Solve the following system of equations by the substitution method. 3x = y + 6 x = 2y - 1

Final answer:

Two valid sequences that insert five numbers between 1/27 and 27 to form a geometric sequence are 1/27, 1/9, 1/3, 1, 3, 9, 27 using a common ratio of 3, and 1/27, -1/9, 1/3, -1, 3, -9, 27 using a common ratio of -3.

Explanation:

To insert five numbers between 1/27 and 27 to form a geometric sequence, we first identify that in a geometric sequence, each term after the first is found by multiplying the previous one by a constant called the common ratio (r). We are effectively looking for seven terms in total (including the given 1/27 and 27) which form this sequence.

The formula for the nth term of a geometric sequence is a_n = a_1 × r^(n-1), where a_n is the nth term of the sequence, a_1 is the first term, and n is the term number.

Since we are given the first term (1/27) and the 7th term (27), we can use these to find the common ratio (r), through the equation 27 = (1/27) × r^(7-1), or simplifying, 27 = (1/27) × r^6. Solving for r, we get r = 3 (considering the positive root for practical purposes in a school context).

Thus, the sequence is: 1/27, 1/9, 1/3, 1, 3, 9, 27. However, since it is mentioned there are two answers, another valid sequence can be determined by using the negative root, r = -3. Therefore, another valid sequence is: 1/27, -1/9, 1/3, -1, 3, -9, 27. These two sequences incorporate the geometric sequence properties correctly.

Answer:

The answer is False

Step-by-step explanation:

The  count does no suggest that the normal model is a good fit for sampling the distribution because the questions used for the test and survey is the one of many other question about cheating which implies that if other questions about college life are being used as the survey, the response would probably be that more of the student would not have admitted to cheating. This concept therefore disobeys the normal distribution model which is a bell shaped model and therefore assumes that at an average, the number of students that admitted to cheating in the major courses should be equal.

Answer:

There is sufficient evidence to conclude that child booster seats meet the specific requirement.

Step-by-step explanation:

Sample: 697, 759, 1266, 621, 569, 432

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{4344}{6} = 724[/tex]

Sum of squares of differences = 415616

[tex]S.D = \sqrt{\frac{415616}{5}} = 288.31[/tex]

We are given the following in the question:  

Population mean, μ = 1000 hic

Sample mean, [tex]\bar{x}[/tex] = 724

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, s = 288.31

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 1000\text{ hic}\\H_A: \mu < 1000\text{ hic}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{724 - 1000}{\frac{288.31}{\sqrt{6}} } = -2.344[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 5 degree of freedom } = -2.015[/tex]

Calculation the p-value from table,

P-value = 0.033

Since,                  

Since, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that the measurement is less than 1000 hic.

Thus, there is sufficient evidence to conclude that child booster seats meet the specific requirement.

Answer:

Step-by-step explanation:

Given that X, the mutual fund returns in the 1st quarter of 2013, is

N(0.054, 0.015)

a) P(X>6.8%) = [tex]1-0.8246\\=0.1754[/tex]

b) [tex]P(0<X<0.076)\\\\\\=0.9288-0.0002\\=0.9286[/tex]

c) [tex]P(X>0.01)\\=1-0.0017\\=0.9983[/tex]

d) [tex]P(X<0) = 0.0002[/tex]

A) The expected percentage of returns that are over 6.8% is _17.5_____%

b) The expected percentage of returns that are between 0& and 7.6% is __92.9__%

C) The expected percentage of returns that are more than 1% is _99.8___%

D) The expected percentage of returns that are less than 0% is _0.02___%

Answer:

C.

Step-by-step explanation:

Hypothesis testing procedure:

Hypothesis:

The null hypothesis will be the variances of sales of two musical stores are equal and the alternative hypothesis will be the variances of sales of two musical stores are not equal

Level of significance: alpha=0.05

Test statistic: F=variance A/variance B=(30)^2/(20)^2=900/400=2.25

P-value: p=0.107

As the alternative hypothesis mentioned that the variances are not equal this leads to two tailed test. so the p-value is calculated using excel function 2*F.DIST.RT(2.25,24,15).

Conclusion:

The p-value seems to exceed the alpha=0.05 and this depicts that the null hypothesis should not be rejected.

At 95% confidence, the null hypothesis should not be rejected. The correct answer is option c. should not be rejected.

Step 1

To test whether the variances of the sales at music stores A and B are equal, we perform an F-test. The null hypothesis [tex](\(H_0\))[/tex] states that the variances are equal, while the alternative hypothesis [tex](\(H_a\))[/tex] states that they are not equal.

The F-statistic is calculated as the ratio of the larger sample variance to the smaller sample variance:

[tex]\[ F = \frac{s_1^2}{s_2^2} \][/tex]

Where [tex]\( s_1^2 \)[/tex] is the variance of store A and [tex]\( s_2^2 \)[/tex] is the variance of store B.

Step 2

In this case, the F-statistic is [tex]\( \frac{30^2}{20^2} = \frac{900}{400} = 2.25 \)[/tex].

Using a significance level of 0.05 and degrees of freedom (df) as [tex]\( n_1 - 1 \)[/tex] and [tex]\( n_2 - 1 \)[/tex], we compare this value to the critical F-value from the F-distribution table.

Since the calculated F-statistic of 2.25 is less than the critical F-value, we fail to reject the null hypothesis.

Therefore, at 95% confidence, the correct answer is c. should not be rejected.

Answer:

c. Interval or ratio

Step-by-step explanation:

There exists the following measurement scales:

Nominal: A variable is linked to a number. For example, Buffalo Bills players, 27 is Tre'Davious White, 49 Tremanine Edmunds, and then on...

Ordinal: Ranks the intensity of something. For example, grading some pain on a 1 to 10 scale.

Interval or ratio: Represents quantity and has an equality of units. One example is the rates of unemployment.

Descriptive: Tries to attribute qualities to quantitative data. For example, rates of unemployment being classified as very low, low, medium, and then on...

So the correct answer is:

c. Interval or ratio

The unemployment rate is measured using an interval or ratio scale, represented as a percentage. Various methods and measures are considered for accurate calculation, though it has limitations in fully capturing unemployment's societal impact.

The reported unemployment rate of 5.5% of the population uses an interval or ratio scale for measurement. This type of scale is used because the unemployment rate is a percentage that represents a proportion of the population. The measurement of unemployment involves a ratio of two quantities: the number of unemployed individuals and the total labor force. Different methods such as Labor Force Sample Surveys, Official Estimates, Social Insurance Statistics, and Employment Office Statistics are used to calculate this figure. Moreover, the U.S. Bureau of Labor Statistics employs six different measures (U1 - U6) to capture various aspects of unemployment. While the rate is informative, there are shortcomings in how it represents the real impact on society, as it does not account for underemployment or those who have stopped looking for work. Understanding these statistics is crucial as unemployment has significant economic and social consequences, such as increasing inequality and potentially leading to civil unrest.

Answer:The margin of error is 0.01323 and 95% confidence interval is (0.674,0.73).

Step-by-step explanation:

Since we have given that

p = 0.70

n = 1200

We need to find the margin of error;

Margin of error would be

[tex]\sqrt{\dfrac{p(1-p)}{n}}\\\\=\sqrt{\dfrac{0.7\times 0.3}{1200}}\\\\=0.01323[/tex]

At 95% confidence level, α = 1.96

so, 95% confidence interval would be

[tex]p\pm z\times 0.01323\\\\=0.7\pm 1.96\times 0.01323\\\\=0.7\pm 0.02593\\\\=(0.7-0.02593,0.7+0.02593)\\\\=(0.674,0.73)[/tex]

Hence, the margin of error is 0.01323 and 95% confidence interval is (0.674,0.73).

Answer:

E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4

Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24

Step-by-step explanation:

Previous concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]

For this distribution the expected value is the same parameter [tex]\lambda[/tex]

[tex]E(X)=\mu =\lambda[/tex]

Other equations useful:

Let X and Y random variables

E(X) =E[E(X|Y)] (conditional expectation)

Var(X)=E[Var(X|Y)]+Var[E(X|Y)] (Total variance)

Solution to the problem

Let A the random variable that represent the number of winter storms next year

B a binary variable, B=1 if the next year is a good year and B=0 in the other case. Then we have this:

E(A|B=1) = 4  and E(A|B=0)=5

We can use the propoerties of conditional expectation like this:

E(A)= E[E(A|B)]= E(A|B=1)P(B=1) +E(A|B=0)P(B=0)

E(A)= E[E(A|B)]= 4*0.6 +5*0.4 =4.4

And we can use also the properties for conditional variance we have the following values:

Var(A|B=1)=4 Var(A|B=0)=5, by the propertis of the Poisson distribution

And then the conditional variance is givne by:

[tex]Var[E(A|B)]= E(A|B=1)^2 P(B=1) +E(A|B=0)^2 P(B=0)[/tex]

And if we replace we got:

[tex]Var[E(A|B)]= 4^2 *0.6 +5^2 *0.4 =19.6[/tex]

And we have also that the expected value for the conditional variance is given by:

E[Var(A|B)]= 4*0.6 +5*0.4 =4.4

And then finally the variance for the random variable A is given by:

Var(A)= E[Var(A|B)] +Var[E(X|Y)]]=4.4+19.6=24

The expected value and variance in the number of storms are 4.4 and 4.4 respectively.

To calculate the expected number of winter storms, we can use the probability of a good or bad year along with their respective average storms. The expected value is then given by:

E(X) = (0.6 * 4) + (0.4 * 5) = 2.4 + 2 = 4.4 storms.

For the variance, since the variance of a Poisson distribution is equal to its mean, we have:

Variance in a good year = 4

Variance in a bad year = 5

The overall variance is a weighted average:

V(X) = (0.6 * 4) + (0.4 * 5) = 2.4 + 2 = 4.4

Using the same formula from your other question:

A = P x (r/12(1+r)^t)/ (1+r/12)^t -1)

A = 222000 x (0.08/12(1+0.08/12)^300 / (1+0.08/12)^300 - 1)

A = $1,713.43 per month

The requirements necessary for a normal probability distribution to be a standard normal probability​ distribution is that The mean and standard deviation have the values of mu equals 1and sigma equals 1. Hence the correct answer is A.

A probability distribution is one whose mean data points are symmetric. That is, most values from such a distribution cluster around the mean.

Another name for Normal Probability Distribution is Gaussian Distribution.

Learn more about Normal Probability Distribution at:

https://brainly.com/question/6476990

Final answer:

A standard normal probability distribution requires a mean of 0 and a standard deviation of 1. The correct answer is B, which reflects these necessary conditions for standardization, allowing for the comparison of z-scores across different distributions.

Explanation:

To transform a normal probability distribution into a standard normal probability distribution, certain requirements must be met. Specifically, the distribution must have a mean (mu) of 0 and a standard deviation (sigma) of 1. Among the given options, the correct answer is B, where the mean and standard deviation have the values of mu equals 0 and sigma equals 1.

This standardization process allows any normal distribution to be compared on a common scale, and it is fundamental for calculating z-scores, which indicate how many standard deviations an element is from the mean.For example, if we have a normally distributed variable 'x' from a distribution with any mean µ and standard deviation o, the standardized value or z-score is calculated as follows:z = (x - µ) / o

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

Step-by-step explanation:

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]\frac{\sigma}{\sqrt{n}}[/tex].

Normal probability distribution

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

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

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

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

In this problem, we have that:

[tex]\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10[/tex].

What is the probability that the average amount billed on the sample bills is greater than $500?

This probability is 1 subtracted by the pvalue of Z when [tex]X = 500[/tex]. So

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

[tex]Z = \frac{500 - 465}{10}[/tex]

[tex]Z = 3.5[/tex]

[tex]Z = 3.5[/tex] has a pvalue of 0.9998.

So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.

The probability that the average amount billed in a sample of 900 bills is greater than $500 is approximately 0.0002.

Given the mean amount billed is $465

standard deviation of $300,

sample size of 900,

the probability that the average amount billed exceeds $500.

First, we need to calculate the standard error of the mean (SEM):

[tex]SEM = \(\frac{\sigma}{\sqrt{n}}\)[/tex]

Substituting the given values:

[tex]SEM = \frac{300}{\sqrt{900}} = \frac{300}{30} = 10[/tex]

Now, we can use the Z-score formula to find the probability.

The Z-score is calculated as follows:

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

(X = 500),

mu = 465

SEM=10

[tex]Z = \frac{500 - 465}{10} = 3.5[/tex]

Using Z-tables or a standard normal distribution calculator, we can find the probability corresponding to a Z-score of 3.5:

P(Z > 3.5) ≈ 0.0002

Therefore, the probability that the average amount billed in a sample of 900 bills is greater than $500 is approximately 0.0002 (rounded to four decimal places).

Answer:

a)[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

b)

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Quality management        Excellent      Good     Fair    Total

Excellent                                40                35         25       100

Good                                      25                35         10         70

Fair                                         5                   10          15        30

Total                                       70                 80         50       200

Part a

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the two categorical variables

H1: There is association between the two categorical variables

The level of significance assumed for this case is [tex]\alpha=0.05[/tex]

The statistic to check the hypothesis is given by:

[tex]\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{70*100}{200}=35[/tex]

[tex]E_{2} =\frac{80*100}{200}=40[/tex]

[tex]E_{3} =\frac{50*100}{200}=25[/tex]

[tex]E_{4} =\frac{70*70}{200}=24.5[/tex]

[tex]E_{5} =\frac{80*70}{200}=28[/tex]

[tex]E_{6} =\frac{50*70}{200}=17.5[/tex]

[tex]E_{7} =\frac{70*30}{200}=10.5[/tex]

[tex]E_{8} =\frac{80*30}{200}=12[/tex]

[tex]E_{9} =\frac{50*30}{200}=7.5[/tex]

And the expected values are given by:

Quality management        Excellent      Good     Fair       Total

Excellent                                35              40          25         100

Good                                      24.5           28          17.5        85

Fair                                         10.5            12           7.5         30

Total                                       70                 80         65        215

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(3-1)(3-1)=4[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

Part b

We can find the probabilities that Quality of Management and the Reputation of the Company would be the same like this:

Let's define some notation first.

E= Quality Management excellent     Ex=Reputation of company excellent

G= Quality Management good     Gx=Reputation of company good

F= Quality Management fait     Ex=Reputation of company fair

P(EΛ Ex) =40/215=0.186

P(GΛ Gx) =35/215=0.163

P(FΛ Fx) =15/215=0.0697

If we have dependence then the conditional probabilities would be higher values.

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Answer:

Brownian Motion- The usual model for the time-evolution of an asset price S(t) is given by the geometric Brownian motion.

Now the geometric Brownian motion is represented by the following stochastic differential equation:

To solve the problem now we have the been Data provided:

μ= 0.12,

σ=0.24,

Step-by-step explanation:

Step A:

we have, the variables of Black Scholes Model, by putting the values of variables available, we get:

Next is, "r" the risk free rate,

Step B:

We now need to calculate the parameter d₂ of the Black Scholes Model. .

Step C:

As step C is done on excel for further calculations so, do use it if you are solving it on computer.

Final answer:

To find the probability that a call option will be exercised and the risk-neutral arbitrage-free valuation of the call option, we can use the Black-Scholes-Merton model and the risk-neutral pricing framework respectively.

Explanation:

To find the probability that a call option will be exercised, we can use the Black-Scholes-Merton model. In this case, we have:

The current price of the security (S) = $40

The strike price of the option (K) = $42

The time to expiration (T) = 4 months (or 0.33 years)

The risk-free interest rate (r) = 8% (or 0.08)

The volatility of the security (σ) = 0.24

Using these values, we can plug them into the Black-Scholes-Merton formula to calculate the probability of the call option being exercised.

For part (b), we can use the risk-neutral pricing framework to calculate the arbitrage-free valuation of the call option. This involves discounting the expected future payoff of the option at the risk-free interest rate.

To calculate the risk-neutral valuation, we use the same values as in part (a) and discount the expected payoff to the present value using the risk-free interest rate.

Answer:

The x axis in the function represents, the number of hours after 7:00 A.M. , the person reaches her car. The person reaches the car at 1:00 P.M.

Step-by-step explanation:

The x axis denotes the no. of hours and and the y axis denotes the distance from the car.

X Intercept is a point where the line intersects the X axis, we can easily notice the fact that at that point, y=0 ie. The person has reached his/her respective car.

The line intersects x at 6.

Therefore, a total of 6 hours are taken from the beginning of the hike.

Thus, the person reaches the car at 1:00 P.M.

Answer:

The person will take 6 hours to get back to her car.

Step-by-step explanation:

Answer:

The firm's sample size must be of at least 171 adults.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the length of the sample.

In this problem, we have that:

[tex]M = 60, \sigma = 400[/tex]. So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]60 = 1.96*\frac{400}{\sqrt{n}}[/tex]

[tex]60\sqrt{n} = 784[/tex]

[tex]\sqrt{n} = 13.07[/tex]

[tex]n = 170.7[/tex]

The firm's sample size must be of at least 171 adults.

Answer:

Option D) Yes, because the test statistic is -2.01

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 30 pound

Sample mean, [tex]\bar{x}[/tex] = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s =  2 pounds

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30\text{ pounds}\\H_A: \mu < 30\text{ pounds}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{29.1 - 30}{\frac{2}{\sqrt{20}} } = -2.012[/tex]

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = -1.729[/tex]

Since,                    

[tex]t_{stat} < t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. Thus, there were enough evidence to conclude that the fishing line breaks with an average force of less than 30 pounds.

Option D) Yes, because the test statistic is -2.01

Answer:

Option c. 17%

Step-by-step explanation:

Data provided in the question:

Amount paid on food last year = 12% of Annual after tax income

Amount paid on food for the current year = 10% of Annual before tax income

Now,

Let the after tax income be 'x'

and tax be 'y'

Therefore,

Income before tax = x + y

Amount paid on food = 12% of x

According to the question

12% of x = 10% of  (x + y)

or

0.12x = 0.10 (x + y)

or

1.2x - x = y

0.2x = y

or

x = 5y

Thus,

percent of the family's annual before-tax income that was paid for tares last year

= [Tax ÷ Income before tax] × 100%

= [ y ÷ ( x + y )] × 100%

=  [ y ÷ ( 5y + y )] × 100%

or

= 0.167 × 100% ≈  17%

Hence,

Option c. 17%

Answer:

[tex]work \ done= 45937.5[/tex]

Step-by-step explanation:

Work done is given by

[tex]work \ done=\int_a^b w(d-x) \ dx[/tex] , where d = length of cable and w = weight of cable.

Here, d = 175 ft and w = 3 lb/ft

Now, [tex]work \ done=\int_0^{175} 3(175-x) \ dx[/tex]

[tex]work \ done= 3\left [175x-\frac{x^2}{2}  \right ]_0^{175}[/tex]

[tex]work \ done= 3\left [175^2-\frac{175^2}{2}  \right ][/tex]

[tex]work \ done= 3\cdot \frac{175^2}{2}[/tex]

[tex]work \ done= 45937.5[/tex]

Answer:

The probability that can be applied to the given situation is Poisson distribution

Step-by-step explanation:

The probability that can be applied to the given situation is Poisson distribution because this distribution applied when the duration of occurrence of an event is known. In the given question laws have occurred every 100 feet therefore we have the number of events that have occurred. Moreover, Poisson distribution predicts the probability of events in fixed time interval

The probability distribution that has the greatest chance of applying to the number of blemishes or flaws occurring in each 100 feet of material in the textile industry is the Poisson distribution.

The probability distribution that has the greatest chance of applying to this situation is the Poisson distribution.

The Poisson distribution is commonly used to model the number of events that occur in a fixed interval of time or space. In this case, the manufacturer is interested in the number of blemishes or flaws occurring in each 100 feet of material, which can be seen as events occurring in a fixed space.

The Poisson distribution is appropriate when the events occur randomly and independently, and the average rate of occurrence is known. It allows for both discrete and continuous distributions.

https://brainly.com/question/33722848

#SPJ3

Answer:

[tex]P(X<0.1)= 5.5x10^{-5}[/tex]

Step-by-step explanation:

Previous concepts

Beta distribution is defined as "a continuous density function defined on the interval [0, 1] and present two parameters positive, denoted by α and β, both parameters control the shape. "

The probability function for the beta distribution is given by:

[tex] P(X)= \frac{x^{\alpha-1} (1-x)^{\beta -1}}{B(\alpha,\beta)}[/tex]

Where B represent the beta function defined as:

[tex]B(\alpha,\beta)= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}[/tex]

Solution to the problem

For our case our random variable is given by:

[tex] X \sim \beta (\alpha=5, \beta =2)[/tex]

We can use the following R code to plot the distribution for this case:

> x=seq(0,1,0.01)

> plot(x,dbeta(x,5,2),main = "Beta distribution a=5, b=2",ylab = "Probability")

And we got as the result the figure attached.

And for this case we want this probability, since we want the probability that she has at most 10% or 0.1 change of winning:

[tex]P(X<0.1)[/tex]

And we can find this probability with the following R code:

> pbeta(0.1,5,2)

[1] 5.5e-05

And we got then this : [tex]P(X<0.1)= 5.5x10^{-5}[/tex]

Answer:

The volume is [tex]V=\frac{64}{15}[/tex]

Step-by-step explanation:

The General Slicing Method is given by

Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is

[tex]V=\int\limits^b_a {A(x)} \, dx[/tex]

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is

The key observation is that the width is the distance between the upper bounding curve [tex]y = 2 - x^2[/tex] and the lower bounding curve [tex]y = x^2[/tex]

The width of each square is given by

[tex]w=(2-x^2)-x^2=2-2x^2[/tex]

This means that the area of the square cross section at the point x is

[tex]A(x)=(2-2x^2)^2[/tex]

The intersection points of the two bounding curves satisfy [tex]2 - x^2=x^2[/tex], which has solutions x = ±1.

[tex]2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}[/tex]

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

[tex]V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}[/tex]

Answer:

The relation is antisymmetric and transitive

Step-by-step explanation:

Let a,b,c be elements of the set of all people.

1) Let a be a person who is 20 years old. aRa means that this person is younger than themselves, which it's false because 20<20 is false. Then R is not reflexive.

2) Let a be a person who is 20 years old and b a person who is 30 years old. Then a is younger than b, that is, aRb.

However, it is not true that b is younger than a, as 30<20 is false, therefore bRa is false and R is not symmetric.

3) Suppose that aRb, so that a is younger than b. Then, b is not younger than a. If n denotes the age of a and m denotes the age of b, we have that n<m which implies that m<n is false. Then bRa is false, thus R is antisymmetric.

4) Suppose that aRb and bRc. Let n,m,p denote the ages of a,b,c respectively. Then n<m and m<p (a is younger than b and b is younger than c), and by transitivity of the ordering of numbers, n<p, that is, a is younger than c. Thus aRc, and R is transitive.

Answer: The function is constant from x = -2 to x=1

Final answer:

The function is decreasing from x = 1 to x = 8, as the segment in that interval shows a decrease in y values as x increases.

Explanation:

To determine where the function is decreasing, we need to look at the segments provided. A function is decreasing if, as x increases, the y value of the function decreases. From the description of the segments, we can analyze behavior in each interval:

Therefore, the function is decreasing from x = 1 to x = 8.

Answer:

Step-by-step explanation:

Answer: The number of scores between 21 and 25 is 2872

Step-by-step explanation:

Since the test scores are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = test scores

u = mean test score

s = standard deviation

From the information given,

u = 23

s = 4

We want to find the probability test scores between 21 points and 25. It is expressed as

P(21 lesser than or equal to x lesser than or equal to 25)

For x = 21,

z = (21 - 23)/4 = - 0.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.30854

For x = 25,

z = (25 - 23)/4 = 0.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.69146

P(21 lesser than or equal to x lesser than or equal to 25)

= 0.69146 - 0.30854 = 0.38292

The number of scores between 21 and 25 would be

0.38292 × 7500 = 2872

Answer: c. –1 and .1587

Step-by-step explanation:

As per given , we have

Null hypothesis : [tex]H_0 : p= 0.35[/tex]

Alternative hypothesis :  [tex]H_1 : p< 0.35[/tex]

Since Alternative hypothesis is left-tailed ,so the test must be a left tailed test .

Z -Test statistic for proportion = [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

, where p= population proportion

[tex]\hat{p}[/tex] = sample proportions

n= Sample size.

Let x be the number of successes.

For n= 40 and x= 11

[tex]\hat{p}=\dfrac{x}{n}=\dfrac{11}{40}=0.275[/tex]

Then ,  [tex]z=\dfrac{0.275-0.35}{\sqrt{\dfrac{0.35(1-0.35)}{40}}}[/tex]

[tex]z=\dfrac{-0.075}{\sqrt{0.0056875}}[/tex]

[tex]z=\dfrac{-0.075}{0.075415515645}[/tex]

[tex]z=-0.99449031619\approx-1[/tex]

By using z-table ,

P-value for left-tailed test = P(z<-1)= 1-P(z<1)    [∵ P(Z<-z)= 1-P(Z<z) ]

= 1-0.8413

=0.1587

Hence, the -score and P-value are  –1 and 0.1587 .

So the correct option is c. –1 and .1587

Answer:

Student needs pens= n = 27.04

Rounding off with upper floor function ⇒ n =28

Rounding off with lower floor function ⇒ n =27

Step-by-step explanation:

Given that lifetime of each pen is a exponential random variable with mean 1 week.

Let [tex]S_{n}[/tex] be total sum of lifetime of n pens.

So mean of [tex]S_{n}[/tex] = μ = n.1

Standard deviation of [tex]S_{n}[/tex] =[tex]\sigma=\sqrt{n}[/tex]

Probability that he doesnot run out of pens= 0.99

Considering Sn be sum of n lifetimes, using central limit theorem

[tex]\frac{S_{n}-n}{\sqrt{n}}\approx N(0,1)\\\\P(S_{n}>15)=[P(\frac{S_{n}-n}{\sqrt{n}})>\frac{15-n}{\sqrt{n}}]\\ 1-\phi(\frac{15-n}{\sqrt{n}})=\phi(-(\frac{15-n}{\sqrt{n}}))=0.99\\[/tex]

From table of standard normal distribution

[tex]\frac{15-n}{\sqrt{n}}=-2.3263\\15-n=-2.3263\sqrt{n}\\n-2.3263-15\sqrt{n}[/tex]

Solving the quadratic Equation in variable x we get

n=27.04

The required condition for a discrete probability function is that the sum of the probabilities for all possible outcomes, represented by '∑f(x)', must equal 1.

The correct answer to this question is d. ∑f(x) = 1 for all values of x. This is a required condition for a discrete probability function. In probability theory, a probability mass function (PMF), also known as a discrete probability function, provides the probabilities of discrete random variables. The function gives the probability that a discrete random variable is exactly equal to some value. The terms 'probability distribution function' and 'probability function' have also been used to denote the same concept.

As a basic rule, the sum of the probabilities for all possible outcomes (x-values) in a discrete probability function should equal 1. This is because these probabilities together represent the entire possible outcome set for your discrete random variable, which accounts for all possibilities and hence should total to 100% or, in decimal form, 1.

https://brainly.com/question/33443936

#SPJ3

Final answer:

The correct condition for a discrete probability function is that the sum of all the probabilities must equal 1, which is represented as Σf(x) = 1 for all values of x. The correct option is d.

Explanation:

The required condition for a discrete probability function is that the sum of all probabilities must equal 1, which is known as the normalization condition. This implies that when we list all possible outcomes or events that a random variable can take, the sum of their probabilities must be 1. This condition is represented as Σf(x) = 1 for all values of x. Therefore, the correct option is d. Σf(x) = 1 for all values of x.

Answer:

Alternative hypothesis:[tex]\mu_1 -\mu_2 \neq 0[/tex]

Or in the alternative way would be:

 Alternative hypothesis:[tex]\mu_1 \neq \mu_2 [/tex]

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

On this case the claim that they want to test is: "The means for the two groups (American Idol and 60 Minutes) is the same". So we want to check if we have significant differences between the two means, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.

Null hypothesis:[tex]\mu_1 -\mu_2 = 0[/tex]

This null hypothesis can be expressed like this:

Null hypothesis:[tex]\mu_1 = \mu_2 [/tex]

Alternative hypothesis:[tex]\mu_1 -\mu_2 \neq 0[/tex]

Or in the alternative way would be:

Alternative hypothesis:[tex]\mu_1 \neq \mu_2 [/tex]

Final answer:

Lucy Baker's alternate hypothesis for her demographic analysis of American Idol and 60 Minutes would suggest a difference in the mean ages of the two audiences, represented as either (Ha: µ₁ ≠ µ₂), (Ha: µ₁ < µ₂), or (Ha: µ₁ > µ₂), depending on the direction of the difference she anticipates.

Explanation:

Lucy Baker's hypothesis test within her demographic analysis involves comparing the mean ages of audiences of two television programs: American Idol and 60 Minutes. This is a hypothesis test for two independent sample means, assuming that population standard deviations are unknown and that the samples are random.

Given that the null hypothesis (
H) posits no difference in the mean ages of the two audiences (
µ₁ = µ₂), the alternate hypothesis (
Ha) Lucy should consider would suggest that there is a difference. Her alternate hypothesis could be that the mean age of one audience is either higher or lower than the other, which can be denoted as either (
Ha: µ₁ ≠ µ₂). However, if she has a specific direction in mind (e.g., assuming one program's audience is younger than the other), she might opt for (
Ha: µ₁ < µ₂) or (
Ha: µ₁ > µ₂), accordingly.

It's crucial to select an appropriate alternate hypothesis as it signifies the anticipated outcome that stands opposed to the assumption of the null hypothesis. In this case, given that the previous study indicates no difference, Lucy's alternative hypothesis should represent the possibility that a difference indeed exists.

Answer:  c)[50,60]

Step-by-step explanation:

The Empirical rule says that , About 68% of the population lies with the one standard deviation from the mean (For normally distribution).

We are given that , The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches.

Then by Empirical rule, about 68% of the heights of students lies between one standard deviation from mean.

i.e. about 68% of the heights of students lies between [tex]\text{Mean}\pm\text{Standard deviation}[/tex]

i.e. about 68% of the heights of students lies between [tex]55\pm5[/tex]

Here, [tex]55\pm5=(55-5, 55+5)=(50,60)[/tex]

i.e.  The required interval that contains the middle 68% of the heights. = [50,60]

Hence, the correct answer is c) (50,60)

Final answer

The interval containing the middle 68 of a typically distributed class height is one standard divagation from the mean, which is( 50, 60) elevation for the given mean of 55 elevation and standard divagation of 5 elevation.

Explanation

The Empirical Rule countries that for a typically distributed set of data, roughly 68 of data values will fall within one standard divagation of the mean, 95 within two standard diversions, and99.7 within three standard diversions. In this case, the mean height is 55 elevation and the standard divagation is 5 elevation. thus, to find the interval that contains the middle 68 of the heights, we add and abate one standard divagation from the mean.

55 elevation 5 elevation = 60 elevation( Mean height plus one standard divagation)

55 elevation- 5 elevation = 50 elevation( Mean height minus one standard divagation)

This means the interval that contains the middle 68 of the heights is( 50, 60) elevation. Hence, the correct answer is option( c).

Answer:

The probability that she will escape detection for at least three years is P=0.343.

Step-by-step explanation:

If Jody has a 30% chance each year of having her tax returns audited, she also has 70% chance each year of escaping detection.

The probability of this happening 3 years in a row is:

[tex]P_n=q^n=0.7^3=0.343[/tex]

Answer:

108

Step-by-step explanation: