Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
Explanation:The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.
The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.
In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
Learn more about Empirical Rule here:https://brainly.com/question/35669892
#SPJ3
Department 1 of a two department production process shows: Units Beginning Work in Process 9900 Ending Work in Process 49000 Total units to be accounted for 180200 How many units were transferred out to Department 2?
Answer:
131200 Units
Step-by-step explanation:
Now, in the production process:
Units to be accounted for= Beginning Work in Process+Work Started
Units Accounted for=Ending Work in Process + Completed and Transferred out
Units to be accounted for = Units Accounted for
Beginning Work in Process =9900 Ending Work in Process = 49000 Total units to be accounted for= 180200
Since
Units to be accounted for = Units Accounted for
Then:
Units Accounted for=Ending Work in Process + Completed and Transferred out
180200=49000+Completed and Transferred out
Completed and Transferred out=180200-49000
=131200
131200 units were transferred to Department 2.
The number of units transferred out to Department 2 can be found by subtracting the Ending Work in Process from the Total units to be accounted for. In this case, that would be 180200 - 49000, which equals 131200 units.
Explanation:In the scenario described, the number of units transferred out to Department 2 can be calculated by deducting the Ending Work in Process from the Total units to be accounted for. This is because the Total units to be accounted for includes all units at the start (Beginning Work in Process), all units processed during the period, and all units still in process at the end (Ending Work in Process). So, the units transferred to Department 2 were processed by Department 1 but are not part of Department 1's ending work in process.
Hence, using the provided numbers:
Total units to be accounted for = 180200Ending Work in Process = 49000
Units transferred out to Department 2 = Total units to be accounted for - Ending Work in Process = 180200 - 49000 = 131200
Learn more about Business Operations here:https://brainly.com/question/33727098
#SPJ3
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
7.3 6.8 8.6 9.7 11.6 9.7 6.8 7.7 11.8 7.4 8.1 8.7 6.3 7.9 7.0 7.9 7.8 6.5 6.3 7.0 7.5 7.7 7.2 11.3 9.0 10.7 5.1
(a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. [Hint: Σxi = 219.4.] (Round your answer to three decimal places.) State which estimator you used.
a) s b) x c) p? d) x tilde e) s / x
(b) Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50%. State which estimator you used.
a) s b) x c) p? d) x tilde e) s / x
(c) Calculate a point estimate of the population standard deviation σ. [Hint: Σxi2 = 1858.92.] (Round your answer to three decimal places.)
Interpret this point estimate.
a. This estimate describes the center of the data.
b. This estimate describes the bias of the data.
c. This estimate describes the linearity of the data.
d. This estimate describes the spread of the data.
Answer:
Step-by-step explanation:
Hello!
The variable of interest is X: flexural strength of concrete beams.
Data:
7.3 6.8 8.6 9.7 11.6 9.7 6.8 7.7 11.8 7.4 8.1 8.7 6.3 7.9 7.0 7.9 7.8 6.5 6.3 7.0 7.5 7.7 7.2 11.3 9.0 10.7 5.1
a.
The point estimate of the mean value is the sample mean. You calculate it using the following formula:
X[bar]= ∑X/n= 219.40/27= 8.1259≅ 8.13
b)x
b.
The point estimate that divides the sample in 50%-50% is the measure of central tendency called Median(Me).
To reach the median you have to calculate it's position (PosMe)
For uneven samples PosMe= (n+1)/2= (27+1)/2= 14 ⇒ This means that the Median is the 14th observation of the data set.
So next is to order the data from lower to heighest and identify the value in the 14th position:
5,1 ; 6,3; 6,3; 6,5 ; 6,8; 6,8; 7; 7; 7,2; 7,3; 7,4 ; 7,5 ; 7,7 ; 7,7 ; 7,8 ; 7,9 ; 7,9 ; 8,1 ; 8,6 ; 8,7 ; 9 ; 9,7 ; 9,7 ; 10,7 ; 11,3 ; 11,6 ; 11,8
Me= 7.7
d) x tilde
c.
The point estimate of the population standard deviation is the sample standard deviation. The standard deviation is the square root of the variance, so the first step is to calculate the sample variance:
[tex]S^2= \frac{1}{n-1}*[sumX^2-\frac{(sumX)^2}{n} ][/tex]
∑X²= 1858.92
[tex]S^2= \frac{1}{26}*[1858.92-\frac{219.40^2}{27} ] = 2.926[/tex]
S= √2.926= 1.71
The standard deviation is a measurement of variability, it shows how to disperse the data is concerning the sample mean.
d. This estimate describes the spread of the data.
I hope it helps!
(a) Point estimate of mean strength: [tex]\( \bar{x} \) ≈ 8.118 MPa.[/tex] (d)
(b) Point estimate of separating strength: 7.7 MPa.
(d) (c) d) Spread of data.
(a) To calculate the point estimate of the mean value of strength, we'll use the sample mean estimator, denoted as [tex]\( \bar{x} \)[/tex].
Step 1:
Sum the data points:[tex]\( \sum{x_i} = 219.4 \)[/tex]
Step 2:
Divide the sum by the number of data points (n = 27): [tex]\( \bar{x} = \frac{\sum{x_i}}{n} = \frac{219.4}{27} \approx 8.118 \)[/tex] (rounded to three decimal places).
So, the point estimate of the mean strength is approximately 8.118 MPa.
(b) To find the strength value that separates the weakest 50% from the strongest 50%, we'll use the sample median estimator, denoted as [tex]\( \tilde{x} \).[/tex]
Step 1:
Arrange the data in ascending order.
Step 2:
Since there are 27 data points, the median will be the value of the 14th data point. Counting from the lowest value, the 14th value is 7.7.
So, the point estimate of the separating strength value is 7.7 MPa.
(c) To estimate the population standard deviation, we'll use the sample standard deviation estimator, denoted as (s).
Step 1:
Use the formula[tex]\( s = \sqrt{\frac{\sum{x_i^2} - \frac{(\sum{x_i})^2}{n}}{n-1}} \).[/tex]
Step 2:
Substitute the given values:[tex]\( s = \sqrt{\frac{1858.92 - \frac{(219.4)^2}{27}}{26}} \approx 1.624 \)[/tex](rounded to three decimal places).
This estimate describes the spread of the data.
So, the answers are:
(a) d)[tex]\( \bar{x} \)[/tex]
(b) d)[tex]\( \tilde{x} \)[/tex]
(c) d) This estimate describes the spread of the data.
A recent study reported that 28% of the residents of a particular community lived in poverty. Suppose a random sample of 300 residents of this community is taken. We wish to determine the probability that 33% or more of our sample will be living in poverty. Complete parts (a) and (b) below.
Before doing any calculations, determine whether this probability is greater than 50% or less than 50%. Why? The answer should be less than 50%, because the resulting z-score will be negative and the sampling distribution is approximately Normal. The answer should be greater than 50%, because 0.24 is greater than the population proportion of 0.20 and because the sampling distribution is approximately Normal. The answer should be less than 50%, because 0.24 is greater than the population proportion of 0.20 and because the sampling distribution is approximately Normal. The answer should be greater than 50%, because the resulting z-score will be positive and the sampling distribution is approximately Normal. Calculate the probability that 24% or more of the sample will be living in poverty. Assume the sample is collected in such a way that the conditions for using the CLT are met. P (p ge 0.24) = (Round to three decimal places as needed.)
Answer:
a) The answer should be less than 50%, because 0.33 is greater than the population proportion of 0.28 and because the sampling distribution is approximately Normal.
b) P(x ≥ 0.33) = 2.68% = 0.027 to 3 d.p
Step-by-step explanation:
a) The required probability is P(x ≥ 0.33)
The population proportion of people living in poverty has already been given as 0.28, So, whatever the standard deviation of the distribution of sample means is, the sample proportion of 0.33 is more than the population proportion, So, it gives a z-score that is greater than 0. The probability from the z-score of the mean (0) to the end of distribution is exactly 50%; So, a Probability that only covers from a particular positive z-score to the end of the distribution will definitely be less than 50%.
So, because 33% is more than population proportion of 28%, and the sampling distribution is approximately normal, the probability of 33% or more of the sample living in poverty is less than 50%.
b) P(x ≥ 0.33)
The sample mean = population mean
μₓ = μ = 0.28
Standard deviation of the distribution of sample means = √[p(1-p)/n] (this is possible due to the central limit theorem for n greater than 30)
σₓ = √[(0.28×0.72)/300] = 0.0259
We then normalize 0.33
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (0.33 - 0.28)/0.0259 = 1.93
To determine the probability of 33% or more of the sample size is living in poverty.
P(x ≥ 0.33) = P(z ≥ 1.93)
We'll use data from the normal probability table for these probabilities
P(x ≥ 0.33) = P(z ≥ 1.93) = 1 - P(z < 1.93)
= 1 - 0.9732 = 0.0268 = 2.68%
It is indeed way less than 50%.
Hope this Helps!!!
The answer to whether the probability is greater than 50% or less than 50% is; less than 50% because 0.33 is greater than the population proportion
What is the probability of the normal distribution?A) We want to determine the probability that 33% or more of our sample will be living in poverty and this is expressed as P(x ≥ 0.33)
Population proportion is; p = 0.28.
Formula for z-score is;
z = (x' - μ)/σ
Since our sample proportion is greater than our population proportion, it means the z-score will be greater than 0
Finally, due to the fact that 33% is more than population proportion of 28%, and the sampling distribution is approximately normal, the probability of 33% or more of the sample living in poverty will definitely be less than 50%.
b) We want to now calculate the above probability;
P(x ≥ 0.33)
The sample mean will be equal to population mean as;
μₓ = μ = 0.28
Standard deviation of the distribution of sample means is;
σₓ = √(p(1 - p)/n)
σₓ = √((0.28×0.72)/300)
σₓ = 0.0259
Thus, z-score is;
z = (x - μ)/σ
z = (0.33 - 0.28)/0.0259
z = 1.93
Thus, the of 33% or more of the sample size is living in poverty is;
P(x ≥ 0.33) = P(z ≥ 1.93) = 1 - P(z < 1.93)
P(x ≥ 0.33) = 1 - 0.9732
P(x ≥ 0.33) = 0.0268
P(x ≥ 0.33) = 0.027
It is indeed way less than 50%.
Read more about normal distribution at; https://brainly.com/question/4079902
The research question below describes a relationship between two quantitative variables. Which variable should be plotted on the horizontal X-axis? Is the sales price of a townhouse in San Francisco related to the number of square feet in the townhouse?
Answer:
number of square feet in the townhouse.
Step-by-step explanation:
When the relationship of two quantitative variables is assessed through plot then the scatter plot is made. Scatter plot shows relationship between two qualitative variables by showing dependent variable on y axis and independent variable on x axis.
When the relation between sale price of house and number of square feet in house is assessed then the sale price is dependent variable and square feet in house. Sale price is dependent variable because the price of house depends on the number of square feet of house. Thus, number of square feet in the town house will be plotted on the horizontal X-axis.
Use seq() to generate a sequence 1, 3, ..., 27. (b) Use log() to generate a new sequence where each element is log-transformed from the sequence in (a). (c) Remove the second to fifth elements in the resulting sequence in (b). (d) Use length() to obtain the length of the resulting sequence in (c). (e) Sort the resulting sequence in (c) from high to low using sort().
Answer:
Step-by-step explanation:
sequence=seq(1,27, by=3)
new_seq=c()
for (i in sequence){
n_seq.append(log(i))
}
new_seq<- n_seq[-(2:5)]
seq_length=length(new_seq)
sorted_seq=sort(new_seq,decreasing=TRUE)
A super has a key ring with 12 keys. He has forgotten which one opens the apartment he needs to enter. What is the probability he is able to enter before reaching the 4th key?
Answer:
Therefore, the probability is P=1/4.
Step-by-step explanation:
We know that a super has a key ring with 12 keys. He has forgotten which one opens the apartment he needs to enter.
We conclude that each key has an equal probability of opening an apartment. Since there are 12 keys, it follows that the probability for each key is equal,
p = 1/12.
We calculate the probability he is able to enter before reaching the 4th key. So he will try three keys by then. We get:
[tex]P=p+p+p\\\\P=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\\\\P=\frac{3}{12}\\\\P=\frac{1}{4}[/tex]
Therefore, the probability is P=1/4.
Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. Summary statistics yield Upper X overbarequals180.975 and Sequals143.042. Calculate a 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments.
Answer:
[tex]180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370[/tex]
[tex]180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580[/tex]
So on this case the 95% confidence interval would be given by (61.370;300.580)
Step-by-step explanation:
Previous concepts
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".
Solution to the problem
[tex]\bar X=180.975[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=143.042[/tex] represent the sample standard deviation
n=8 represent the sample size
The confidence interval on this case is given by:
[tex]\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}} [/tex] (1)
We can find the degrees of freedom and we got:
[tex] df = n-1= 8-1=7[/tex]
The next step would be find the value of [tex]\t_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]
Using the t table with df =7, excel or a calculator we see that:
[tex]t_{\alpha/2}=2.365[/tex]
Since we have all the values we can replace:
[tex]180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370[/tex]
[tex]180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580[/tex]
So on this case the 95% confidence interval would be given by (61.370;300.580)
A new article reported that college students who have part-time jobs work an average of 15 hour per week. The staff of a college from for newspaper thought that the average might be different from 15 hours per week for their college. Data were collected on the number of hours worked per week for a random sample of students at the college who have part-time jobs. The data were used to test the hypotheses H_o: mu = 15 H_a: mu notequalto 15. where mu is the mean number of hours worked per week for all students at the college with part-time jobs. The results of the test are shown in the table below. Assuming all conditions for inference were met, which of the following represents a 95 percent confidence interval for mu?
(A) 13.755 plusminus 0.244
(B) 13.755 plusminus 0.286
(C) 13.755 plusminus 0.707
(D) 13.755 plusminus 1.245
E) 13, 755 plusminus 1.456
Answer:
E (13.755 ± 1.456)
Step-by-step explanation:
with 95% confidence interval and df=25
we can find t score on the t table : t*=2.06
(because the table has already provided us standard error , we DON'T have to calculate it by using σsample=S/[tex]\sqrt{n}[/tex]) (standard error = standard deviation)
so the interval should be 13.755± 2.06(0.707)= 13.755± 1.456
The 95% confidence interval for [tex]\(\mu\)[/tex] is e. [tex]\(13.755 \pm 1.456\)[/tex], based on a sample mean of 13.755, standard error of 0.707, and 25 degrees of freedom.
To construct a 95% confidence interval for the mean [tex]\(\mu\)[/tex], we need to use the sample mean, the standard error, and the appropriate critical value for a t-distribution.
Given data:
- Sample mean [tex](\(\bar{x}\))[/tex]: 13.755
- Standard error (SE): 0.707
- Degrees of freedom (df): 25
For a 95% confidence interval with 25 degrees of freedom, we need the critical value [tex]\( t^* \).[/tex] We can find [tex]\( t^* \)[/tex] using a t-table or statistical software.
For 25 degrees of freedom, the critical value [tex]\( t^* \)[/tex] for a 95% confidence interval is approximately 2.064.
The formula for the confidence interval is:
[tex]\[\bar{x} \pm t^* \times \text{SE}\][/tex]
Substitute the values:
[tex]\[13.755 \pm 2.064 \times 0.707\][/tex]
Calculate the margin of error:
[tex]\[2.064 \times 0.707 \approx 1.459\][/tex]
Thus, the 95% confidence interval is:
[tex]\[13.755 \pm 1.459\][/tex]
So, the correct answer is:
e. [tex]\[\boxed{13.755 \pm 1.456}\][/tex]
Complete Question:
18. Brown eyes (B) are dominant to blue eyes (b). The presence of freckles (F) is dominant to the absence of freckles (f). What is the probability that two people who are heterozygous for both traits will have a child that has blue eyes and no freckles?
Answer:
For bbff we have only 6.3% probability
Step-by-step explanation:
If the parents are heterozygous for both traits, them they are represented by:
BbFf × BbFf
Parent 1: BbFf
Parent 2: BbFf
We have to find the percentage of occurence of bb × ff, which is a child that has blue eyes and no freckles, with no dominant factor.
By distributing the possibilities in a Punnett square, vide picture. We have the following possibilities:
Genotype Count Percent
bBfF 4 25
BBfF 2 12.5
bBFF 2 12.5
bBff 2 12.5
bbfF 2 12.5
BBFF 1 6.3
BBff 1 6.3
bbFF 1 6.3
bbff 1 6.3
For bbff we have only 6.3% probability
Two balanced dice are rolled. Let X be the sum of the two dice. Obtain the probability distribution of X (i.e. what are the possible values for X and the probability for obtaining each value?). Check that the probabilities sum to one.
1) What is the probability for obtaining X \geq 8?
2) What is the average value of X?
Answer:
Step-by-step explanation:
Each die has faces numbered from 1 to 6. Hence, the sums range from 2 to 12.
There are 6 × 6 = 36 possible combinations of the dice.
The sums, their possible combinations and their probabiities are as below:
2 = 1+1 1/36
3 = 1+2 = 2+1 2/36= 1/18
4 = 1+3 = 2+2 = 3+1 3/36= 1/12
5 = 1+4 = 2+3 = 3+2 = 4+1 4/36= 1/9
6 = 1+5 = 2+4 = 3+3 = 4+2 = 5+1 5/36
7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 6/36= 1/6
8 = 2+6 = 3+5 = 4+4 = 5+3 = 6+2 5/36
9 = 3+6 = 4+5 = 5+4 = 6+3 4/36= 1/9
10 = 4+6 = 5+5 = 6+4 3/96= 1/12
11 = 5+6 = 6+5 2/36= 1/18
12 = 6+6 1/36
Sum of probabilities = 1/36 + 1/18 + 1/12 + 1/9 + 5/36 + 1/6 + 5/36 + 1/9 + 1/12 + 1/18 + 1/36 = 1
1) [tex]P(X\ge 8) = 5/36+1/9+1/12+1/18+1/36=15/36[/tex]
2) The average value of X = 1/36*(2+12) + 1/18*(3+11) + 1/12*(4+10)+ 1/9(5+9)+ 5/36(6+8)+ 1/6(7) = 14*(5/6)+7/6 = 77/6
Which of the following meets the requirements of a stratified random sample? Multiple Choice A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six people who volunteer for the sample. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six people chosen at random, without regard to age. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six males chosen at random, without regard to age. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include two people chosen at random under the age of 25 and four people chosen at random over 25.
Answer:
Correct option is D.
Step-by-step explanation:
Random sampling implies the selection of of values or individuals in a random pattern.
A stratified random sampling is a sampling method where:
First divide the entire population into homogeneous subgroups, known as strata.Then take a random sample form each of the strata such that the sample size is proportional to the size of the strata.In this case the it is provided that the population consists of 30 members, 10 under the age of 25 and 20 over the age of 25.
So the stratas are:
10 members under the age of 25.20 members over the age of 25.Now to a random sample of size 2 is selected from strata 1 and a random sample of size 4 is selected from strata 2.
This forms a stratified random sample.
Correct option is:
A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include two people chosen at random under the age of 25 and four people chosen at random over 25.
Final answer:
The correct choice for a stratified random sample is selecting two people randomly under the age of 25 and four people randomly over 25, as it ensures that the sample is proportional and representative of the two age groups within the population.
Explanation:
To determine which option meets the requirements of a stratified random sample, let's first define it. A stratified random sample divides the population into separate groups, known as strata, and a random sample is taken from each group. The key point here is that the sample from each stratum is proportional to the size of the stratum within the whole population, ensuring the sample is representative.
Given the multiple-choice options provided:
Sampling based on volunteers is a convenience sample, not stratified.Choosing people at random without regard to age is simple random sampling.Selecting only males is not stratified, as it does not account for another relevant characteristic, such as age.Selecting two people randomly under the age of 25 and four people randomly over 25 is the correct choice, as it ensures the sample is representative of the two age groups within the population.This type of sampling accounts for the different proportions of the sub-groups (under and over 25) in the population, which is a key aspect of stratified sampling.
find the diameter of a circle with an area of 74 square millimeters
Answer:
The diameter of the circle is 9.7mm
Step-by-step explanation:
The area of a circle is given by the following formula:
[tex]A_{c} = \pi r^{2}[/tex]
In which r is the radius.
In this problem, we have that
[tex]A_{c} = 74 mm^{2}[/tex]
So
[tex]74 = \pi r^{2}[/tex]
[tex]r = \sqrt{\frac{74}{\pi}}[/tex]
[tex]r = 4.85mm[/tex]
The diameter is twice the radius:
So
[tex]d = 2r = 2*4.85 = 9.7[/tex]
The diameter of the circle is 9.7mm
Final answer:
To find the diameter of a circle with an area of 74 square millimeters, use the area formula A = πr², solve for r, then double the result to get the diameter. The estimated diameter is approximately 9.71 mm.
Explanation:
The question asks us to find the diameter of a circle with an area of 74 square millimeters. We use the formula for the area of a circle, which is A = πr² where A is the area and r is the radius. To find the diameter, we need to first solve for the radius, and then multiply by 2 since the diameter is twice the length of the radius.
Steps to Find the Diameter:
Write down the area formula: A = πr².Rearrange the formula to solve for the radius: r = √(A/π).Substitute the given area of 74 mm² into the formula and solve for r.Once the radius is found, multiply by 2 to get the diameter: D = 2r.Use a calculator to compute the values, remembering to square the calculation's output back to the same number of significant figures as the original data.Assuming π is approximately 3.14, we can estimate:
A = 74 mm² = πr²
r = √(74/π) mm
r ≈ √(74/3.14) mm ≈ √(23.567) mm ≈ 4.855 mm
The radius is approximately 4.855 mm, so the diameter would be 2 × 4.855 mm ≈ 9.71 mm.
A company services home air conditioners. It is known that times for service calls follow a normal distribution with a mean of 60 minutes and a standard deviation of 10 minutes. A random sample of eight service calls is taken. What is the probability that exactly two of them take more than 68.4 minutes
Answer:
[tex]P(y = 2) = 0.294[/tex]
Step-by-step explanation:
If the times for service calls follow a normal distribution with mean μ = 60 minutes and standar deviation σ = 10 minutes, then we can calculate the probability that one call take more than 68.4 minutes, thus:
P(x>68.4) = 1 - P(X<68.4)
We standardize as follow
[tex]z = \frac{x-\mu}{\sigma}[/tex]
[tex]z = \frac{68.4 - 60}{10} = 0.84[/tex]
Since the table of the distribution normal, we obtain:
1- P(z <0.84) = 1 - 0.7995 = 0.2005
Therefore:
P(x>68.4) = 0.2005
Now if you select random eight service calls, the number of calls takes more than 68.4 minutes is a binomial variable with
p = 0.2005
n = 8
And we need calculate this probability
[tex]P(y = 2) = (8C2)(0.2005)^{2}(1-2005)^{8-2}[/tex]
[tex]P(y = 2) = 0.294[/tex]
To find the probability that exactly two of the service calls take more than 68.4 minutes, we can use the properties of the normal distribution and the binomial distribution. First, we find the probability of a single service call taking more than 68.4 minutes using the z-score. Then, we use the binomial distribution to find the probability of exactly two service calls taking more than 68.4 minutes in a random sample of eight service calls.
Explanation:To solve this problem, we will use the properties of the normal distribution. We know that the mean service time is 60 minutes and the standard deviation is 10 minutes. We need to find the probability that exactly two of the service calls take more than 68.4 minutes. First, we need to find the probability that a single service call takes more than 68.4 minutes. Using the z-score formula, we find that the z-score for 68.4 minutes is (68.4 - 60) / 10 = 0.84. Looking up this value in the standard normal distribution table, we find that the probability of a service call taking more than 68.4 minutes is approximately 0.2019. Since we have a random sample of eight service calls, we can use the binomial distribution to find the probability of exactly two service calls taking more than 68.4 minutes. The formula for this is P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where n is the sample size, k is the number of successes, p is the probability of success, and C(n, k) is the binomial coefficient. Plugging in the values from our problem, we have P(X = 2) = C(8, 2) * 0.2019^2 * (1-0.2019)^(8-2). Evaluating this expression, we find that the probability of exactly two service calls taking more than 68.4 minutes is approximately 0.2262.
Learn more about Normal distribution here:https://brainly.com/question/34741155
#SPJ11
Consider three independent rolls of a fair six-sided die. (a) What is the probability that the sum of the three rolls is 11? (b) What is the probability that the sum of the three rolls is 12? (c) In the seventeenth century, Galileo explained the experimental observation that a sum of 10 is more frequent than a sum of 9, even though both 10 and 9 can be obtained in six distinct ways. Can you retrace Galileo’s thinking?
Answer:
a) 1/8 b) 1/12 c) probability of obtaining sum of 9 in a single roll of die without having the same number repeated twice is less than than of sum of 10
Step-by-step explanation:
Probability of any digit in a single roll = 1/6
a) sum of 11 will be obtained if each roll has the following result
1 and 4 and 6, 1 and 5 and 5, 1 and 6 and 4, 4 and 1 and 6, 4 and 6 and 1, 6 and 4 and 1, 6 and 1 and 4, 5 and 1 and 5, 5 and 5 and 1, 2 and 5 and 4, 2 and 4 and 5, 4 and 2 and 5, 4 and 5 and 2, 5 and 2 and 4, 5 and 4 and 2, 2 and 6 and 3, 2 and 3 and 6, 3 and 2 and 6, 3 and 6 and 2, 6 and 3 and 2, 6 and 2 and 3, 3 and 3 and 5, 3 and 5 and 3, 5 and 3 and 3, 3 and 4 and 4, 4 and 3 and 3, 3 and 4 and 3
27(1/6 × 1/6 × 1/6)= 1/8
b) 2 and 4 and 6, 2 and 5 and 5, 4 and 2 and 6, 4 and 6 and 2, 6 and 4 and 2, 6 and 2 and 4, 5 and 2 and 5, 5 and 5 and 2, 2 and 6 and 4, 3 and 3 and 6, 3 and 6 and 3, 6 and 3 and 3, 3 and 4 and 5, 3 and 5 and 4, 4 and 3 and 5, 4 and 5 and 3, 5 and 3 and 4, 5 and 4 and 3
18(1/6 × 1/6 × 1/6)= 1/12
c) six ways of obtaining 10: 3+1+6, 3+2+5, 3+3+4, 4+2+4,4+5+1, 6+2+2
six ways of obtaining 9: 1+2+6, 1+3+5, 1+4+4, 2+2+5, 2+3+4,3+3+3
To get 10, there are only two ways of repeating a number out of 6 ways. To get 9, there are three ways of repeating a number out of 6 ways
The probability of sums from rolls of dice is based on the possible combinations of outcomes, where each side of a dice has a 1/6 chance. For three independent rolls, specific sums like 11 or 12 would require identifying all possible combinations that achieve these sums. Galileo's observation relates to the unequal probabilities of different number combinations despite having the same number of combinations for sums of 9 and 10.
Explanation:The probabilities for the sums of dice rolls are calculated based on the number of ways those sums can be achieved using the possible outcomes of each individual roll. For a fair six-sided die, the probability of rolling any given number is 1/6.
(a) To find the probability that the sum of three rolls is 11, we would identify all the combinations of rolls that result in that sum and calculate the likelihood of each. Possible combinations for a sum of 11 include (5,5,1), (5,4,2), (6,4,1), etc. Each of the three dice has a 1/6 chance of landing on a specified number, and since each roll is independent, the probabilities are multiplied.
(b) Similarly, for a sum of 12, possible combinations include (4,4,4), (6,5,1), (6,4,2), etc.
(c) As for Galileo's observation about sums of 10 being more frequent than 9, it is important to note that while both sums have six distinct combinations, some combinations are more likely than others because rolls are independent and not all number combinations are equally probable. This leads to a higher overall probability for a sum of 10.
You bike
2
miles
miles the first day of your training,
2.5 miles the second day,
3.5 miles the third day, and
5.5 miles the fourth day. If you continue this pattern, how many miles do you bike the seventh day?
Answer:
7.9
Step-by-step explanation:
A survey is conducted of students enrolled during both the Winter 2013 and Winter 2014 semesters. 12% of students got the flu during Winter 2013. 18% of students came got the flu during Winter 2014. 5% of students got the flu during both years. What is the probability that a randomly selected survey participant didnt?
Answer:
The probability that a randomly selected survey participant didn't got the flew = 0.75
Step-by-step explanation:
Students got the flu during Winter 2013 = 12 %
Students got the flu during Winter 2014 = 18 %
Students got the flu during both years = 5 %
Students got the flu at least in one year = 12 + 18 - 5 = 25 %
Students didn't got the flew = 100 - 25 = 75 %
⇒ The probability that a randomly selected survey participant didn't got the flew = [tex]\frac{75}{100}[/tex]
⇒ 0.75
This is the probability that a randomly selected survey participant didn't got the flew.
A group of students bakes 100 cookies to sell at the school bake sale. The students want to ensure that the price of each cookie offsets the cost of the ingredients. If all the cookies are sold for $0.10 each, the net result will be a loss of $4. If all the cookies are sold for $0.50 each. The students will make a $36 profit. First, write the linear function p(x) that represents the net profit from selling all the cookies, where x is the price of each cookie. Then, determine how much profit the students will make if they sell the coolies for $0.60 each. Explain. Tell how your answer is reasonable.
Answer:
(a) [tex]p=100x-14[/tex]
(b) [tex]p=\$46[/tex]
Step-by-step explanation:
Linear Modeling
It consists of finding an equation of a line that fits the conditions of a certain situation in real life. We'll use a linear model for the cookies of the students.
(a) We know that we have a total of n=100 cookies. If sold for $0.10 each, they lose $4. We have an initial condition (x,p) = (0.10,-4), where x is the price of each cookie and p(x) is the net profit from selling all the cookies. The second conditions are that when then the price is $0.50 each, there is a positive profit of $36, which is a second point (0.5,36). That is enough to build the linear function, that can be found by
[tex]\displaystyle p-p_1=\frac{p_2-p_1}{x_2-x_1}(x-x_1)[/tex]
[tex]\displaystyle p+4=\frac{36+4}{0.5-0.1}(x-0.1)[/tex]
Reducing
[tex]p=100x-14[/tex]
(b) If the students sell the cookies for x=0.60 each, the profit will be
[tex]p=100(0.6)-14=46[/tex]
[tex]p=\$46[/tex]
It's a reasonable answer because we have found that increasing the price, the profit will increase also. The model doesn't have any restriction for the price
A developmental psychologist is i nterested in assessing t he "" emotional i ntelligence"" o f co llege s tudents. The e xperimental des ign ca lls f or a dministering a questionnaire that measures emotional intelligence to a sample of 100 undergraduate student volunteers who are enrolled in an introductory psychology course currently b eing t aught at her u niversity. A ssume t his is the only sample being used for this study and discuss the adequacy of the sample
Answer:
In terms of size it is adequate . Sample isn't randomly selected
Step-by-step explanation:
In order for the sampling ot be adequate the sample size must be large enough which in this case it is. The samples must also be randomly selected. Here, the sample includes students from one course only. In order for the study to represent whole population that is college students, students from other courses must also be included. Junior as well as senior students must also be part of this study.
The number of square feet per house are normally distributed with a population standard deviation of 137 square feet and an unknown population mean. A random sample of 19 houses is taken and results in a sample mean of 1350 square feet. Find the margin of error for a 80% confidence interval for the population mean. z0.10z0.10z0.05z0.05z0.025z0.025z0.01z0.01z0.005z0.005 1.2821.6451.9602.3262.576 You may use a calculator or the common z values above. Round the final answer to two decimal places.
Answer:
The MOE for 80% confidence interval for μ is 5.59.
Step-by-step explanation:
The random variable X is defined as the number of square feet per house.
The random variable X is Normally distributed with mean μ and standard deviation σ = 137.
The margin of error for a (1 - α) % confidence interval for population mean is:
[tex]MOE=z_{\alpha /2}\times\frac{\sigma}{\sqrt{n}}[/tex]
Given:
n = 19
σ = 137
[tex]z_{\alpha /2}=z_{0.20/2}=z_{0.10}=1.282[/tex]
Compute MOE for 80% confidence interval for μ as follows:
[tex]MOE=1.282\times\frac{137}{\sqrt{19}}=1.282\times4.36=5.58952\approx5.59[/tex]
Thus, the MOE for 80% confidence interval for μ is 5.59.
In the following equations, based on the variable costing income statement, identify the items designated by X: a. Net Sales – X = Manufacturing Margin b. Manufacturing Margin – X = Contribution Margin c. Contribution Margin – X = Income from Operations a. b. c.
Answer:
A = Total cost
B = Variable cost
C = Fixed cost
Step-by-step explanation:
A - Manufacturing cost includes all the cost directly and/or indirectly involved in the production process. It includes the variable component and fixed component, which represent the total cost. Hence, to arrive at the manufacturing margin, the total cost is subtracted from the net sales. The results us the manufacturing margin.
B - Contribution margin only includes and ultimately works with the individual variable element of cost, unlike the manufacturing margin. Hence, to arrive at the contribution margin, the variable cost element is subtracted from the manufacturing margin.
C - To arrive at the income from operations, the fixed cost element is subtracted from the contribution margin. With this, our income from operations can be derived.
Graduation. It's believed that as many as 25% of adults over 50 never graduated from high school. We wish to see this percentage is the same among the 25 to 30 age group.
a. How many of this younger age group must we survey in order to estimate the proportion of non-grads to within 6% with 90% confidence?
b. Suppose we want to cut the margin of error to 4%. What' s the necessary sample size?
c. What sample size would produce a margin of error of 3%?
Answer:
a) [tex]n=\frac{0.25(1-0.25)}{(\frac{0.06}{1.64})^2}=140.08[/tex]
And rounded up we have that n=141
b) [tex]n=\frac{0.25(1-0.25)}{(\frac{0.04}{1.64})^2}=315.19[/tex]
And rounded up we have that n=316
c) [tex]n=\frac{0.25(1-0.25)}{(\frac{0.03}{1.64})^2}=560.33[/tex]
And rounded up we have that n=561
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
Part a
The estimated proportion for this case is [tex]\hat p =0.25[/tex]
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.06[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.25(1-0.25)}{(\frac{0.06}{1.64})^2}=140.08[/tex]
And rounded up we have that n=141
Part b
[tex]n=\frac{0.25(1-0.25)}{(\frac{0.04}{1.64})^2}=315.19[/tex]
And rounded up we have that n=316
Part c
[tex]n=\frac{0.25(1-0.25)}{(\frac{0.03}{1.64})^2}=560.33[/tex]
And rounded up we have that n=561
After recording the pizza delivery times for two different pizza shops you conclude one pizza shop has a mean delivery of 45 minutes with a standard deviation of 5 minutes. the other shop has a mean delivery time of 44 minutes with a standard deviation of 19 minutes. interpret these figures. if you like pizzas from both shops equally well, which one would you order from?Why?
A. the means are nearly equal but the variation is significantly lower for the second shop than the first
B. the variations are nearly equal but the mean is greater for the first shop than for the second
C. both means and the variations are nearly equal
D. the means are nearly equal but the variations is significantly greater for the second shop than the first.
If you liked the pizzas from both shops equally well which one would you order from?why?
A. choose the second shop. the delivery time is more reliable because it has a larger standard deviation.
B. choose the second shop. the delivery time is more reliable because it has a larger mean.
C. choose the first shop. the delivery time is more reliable because it has a lower standard deviation.
D. choose the first shop. the delivery time is more reliable because it has a lower mean.
Answer:
C
Step-by-step explanation:
lolololpygdffffygccffghhjjjj
An internet service provider uses 50 modems to serve the needs of 1000 customers. It is estimated that at a given time. each customer will need a connection with probability 0.01, independent of the other customers.
What is the probability mass function of the number of modems in use at the given time?
Answer:
The probability mass function of the number of modems in use at the given time is:
[tex]P (X=x)={1000\choose x}(0.01)^{x}(1-0.01)^{1000-x};\ x=0, 1, 2, ...49[/tex]
Step-by-step explanation:
Let the random variable X = number of modems in use.
The internet provider uses 50 modems.
The number of customers served by the internet user is, n = 1000.
The probability that a customer will require an internet connection is, p = 0.01.
It is provided that the customers are independent of each other.
The random variable X satisfies all the properties of a Binomial distribution with parameters n = 1000 and p = 0.01.
The probability mass function of a Binomial distribution is:
[tex]P (X=x)={n\choose x}(p)^{x}(1-p)^{n-x};\ x=0, 1, 2, ...49[/tex]
Then the probability mass function of the number of modems in use at the given time is:
[tex]P (X=x)={1000\choose x}(0.01)^{x}(1-0.01)^{1000-x};\ x=0, 1, 2, ...49[/tex]
The question is about calculating the probability mass function of a binomial distribution. Given a fixed number of independent trials (1000 customers), with a constant probability of success (0.01), the pmf is given by P(X = k) = C(n, k) * (p)^k * (1-p)^(n-k), where n=1000, p=0.01, and k varies from 0 to 50.
Explanation:The situation described is a case of a binomial distribution. This is because there are a fixed number of independent trials (1000 customers attempting to connect), and each trial has two possible outcomes (the customer needs a connection or does not need a connection), with the probability of success (need a connection) being constant (0.01).
The probability mass function (pmf) of a binomial distribution is given by:
P(X = k) = C(n, k) * (p)^k * (1-p)^(n-k)
where:
- P(X = k) is the probability that exactly k trials are successful
- n is the total number of trials (here, 1000)
- k is the number of successful trials (modems in use)
- p is the probability of success on a single trial (here, 0.01)
- C(n, k) is the combination of n items taken k at a time.
To find the pmf of the number of modems in use, substitute the given values into the pmf formula for each possible value of k (0 to 50, because the ISP has only 50 modems). For each k, the result is the probability that k modems are in use at the given time.
Learn more about Binomial Distribution here:https://brainly.com/question/33656163
#SPJ3
A random sample of 16 students selected from the student body of a large university had an average age of 25 years. We want to determine if the average age of all the students at the university is significantly different from 24. Assume the distribution of the population of ages is normal with a standard deviation of 2 years.At a .05 level of significance, it can be concluded that the mean age is _____. a. significantly less than 25 b. not significantly different from 24 c. significantly less than 24 d. significantly different from 24
Answer:
Option D) significantly different from 24
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 24
Sample mean, [tex]\bar{x}[/tex] = 25
Sample size, n = 16
Alpha, α = 0.05
Population standard deviation, σ = 2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 24\text{ years}\\H_A: \mu \neq 24\text{ years}[/tex]
We use Two-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{25 - 24}{\frac{2}{\sqrt{16}} } = 2[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]
Since,
The calculated z statistic does not lie in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Thus, it can be concluded that the mean age is
Option D) significantly different from 24
Which equation has a solution of x = 13?
A. x+3= 10
B. x-10=3
C. 1 / 2 = 26
D.X+ 12 = 1
Answer:
B
Step-by-step explanation:
we can plug 13 into each equation
A. 13+3 =10
16≠10, so A is wrong
B. 13-10=3
3=3, B is correct
C, is definitely wrong
D. 13+12=1
25≠1
An insurance company has 1,000 policies on men of age 50. The company estimates that the probability that a man of age 50 dies within a year is .01. Estimate the number of claims that the company can expect from beneficiaries of these men within a year.
Answer:
the expected number of claims within a year is 10
Step-by-step explanation:
since the life expectancy of each man is independent from others, then the random variable X= number of man of 50 years who die within a year , follows a binomial distribution. Thus the expected value of X is given by
E(X) = n*p
where
p= probability that a man of age 50 dies within a year = 0.01
n = number of policies on men of age 50 = 1000
replacing values
E(X) = n*p = 0.01 * 1000 = 10
therefore the expected number of claims within a year is 10
Using the binomial distribution, it is found that the estimate of the number of claims that the company can expect from beneficiaries of these men within a year is of 10.
For each men, there are only two possible outcomes, either there is a claim(they die), or there is not a claim. The probability of a men dying is independent of any other men, hence, the binomial distribution is used to solve this question.
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability, and has expected value given by:
[tex]E(X) = np[/tex]
In this problem:
There are 1000 policies, hence [tex]n = 1000[/tex].Each of them has a 0.01 probability of dying, hence causing a claim, which means that [tex]p = 0.01[/tex].Then:
[tex]E(X) = 1000(0.1) = 10[/tex]
The expected number of claims is of 10.
To learn more about the binomial distribution, you can take a look at https://brainly.com/question/25642476
When designing an experiment to study tree growth, the following four treatments are used: none, irrigation only, fertilization only, irrigation and fertilization. A row of 10 trees is used in the study and each tree is randomly assigned to one of the four treatments. How many different treatment arrangements are possible
Answer:
286 different treatment arrangements
Step-by-step explanation:
This is a combination problem.
We want to share 10 trees amongst 4 experimental groups.
13C3 = 286 different treatment arrangements.
There are 286 different treatment arrangements of trees possible.
What is tree treatment ?The tree treatment includes the medical aid of trees in the growth and development such as fertilizers and various chemical that allows for their development.
As per the question the experiment was conducted to study the tree's growth and the treatments were used that included irrigation, fertilization. The row of 10 trees was used in the study and each tree was randomly taken.
This is a combination problem. About 10 trees are given that have 4 experimental groups. Thus the 13 C3 = 286 different treatment arrangements.
Find out more information about the treatments.
brainly.com/question/13648997.
A researcher collected a sample of 938 low-income women and asked about their material well-being. The researcher was interested in the effects of two independent variables:
current employment status (no, yes), and if they had received cash welfare assistance in the past (no, yes).
More specifically, the researcher hypothesized that the effect of current employment on material well-being would be different depending on whether the women had received cash for welfare in the past.
What type of statistical analysis is appropriate for the researcher’s hypothesis?
a) ANOVA
b) ANCOVA 2-Way
c) ANOVA 2-Way
d) ANCOVA
Final answer:
The correct statistical analysis for assessing the effects of two independent variables on a single outcome and understanding interactions between the variables is c) ANOVA 2-Way.
Explanation:
The appropriate statistical analysis for the researcher’s hypothesis, which involves understanding the effects of two independent variables (current employment status and prior cash welfare assistance) on a dependent variable (material well-being) is a two-way ANOVA. This statistical method is used to analyze the impact of two independent categorical variables on a continuous outcome and to interact between these variables.
A two-way ANOVA, also known as factorial ANOVA, is particularly useful in this case as the researcher is interested not just in the separate effect of each variable, but also in the potential interaction between current employment status and the experience of receiving cash welfare in the past. This interaction term allows the researcher to understand if the effect of employment status on material well-being differs based on whether the women had received cash welfare.
Therefore, the correct answer to the question is: c) ANOVA 2-Way
A plane delivers two types of cargo between two destinations. Each crate of cargo I is 9 cubic feet in volume and 187 pounds in weight, and earns $30 in revenue. Each crate of cargo II is 9 cubic feet in volume and 374 pounds in weight, and earns $45 in revenue. The plane has available at most 540 cubic feet and 14,212 pounds for the crates. Finally, at least twice the number of crates of I as II must be shipped. Find the number of crates of each cargo to ship in order to maximize revenue. Find the maximum revenue. crates of cargo I crates of cargo II maximum revenue $
Answer:
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
Step-by-step explanation:
Given that a plane delivers two types of cargo between two destinations
Crate I Crate II
Volume 9 9
Weight 187 374
Revenue 30 45
Let X be the no of crate I and y that of crate II
Then
[tex]9x+9y\leq 540\\187x+374y\leq 14212\\x\geq 2y[/tex]
Simplify these equations to get
[tex]x+y\leq 60\\x+2y\leq 76\\x\geq 2y[/tex]
Solving we get
[tex]y\leq 16\\x\leq 46 and x\geq 32\\32\leq x\leq 46[/tex]
REvenue = 30x+45y
The feasible region would have corner points as (60,0) or (32,16) or (46,16)
Revenue for (60,0) = 1800
(32,16) = 1680
(46,16)=2100
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. The problem can be solved using linear programming techniques to find the optimal solution.
Explanation:To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. Let's assume the number of crates of cargo I is x and the number of crates of cargo II is y.
Based on the given information, the constraints for the problem are:
Volume constraint: 9x + 9y ≤ 540Weight constraint: 187x + 374y ≤ 14,212Relationship constraint: x ≥ 2yTo find the maximum revenue, we need to maximize the objective function: Revenue = 30x + 45y.
The problem can be solved using linear programming techniques, such as graphical or simplex method, to find the optimal solution. However, since these methods require plotting and iterations, the detailed calculations are beyond the scope of this response. The optimal solution will provide the values of x and y, which can be used to determine the maximum revenue.
Learn more about Optimizing revenue for cargo shipment here:https://brainly.com/question/34408399
#SPJ3
A certain lottery has 38 numbers. In how many different ways can 4 of the numbers be selected? (Assume that order of selection is not important.)
Answer:
There are 73815 ways of selecting 4 of the lottery numbers.
Step-by-step explanation:
The total amount of ways to pick 4 numbers from a set of 38 ignoring the order is given by the combinatorial number of 38 with 4, denoted by [tex] {38 \choose 4} [/tex] , which is
[tex] {38 \choose 4} = \frac{38!}{4!(38-4)!} = 73815 [/tex]
Therefore, there are 73815 ways of picking 4 of the numbers.
Answer:
We conclude that have 73815 a different ways.
Step-by-step explanation:
We know that a certain lottery has 38 numbers. We assume that order of selection is not important. We calculate how many different ways can 4 of the numbers be selected.
So out of 38 numbers we choose 4 numbers. We get:
[tex]C_4^{38}=\frac{38!}{4!(38-4)!}=73815[/tex]
We conclude that have 73815 a different ways.