Answer:
The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
All songs
Mean 215 seconds, standard deviation 35 seconds
Sample
49
Mean 215, standard deviation [tex]s = \frac{35}{\sqrt{49}} = 5[/tex]
The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.
Answer:
The sample mean would be:
[tex]\mu_{\bar X} = 215 seconds[/tex]
And the deviation:
[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]
Step-by-step explanation:
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
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
We know the following info for the random variable X who represent the duration
[tex]\mu = 215, \sigma=35[/tex]
For this case we select a sampel size of n =49>30. So we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The sample mean would be:
[tex]\mu_{\bar X} = 215 seconds[/tex]
And the deviation:
[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]
what is the h?
h/6 = 20/24
Answer:
h = 5
Step-by-step explanation:
[tex] \frac{h}{6} = \frac{20}{24} \\ 24 \times h = 20 \times 6 \\ 24h = 120 \\ h = \frac{120}{24} \\ h = 5[/tex]
State the Pythagorean Theorem in your own words
Answer:
A ^2 + B^2 =C^2. The Pythagorean Theorem is a statement about triangles containing a right angle
Step-by-step explanation:
Darcia made a snack mix using the following recipe. 1 ¼ cups granola, ¾ cup peanuts, ½ cup raisins, ¼ cup chocolate chips. If she used 3 cups of granola and wanted to keep the ingredients proportional, how much would she use of each of the other ingredients.
Answer:
2345
Step-by-step explanation:
because3456
A college entrance exam company determined that a score of 23 on the mathematics portion of the exam suggests mat a student is ready for college-level mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 200 students who completed this core set of courses results n a mean math score of 23.6 on the college entrance exam with a standard deviation of 3.2. Do these results suggest mat students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 23 on the math portion of the exam? Complete parts a through d below. a. State the appropriate null and alternative hypotheses. Choose the correct answer below i. H_0: mu = 23.6 versus H_1 = mu notequalto 23.6 ii.H_0: mu = 23.6 versus H_1 = mu > 23.6 iii.H_0: mu = 23.6 versus H_1 = mu < 23.6 iv. H_0: mu = 23.6 versus H_1 = mu > 23 v. H_0: mu = 23.6 versus H_1 = mu < 23 b. Verify that the requirements to perform the test using the t-distribution are satisfied. Is the sample obtained using simple random sampling or from a randomized experiment? i. Yes, because only high school students were sampled. ii. No, because not all students complete the courses.iii. No, because only high school students were sampled.iv. Yes, because the students were randomly sampled. Is the population from which the sample is drawn normally distributed or is the sample size, n, large (n Greaterthanorequalto 30)? i. No, neither of these conditions are true ii. Yes, the sample size is larger man 30 iii. Yes, the population is normally distributed It is impossible to determine using the given information c. Are the sampled values independent of each other? i. Yes, because each student's test score does not affect other students' test scores ii. No. because students from the same class will affect each other's performance iii. Yes, because the students each take their own tests iv. No, because every student takes the same test d. Use the P-value approach at the alpha = 0 10 level of significance to test the hypotheses. (Round to three decimal places as needed) State the conclusion for the test Choose the correct answer below i. Do not reject the null hypothesis because the P-value is greater than the alpha = 0 10 level of significance ii. Reject the null hypothesis because the P-value a less than the alpha = 0 10 level of significance iii. Do not reject (he null hypothesis because the P value is less than the alpha = 0 10 level of Significance. iv. Reject the nun hypothesis because the P-value greater than the alpha = 010 level of significance e. Write a conclusion based on the results. Choose the correct answer below. i. There is sufficient evidence to conclude that the population mean is greater than 23. ii. There is sufficient evidence to conclude that the population mean is less than 23 iii.There is not sufficient evidence to conclude that the population mean is greater than 23 iv. There is not sufficient evidence to conclude that the population mean is less man 23.
Answer:
a) The null hypothesis is represented as
H₀: μ ≤ 23
The alternative hypothesis is given as
Hₐ: μ > 23
b) Check the Explanation
The conditions for a t-test to be performed are satisfied or not?
- Yes, because the students were randomly sampled.
- Yes, the sample size is larger man 30.
And the central limit theorem allows us to approximate that the random sample obtained from the population is a normal distribution.
c) Are the sampled values independent of each other?
Yes, because each student's test score does not affect other students' test scores.
d) p-value obtained = 0.004
Reject the null hypothesis because the P-value a less than the alpha = 0.10 level of significance
e) There is sufficient evidence to conclude that the population mean is greater than 23.
Step-by-step explanation:
For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, we want to check if results suggest that students who complete the core curriculum are ready for college-level mathematics.
The only condition to be ready for college is scoring above 23.
So, the null hypothesis would be that the mean of test scores of students that complete core curriculum is less than or equal to 23. That is, there isn't significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics.
And the alternative hypothesis would be that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.
Mathematically
The null hypothesis is represented as
H₀: μ ≤ 23
The alternative hypothesis is given as
Hₐ: μ > 23
b) The conditions required before performing t-test.
- The sample should be a random sample
- The dependent variable should be approximately normally distributed.
- The observations are independent of one another.
- The dependent variable should not contain any outliers
All of these conditions are satisfied for our distribution.
c) Are the sampled values independent of each other?
Yes, because each student's test score does not affect other students' test scores.
d) To do this test, we will use the t-distribution because no information on the population standard deviation is known
So, we compute the t-test statistic
t = (x - μ₀)/σₓ
x = sample mean = 23.6
μ₀ = 23
σₓ = standard error = (σ/√n)
where n = Sample size = 200
σ = Sample standard deviation = 3.2
σₓ = (3.2/√200) = 0.226
t = (23.6 - 23) ÷ 0.226 = 2.65
checking the tables for the p-value of this t-statistic
- Degree of freedom = df = n - 1 = 200 - 1 = 199
- Significance level = 0.10
- The hypothesis test uses a one-tailed condition because we're testing only in one direction.
p-value (for t = 2.65, at 0.10 significance level, df = 199, with a one tailed condition) = 0.004348 = 0.004 to 3 d.p.
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.10
p-value = 0.004
0.004 < 0.10
Hence,
p-value < significance level
This means that we reject the null hypothesis, accept the alternative hypothesis and say that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.
e) The result of the p-value obtained is that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.
Hope this Helps!!!
Translate the phrase "4 times as old as she was
33 years ago" into algebraic terms.
Answer:
4 * x - 33
Step-by-step explanation:
four times as old as she was 33 years ago
let x represent how old she is now
4 * x - 33
Four ( 4 ) times ( * ) as old as she was 33 years ago ( x - 33 )
Final answer:
The phrase "4 times as old as she was 33 years ago" can be represented algebraically as the equation x = 4(x - 33), where x is the current age of the person.
Explanation:
To translate the phrase "4 times as old as she was 33 years ago" into algebraic terms, we first denote the current age of the person as a variable, let's call it x. The age of the person 33 years ago would then be expressed as x - 33. To articulate that someone is now four times as old as they were 33 years ago, we create the algebraic expression 4(x - 33), which represents the current age being four times the age 33 years prior.
Example:
If we want to set up an equation where someone's current age is four times their age from 33 years ago, it would look like this:
Let x = current age
Then x - 33 = age 33 years ago
The phrase translates to the equation x = 4(x - 33)
By multiplying both sides by the same factor, we can manipulate this equation if necessary to solve for x. As an example, multiplying both sides by 1 does not change the equation, but it can sometimes be helpful in other contexts to multiply both sides to simplify or to get rid of fractions.
The manager of the toy store at the mall is curious about why nobody is buying the new robot toys. He's very annoyed because he paid a lot of money for inventory, and wants to know how to get the merchandise sold. What is the best method of sampling for him to get his answers?
A.
use a convenience sample of customers in other toy stores
B.
survey customers as they leave the store
C.
survey the toy store employees
D.
random sampling of people walking in the mall
Answer: B
Step-by-step explanation: I believe it's B because if you're able to get the opinions of others about your store. You'll be able to see whats the customers are more interested in, so therefore be able to supply your store with stuff people are more intrigued in. (Let me know if I'm wrong, have a good day)
A consumer organization estimates that over a 1-year period 17% of cars will need to be repaired once, 10% will need repairs twice, and 2% will require three or more repairs. If you own two cars, what is the probability that a) neither will need repair? b) both will need repair? c) at least one car will need repair?
Answer:
a) 50.41% probability that neither will need repair.
b) 8.41% probability that both will need repair.
c) 49.59% probability that at least one car will need repair.
Step-by-step explanation:
For each car, there are only two possible outcomes. Either it will need repairs, or it will not need repairs. The probability of a car needing repairs is independent of other cars. 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.
17% of cars will need to be repaired once, 10% will need repairs twice, and 2% will require three or more repairs.
This means that [tex]p = 0.17+0.1+0.02 = 0.29[/tex]
Two cars:
This means that [tex]n = 2[/tex]
a) neither will need repair?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2,0}.(0.29)^{0}.(0.71)^{2} = 0.5041[/tex]
50.41% probability that neither will need repair.
b) both will need repair?
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.29)^{2}.(0.71)^{0} = 0.0841[/tex]
8.41% probability that both will need repair.
c) at least one car will need repair?
Either none will need repair, or at least one will. The sum of these probabilities is 100%.
From a)
50.41% probability that neither will need repair.
p = 100 - 50.41 = 49.59%
49.59% probability that at least one car will need repair.
Brawdy Plastics, Inc., produces plastic seat belt retainers for General Motors at the Brawdy Plastics plant in Buffalo, New York. After final assembly and painting, the parts are placed on a conveyor belt that moves the parts past a final inspection station. How fast the parts move past the final inspection station depends upon the line speed of the conveyor belt (feet per minute). Although faster line speeds are desirable, management is concerned that increasing the line speed too much may not provide enough time for inspectors to identify which parts are actually defective. To test this theory, Brawdy Plastics conducted an experiment in which the same batch of parts, with a known number of defective parts, was inspected using a variety of line speeds.
The following data were collected. If required, enter negative values as negative numbers.
Line Speed Devective number
20 23
20 21
30 19
30 16
40 15
40 17
50 14
50 11
a. Select a scatter diagram with the line speed as the independent variable.b. What does the scatter diagram developed in part (a) indicate about the relationship between the two variables?
Scatter plot shows negative correlation; regression equation: Defective Parts = [tex]27.5 - 0.3 \times \text{Line Speed}[/tex]; 20 defects predicted at 25 fpm.
a. The scatter diagram with the line speed as the independent variable is displayed. It shows the number of defective parts found at various line speeds.
(DIAGRAM IS GIVEN BELOW)
b. The scatter diagram indicates that there is a negative relationship between the two variables. As the line speed increases, the number of defective parts found decreases. This suggests that at higher speeds, perhaps the inspectors are not able to identify all the defective parts due to the speed at which the parts are moving past the inspection station.
c. Using the least squares method to develop the estimated regression equation with line speed as the independent variable (to 1 decimal), we get:
Defective Parts = [tex]27.5 - 0.3 \times \text{Line Speed}[/tex]
where 27.5 is the y-intercept and -0.3 is the slope of the line.
d. To predict the number of defective parts found for a line speed of 25 feet per minute, we can substitute x = 25 into the regression equation:
Defective Parts = [tex]27.5 - 0.3 \times 25 = 20.0[/tex]
Therefore, the model predicts that for a line speed of 25 feet per minute, there would be 20 defective parts found.
The complete question is here:
Brawdy Plastics, Inc. produces plastic seat belt retainers for General Motors at their plant in Buffalo, New York. After final assembly and painting, the parts are placed on a conveyor belt that moves the parts past a final inspection station. How fast the parts move past the final inspection station depends upon the line speed of the conveyor belt (feet per minute). Although faster line speeds are desirable, management is concerned that increasing the line speed too much may not provide enough time for inspectors to identify which parts are actually defective. To test this theory, Brawdy Plastics conducted an experiment in which the same batch of parts, with a known number of defective parts, was inspected using a variety of line speeds. The following data were collected. If required, enter negative values as negative numbers.
(DATA IS GIVEN BELOW)
a. Select a scatter diagram with the line speed as the independent variable.
b. What does the scatter diagram developed in part (a) indicate about the relationship between the two variables?
c. Use the least squares method to develop the estimated regression equation (to 1 decimal). = + x d. Predict the number of defective parts found for a line speed of 25 feet per minute.
Application of the least squares method results in values of the y intercept and the slope that minimizes the sum of the squared deviations between the
a. observed values of the independent variable and the estimated values of the independent variable
b. actual values of the independent variable and estimated values of the dependent variable
c. observed values of the dependent variable and the estimated values of the dependent variable
d. None of these answers is correct.
Answer:
Копиравать
Step-by-step explanation:
Копиравать ладно
A study is conducted to determine whether sunshine affects depression. Eight individuals are given a questionnaire measuring depression immediately following a run of 10 consecutive days when the sun shone for over 80% of the daylight hours. The same individuals have their depression measured immediately following 10 consecutive days without any sunshine. The following data are collected. The higher the score the greater the depression.
Individuals 1 2 3 4 5 6 7 8
Sunshine 10 12 14 11 12 10 14 15
No Sunshine 20 21 17 14 18 8 18 14
Using the Wilcoxon signed ranks test to evaluate the data, with a = 0.052 tail, Tcrit =
a.- 3
c.33
b.3
d.4
Based on the information given, it should be noted that the Wilcoxon signed rank test will be 3.
From the table, it can be seen that the first tank is assigned to the first absolute value. The second rank is assigned to the second absolute value while the average of the third and the fourth value are given the value of 3.
The sum of the positive ranks will be:
= 1 + 2 = 3
The sum of the negative ranks will be:
= 8 + 7 + 3.5 + 3.5 + 6 + 5 = 33.
Therefore, the Wilcoxon signed-rank will be the minimum of 33 and 3 which is 3.
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Help please with this question
Given:
The given two functions are [tex]f(x)=\frac{4 x^{2}+24 x}{x^{2}-11 x+30}[/tex] and [tex]g(x)=\frac{x-5}{x^{2}}[/tex]
We need to determine the value of R(x)
Value of R(x):
The value of R(x) can be determined by R(x) = f(x) × g(x)
Substituting the values, we get;
[tex]R(x)=\frac{4 x^{2}+24 x}{x^{2}-11 x+30} \cdot \frac{x-5}{x^{2}}[/tex]
Multiplying the fractions, we have;
[tex]R(x)=\frac{\left(4 x^{2}+24 x\right)(x-5)}{\left(x^{2}-11 x+30\right) x^{2}}[/tex]
Let us factor out the common term 4x from the term (4x² + 24x)
Thus, we have;
[tex]R(x)=\frac{4 x(x+6)(x-5)}{\left(x^{2}-11 x+30\right) x^{2}}[/tex]
Let us factor the term (x² -11x + 30), we get;
[tex]R(x)=\frac{4(x+6)(x-5)}{x(x-5)(x-6)}[/tex]
Cancelling the common term, we have;
[tex]R(x)=\frac{4(x+6)}{x(x-6)}[/tex]
Thus, the value of R(x) is [tex]\frac{4(x+6)}{x(x-6)}[/tex]
A round ball has a diameter of 10 inches. What is the approximate volume of the ball?
Answer:
523.333333333
Step-by-step explanation:
Using the formula 4/3πr³:
(4/3)(3.14)(5^3)
Hope this helps :)
−8x+4y=24 What does X and Y equal?
−7x+7y=28
Answer:(x,y) = (-2,2)
Step-by-step explanation:
In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether the proportions have changed, a random sample of 300 students from ABC University was selected. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. Refer to Exhibit 12-4. If the proportions are the same as they were in the past, the expected frequency for the Business College is
Answer: The expected frequency for the Business college is 31.5.
Step-by-step explanation:
Since we have given that
Number of students from ABC university = 300
Number of students from Liberal Arts college = 120
Number of students in Education college = 90
Number of students in Business college = 90
Probability of students in Business college = 35% = 0.35
Probability of students in Liberal arts college = 0.35
Probability of students in Education college = 0.30
So, Expected frequency for the Business College is given by
[tex]0.35\times 90=31.5[/tex]
Hence, the expected frequency for the Business college is 31.5.
Before sending track and field athletes to the Olympics, the U.S. holds a qualifying meet.
The upper box plot shows the top 12men's long jumpers at the U.S. qualifying meet. The lower box plot shows the distances (in meters) achieved in the men's long jump at the2012 Olympic games.
Which pieces of information can be gathered from these box plots?
Choose all answers that apply:
Choose all answers that apply:
(Choice A)
A
The Olympic jumps were farther on average than the U.S. qualifier jumps.
(Choice B)
B
All of the Olympic jumps were farther than all of the U.S. qualifier jumps.
(Choice C)
C
The Olympic jumps vary noticeably more than the U.S. qualifier jumps.
(Choice D)
D
None of the above
2 horizontal boxplots titled U.S. Qualifier and Olympics are graphed on the same horizontal axis, labeled Distance, in meters. The boxplot titled U.S. Qualifier has a left whisker which extends from 7.68 to 7.7. The box extends from 7.7 to 7.89 and is divided into 2 parts by a vertical line segment at 7.74. The right whisker extends from 7.9 to 7.99. The boxplot titled Olympics has a left whisker which extends from 7.7 to 7.83. The box extends from 7.83 to 8.12 and is divided into 2 parts by a vertical line segment at 8.04. The right whisker extends from 8.12 to 8.31. All values estimated.
Answer:
a and c
Step-by-step explanation:
Looking at the pieces of information, the pieces of information that can be gathered from these box plots are:
The Olympic jumps were farther on average than the U.S. qualifier jumps.The Olympic jumps vary noticeably more than the U.S. qualifier jumps.What is Olympics?Olympics actually refers to the major international event which usually features different sports and normally holds once in every four years. It is also known as the Games of the Olympiad.
From the pieces of information given, we can actually conclude that the above options can be gathered from the information.
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Stefan’s neighborhood has a community garden. The garden has 15 equally sized rectangular plots for people to grow fruits and vegetables. Each rectangular plot measures 8 feet by 10 2 5 feet. A rectangle has a base of 8 feet and height of 10 and two-fifths feet. What is the area of each plot? Of the garden? The area of each plot is ft2. The total area of the garden is ft2.
Answer:
top 83 1/5
bottom 1,248
Step-by-step explanation:
i got it right
Answer:
Top 83 1/5
Bottom 1,248
Step-by-step explanation:
Just took it
principal: $1000, annual interest rate: 4.8%, time: 2 yr
The question involves calculating compound interest using a given principal, interest rate, and time period. The calculation is performed using the compound interest formula, which takes into account the principal amount of $1000, an annual interest rate of 4.8%, and a time span of 2 years.
Explanation:The question pertains to calculating compound interest over a certain period using a principal amount, an annual interest rate, and a given time frame. Given a principal of $1000, an annual interest rate of 4.8%, and a time period of 2 years, we can find the amount of interest accrued using the formula:
Interest = Principal imes rate imes time
Compound Interest = Principal imes (1 + interest rate)time
In this case, the formula would be:
Compound Interest = $1000 imes (1 + 0.048)2
Performing this calculation would give us the total amount after 2 years, including both the principal and the interest earned.
Which word describes the slope of the line?
Answer: the answer is zero my bad
Step-by-step explanation:
The smallest angle of a right triangle is 20-degree angle. What is the measure of the medium angle?
Answer:
70 degrees.
Step-by-step explanation:
Hello!
It says that this triangle is a right triangle. That means one angle is 90°. Since there are 180° in a triangle, all we have to do is subtract 90 + 20 from 180.
90 + 20 = 110
180 - 110 = 70.
So the medium angle is 70°.
Hope this helps!
solve for x
5x^2-x-4=0
Answer:
x = -5/4 x=1
Step-by-step explanation:
5x^2 -x -4 =0
Factor
(5x+4) (x-1)=0
Using the zero product property
5x+4 =0 x-1 =0
5x=-4 x=1
x = -5/4 x=1
Answer:
x = -0.8, 1
Step-by-step explanation:
5x² - x - 4 = 0
5x² - 5x + 4x - 4 = 0
5x(x - 1) + 4(x - 1) = 0
(5x + 4)(x - 1) = 0
x = -⅘, 1
Forester Company has five products in its inventory. Information about the December 31, 2018, inventory follows.Product Quantity UnitCost UnitReplacementCost UnitSellingPriceA 1,000 $ 10 $ 12 $ 16 B 800 15 11 18 C 600 3 2 8 D 200 7 4 6 E 600 14 12 13 The cost to sell for each product consists of a 15 percent sales commission. The normal profit percentage for each product is 40 percent of the selling price.Required:1. Determine the carrying value of inventory at December 31, 2018, assuming the lower of cost or market (LCM) rule is applied to individual products.2a. Determine the carrying value of inventory at December 31, 2018, assuming the LCM rule is applied to the entire inventory.2b. Assuming inventory write-downs are usual business practice for Forester, record any necessary year-end adjusting entry.
Answer:
See the image below.
The lower of cost or market rule states that a business must record the cost of inventory at whichever cost is lower – the original cost or its current market price.
To determine the carrying value of inventory under the lower of cost or market (LCM) rule, we compare unit cost and unit replacement cost for each product and choose the lower value. We then multiply these values with their respective quantities and sum them to get the total carrying value. For inventory write-downs, we record an adjustment entry.
Explanation:The carrying value of inventory for Forester Company at the end of December 31, 2018, involves two scenarios. In the first scenario, we apply the lower of cost or market (LCM) rule to individual products (product A through E). For each product, we compare its unit cost and unit replacement cost and choose the lower value. The total carrying value under this scenario is obtained by summing the products of the chosen values and the respective quantities.
In the second scenario, we apply the LCM rule to the entire inventory. Here, we compare the total cost and the total replacement cost of all products, and choose the lesser value as the carrying value of inventory. Inventory write-downs are typical business practice for Forester, in this case, we record an adjusting entry to recognize the loss resulting from decline in the value of inventory.
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Un padre paga la mesada a sus tres hijas de forma que a cada una le corresponde una cantidad
proporcional a su edad. A la mayor, que tiene 20 años, le da 5.000 pesos. ¿Cuánto dará a las otras
dos hijas de 15 años y 8 años de edad?
las otras dos hijas obtendrán 3750 pesos y 2000 pesos.
¿Qué es razón y proporción?Cuando dos números son divisibles se pueden escribir en la forma p:q, cuando dos razones son iguales se dice que son proporcionales.
El padre da dinero a las hijas en la proporción de su edad.
La hija mayor, que tiene 20 años, da 5.000 pesos.
La relación es 5000: 20 = 250:1
la edad de las otras hijas es 15 y 8
Que la hija de 15 años se lleve x pesos
y la hija de 8 años recibe y pesos
por lo que se forma la relación x: 15 y x: 8 que es igual a 250: 1
Comparando las dos proporciones
x/15 = 250/1
X = 15*250
x = 3750 pesos
x/8 = 250/1
x = 2000 pesos
Por lo tanto, las otras dos hijas obtendrán 3750 pesos y 2000 pesos.
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6 of 25
Camden spent a total of $599.38 for 63 meals. How much did he average on each meal?
dollars
coral beef grew 19.5 mm taller. how much did it grow in meters?
1 meter = 1,000 millimeter
We can solve this by making two proportion or millimeter over meter: [tex]\frac{millimeter}{meter}[/tex]
[tex]\frac{19.5 mm}{x meter} = \frac{1,000 mm}{1 m}[/tex]
Now you must solve for the unknown meter under the first proportion:
(1,000)(x meter) = 19.5 * 1
1,000(x meter) = 19.5
x meter = 0.0195 m
19.5 mm = 0.0195 m
Hope this helped! Let me know if you have any further questions
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A cardboard box without a lid is to have a volume of 8,788 cm3. Find the dimensions that minimize the amount of cardboard used. (Let x, y, and z be the dimensions of the cardboard box.)
Answer:
x = y = 26 cm; z = 13 cm
Step-by-step explanation:
The generic solution for the minimum material in an open-top box is that the box is square and half as high as it is wide. It is half a cube of twice the volume.
The dimensions of the square base are ...
∛(2·8788 cm³) = 26 cm
Then the height is half that, or 13 cm.
x = y = 26 cm; z = 13 cm
_____
If you need to see the development, you can use the method of Lagrange multipliers to find the minimum area for the given volume;
area = xy +2(xz +yz)
volume = xyz = 8788
We require each of the partial derivatives of L with respect to x, y, z, and λ to be zero.
L = xy +2(xz +yz) +λ(xyz -8788)
partial with respect to x: 0 = y+2z +λyz
partial with respect to y: 0 = x +2z +λxz
partial with respect to z: 0 = 2x+2y +λxy
partial with respect to λ: 0 = xyz -8788
From the first two equations, we have ...
λ = (y +2z)/(yz) = 1/z +2/y
λ = (x +2z)/(xz) = 1/z +2/x
Equating these expressions for λ, we find ...
1/z +2/y = 1/z +2/x ⇒ x = y
The third equation then tells us ...
λ = (2x +2y)/(xy) = 2/y +2/x
Comparing this to either of the first two expressions for λ, we see ...
1/z +2y = 2/x +2/y ⇒ z = x/2
This is the result we started the answer with:
x = y = 2z = ∛(2·8788 cm³)
Clark and Phil are each running to raise money. The amount of money (y), in dollars, they each raise is based on the distance (x), in miles, they each run. Clark has an initial donation that he has received regardless of how many miles he runs. The graphs model the amount of money each will raise based on the distance they each run. What is the unit rate for the person for whom the amount of money and the number of miles are proportionally related? A) $5.00 per mile B) $7.50 per mile C) $15.00 per mile D) $30.00 per mile
Answer:
Your answer is B
Step-by-step explanation:
i just did the test
Answer:$7.50
Step-by-step explanation:
In a sample of nequals16 lichen specimens, the researchers found the mean and standard deviation of the amount of the radioactive element, cesium-137, that was present to be 0.009 and 0.003 microcurie per milliliter, respectively. Suppose the researchers want to increase the sample size in order to estimate the mean mu to within 0.001 microcurie per milliliter of its true value, using a 95% confidence interval. Complete parts a through c. a. What is the confidence level desired by the researchers? The confidence level is 95. b. What is the sampling error desired by the researchers? The sampling error is 0.001. c. Compute the sample size necessary to obtain the desired estimate. The sample size is nothing. (Type a whole number.)
Answer:
(a) The confidence level desired by the researchers is 95%.
(b) The sampling error is 0.001 microcurie per millilitre.
(c) The sample size necessary to obtain the desired estimate is 36.
Step-by-step explanation:
The mean and standard deviation of the amount of the radioactive element, cesium-137 present in a sample of n = 16 lichen specimen are:
[tex]\bar x=0.009\\s=0.003[/tex]
Now it is provided that the researchers want to increase the sample size in order to estimate the mean μ to within 0.001 microcurie per millilitre of its true value, using a 95% confidence interval.
The (1 - α)% confidence interval for population mean (μ) is:
[tex]CI=\bar x \pm z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]
(a)
The confidence level is the probability that a particular value of the parameter under study falls within a specific interval of values.
In this case the researches wants to estimate the mean using the 95% confidence interval.
Thus, the confidence level desired by the researchers is 95%.
(b)
In case of statistical analysis, during the computation of a certain statistic, to estimate the value of the parameter under study, certain error occurs which are known as the sampling error.
In case of the estimate of parameter using a confidence interval the sampling error is known as the margin of error.
In this case the margin of error is 0.001 microcurie per millilitre.
(c)
The margin of error is computed using the formula:
[tex]MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
*Use a z-table.
Compute the sample size value as follows:
[tex]MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\times s}{MOE}]^{2}[/tex]
[tex]=[\frac{1.96\times 0.003}{0.001}]^{2}[/tex]
[tex]=34.5744\\\approx36[/tex]
Thus, the sample size necessary to obtain the desired estimate is 36.
The sample size necessary to estimate the mean within the desired sampling error can be calculated using the formula: n = (z * σ / E)^2, where n is the sample size, z is the z-score corresponding to the desired confidence level, σ is the standard deviation, and E is the desired sampling error. In this case, the sample size is approximately 11,447.
Explanation:To determine the sample size necessary to estimate the mean within the desired sampling error, we can use the formula:
n = (z * σ / E)^2
Where n is the sample size, z is the z-score corresponding to the desired confidence level (in this case, 1.96 for a 95% confidence level), σ is the standard deviation, and E is the desired sampling error.
Plugging in the values, we get:
n = (1.96 * 0.003 / 0.001)^2
Simplifying, we find that n is approximately 11,447. Therefore, the sample size necessary to obtain the desired estimate is 11,447 (rounded to the nearest whole number).
Learn more about Sample Size Calculation here:https://brainly.com/question/34288377
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In a poll,
327 students voted. Nominee D received two thirds of the votes. How many votes did Nominee D receive?
Answer:
218
Step-by-step explanation:
327*2/3=218
An online vendor requires that customers select a password that is a sequence of upper-case letters, lower-case letters and digits. A valid password must be at least 10 characters long, and it must contain at least one character from each of the three sets of characters. What is the probability that a randomly selected string with exactly ten characters results in a valid password? The alphabet for the strings in the sample space from which the string is drawn is the union of the three sets of characters.
Answer:
What is the probability that a randomly selected string with exactly ten characters results in a valid password? = 0.02836
Step-by-step explanation:
CHECK THE ATTACHMENT BELOOW
Final answer:
The probability of a randomly selected string with exactly ten characters resulting in a valid password can be determined using combinatorics. The probability is given by (62! / (52!*10!)) / (62^10)
Explanation:
To find the probability of a randomly selected string with exactly ten characters resulting in a valid password, we need to determine how many valid passwords exist out of all possible strings of length ten. Since the password must contain at least one character from each of the three sets (upper-case letters, lower-case letters, and digits), we can calculate the probability using combinatorics.
There are 26 upper-case letters, 26 lower-case letters, and 10 digits, so the total number of characters in the union of the three sets is 26 + 26 + 10 = 62.
The probability of selecting a valid password is then:
P(valid password) = (Number of valid passwords) / (Total number of passwords)
For a valid password, the first character can be any of the 62 characters, the second character can be any of the remaining 61 characters, and so on. Therefore, the number of valid passwords is: 62 * 61 * 60 * ... * 53.
Since there are 10 characters in a valid password, the number of valid passwords is:
62 * 61 * 60 * ... * 53 = 62! / (52!*10!).
The total number of passwords of length 10 is: 62^10.
Therefore, the probability of a randomly selected string with exactly ten characters resulting in a valid password is:
P(valid password) = (62! / (52!*10!)) / (62^10).
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is believed that the machine is underfilling of overfilling the bags. A 61 bag sample had a mean of 424 grams with a standard deviation of 26. Assume the population is normally distributed. A level of significance of 0.01 will be used. Specify the type of hypothesis test.
Answer:
[tex]t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202[/tex]
[tex]p_v =2*P(t_{(60)}>1.202)=0.234[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.
Step-by-step explanation:
Data given and notation
[tex]\bar X=424[/tex] represent the sample mean
[tex]s=26[/tex] represent the sample standard deviation
[tex]n=61[/tex] sample size
[tex]\mu_o =420[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is different from 420 or not, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 420[/tex]
Alternative hypothesis:[tex]\mu \neq 420[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=61-1=60[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(60)}>1.202)=0.234[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.