  

And the 95% confidence interval would be given (0.00129;0.0787).  

We are confident at 95% that the difference between the two proportions is between [tex]0.00129 \leq p_B -p_A \leq 0.0787[/tex]

And as we can see the confidence interval for the difference on this case not contains the 0.

The largest intervals on which the existence and uniqueness theorem guarantees a unique solution for the given initial conditions.

The interval a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution depends on the initial conditions. For the given initial conditions:

Answer:

answers and explanation is shown in the attachment

Step-by-step explanation:

The detailed steps and necessary substitution is as shown in the attached files.

Answer:

[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{n} = \frac{1220750}{500}=2441.5[/tex]

[tex] s= \sqrt{\frac{N \sum x^2 f -[\sum xf]^2}{N(N-1}}= \sqrt{\frac{500*3408029125 -[1220750]^2}{50*49}}=9341.2405[/tex]

Step-by-step explanation:

In order to find the mean and standard deviation we can create the following table:

Limits            Frequency(f)      x(midpoint)      x*f           x^2 *f

__________________________________________________

0-499                   9                   249.5         2245.5    560252.3

500-999              13                  749.5         9743.5     7302753

1000-1499           33                 1249.5       41233.5    51521258.25

1500-1999           115                1749.5       201192.5   351986278.8

2000-2499         125               2249.5      281187.5   632531281.3

2500-2999          81                2749.5      222709.5  612339770.3

3000-3499          47                3249.5      152726.5   496284761.8

3500-3999          45                3749.5      168727.5   632643761.3

4000-4499          22                4249.5      93489       397281505.5

4500-4999          10                 4749.5       47495      225577502.5

_____________________________________________________

Total                    500                               1220750     3408029125

We can calculate the mean with the following formula:

[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{n} = \frac{1220750}{500}=2441.5[/tex]

And the standard deviation would be given by:

[tex] s= \sqrt{\frac{N \sum x^2 f -[\sum xf]^2}{N(N-1}}= \sqrt{\frac{500*3408029125 -[1220750]^2}{50*49}}=9341.2405[/tex]

Answer:

A or C

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

32

Step-by-step explanation:

32 X 2 = 64. 64 - 9 = 55.

Answer:

There is a 33.67% probability that exactly one of them is defective.

Step-by-step explanation:

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

Here, we can have different formats. For example, D-ND-ND is the same as ND-D-ND, that is, the ordering is not important. So we use the combinations formula.

Combinations formula:

[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

One defective(one from a set of 55) and two non defective(two from a set of 45). So

[tex]D = C_{55,1}*C_{45,2} = \frac{55!}{54!1}*\frac{45!}{43!2!} = 55*45*22 = 54450[/tex]

Total outcomes

Three from a set of 100. So

[tex]T = C_{100,3} = \frac{100!}{97!3!} = 161700[/tex]

What is the probability that exactly one of them is defective

[tex]P = \frac{D}{T} = \frac{54450}{161700} = 0.3367[/tex]

There is a 33.67% probability that exactly one of them is defective.

Answer:

As proved in the attached files

Step-by-step explanation:

The detailed step and mathematical derivation and manipulation is as shown in the attachment.

Answer:

a) 6.68%

b) 0.135%

c) 43.32%

d) 1.7

e) -2.5

Step-by-step explanation:

a)

P(X>615)=P(z>(615-540)/50)

P(X>615)=P(z>1.5)

P(X>615)=P(0<z<∞)-P(0<z<1.5)

P(X>615)=0.5-0.4332

P(X>615)=0.0668

The percentage of students have an SAT math score greater than 615 is 6.68%

b)

P(X>690)=P(z>(690-540)/50)

P(X>690)=P(z>3)

P(X>690)=P(0<z<∞)-P(0<z<3)

P(X>690)=0.5-0.49865

P(X>690)=0.00135

The percentage of students have an SAT math score greater than 690 is 0.135%

c)

P(465<X<540)=P((465-540)/50<z<(540-540)/50))

P(465<X<540)=P(-1.5<z<0)

P(465<X<540)=0.4332

The percentage of students have an SAT math score between 465 and 540 is 43.32%

d)

z=(625-540)/50

z=85/50

z=1.7

The z-score for student with an SAT math score of 625 is 1.7.

e)

z=(415-540)/50

z=-125/50

z=-2.5

The z-score for a student with an SAT math score of 415 is -2.5.

Answer:

See below

Step-by-step explanation:

a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.

b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.

c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.

d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:

a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b

Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.

e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.

f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.

g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.

Answer:

a) y= y₀+vy*t , x= x₀+vx*t

b) they are closest at t= 29.23 s

c) r min = 63.79 ft

Step-by-step explanation:

a) denoting v as velocities and "₀" as initial conditions , then the position of Sven is given by the coordinate (0,y) where

y= y₀+vy*t

and the position of Rudyard is given by the coordinate (x,0) where

x= x₀+vx*t

b) the distance r between Sven and Rudyard  is given by

r²=x²+y²

the distance will be minimum when the derivative of r with respect to the time is 0 . Then taking the derivative of the equation above

2*r*dr/dt = 2*x*dx/dt + 2*y*dy/dt

since dx/dt= vx and dy/dt= vy , then

r*dr/dt = x*vx+ y*vy

dr/dt = (x*vx+ y*vy)/r

assuming that r cannot be 0 , then

dr/dt =0 → x*vx+ y*vy = 0

(x₀+vx*t)*vx + (y₀+vy*t)*vy = 0

-(x₀*vx + y₀*vy) = (vx²+vy²)*t

t= -(x₀*vx + y₀*vy)/(vx²+vy²)

replacing values

t= -(x₀*vx + y₀*vy)/(vx²+vy²) = -[ 140 ft*(-6ft/s) + (-170 ft)*4 ft/s]/[ (-6ft/s)²+ (4 ft/s)²] = 29.23 s

then they are closest at t= 29.23 s

and the minimum distance will be

x = x₀ + vx*t = 140 ft+(-6ft/s)*29.23 s = -35.38 ft

y= y₀+vy*t = (-170 ft)+ 4 ft/s*29.23 s = -53.08 ft

r min = √(x²+y²)= 63.79 ft

r min = 63.79 ft

Note

to prove our assumption that r is not 0 , then x and y should be 0 at the same time. thus

0= y₀+vy*t → t = (-y₀)/vy = -140 ft/(-6ft/s) = 26.33 s

0= x₀+vx*t → t= (-x₀)/vx = -(-170 ft)/4 ft/s = 42.5 s

then r is never 0

Answer:

(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2a) The percentage of results more than 45 is 79.67%.

(2b) The percentage of results less than 85 is 91.77%.

(2c) The percentage of results are between 75 and 90 is 15.58%.

(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.

Step-by-step explanation:

(1)

Let Y = the time taken to deliver a pizza.

The random variable Y follows a Uniform distribution, U (20, 60).

The probability distribution function of a Uniform distribution is:

[tex]f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.[/tex]

Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:

[tex]P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70[/tex]

Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2)

Let X = results of a certain blood test.

It is provided that the random variable X follows a Normal distribution with parameters [tex]\mu = 60[/tex] and [tex]s = 18[/tex].

The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.

The z-scores follows a Standard normal distribution, N (0, 1).

(a)

Compute the probability that the results are more than 45 as follows:

[tex]P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z<0.833)=0.7967[/tex]

The percentage of results more than 45 is: [tex]0.7967\times100=79.67\%[/tex]

Thus, the percentage of results more than 45 is 79.67%.

(b)

Compute the probability that the results are less than 85 as follows:

[tex]P(X<85)=P(\frac{X-\mu}{\sigma}< \frac{85-60}{18})=P(Z<1.389)=0.9177[/tex]

The percentage of results less than 85 is: [tex]0.9177\times100=91.77\%[/tex]

Thus, the percentage of results less than 85 is 91.77%.

(c)

Compute the probability that the results are between 75 and 90 as follows:

[tex]P(75<X<90)=P(\frac{75-60}{18}<\frac{X-\mu}{\sigma}< \frac{90-60}{18})\\=P(0.833<Z<1.67)\\=P(Z<1.67)-P(Z<0.833)\\=0.9525-0.7967\\=0.1558[/tex]

The percentage of results are between 75 and 90 is: [tex]0.1558\times100=15.58\%[/tex]

Thus, the percentage of results are between 75 and 90 is 15.58%.

(d)

Compute the probability that the results are between 20 and 100 as follows:

[tex]P(20<X<100)=P(\frac{20-60}{18}<\frac{X-\mu}{\sigma}< \frac{100-60}{18})\\=P(-2.22<Z<2.22)\\=P(Z<2.22)-P(Z<-2.22)\\=0.9868-0.0132\\=0.9736[/tex]

Then the probability that the results outside the range 20 to 100 is: [tex]1-0.9736=0.0264[/tex].

The percentage of results outside the range 20 to 100 is: [tex]0.0264\times100=2.64\%[/tex]

Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.

We can define two variables: S represents Sophia's hours, and L represents London's hours. The first equation representing total combined work time is S + L = 11 hours. The second equation, representing total shirts completed, is 40S + 25L = 365 shirts.

Let's define our variables as follows: S stands for the hours Sophia worked and L for the hours London worked. Since Sophia and London worked a total of 11 hours, we can write the first equation as S + L = 11. Each worker's production rate times the hours they worked will total the shirts they produced. So, our second equation is 40S + 25L = 365. These two equations can be used to find the number of hours Sophia and London worked respectively.

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