Answer:
43.62% probability that it will take more than an additional 34 minutes
Step-by-step explanation:
To solve this question, we need to understand the exponential distribution and the conditional probability formula.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
Conditional probability formula:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Taking more than 24 minutes.
Event B: Taking ore than 24+34 = 58 minutes.
P(A)
More than 24, use the exponential distribution.
Mean of 41, so [tex]m = 41, \mu = \frac{1}{41} = 0.0244[/tex]
[tex]P(A) = P(X > 24) = e^{-0.0244*24} = 0.5568[/tex]
Intersection:
More than 24 and more than 58, the intersection is more than 58. So
[tex]P(A \cap B) = P(X > 58) = e^{-0.0244*58} = 0.2429[/tex]
Then:
[tex]P(B|A) = \frac{0.2429}{0.5568} = 0.4362[/tex]
43.62% probability that it will take more than an additional 34 minutes
In the growth equation y = 3(1.2)^5 what is the rate of increase?
The rate of increase is 1.2 as it represents b in the equations A+b power of x+c then +D.
Answer:
y=7.46496
Step-by-step explanation:
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.
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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For a healthy human, a body temperature follows a normal distribution with Mean of 98.2 degrees Fahrenheit and Standard Deviation of 0.26 degrees Fahrenheit. For an individual suffering with common cold, the average body temperature is 100.6 degrees Fahrenheit with Standard deviation of 0.54 degrees Fahrenheit. Simulate 10000 healthy and 10000 unhealthy individuals and answer questions 14 to 16. 14. If person A is healthy and person B has a cold, which of the events are the least likely
Complete Question:
For a healthy human, a body temperature follows a normal distribution with Mean of 98.2 degrees Fahrenheit and Standard Deviation of 0.26 degrees Fahrenheit. For an individual suffering with common cold, the average body temperature is 100.6 degrees Fahrenheit with Standard deviation of 0.54 degrees Fahrenheit. Simulate 10000 healthy and 10000 unhealthy individuals and answer questions 14 to 16.
14. If person A is healthy and person B has a cold, which of the events are the most likely? Pick the closest answer.
a. Person B will have higher temperature than 101 degrees.
b. Person A will have temperature higher than 99 degrees
c. Person B will have temperature lower than 100 degrees
d. Person A will have temperature lower than 97.5 degrees
15. What would be a range [A to B], which would contain 68% of healthy individuals? Pick the closest answer.
a. Between 97.9 and 98.46
b. Between 99.5 and 101.6
c. Between 100.06 and 101.14
d. Between 100.1 and 102.2
16. What is the approximate probability that a randomly picked, unhealthy individual (one with the cold) would have body temperature above 101 degrees Fahrenheit? Pick the closest answer.
a. About 10%
b. About 15%
c. About 22%
c. About 34%
Answer:
14. option a
15. option a
16. option a
explanation:
14 option (a)
person B have higher temperature than 101 degrees,
(b) cannot be true if person a has higher temperature than 99 then he has cold most likely and as mentioned by the standard dev and mean of healthy people
(c) AND (d) also cannot be true as they also do not satisfy the given standard dev and mean
15 option (a) because of the A to B range and not B to A range ....... as the healthy person's standard dev is 0.26. Hence the 68% data will lie in 97.94 - 98.46
but, if it was B to A, then the (c) option will be true
16 option (a) because most of the unhealthy individuals above the 0.54 standard dev will come somewhat near 90%
Which word describes the slope of the line?
Answer: the answer is zero my bad
Step-by-step explanation:
6 of 25
Camden spent a total of $599.38 for 63 meals. How much did he average on each meal?
dollars
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
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
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 number between 0 and 1, expressed as one number over another, is called a _ _ _ _ _ _ _ _.
Answer:
Fraction
Step-by-step explanation:
Answer:
fraction
Step-by-step explanation:
pls mark brainliest
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.
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.
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]
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 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!!!
A random sample of 340 people in Chicago showed that 66 listened to WJKT-1450, a radio station in South Chicago Heights. Based on this sample information, what is the point estimate for the proportion of people in Chicago that listen to WJKT-1450? Question 29 options: 1450 340 About 0.194 66
Answer:
Point of estimate of the true Proportion 'p' = 0.1941
Step-by-step explanation:
Explanation:-
Given data a random sample of 340 people in Chicago showed that 66 listened to WJKT-1450, a radio station in South Chicago Heights.
Given sample size 'n'= 340
Point of estimate of the true Proportion P = [tex]\frac{x}{n}[/tex]
Properties of Estimation
An estimator is not expected to estimate the Population parameter without error.
An estimator should be close to the true value of un-known parameter.
Point of estimate of the true Proportion = [tex]p = \frac{66}{340}= 0.1941[/tex]
Conclusion:-
Point of estimate of the true Proportion = 0.1941
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]
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).
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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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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)
For a community service project, of the students in a class bring paper towels 3/4 2/5 of the students bring towels and soap What fraction of the students who bring paper towels also bring soap ? es
Step-by-step explanation:
For a community service project, the no of the students in a class bring paper towels = [tex]\frac{3}{4}[/tex]
The students bring towels and soap = [tex]\frac{2}{5}[/tex]
The students who bring paper towels = [tex]\frac{3}{4}[/tex]
The fraction of students who bring paper towels also bring soap = [tex]\frac{2}{5}[/tex]
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
What is the g?
-15 - g/3 = -5
Answer:
g = - 30
Step-by-step explanation:
[tex] - 15 - \frac{g}{3} = - 5 \\ - \frac{g}{3} = - 5 + 15 \\ - \frac{g}{3} = 10 \\ - g = 10 \times 3 \\ - g = 30 \\ \huge \red{ \boxed{g = - 30}}[/tex]
The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average
percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that
the percentage of SiO2 in a sample is normally distributed with (sigma) = 0.3 and that x-bar = 5.25.
a) Does this indicate conclusively that the true average percentage differs from 5.5?
b) If the true average percentage is (mu) = 5.6 and a level a = 0.01 test based on n = 16 is used,
what is the probability of detecting this departure from H0?
Answer:
A. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.
B. P(Z<-4.93333) = 0.0000
C. n = 5945 samples
Step-by-step explanation:
A)
The rightnull and alternative hypotheses are given as below:
Null hypothesis H0: µ = 5.5
Alternative hypothesis Ha: µ ≠ 5.5
Conducting the one sample z test for population mean.
Test statistic formula is given as below:
Z = (Xbar - µ)/[σ/√(n)]
Given,
Xbar = 5.23
µ = 5.5
σ = 0.30
n = 16
Z = (5.23 – 5.5) / [ 0.30 /√(16)]
Z = -3.6000
P-value = 0.0003
α value = 0.05
P-value < α value
So, we reject the null hypothesis. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.
B
From the information given
Xbar = 5.23
µ = 5.6
σ = 0.30
n = 16
α value = 0.01
Z = (Xbar - µ)/[σ/√(n)]
Z = (5.23 - 5.6)/(0.30/√(16))
Z = -4.93333
P(Z<-4.93333) = 0.0000
Required probability = 0.0000
C
We are given α =0.01
Therefore, critical Z value = 2.57
E = 0.01
σ = 0.30
n = (Z*σ/E)^2
n = (2.57*0.30/0.01)^2 = 5944.41
n = 5945 samples
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:
The diameter of the sun is 1.391 million kilometers. Represent this number in scientific notation.
Answer:
1391000000
Step-by-step explanation:
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
~ Just a girl in love with Shawn Mendes
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.
Learn more about Olympic Games on https://brainly.com/question/10409012
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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 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.
Learn more about data on:
https://brainly.com/question/19243813