A Common Measure of Student Achievement, The National Assessment of Educational Progress (NAEP) is the only assessment that measures what U.S. students know and can do in various subjects across the nation, states, and in some urban districts. Also known as The Nation's Report Card, NAEP has provided important information about how students are performing academically since 1969. NAEP is a congressionally mandated project administered by the National Center for Education Statistics (NCES) within the U.S. Department of Education and the Institute of Education Sciences (IES). NAEP is given to a representative sample of students across the country. Results are reported for groups of students with similar characteristics (e.g., gender, race and ethnicity, school location), not individual students. National results are available for all subjects assessed by NAEP. In the most recent year, The NAEP sample of 1077 young women had mean quantitative score of 275. Individual NAEP scores have a Normal distribution with standard deviation of 60. (This is indicating that the population standard deviation σσ is 60) Find a 99% Confidence Interval for the mean quantitative scores for young women. a) Check that the normality assumptions are met. ? b) What is the 99% confidence interval for the mean quantitative scores for young women? ____ ≤μ≤ _____ c) Interpret the confidence interval obtained in previous question

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let [tex]\mu[/tex] denotes the average hours of sleep per night.

As per given , we have

[tex]H_0:\mu=7\\H_a:\mu<7[/tex]

, since [tex]H_a[/tex] is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : [tex]\overline{x}=7.24[/tex]

Sample  standard deviation : s= 1.93

Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

i.e.  [tex]t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58[/tex]

For significance level [tex]\alpha=0.05[/tex] and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= [tex]t_{\alpha, df}=t_{0.05,21}=- 1.7207[/tex]

Decision : Since the test statistic value (0.58) > critical value  1.7207, it means we are failed to reject the null hypothesis .

[Note : When [tex]|t_{cal}|>|t_{cri}|[/tex], then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

To determine if LTCC Intermediate Algebra students get less than seven hours of sleep per night, a one-sample t-test can be conducted.

To determine if LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average, we can conduct a one-sample t-test. The null hypothesis (H0) is that the mean number of hours of sleep is at least seven, while the alternative hypothesis (Ha) is that the mean number of hours of sleep is less than seven. With a significance level of 5%, we compare the t-statistic calculated from the sample data to the critical t-value from the t-distribution table. If the calculated t-statistic is less than the critical t-value, we reject the null hypothesis and conclude that LTCC Intermediate Algebra students get less than seven hours of sleep on average.

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Final answer:

The best point estimate for the average number of credit hours per semester for all students at the local college is 9.1.

Explanation:

The best point estimate for the average number of credit hours per semester for all students at the local college can be found using the sample mean. In this case, the study found that students earned an average of 9.1 credit hours per semester based on a sample of 96 students. Therefore, the best point estimate for the average number of credit hours per semester for all students at the local college is 9.1.

8x^4y – 16x^2y^2=

=8x^4y-16x^4y

=-8x^4y

=-8*(x^4y)

Answer:

8\,x^2\,y\,(x-\sqrt{2} )(x+\sqrt{2} )

Step-by-step explanation:

Let's start by extracting all common factors from the two terms of this binomial. These common factors are: 8, [tex]x^2[/tex], and [tex]y[/tex].

The extraction renders:

[tex]8\,x^2\,y\,(x^2-2)[/tex]

In the real number system, the binomial in parenthesis can still be factored out considering that 2 is the perfect square of [tex]\sqrt{2}[/tex], that is:

[tex]2=(\sqrt{2} )^2[/tex]

We can then forwards with the factoring of this binomial using the factorization of a difference of squares as:

[tex](x^2-2) = (x^2-(\sqrt{2} )^2)=(x-\sqrt{2} )(x+\sqrt{2} )[/tex]

Thus giving the complete factorization as:

[tex]8\,x^2\,y\,(x-\sqrt{2} )(x+\sqrt{2} )[/tex]

To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. The 99% confidence interval for the given sample mean (x = 880) and sample standard deviation (s = 5) is (877.371, 882.629). Assuming a different sample standard deviation, the 99% confidence intervals are (874.24, 885.76) for s = 10 and (868.48, 891.52) for s = 20. The confidence interval width increases as the sample standard deviation increases.

To construct a confidence interval for the population mean with an unknown standard deviation, we use Student's t-distribution. First, we find the value of the t-statistic for the given confidence level and degrees of freedom. For a 99% confidence level and a sample size of 24, the degrees of freedom is 23. Using Appendix D or a t-table, we find the critical t-value to be approximately 2.819.

(a) To construct a confidence interval with the given sample mean (x = 880) and sample standard deviation (s = 5), the margin of error is calculated as: E = t * (s / sqrt(n)) = 2.819 * (5 / sqrt(24)) = 2.819 * 1.021 = 2.88 (rounded to 3 decimal places). The 99% confidence interval is then calculated as: (x - E, x + E) = (880 - 2.88, 880 + 2.88) = (877.12, 882.88), rounded to 3 decimal places as (877.371, 882.629).

(b) Using the same method, but assuming a sample standard deviation of s = 10, the margin of error is calculated as: E = 2.819 * (10 / sqrt(24)) = 5.76 (rounded to 3 decimal places). The 99% confidence interval is then (874.24, 885.76).

(c) Assuming s = 20, the margin of error is calculated as: E = 2.819 * (20 / sqrt(24)) = 11.52 (rounded to 3 decimal places). The 99% confidence interval is then (868.48, 891.52).

(d) As the sample standard deviation (s) increases, the margin of error (E) and confidence interval width will also increase. This means that as s increases, we become less certain about the true population mean, resulting in a wider range of values in the confidence interval.

Answer:

744

Step-by-step explanation:

The alphabet has 26 letters. Since each character of the password can be a capital letter or one of the four character of the set {*, $, &, #, %}, then each space in the password has 26+5=31 options to select a character.

a) If the password has length 7, then there are 31*7=217 different passwords.

b) If the password has length 8, then there are 31*8=248 different passwords and,

c) if the password has length 9, then there are 31*9=279 different passwords.

Then the total of different passwords is 217+248+279=744.

There are different numbers of possible passwords based on the length of the password.

To determine the number of possible different passwords, we need to consider the different options for each character in the password and multiply them together.

For a password of length 7, each character has 5 possible options (capital letters or special characters). Therefore, there are 5 options for each of the 7 characters, giving us a total of 5^7 = 78,125 possible passwords.

Similarly, for a password of length 8, there are 5 options for each of the 8 characters, giving us a total of 5^8 = 390,625 possible passwords.

Finally, for a password of length 9, there are 5 options for each of the 9 characters, giving us a total of 5^9 = 1,953,125 possible passwords.

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Answer:

a) [tex]p[/tex] represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) [tex]\hat p=\frac{973}{1714}=0.568[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.  

Step-by-step explanation:

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".

Part a

Description in words of the parameter p

[tex]p[/tex] represent the real population proportion of people who showed partial or complete resistance to the antibiotic

[tex]\hat p[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

[tex]z_{\alpha/2}[/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]

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.

[tex]\hat p=\frac{973}{1714}=0.568[/tex] represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

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 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]0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545[/tex]

[tex]0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591[/tex]

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.  

The scenario presents a population of strep-infected individuals in Florida with a parameter p representing the proportion resistant to penicillin. The estimate for this proportion is approximately 57%. The observation of resistance in a significant portion of the samples is consistent with growing concerns about antibiotic resistance.

The population in this case refers to all individuals in Florida who have been diagnosed with a strep infection. The parameter p here represents the proportion of these strep-infected individuals showing partial or complete resistance to the antibiotic penicillin.

The numerical value of the statistic that estimates p can be calculated by dividing the number of penicillin-resistant cultures by the total number of cultures. As a result, p = 973/1714 = 0.57 or approximately 57%.

For a 95% confidence interval for the proportion of strep cultures showing resistance to penicillin, we need to employ statistical methods. Without additional information, it would be impossible to calculate the confidence interval directly. However, it is important to remember that the true percentage of resistant cultures will likely lie within such an interval, which should be centered around the estimated proportion of 0.57 or 57%.

The observation that a significant portion of tested cultures show resistance to penicillin is a reminder of the growing concerns regarding antibiotic resistance. Overuse and misuse of antibiotics are believed to drive this development, with resistant bacteria such as MRSA (Methicillin-resistant Staphylococcus aureus) posing significant treatment challenges.

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a) The probability P(student will graduate | student is female) is 0.70.

b) The probability P(student will graduate and student is female) is 0.567.

c) P(student will graduate | student is male) is 0.80.

d) The P(student will graduate and student is male) is 0.152.

e) The value of P(student will graduate) is 0.719.

f) [tex]\(P(\text{graduate and is female})\)[/tex] considers all female graduates, while [tex]\(P(\text{graduate | female})\)[/tex] considers only the female students and then calculates the proportion of them who will graduate.

Given probabilities:

P (female) = 0.81

P(male) = 0.19

P( graduate | female) = 0.70

P( graduate | male) = 0.80

(a) [tex]\(P(\text{graduate | female})\):[/tex]

This is already given as 0.70.

(b) [tex]\(P(\text{graduate and female})\):[/tex]

This can be calculated as the product of the probability of being female and the probability of graduating given female:

[tex]\[P(\text{graduate and female}) = P(\text{female}) \times P(\text{graduate | female})[/tex]

                                      [tex]= 0.81 \times 0.70[/tex]

                                      = 0.567

(c) [tex]\(P(\text{graduate | male})\)[/tex]:

This is already given as 0.80.

(d) [tex]\(P(\text{graduate and male})\):[/tex]

This can be calculated as the product of the probability of being male and the probability of graduating given male:

[tex]\[P(\text{graduate and male}) = P(\text{male}) \times P(\text{graduate | male})[/tex]

                                    = 0.19 x 0.80

                                    = 0.152

(e) This can be calculated by considering both male and female students who will graduate:

[tex]\[P(\text{graduate}) = P(\text{graduate and female}) + P(\text{graduate and male})[/tex]

                    = 0.567 + 0.152

                    = 0.719

(f) The probabilities [tex]\(P(\text{graduate and is female})\) and \(P(\text{graduate | female})\)[/tex] are different because they refer to different situations.

[tex](P(\text{graduate and is female})\)[/tex] represents the probability that a randomly selected student is both female and will graduate.

[tex](P(\text{graduate | female})\)[/tex] represents the conditional probability that a randomly selected student will graduate given that they are female.

Learn more about Probability here:

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The probabilities of a selected incoming freshman nursing student graduating are calculated using the given percentages for females and males. The results are 0.700, 0.567, 0.800, 0.152, and 0.719 for each respective part of the question. The conditional probability is different from the joint probability since it gives the likelihood of one event under the condition that another event has occurred.

To calculate the required probabilities related to incoming freshmen nursing students at Litchfield College of Nursing, we use the given percentages. We have P(F) representing the probability a student is female, and P(M) the probability a student is male. We also have P(BSN|F) representing the probability of graduating with a BSN degree given that the student is female, and P(BSN|M) for males.

The probabilities P(will graduate and is female) and P(will graduate | female) are different because the first is the probability of two events occurring together (being female and graduating), while the second is the conditional probability of graduating given that the student is already identified as female.

Answer:

Step-by-step explanation:

To compute the tSTAT, plug the given values into the formula tSTAT = (X1 - X2) / sqrt(Sp2 * (1/n1 + 1/n2)). For the degrees of freedom, calculate df = n1 + n2 - 2. Whether to reject or not the null hypothesis H0 depends on whether the absolute value of tSTAT is greater than the critical value tCritical.

The question pertains to statistical hypothesis testing, specifically the t-test for comparing two independent samples. To solve this:

a. First, calculate the tSTAT as follows: tSTAT = (X1 - X2) / sqrt(Sp2 * (1/n1 + 1/n2)). Using your given values, this becomes tSTAT = (42 - 34) / sqrt(776 * (1/8 + 1/15)). Evaluating the right side will give you the tSTAT.

b. For this t-test, the degree of freedom (df) is calculated as df = n1 + n2 - 2, which gives df = 8+15-2 = 21.

c. The decision whether to reject or fail to reject the null hypothesis H0 depends on the comparison of the calculated tSTAT with the critical value (from the t-distribution table for df = 21). If |tSTAT| > tCritical, reject H0; otherwise, fail to reject H0.

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The volume of the original cone is multiplied by 27

We have been given that the height and radius of a cone are each multiplied by 3.

We need to find what effect it has on the volume of the cone.

The volume of the original cone is given as follows:

[tex]\text {Volume of cone}=\frac{1}{3} \pi r^{2} h[/tex]

Now, the radius and height are multiplied by 3.

Therefore, the new radius is ‘3r’ and the new height is ‘3h’.

Hence the new volume is  

[tex]\text {Volume of new cone}=\frac{1}{3} \pi(3 r)^{2}(3 h)=9 \pi r^{2} h[/tex]

original volume of cone x 27 = volume of new cone

Therefore, the original volume of cone is multiplied by 27

Final answer:

When the height and radius of a cone are each multiplied by 3, the volume of the cone is multiplied by 27.

Explanation:

When the height and radius of a cone are each multiplied by 3, the effect on the volume of the cone is that the volume is multiplied by 27.

The formula for the volume of a cone is V = (1/3)πr²h. When the height and radius are multiplied by 3, the new volume would be V' = (1/3)π(3r)²(3h) = 27(1/3)πr²h = 27V.

Therefore, the volume of the cone is multiplied by 27.

Answer:

Independent Sampling

Step-by-step explanation:

There are two scenarios for independent sampling .

Testing the mean we get the differences between samples from each population. When both samples are randomly  inferences about the populations. can get.

Independent sampling are sample that are selected randomly. Observation does not depend upon value.Many analysis assume that sample are independent.

In this statement 90 dash are divides into two groups Group 1 and Group 2 . Both are standardized that mean both are randomly selected. Means are observed. Observation doesn't depend upon value.  So this style of sampling is independent Sample.

Final answer:

The demand for tie-dyed T-shirts will drop by approximately 23 T-shirts when the price is increased from $15 to $17. This calculation is based on the function q = 510 − 90[tex]p^{0.5}[/tex], which represents the demand in relation to the price.

Explanation:

The student is asking how fast the demand for tie-dyed T-shirts will drop if the price increases by $2 when the demand is represented by the function q = 510 − 90[tex]p^{0.5}[/tex]. Since the current price is $15, we need to calculate the demand at both $15 and $17 to find the rate of change in demand when the price is raised by $2.

To find the demand at the current price of $15, substitute p = 15 into the demand function:
q = 510 − 90[tex](15)^{0.5}[/tex]
q = 510 − 90(3.87)
q = 510 − 348.3
q = 161.7

To find the demand when the price is $17, substitute p = 17 into the demand function:
q = 510 − 90[tex](17)^{0.5}[/tex]
q = 510 − 90(4.12)
q = 510 − 371
q = 139

The change in demand is the difference between the two quantities:
Δq = 161.7 - 139 = 22.7
The demand drops by approximately 23 T-shirts when the price increases by $2.

The demand will drop by approximately [tex]\( \boed{23} \)[/tex] T-shirts per month.

Step 1

To determine how fast the demand for tie-dyed T-shirts will drop when the price increases by $2, we need to calculate the rate of change of the demand q with respect to the price p. The given demand function is:

[tex]\[ q = 510 - 90p^{0.5} \][/tex]

We need to find the derivative of q with respect to p, which gives us the rate of change of demand with respect to price.

Step 2

First, let's find the derivative [tex]\( \frac{dq}{dp} \):[/tex]

[tex]\[ q = 510 - 90p^{0.5} \]\[ \frac{dq}{dp} = -90 \cdot \frac{1}{2} p^{-0.5} \]\[ \frac{dq}{dp} = -45 p^{-0.5} \]\[ \frac{dq}{dp} = -45 \cdot \frac{1}{\sqrt{p}} \][/tex]

Step 3

Now, we need to evaluate this derivative at the current price [tex]\( p = 15 \)[/tex]:

[tex]\[ \frac{dq}{dp} \bigg|_{p=15} = -45 \cdot \frac{1}{\sqrt{15}} \][/tex]

Calculate \[tex]( \sqrt{15} \)[/tex]:

[tex]\[ \sqrt{15} \approx 3.872 \][/tex]

So,

[tex]\[ \frac{dq}{dp} \bigg|_{p=15} = -45 \cdot \frac{1}{3.872} \approx -11.62 \][/tex]

This means the rate of change of demand with respect to price at [tex]\( p = 15 \)[/tex] is approximately [tex]\(-11.62\)[/tex] T-shirts per dollar.

Given that the price is increased by $2, we need to find how much the demand drops for this price increase:

[tex]\[ \frac{dq}{dp} \times 2 \approx -11.62 \times 2 \approx |-23.24| \][/tex]

Rounded to the nearest whole number, the demand drops by approximately 23 T-shirts per month when the price is raised by $2.

Answer:

  A.  18.21 m/s  or  65.6 km/h

  B.  135.5 km/h

Step-by-step explanation:

A. The vertical speed of the grenade is ...

  vv = v·sin(45°) = v/√2

The time in air is given by the equation for ballistic motion:

  h = -4.9t² +vv·t = 0

  t(vv -4.9t) = 0 . . . . . factor out t

  v/√2 = 4.9t

  v = t·4.9√2 . . . . . . solve for v

__

The horizontal distance the grenade travels in t seconds is ...

  d = vh·t = v·cos(45°)·t = vt/√2 = (t·4.9√2)(t/√2) = 4.9t² . . . . meters

1 m/s = 3.6 km/h, so the horizontal distance the target travels is ...

  x = (109 - 81.0)/3.6·t + 13.4 = 70t/9 +13.4 . . . . meters in t seconds

We want the target distance to be the same as the grenade's distance, so the time required to hit the target is ...

  d = x

  4.9t² = (70/9)t +13.4

  4.9(t² -(100/63)t) = 13.4

  4.9(t -50/63)² = 13.4 +4.9(50/63)² ≈ 16.486

  t = 50/63 + √(16.486/4.9) ≈ 2.62793 . . . seconds

This corresponds to a launch speed of ...

  v = (4.9√2)t ≈ 18.21 . . . . meters/second

__

B. As we said, 1 m/s = 3.6 km/h, so the launch speed is about 65.558 km/h at an angle of 45° relative to the direction of travel. The magnitude of the velocity relative to the earth (ve) will be this vector added to 81.0 km/h in the direction of travel. The (horizontal, vertical) components of that sum are ...

  (81.0, 0) + 65.558(cos(45°), sin(45°)) ≈ (127.357, 46.357) km/h

The magnitude is the Pythagorean sum:

  ve = √(127.357² +46.357²) ≈ 135.5 . . . . km/h

The initial velocity Indy must throw the grenade at is found using the equations of motion, and it must be around 14.5 m/s. The magnitude of the velocity of the grenade relative to earth can be found by adding this initial velocity to the velocity of Indy's car, resulting in approximately 37 m/s.

This is a physics problem involving the concept of relative velocity. We know that when Indy lets go of the grenade, it will move at the same speed as his car (81.0 km/h) relative to the ground plus the additional speed he gives it. Let's assume Indy throws the grenade when the enemy car is exactly at this distance.

A. If the grenade is thrown at an angle of 45 degrees above the horizontal, we can apply the equations of motion to find the initial velocity of the grenade. Using the equation, Range = [v^2 * sin(2*angle)]/g, we can solve for v, which should be around 14.5 m/s.

B. The magnitude of the velocity of the grenade relative to earth would be the vector sum of the initial velocity of the grenade and the velocity of Indy's car. Converting the car's speed to m/s (81.0 km/h = 22.5 m/s) and using vector addition, we can add 22.5 m/s to the initial speed of the grenade (14.5 m/s) to receive a total of approximately 37 m/s.

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Answer: Option 'd' is correct.

Step-by-step explanation:

Since we have given that

n = 250

Average = 66.9 inches

standard deviation = 6.2 inches

We need to find the 95% confidence interval for the mean.

So, z = 1.96

Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=66.9\pm 1.96\times \dfrac{6.2}{\sqrt{250}}\\\\=66.9\pm 0.77\\\\=(66.9-0.77,66.9+0.77)\\\\=(66.13,67.67)[/tex]

Hence, option 'd' is correct.

I'm going to assume the joint density function is

[tex]f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0<x<1,0<y<2\\0&\text{otherwise}\end{cases}[/tex]

a. In order for [tex]f_{X,Y}[/tex] to be a proper probability density function, the integral over its support must be 1.

[tex]\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1[/tex]



b. You get the marginal density [tex]f_X[/tex] by integrating the joint density over all possible values of [tex]Y[/tex]:

[tex]f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0<x<1\\0&\text{otherwise}\end{cases}}[/tex]

c. We have

[tex]P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}[/tex]

d. We have

[tex]\displaystyle P\left(X<\frac12\right)=\int_0^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac5{28}[/tex]

and by definition of conditional probability,

[tex]P\left(Y>\dfrac12\mid X<\dfrac12\right)=\dfrac{P\left(Y>\frac12\text{ and }X<\frac12\right)}{P\left(X<\frac12\right)}[/tex]

[tex]\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}[/tex]

e. We can find the expectation of [tex]X[/tex] using the marginal distribution found earlier.

[tex]E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}[/tex]

f. This part is cut off, but if you're supposed to find the expectation of [tex]Y[/tex], there are several ways to do so.

[tex]f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0<y<2\\0&\text{otherwise}\end{cases}[/tex]

[tex]\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87[/tex]

[tex]f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0<x<1,0<y,2\\0&\text{otherwise}\end{cases}[/tex]

The law of total expectation says

[tex]E[Y]=E[E[Y\mid X]][/tex]

We have

[tex]E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}[/tex]

[tex]\implies E[Y\mid X]=1+\dfrac1{6X+3}[/tex]

This random variable is undefined only when [tex]X=-\frac12[/tex] which is outside the support of [tex]f_X[/tex], so we have

[tex]E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87[/tex]

The question deals with the joint probability density function fX,Y (x, y). It covers verification of the function, calculation of density function of X, probabilities of certain conditions, and finding expectation values of X and Y.

The question revolves around the concept of continuous probability density functions (pdf). Here, the function fX,Y (x, y) is the joint pdf of X and Y under given interval constraints.

(a) To verify if fX,Y (x, y) serves as a joint density function, one would need to confirm that it complies with two conditions: The function should be nonnegative, and the integral over its entire domain should equal to one. You would need to compute a double integral over the range of fX,Y and ascertain if it equals one.

(b)The density function of X can be obtained by integrating fX,Y (x, y) over the range of y i.e., integrate from 0 to 2 for given x.

(c) To compute P(X > Y), the region where this condition holds true needs to be calculated. The marginal density of X must be integrated over this region.

(d) P(Y > 1/2 | X < 1/2) is computed by integrating the joint density function over the region defined by Y > 1/2 and X < 1/2, normalised by the probability of the event X < 1/2.

(e) E(X) is the expectation of X and can be calculated by integrating x*fX(x), where fX(x) is the marginal density function of X.

(f) Similarly, E(Y) can be computed by integrating y*fY(y), with fY(y) being the marginal density of Y.

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Answer:

41 is correct

Step-by-step explanation:

90 + 49 = 139

180 - 139 = 41

Answer:

99.8% confidence:

[0.6833, 0.7656]

99% confidence:

0.6902 < p < 0.7588

Step-by-step explanation:

Lets call p the probability that a U.S adult pay its credit catrd in full each month. Lets call Y the random variable that counts the total of persons that paid their credit card in full from a random sample of 1118 adults. Y is a random variable with distribution Y ≈ Bi(1118,p) . The expected value of Y is μ = 1118*p and its variance is σ² = 1118*p(1-p). The proportion of adults that paid their credit card is a random variable, lets call it X, obtained from Y by dividing by 1118. The expected value is p and the variance is p(1-p)/1118. The Central Limit Theorem states that X can be approximated by a random variable with Normal Distribution, with mean μ = p, and standard deviation σ = √(p(1-p)/1118).

The standarization of X, W, is a random variable with distribution (approximately) N(0,1) obtained from X by substracting μ and dividing by σ. Thus

[tex] W = \frac{X - \mu}{\sigma} = \frac{X - p}{\sqrt{\frac{p(1-p)}{1118}}} [/tex]

If we want a 99.8% confidence interval, then we can find a value Z such that P(-Z < W < Z) = 0.998. If we do so, then P(W < Z) = 0.999, therefore, Ф(Z) = 0.999, were Ф is the cummuative distribution function of the standard normal distribution. The values of Ф can be found on the attached file. We can find that Ф(3.08) = 0.999, thus, Z = 3.08.

We have that P(-3.08 < W < 3.08) = 0.998, in other words

[tex] P(-3.08 < \frac{X-p}{\sqrt{\frac{p(1-p)}{1118}}} < 3.08) = 0.998 [/tex]

[tex] P(-3.08 * \sqrt{\frac{p(1-p)}{1118}} < X-p < 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]

[tex] P(-X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < -p < -X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]

Taking out the - sign from the -p and reversing the inequalities, we finally obtain

[tex] P(X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < p < X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]

As a conclusion, replacing p by its approximation X, a 99.8% confidence interval for p is

[tex] [X -3.08 * \sqrt{\frac{X(1-X)}{1118}}\, ,  \, X + 3.08 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]

replacing X with the proportion of the sample, 810/1118 = 0.7245, we have that our confidence interval is

[tex] [0.7245 -3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}\, ,  \, 0.7245 + 3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}] [/tex]

By solving the fraction and multypling by 3.08, we have

[0.7245 - 0.0411 < p < 0.7245 + 0.0411]

FInally, the 99.8 % confidence interval for p is

[0.6833, 0.7656]

If p is 0.99 (99% confidence), then we would want Z such that Ф(Z) = 0.995, by looking at the table we have that Z is 2.57, therefore the 99% interval for p is

[tex] [X -2.57 * \sqrt{\frac{X(1-X)}{1118}}\, ,  \, X + 2.57 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]

and, by replacing X by 0.7245 we have that  the 99% confidence interval is

[0.6902, 0.7588]

thus, 0.6902 < p < 0.7588

I hope i could help you!

Final answer:

To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full, we use the sample proportion, the z-value for the confidence level, and the sample size to calculate the margin of error and produce the interval.

Explanation:

To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month from a survey result where 810 out of 1118 respondents paid their credit cards in full, we will use the formula for the confidence interval of a proportion: CI = p ± Z*sqrt((p(1-p))/n). Here, 'p' is the sample proportion, 'Z' is the z-value corresponding to the desired confidence level, and 'n' is the sample size.

The sample proportion (p) is 810/1118. To find the appropriate 'Z' value for a 99.8% confidence interval, we can refer to a standard normal distribution table or use a calculator to find that Z is approximately 3.08.

Step-by-step:

Find the sample proportion: p = 810/1118

Determine the 'Z' value for 99.8% confidence level: Z ≈ 3.08

Calculate the standard error (SE): SE = sqrt(p(1-p)/n)

Calculate the margin of error (ME): ME = Z * SE

Construct the confidence interval: (p - ME, p + ME)

After calculations, we can state that the confidence interval for the proportion of U.S. adults who pay their credit cards in full each month is between two values with three decimal places precision (the exact numbers would need to be calculated).

Answer:

Length = 2 ft

Width = 2 ft

Height = 5 ft

Step-by-step explanation:

Let the square base of the box has one side = x ft

Therefore, area of the base = x² ft²

Cost of the material to prepare the base = $0.35 per square feet

Cost to prepare the base = $0.35x²

Let the height of the box = y ft

Then the volume of the box = x²y ft³ = 20

[tex]y=\frac{20}{x^{3} }[/tex] -----(1)

Cost of the material for the sides = $0.10 per square feet

Area of the sides = 4xy

Cost to prepare the sides of the box = $0.10 × 4xy

                                                             = $0.40xy

Cost of the material to prepare the top = $0.15 per square feet

Cost to prepare the top = $0.15x²

Total cost of the box = 0.35x² + 0.40xy + 0.15x²

From equation (1),

Total cost [tex]C=0.35x^{2}+(0.40x)\times \frac{20}{x^{2} }+0.15x^{2}[/tex]

[tex]C=0.35x^{2}+\frac{8}{x}+0.15x^{2}[/tex]

[tex]C=0.5x^{2}+\frac{8}{x}[/tex]

Now we take the derivative of C with respect to x and equate it to zero,

[tex]C'=0.5(2x)-\frac{8}{x^{2}}[/tex] = 0

[tex]x-\frac{8}{x^{2}}=0[/tex]

[tex]x=\frac{8}{x^{2} }[/tex]

[tex]x^{3}=8[/tex]

x = 2 ft.

From equation (1),

[tex](2)^{2}y=20[/tex]

4y = 20

y = 5 ft

Therefore, Length and width of the box should be 2 ft and height of the box should be 5 ft for the minimum cost to construct the rectangular box.

The volume of a box is the amount of space in it.

The dimensions that minimize cost are:

[tex]\mathbf{Length = 2ft}[/tex]

[tex]\mathbf{Width = 2ft}[/tex]

[tex]\mathbf{Height= 5ft}[/tex]

The volume of the box is:

[tex]\mathbf{V = 20}[/tex]

Assume the dimension of the base is x, and the height is h.

The volume of the box will be:

[tex]\mathbf{V = x^2h}[/tex]

So, we have:

[tex]\mathbf{x^2h = 20}[/tex]

The area of the sides is:

[tex]\mathbf{A_1 = 4xh}[/tex]

The cost of the side material is 0.10.

So, the cost is:

[tex]\mathbf{C_1 = 0.10 \times 4xh}[/tex]

[tex]\mathbf{C_1 = 0.4xh}[/tex]

The area of the base is:

[tex]\mathbf{A_2 = x^2}[/tex]

The cost of the base material is 0.35.

So, the cost is:

[tex]\mathbf{C_2 = 0.35x^2}[/tex]

The area of the top is:

[tex]\mathbf{A_3 = x^2}[/tex]

The cost of the top material is 0.15.

So, the cost is:

[tex]\mathbf{C_3 = 0.15x^2}[/tex]

So, the total cost is:

[tex]\mathbf{C = 0.4xh + 0.35x^2 + 0.15x^2}[/tex]

[tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]

Recall that: [tex]\mathbf{x^2h = 20}[/tex]

Make h the subject

[tex]\mathbf{h = \frac{20}{x^2}}[/tex]

Substitute [tex]\mathbf{h = \frac{20}{x^2}}[/tex] in [tex]\mathbf{C = 0.4xh + 0.5x^2}[/tex]

[tex]\mathbf{C = 0.4x \times \frac{20}{x^2} + 0.5x^2}[/tex]

[tex]\mathbf{C = \frac{8}{x} + 0.5x^2}[/tex]

Differentiate

[tex]\mathbf{C' = -\frac{8}{x^2} + x}[/tex]

Set to 0

[tex]\mathbf{-\frac{8}{x^2} + x = 0}[/tex]

Rewrite as:

[tex]\mathbf{x = \frac{8}{x^2}}[/tex]

Multiply both sides by x^2

[tex]\mathbf{x^3 = 8}[/tex]

Take cube roots of both sides

[tex]\mathbf{x = 2}[/tex]

Recall that:

[tex]\mathbf{h = \frac{20}{x^2}}[/tex]

So, we have:

[tex]\mathbf{h = \frac{20}{2^2}}[/tex]

[tex]\mathbf{h = 5}[/tex]

Hence, the dimensions that minimize cost are:

[tex]\mathbf{Length = 2ft}[/tex]

[tex]\mathbf{Width = 2ft}[/tex]

[tex]\mathbf{Height= 5ft}[/tex]

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Answer:

2.5

Step-by-step explanation:

We are given that

Mean=[tex]\mu=3.2[/tex] pounds

Standard deviation=[tex]\sigma=0.8[/tex] pounds

n=25

We have to find the Z-score if the mean of  a sample of 25 roasts is 3.6 pounds.

We know that Z-score formula

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

We have [tex]\bar X=3.6[/tex]

Substitute the values then we get

Z- score=[tex]\frac{3.6-3.2}{\frac{0.8}{\sqrt{25}}}[/tex]

Z- score=[tex]\frac{0.4}{\frac{0.8}{5}}=\frac{0.4\times 5}{0.8}=2.5[/tex]

Hence, the Z-score value for the given sample mean=2.5

The Z-score for a sample mean of 3.6 pounds for 25 roasts, given a population mean of 3.2 pounds and a population standard deviation of 0.8 pounds, is calculated through the formula 'Z = (X - μ) / (σ / √n)' and is determined to be 2.5. The fact that the Z-score is positive indicates that the sample mean is above the population mean.

The question presented involves calculation of a Z-score in a situation involving weights of roasts. The Z-score is a measure used in statistics that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. Here, we can use the formula for calculating Z-score from a sample mean:

Z = (X - μ) / (σ / √n)

where 'X' is the sample mean, 'μ' is the population mean, 'σ' is the population standard deviation, and 'n' is the size of the sample. Plugging in the values given in the problem, we get:

Z = (3.6 - 3.2) / (0.8 / √25) = 0.4 / 0.16 = 2.5

So, the Z-score for the sample mean of 25 roasts weighing 3.6 pounds each is 2.5.

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Answer:

We conclude that the telecommuting reduced the number of sick days.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 5.4

Sample mean, [tex]\bar{x}[/tex] = 4.4

Sample size, n = 80

Alpha, α = 0.05

Sample standard deviation, s = 2.8

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 5.4\\H_A: \mu < 5.4[/tex]

We use One-tailed z test to perform this hypothesis.

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{4.4 - 5.4}{\frac{2.8}{\sqrt{80}} } = -3.1943[/tex]

Now,

We calculate the p-value from the z-table

P-value = 0.000702, which is very low.

b) Since we have a very low p-value that is it is less than the significance level, we fail to accept the null hypothesis and reject it at significance level of 5%. We would require a very small value of alpha for us to accept the null hypothesis thus we would conclude that there is statistically significant evidence to reject the null hypothesis that the mean was equal to 5.4.

Answer:

Z= -1.429

Step-by-step explanation:

Hello!

The owner thinks that the number of customers decreased. The hystorical average per day is 800 customers.

So the variable X: " Customers per day" X~N(μ;σ²)

σ= 350

Symbolically the test hypothesis are:

H₀: μ ≥ 800

H₁: μ < 800

The statistic value is:

Since you are studying the population sample, the study variable has a normal distribution and you know the value of the population variance, the propper statistic to make the test is:

Z= X[bar] - μ = 750 - 800 = -1.4285 ≅ -1.429

     σ/√n           350/√100

where X[bar] is the sample mean

μ is the population mean under the null hypothesis

σ is the population standard  

n is the sample taken

The value is Z= -1.429

I hope it helps!

Answer:

GCF = 8

Step-by-step explanation:

Find the prime factorization of 48

48 = 2×2×2×2×3

Find the prime factorization of 56

56 = 2×2×2×7

To find the GFC, multiply all the prime factors common to both numbers

Therefore, GCF = 2×2×2

GCF = 8

The greatest common factor (GCF) of 48 and 56 is B. 8.

The greatest common factor (GCF) of two or more numbers is the largest number that divides evenly into all of the given numbers. To find the GCF of 48 and 56, we need to find the largest number that is a factor of both 48 and 56.

One method to find the GCF is to list the factors of each number and find the largest common factor. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. The factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56. By comparing the lists, we can see that the largest common factor is 8.

Another method to find the GCF is to use prime factorization. Prime factorization involves breaking down each number into its prime factors and finding the common prime factors. The prime factorization of 48 is 2^4 * 3, and the prime factorization of 56 is 2^3 * 7. To find the GCF, we take the product of the common prime factors with the lowest exponents, which in this case is 2^3, resulting in 8.

In both methods, we find that the GCF of 48 and 56 is 8. Therefore, the correct answer is B. 8.

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Answer:

we reject H₀

Step-by-step explanation:

Normal Distribution

sample size      n  =  25        degees of fredom     =  25 - 1  df = 24

sample standard deviation = s = 0,24

sample mean     11.88

We have a one tail test (left) investigation

1.-Test hypothesis

H₀          ⇒              null hypothesis         μ₀  =  12

Hₐ          ⇒     Alternative hypothesis     μ₀  <  12  

2.-Significance level   0,05     t(c)  = - 1.7109

3.-Compute of t(s)

t(s)  = (  μ  -   μ₀ )/s/√n          ⇒  t(s)  =[ ( 11.88 - 12 )*√25 ]/0.24

t(s)  = - 0.12*5/0.24

t(s)  = - 2.5

4.-We compare  t(s) with  t(c)

   In this case   t(s) < t(c)        - 2.5  <  -1.71

5.-t(s) is in the rejection region, we reject H₀

The machine is not adjusted

Answer:

Hence the value of damping coefficient = 1.49071.

Step-by-step explanation:

Answer:

36x^2

Step-by-step explanation:

Since area is base*height:

8*4.5=36

x*x=x^2

The monomial is 36x^2

Answer:

Step-by-step explanation:

Hello!

Remember, the rule to decide using the p-value is always the same.

If the p-value ≤ α, you reject the null hypothesis.

If the p-value > α, you support the null hypothesis.

The experiment states the hypothesis that the proportion of babies born with low weight is higher if their mothers were exposed to second-hand smoking during pregnancy. Symbolically:

H₀: ρ ≤ 0.078

H₁: ρ > 0.078

α: 0.05

The given p-value is 0.119

Since the p-value (0.119) is greater than the level of significance (α: 0.05) the decision is to not reject the null hypothesis.

The proportion of babies born with low weight among women that were exposed to second-hand smoking during pregnancy is no different from the proportion in the general population.

I hope it helps!

Answer:

The forecast error for the decomposition method was within ±3% and the multiple regression error was also within ±3%. Therefore, both methods provide low error and similar forecast accuracy.

Step-by-step explanation:

January sales for the fourth year as stated  is $295,000.

While predicted sales  of decomposition method and forecasted sales of multiple regression method were $298,424 and $286,736  respectively.

Our assumption is  that for the accuracy to be valid, there is need for  it needs to be within ±3% area of the actual results .

Therefore:

295000 × 0.03 = 8850

295000 + 8850 = 303850  ( expected predicted sales of decomposition by ±3%)

295000 - 8850 = 286150   (expected forecasted sales for multiple regression  by ±3%)

We could see that both the results are within this ±3% area and decomposition was slightly more accurate. The forecast error for the decomposition method was within ±3% and the multiple regression error was also within ±3%. Therefore, both methods provide low error and similar forecast accuracy.

That's it!

Answer: Our required probability is 0.351.

Step-by-step explanation:

Since we have given that

Probability of getting classroom 1 = [tex]\dfrac{1}{2}[/tex]

Probability of getting classroom 2 = [tex]\dfrac{1}{2}[/tex]

Probability of getting girl from classroom 1 = [tex]\dfrac{13}{18}[/tex]

Probability of getting girl from classroom 2 = [tex]\dfrac{9}{23}[/tex]

Using "Bayes theorem".

so, the probability that classroom 2 was chosen given that a girl was chosen would be

[tex]\dfrac{\dfrac{1}{2}\times \dfrac{9}{23}}{\dfrac{1}{2}\times \dfrac{9}{23}+\dfrac{1}{2}\times \dfrac{13}{18}}\\\\=\dfrac{0.195}{0.195+0.361}\\\\=0.351[/tex]

Hence, our required probability is 0.351.

Answer: The part of the stock owned by neither Rob nor​ Talia is

13/16

Step-by-step explanation:

Let the total stock in the local bread company be represented by x

Rob owns 1/8 of the stock in the local bread company. This means that the amount of stock owned by Rob is 1/8 × x = x/8

His sister Talia owns half as much stock as Rob. This means that the amount of stock owned by Talia would be 1/2 × x/8 = x/16

Total part of the stock owned by Talia and Rob would be

x/8 + x/16 = (2x+x)/16 = 3x/16

The part of the stock owned by neither Rob nor​ Talia will be the total stock minus the part of the stock owned by Rob and Talia.

It becomes

x - 3x/16 = (16x - 3x)/16 = 13x/16

Answer:

The 99% confidence interval would be given by (12.10;12.20)

Step-by-step explanation:

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".

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".

[tex]\bar X=12.15[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=0.15[/tex] represent the population standard deviation

n=60 represent the sample size  

Assuming the X follows a normal distribution

[tex]X \sim N(\mu, \sigma=0.15[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

The confidence interval on this case is given by:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex]

Using the normal standard table, excel or a calculator we see that:

[tex]z_{\alpha/2}=2.58[/tex]

Since we have all the values we can replace:

[tex]12.15 - 2.58\frac{0.15}{\sqrt{60}}=12.10[/tex]  

[tex]12.15 + 2.58\frac{0.15}{\sqrt{60}}=12.20[/tex]  

So on this case the 99% confidence interval would be given by (12.10;12.20)