Answer:
d. It can be biased by one or two extremely small or large values
Step-by-step explanation:
Specially in a small sample, a single measure can cause a large difference.
For example, you are selling yourself as a tutor, you have five students. 4 of them got good grades, but the other one got 0. The arithmetic mean is not going to be kind to your averages, in virtue of the outlier.
So the correct answer is:
d. It can be biased by one or two extremely small or large values
The arithmetic mean's disadvantage is that extreme values or outliers in the data set can significantly skew the mean. This makes the mean a less accurate reflection of the central tendency or 'average' of the data.
Explanation:The correct answer to your question is "d. It can be biased by one or two extremely small or large values." The arithmetic mean, often simply called the 'mean', is a type of average most commonly used. It is calculated by adding all numbers in a set and then dividing by the quantity of numbers. While useful, it has a significant disadvantage in its sensitivity to extreme values, also referred to as 'outliers'. If there are one or two extremely large or small numbers in the data set, these can drastically affect the calculated mean, thus not accurately representing the central tendency of the overall data.
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Slope intercept form that passes through the given point and is parallel to the given line (2,-1) and y=2x+2
Answer:
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)
The equation of the given line is
y=2x+2
Comparing with the slope intercept form, slope = 2
If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (2,-1) is 2
To determine the intercept, we would substitute m = 2, x = 2 and y = -1 into y = mx + c. It becomes
- 1 = 2×2 + c = 4 + c
c = - 1 - 4 = - 5
The equation becomes
y = 2x - 5
. You have a bag which contains 100 coins. 99 of these coins are normal and fair having the two sides head and tail. But, one coin is fake and has two heads! You pick a coin randomly out of the bag without checking whether it is a normal coin or the fake one. Suppose that each coin is equally likely to be picked. You throw the picked coin n N times in a row (still without checking whether you have a regular coin or not) and observe the coin lands heads n times. What is the probability that you have picked the fake coin?
Answer:I am not very sure but I think [tex]\frac{1}{51.5}[/tex]
Step-by-step explanation:
The probability of picking the fake coin given it lands heads [tex]\(n\)[/tex] times is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]
To find the probability that you have picked the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row, we can use Bayes' theorem.
Let [tex]\(F\)[/tex] be the event of picking the fake coin, and [tex]\(H\)[/tex] be the event of getting heads [tex]\(n\)[/tex] times in a row.
We are asked to find [tex]\(P(F|H)\)[/tex], the probability of picking the fake coin given that [tex]\(n\)[/tex] heads are observed in a row.
By Bayes' theorem:
[tex]\[ P(F|H) = \frac{P(H|F) \times P(F)}{P(H)} \][/tex]
We know:
[tex]- \(P(H|F) = 1\)[/tex] (since the fake coin has two heads).
[tex]- \(P(F) = \frac{1}{100}\)[/tex] (since there is only one fake coin out of 100).
[tex]- \(P(H)\)[/tex] is the probability of getting [tex]\(n\)[/tex] heads in a row, which is the same for both the fake and normal coins. This is [tex]\(0.5^n\).[/tex]
Now, substituting the values:
[tex]\[ P(F|H) = \frac{1 \times \frac{1}{100}}{0.5^n} \][/tex]
[tex]\[ P(F|H) = \frac{1}{100 \times 0.5^n} \][/tex]
[tex]\[ P(F|H) = \frac{1}{100 \times 2^n} \][/tex]
So, the probability of picking the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]
Which of the following conditions must be met if you want to use z-procedures to construct a confidence interval for a population proportion?
A. The sample size must be at least 30 (some texts say 50).
B The population must be at least 10 times as large as the sample (some texts say 20).
C. Both npˆ and n(1 − pˆ) must be at least 10 (some texts say greater than 5).
To use z-procedures for a confidence interval for a population proportion, three conditions must be met. The sample size must be large (at least 30 or 50), the population must be at least 10 or 20 times as large as the sample, and both npˆ and n(1 - pˆ) must be at least 10, ensuring the sampling distribution approximates to normal.
Explanation:
The process of constructing a confidence interval for a population proportion through z-procedures demands certain conditions to be met. These conditions directly shape the accuracy of the findings. There are three key requirements:
The sample size must be large enough, often at least 30 or 50 as per certain textbooks, to ensure the central limit theorem applies and the distribution approximates normal. However, this condition is more relevant for mean-based estimates and less for proportions.The population should be large enough compared to the sample. According to some literature, the population should be at least 10 or 20 times larger than the sample. This condition is known as the '10% rule' for categorical data, which aims to limit the effects of potential sample dependency.Lastly, the quantities of both npˆ and n(1 - pˆ) should be at the minimum of 10 (some sources claim more than 5). This requirement is crucial to validate that the sampling distribution for the sample proportion is approximately normal.Learn more about z-procedures here:https://brainly.com/question/34396416
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The correct condition for using z-procedures to construct a confidence interval for a population proportion is:
C. Both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] must be at least 10 (some texts say greater than 5).
To use z-procedures to construct a confidence interval for a population proportion, we typically rely on the normal approximation to the binomial distribution. For this approximation to be valid, both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] should be sufficiently large.
- [tex]\( n \)[/tex] represents the sample size.
- [tex]\( \hat{p} \)[/tex] represents the sample proportion.
Both conditions ensure that the sampling distribution of the sample proportion is approximately normal.
1. [tex]\( np \hat{p} \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected successes in the sample.
2.[tex]\( n(1 - \hat{p}) \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected failures in the sample.
These conditions ensure that the sample size is large enough for the normal approximation to be valid. They are more precise than A and B, which are common rules of thumb but not absolute requirements for using z-procedures.
Do seasons influence the type of pet people adopt? A researcher is interested in finding out whether thereis a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall)during which such pet would be adopted. He randomly selects 200 adoption files from the humane society andrecords the season and pet type for each of the 200 adoptions.what kind of test scenario is this?A. Simple linear regressionB. One-sample t-test for a population meanC. Paired t-test for a population mean differenceD. Two-sample t-test for the comparison of two population meansE. One-sample Z-test for a population proportionF. Two-sample Z-test for the comparison of two population proportionsG. ANOVA for comparing many population meansH. Chi-squared test of goodness of fitI. Chi-squared test of independenceJ. Chi-squared test of homogeneityK. One-sample Binomial test for a population proportion
Answer:
Answer = I. Chi-squared test of independence
Step-by-step explanation:
The test of independence uses the contingency table format. The analysis of a two-way contingency table helps to answer the question whether two variables are related or independent of each other. Thus, the chi-square statistic measures how much the observed frequencies differ from the expected frequencies when variables are independent. The first procedure is to state the null and alternative hypothesis
H0: No relationship exists between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted
H1: There is a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted
To be able to decide whether to reject the null hypothesis or not, we need to compare the calculated test statistic with its chi-square table or critical value using a given level of significance and degree of freedom and then decide using the decision rule (Reject H0 if the calculated chi-square statistic is greater than the critical value otherwise, accept H0)
The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. Each can holds a maximum of 12.50 ounces of soda. Every can that has more than 12.50 ounces of soda poured into it causes a spill and the can must go through a special cleaning process before it can be sold. What is the probability that a randomly selected can will need to go through this process? A) .1587 B) .6587 C) .8413 D) .3413
Answer: A) .1587
Step-by-step explanation:
Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.
i.e. [tex]\mu=12.30[/tex] and [tex]\sigma=0.20[/tex]
Let x denotes the amount of soda in any can.
Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.
Then, the probability that a randomly selected can will need to go through the mentioned process = probability that a randomly selected can has more than 12.50 ounces of soda poured into it =
[tex]P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587[/tex]
Hence, the required probability= A) 0.1587
Final answer:
The probability that a randomly selected can will need a special cleaning process because it holds more than 12.50 ounces of soda is calculated using the Z-score. The correct answer is A) 0.1587.
Explanation:
The question asks us to calculate the probability that a randomly selected can will require a special cleaning process if it holds more than 12.50 ounces of soda, given a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. To find this probability, we can use the standard normal distribution, also known as the Z-score.
First, we calculate the Z-score for 12.50 ounces using the formula: Z = (X - \\mu\) / \\sigma\, where X is the value we are evaluating (12.50 ounces), \mu is the mean (12.30 ounces), and \sigma is the standard deviation (0.20 ounce). Plugging in the values gives us Z = (12.50 - 12.30) / 0.20 = 1. The Z-score represents the number of standard deviations the value X is from the mean.
Next, to find the probability that a can will overflow (have more than 12.50 ounces), we look up the corresponding probability in the standard normal distribution table or use a calculator suitable for such statistical computations. The probability corresponding to a Z-score of 1 is approximately 0.8413. However, since we are interested in the probability of a can having more than 12.50 ounces, we need the area to the right of the Z-score, which is 1 - 0.8413 = 0.1587. Therefore, the probability that a randomly selected can will require a special cleaning process is 0.1587.
The correct answer to the question is A) 0.1587.
A BINGO card is a 5 × 5 grid. The center square is a free space and has no number. The first column is filled with five distinct numbers from 1 to 15, the second with five numbers from 16 to 30, the middle column with four numbers from 31 to 45, the fourth with five numbers from 46 to 60, and the fifth with five numbers from61 to 75.
Since the object of the game is to get five in a row horizontally, vertically, or diagonally, the order is important.
How many BINGO cards are there?
Answer:
[tex]5.52*10^{26}[/tex]
Step-by-step explanation:
For the columns with 5 slots, there are 15 distinct options to fill in. The number of ways to fill them in is
15 * 14 * 13 * 12 * 11 = 360360 ways (order matters)
For the column with 4 slots and 15 options. The number of ways to fill them in is
15 * 14 * 13 * 12 = 32760 ways (order matters)
Since a bingo card has 4 columns of 5 slots and 1 column with 4 slots, the total number of combination there is:
[tex]360360^4*32760 \approx 5.52*10^{26}[/tex] (order matters)
A teacher finds that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%; F, 5%. At the end of a randomly selected semester, the following grades were recorded. Calculate the chi-square test statistic chi squared used to determine if the grade distribution for the department is different than expected. Use alpha equals 0.01 .
Grade A B C D F
Number 42 36 60 8 14
(A) 3.41
(B) 5.25
(C) 6.87
(D) 4.82
Answer:5.25
Step-by-step explanation:
Intelligence quotas on two different tests are normally distributed. Test A has a mean of 100 and a standard deviation of 13. Test B has a mean of 100 and a standard deviation of 18. Usez-scores to determine which person has the higher IQ: an individual who scores 123 on Test A or an individual who scores 121 on Test B. Which individual has the higher IQ?
Answer:
The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The individual with the higher IQ is the one with the higher z-score. So
Test A has a mean of 100 and a standard deviation of 13. An individual who scores 123 on Test A.
So [tex]\mu = 100, \sigma = 13, X = 123[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{123 - 100}{13}[/tex]
[tex]Z = 1.77[/tex]
Test B has a mean of 100 and a standard deviation of 18. An individual who scores 121 on Test B.
So [tex]\mu = 100, \sigma = 18, X = 121[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{121 - 100}{18}[/tex]
[tex]Z = 1.17[/tex]
The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.
You have a weekend job selecting speed limit signs to put at road curves. The speed limit is determined by the radius of the curve and the bank angle. Part A For a turn of radius 280 m and a 7.0∘ bank angle, what speed limit should you post so that a car traveling at that speed negotiates the turn successfully even when the road is wet and
Final answer:
To determine the speed limit for a banked road curve, use the formula: speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle)). For a curve with a radius of 280 m and a bank angle of 7.0°, the speed limit should be approximately 23.8 m/s.
Explanation:
When determining the speed limit for a banked road curve, we need to consider the radius of the curve and the bank angle. In this case, the radius of the curve is 280 m and the bank angle is 7.0°.
To calculate the speed limit, we can use the formula:
speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle))
Substituting the given values into the formula, we get:
speed limit = square root of (280 * 9.8 * tan(7.0))
Using a calculator, we find that the speed limit should be approximately 23.8 m/s.
Calculate the volume of the solid of revolution generated by revolving the region bounded by the parabolas y 2 = 2 (x − 3) and y 2 = x about y = 0.
Answer:
[tex]9\pi[/tex]
Step-by-step explanation:
given are two parabolas with vertex as (3,0) and (0.0)
[tex]y^2 =2(x-3)\\y^2 =x[/tex]
These two intersect at x=6
Volume of II curve rotated about x axis - volume of I curve rotated about x axis = Volume of solid of revolution
For the second curve limits for x are from 0 to 6 and for I curve it is from 3 to 6
V2 =\pi [tex]\int\limits^6_0 {y^2} \, dx \\=\pi\int\limits^6_0 {x} \, dx\\= \pi\frac{x^2}{2} \\=18\pi[/tex]
V1 =[tex]\pi \int\limits^6_3 {y^2} \, dx\\\pi \int\limits^6_3 {2x-6} \, dx\\=\pi(x^2-6x)\\= \pi[(36-9)-6(6-3[)\\= (27-18)\pi\\=9\pi[/tex]
Volume of solid of revolutin = V2-V1 = [tex]9\pi[/tex]
Andy once heard about a car crash victim who died because he was pinned in the wreckage by a seat belt he could not undo. As a result, Andy refuses to wear a seat belt when he rides in a car. How would you explain to Andy the fallacy behind relying on this anecdotal evidence?
Answer:
You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.
Step-by-step explanation:
You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.
Here Andy is basing himself on one anecdotal evidence he heard. So a sample size of 1. If he were to base his opinion from a large sample size, it is highly likely that 99% of the people were saved by the seatbelt and 1% got hurt. Since Andy only heard the anecdotal evidence, he may be basing himself on a very unlikely event.
Andy's refusal to wear a seat belt based on a single incident is an example of reliance on anecdotal evidence, a logical fallacy. This means he is making a general rule based on a singular event, which is misleading as it does not take into account the reality that seat belts vastly improve safety in car crashes. It's important to base decisions on sound evidence rather than isolated incidents.
Explanation:Andy is relying on anecdotal evidence, which is a logical fallacy. A logical fallacy is a flaw in reasoning that makes an argument invalid. In this case, Andy has heard a story about a particular event, and is applying it as a universal rule.
However, just because a seatbelt caused harm in a single incident does not mean that seat belts are generally harmful or risky. According to various studies, your chance of being killed or seriously injured in a car crash is much higher if you are not wearing a seatbelt.
Therefore, it is illogical to avoid wearing seat belts based on a single unfortunate incident. It's important to remember that anecdotal evidence can be misleading, as it's prone to bias and doesn't take into account the full range of possibilities and outcomes.
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A company produces regular and deluxe ice cream at three plants. Per hour of operation,
Plant A produces 40 gallons of regular ice cream and 20 gallons of deluxe ice cream.
Plant H produces 20 gallons of regular ice cream and 40 gallons of deluxe ice cream,
and Plant M produces 40 gallons of regular and 40 gallons of deluxe.
It costs $60 per hour to operate Plant A, $92 per hour to operate Plant H, and $140 per hour to operate Plant M. The company must produce at least 350 gallons of regular ice cream and at least 240 gallons of deluxe ice cream each day.To minimize the cost of production, Plant A should operate for ___ hours per day, Plant H should operate for ___ hours per day, and Plant M should operate for ______ hours per day. (Round to the nearest tenth as needed.)
Answer:
Answer explained below
Step-by-step explanation:
Let a be the working hours of plant A.
Let b be the working hours of plant H.
Let c be the working hours of plant M.
We have to minimize , Z = 60a + 92b + 140c
subject to constraint , 40a + 20b + 40c >=350
20a + 40b + 40c >=240
where a>=0 , b>=0 ,c>=0
So ,by solving this , a = 7.67 , b= 2.17 , c = 0
So ,working hours of plant A = 8 hrs
working hours of plant H = 3 hrs
working hours of plant M = 0 hrs
Answer:
A = 8hrs
H = 3hrs
M = 0hrs
Step-by-step explanation:
Let a be the working hours of plant A.
Let b be the working hours of plant H.
Let c be the working hours of plant M.
∴ Hours of plants A,H,M are a,b,c respectively.
We have to minimize the cost of production, Z = 60a + 92b + 140c
Regular ice cream: 40a + 20b + 40c = 350
Deluxe ice cream: 20a + 40b +40c =240
Where a>=0, b>=0, c>=0
So, by solving this, a= 7.67, b= 2.77,c= 0
So working hours of plant A = 8 hrs
Working hours of plant H = 3 hrs
Working hours of plant M = 0 hrs
If the p-value for a hypothesis test is 0.07 and the chosen level of significance is α = 0.05, then the correct conclusion is to ____________________. A. reject the null hypothesis B. not reject the null hypothesis if σ = 10 C. reject the null hypothesis if σ = 10 D. reject the null hypothesis
The correct conclusion is to not reject the null hypothesis.
Explanation:If the p-value for a hypothesis test is 0.07 and the chosen level of significance is α = 0.05, then the correct conclusion is to not reject the null hypothesis.
When conducting a hypothesis test, the p-value represents the probability of obtaining the observed data or more extreme values assuming the null hypothesis is true. If the p-value is greater than the significance level (α), which is the threshold for rejecting the null hypothesis, we fail to reject the null hypothesis.
In this case, the p-value of 0.07 is greater than the significance level of 0.05. Therefore, we cannot reject the null hypothesis.
A sample of eight workers in a clothing manufacturing company gave the following figures for the amount of time(in minutes) needed to join a collar to a shirt11 13 14 10 9 16 11 12Construct a 95% confidence interval for the true mean amount of time needed to join a collar.
Answer:
[tex]10.108 < \mu < 13.892[/tex]
Step-by-step explanation:
1) 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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
We have the following distribution for the random variable:
[tex]X \sim N(\mu , \sigma=0.45)[/tex]
And by the central theorem we know that the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
2) Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=12[/tex]
The sample deviation calculated [tex]s=2.268[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=8-1=7[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that [tex]t_{\alpha/2}=2.36[/tex]
Now we have everything in order to replace into formula (1):
[tex]12-2.36\frac{2.268}{\sqrt{8}}=10.108[/tex]
[tex]12+2.36\frac{2.268}{\sqrt{8}}=13.892[/tex]
So on this case the 95% confidence interval would be given by (10.108;13.892)
[tex]10.108 < \mu < 13.892[/tex]
An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route canbe justified only if it reduces the mean travel time by more than 10 minutes.a. If u 1 is the mean for the scenic route and u 2 for the nonscenic route. what hypotheses should be tested?b. If u 1 is the mean for the nonscenic route and u Zfor the scenic route, what hypotheses should be tested?
Final answer:
The null and alternative hypotheses for testing the mean travel time on scenic and nonscenic routes.
Explanation:
a. In this case, the hypotheses to be tested are:
Null hypothesis (H0): The mean travel time for the nonscenic route is not shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 ≤ 10.
Alternative hypothesis (Ha): The mean travel time for the nonscenic route is shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 > 10.
b. If u1 is the mean for the nonscenic route and u2 for the scenic route, the hypotheses to be tested would be:
Null hypothesis (H0): The mean travel time for the scenic route is not shorter than 10 minutes compared to the nonscenic route, or in other words, u2 - u1 ≤ 10.
Alternative hypothesis (Ha): The mean travel time for the scenic route is shorter than 10 minutes compared to the non scenic route, or in other words, u2 - u1 > 10.
The correct options are : - (a) The correct hypothesis is 2. [tex]\(\mu_1 - \mu_2 < -10\)[/tex] and (b) The correct hypothesis is 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]
Part (a): [tex]\(\mu_1\)[/tex] is the mean for the scenic route and [tex]\(\mu_2\)[/tex] is the mean for the non-scenic route
The individual decides that the non-scenic route is justified only if it reduces the mean travel time by more than [tex]10[/tex] minutes. In this context, we need to test whether the mean travel time for the scenic route [tex]\(\mu_1\)[/tex] minus the mean travel time for the non-scenic route [tex](\(\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]
Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = -10\)[/tex]
Alternative Hypothesis[tex](\(H_a\)) : \(\mu_1 - \mu_2 < -10\)[/tex]
So, the correct option for (a) is: 2.[tex]\(\mu_1 - \mu_2 < -10\)[/tex]
Part (b): [tex]\(\mu_1\)[/tex] is the mean for the non-scenic route and [tex]\(\mu_2\)[/tex] is the mean for the scenic route
In this case, we need to test whether the mean travel time for the non-scenic route [tex](\(\mu_1\))[/tex] minus the mean travel time for the scenic route [tex](\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]
Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = 10\)[/tex]
Alternative Hypothesis [tex](\(H_a\)) : \(\mu_1 - \mu_2 < 10\)[/tex]
So, the correct option for (b) is: 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]
The complete Question is
An individual can take either a scenic route to work or a non-scenic route. She decides that use of the non-scenic route can be justified only if it reduces the mean travel time by more than 10 minutes.
(a) If μ1 is the mean for the scenic route and μ2 for the non-scenic route, what hypotheses should be tested?
1. μ1 − μ2 = −10
2. μ1 − μ2 < −10μ1 − μ2 = −10
3. μ1 − μ2 > −10 μ1 − μ2 = −10
4. μ1 − μ2 ≠ −10μ1 − μ2 = 10
5. μ1 − μ2 < 10μ1 − μ2 = 10
6. μ1 − μ2 > 10
(b) If μ1 is the mean for the non-scenic route and μ2 for the scenic route, what hypotheses should be tested?
1. μ1 − μ2 = −10
2. μ1 − μ2 < −10μ1 − μ2 = −10
3. μ1 − μ2 > −10 μ1 − μ2 = −10
4. μ1 − μ2 ≠ −10μ1 − μ2 = 10
5. μ1 − μ2 < 10μ1 − μ2 = 10
6. μ1 − μ2 > 10
Twenty-two concrete blocks were sampled and tested for crushing strength in order to estimate the proportion that were sufficiently strong for a certain application. Eighteen of the 22 blocks were sufficiently strong. Use the small-sample method to construct a 95% confidence interval for the proportion of blocks that are sufficiently strong.
Answer:
The 95% confidence interval with the normla method would be given by (0.657;0.979)
The 95% confidence interval witht he small sample method would be given by (0.784;0.852)
Step-by-step explanation:
1) Notation and definitions
[tex]X=18[/tex] number of blocks that were sufficiently strong
[tex]n=22[/tex] random sample taken
[tex]\hat p=\frac{18}{22}=0.818[/tex] estimated proportion of blocks that were sufficiently strong
[tex]p[/tex] true population proportion of blocks that were sufficiently strong
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]
2) Confidence interval (Normal method)
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 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.657[/tex]
[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.979[/tex]
The 95% confidence interval with the normla method would be given by (0.657;0.979)
3) Confidence interval (Small sample method)
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n} \frac{N-n}{N-1}}[/tex]
But we need to know the total size for the population, and on this case we don't know but let's assumed that N=2000 for example .
If we replace the values obtained we got:
[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.784[/tex]
[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.852[/tex]
The 95% confidence interval witht he small sample method would be given by (0.784;0.852)
The curve c(t) = (cost,sint,t) lies on which of the following surfaces. Enter T or F depending on whether the statement is true or false. (You must enter T or F -- True and False will not work.)___ 1. a plane___ 2. a sphere___ 3. an ellipsoid___ 4. a circular cylinder
Answer:
1. Plane False
2. Sphere False
3. Ellipsoid False
4. Circular cylinder True
Step-by-step explanation:
For this case we have the following curve [tex]C(t) = (cos t , sin t , t[/tex]
And we can express like this the terms for the curve or each component:
[tex] x= cos t, y= sin t , z =t[/tex]
1. Plane False
The general equation for a plane is given by:
a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.
For this case we don't satisfy this since have sinusoidal functions and this equation is never satisfied.
2. Sphere False
The general equation for a sphere is given by:
(x - a)² + (y - b)² + (z - c)² = r²
And for this case if we see our parametric equation again that is not satisfied since we have two cosenoidal functions. And another function z=t
3. Ellipsoid False
The general equation for an ellipsoid is given by:
x^2/a2 + y^2/b2 + z^2/c2 = 1
And for this case again that's not satisfied since we have
[tex]\frac{cos^2 t}{a^2} + \frac{sin^2 t}{b^2}+\frac{t^2}{c^2} \neq 1[/tex]
4. Circular cylinder True
The general equation for a circular cylinder is given by:
[tex]x^2 +y^2 = r^2[/tex]
And if we replace the equations that we have we got:
[tex] cos^2 t + sin^2 t = 1[/tex] from the fundamental trigonometry property.
So then we see that our function satisfy the condition and is the most appropiate option.
The seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row. If there are 30 rows altogether, how many seats are there in all? Select one: O
A. 1470
B. 1635
C. 2940
D. 3270
Answer:
B. 1635
Step-by-step explanation:
We have been given that the seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row.
We can see that the seating order is in form of arithmetic sequence, whose first term is 11 and common difference is 3.
Since there are 30 rows altogether, so we need to find the sum of 30 first terms of the sequence using sum formula.
[tex]S_n=\frac{n}{2}[2a+(n-1)d][/tex], where,
[tex]S_n=[/tex] Sum of n terms,
n = Number of terms,
a = First term,
d = Common difference.
Upon substituting our given value is above formula, we will get:
[tex]S_n=\frac{30}{2}[2(11)+(30-1)3][/tex]
[tex]S_n=15[22+(29)3][/tex]
[tex]S_n=15[22+87][/tex]
[tex]S_n=15[109][/tex]
[tex]S_n=1635[/tex]
Therefore, there are 1635 seats in all and option B is the correct choice.
Answer: the total number of seats is 1635
Step-by-step explanation:
For each additional row after the first row, there are 3 more seats than there are in the previous row. This means that the seats in each row is increasing in arithmetic progression with a common difference of 3. The formula for determining the sum of n terms of an arithmetic sequence is expressed as
Sn = n/2[2a + (n - 1)d]
Where
n represents the number of terms in the sequence.
a represents the first term,
d represents the common difference.
From the information given,
a = 11
n = 30
d = 3
Therefore,
S30 = 30/2[2×11 + (30 - 1)3]
S30 = 15[22 + 29×3]
S30 = 15 × 109= 1635
How do I write the equation of a line in slope intercept form, where slope is -3 and the y-intercept is (0,-10)
Answer:
y= -3x +10
Step-by-step explanation:
you already have your slop so plug it in and your y intercept is 10
The temperature in degrees Celsius on the surface of a metal plate is T(x, y) = 19 − 4x2 − y2 where x and y are measured in centimeters. Estimate the average temperature when x varies between 0 and 2 centimeters and y varies between 0 and 4 centimeters. °C
Answer:
Average value of temperature will be [tex]8.335^{\circ}C[/tex]
Step-by-step explanation:
We have given the temperature in degree Celsius [tex]T(x,y)=19-4x^2-y^2[/tex]
It is given that x varies between 0 to 2
So [tex]0\leq x\leq 2[/tex]
And y varies between 0 and 4
So [tex]0\leq y\leq 4[/tex]
So area between the region A = 4×2 = 8[tex]cm^2[/tex]
Now average temperature is given by
[tex]Average\ value=\frac{1}{A}\int \int T(x,y)dA[/tex]
[tex]Average\ value=\frac{1}{8}\int \int (19-4x^2-y^2)dxdy[/tex]
[tex]=\frac{1}{8}\int \int (19x-4\frac{x^3}{3}-y^2x)dy[/tex]
As limit of x is 0 to 2
[tex]=\frac{1}{8}\int (19\times 2-4\times \frac{2^3}{3}-y^2\times 2)-(19\times 0-4\times \frac{0^3}{3}-y^2\times 0))dy=\frac{1}{8}\int(38-\frac{32}{3}-2y^2)dy[/tex]
=[tex]=\frac{1}{8}(38y-\frac{32}{3}y-\frac{2y^3}{3})[/tex]
As limit of y is 0 to 4
So [tex]Average\ value=\frac{1}{8}(38\times 4-\frac{16}{3}\times 4-\frac{2\times 4^3}{3})-0=8.335^{\circ}C[/tex]
Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert). This simple random sample was obtained at 3:30 pm on a weekday. Use the sample data to construct a 95% confidence interval estimate of the population standard deviation. 62 61 61 57 61 54 59 58 59 69 60 67A.) 4.7 mi/h< <5.6 mi/h
b.) 2.9 mi/h< <6.9 mi/h
c.) 3.1 mi/h< <8.5 mi/h
d.) 1.6 mi/h< <4.9 mi/h
e.) 5.4 mi/h< <9.2 mi/h
Answer:
Option (B) is the correct answer to the following question.
Step-by-step explanation:
Step-1: We have to find the Mean of the series.
The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.
[tex]Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}[/tex] [tex]= 60.67[/tex]
Step-2: We have to find the Standard Deviation.
Let Standard Deviation be x.
Formula of Standard Deviation is: [tex]s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}[/tex]
Put value in formula of Standard Deviation,
[tex]s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}[/tex] = 40.75
Step-3: Then, we have to find the critical value by chi-square.
[tex]X_{1-\alpha/2}^{2}=3.82[/tex]
[tex]X_{1-\alpha/2}^{2}=21.92[/tex]
Then, find the confidence interval which is 95%.
[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9[/tex]
[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9[/tex]
The 95% confidence interval for the population standard deviation, based on the given sample data, is estimated to be approximately between 2.9 mi/h and 6.9 mi/h. Here option B is correct.
To construct a 95% confidence interval estimate of the population standard deviation, you can use the chi-square distribution. The formula for the confidence interval is given by:
[tex]\[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \][/tex]
where:
- n is the sample size,
- s is the sample standard deviation,
- [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] are the critical values from the chi-square distribution for the lower and upper bounds of the confidence interval, respectively, and
- α is the significance level (0.05 for a 95% confidence interval).
Given that the sample data is: 62 61 61 57 61 54 59 58 59 69 60 67, the sample size n is 12, and the sample standard deviation s is approximately 4.75.
Now, you need to find the critical values [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex]. For a 95% confidence interval and df = n-1 = 11, you can consult a chi-square distribution table or use a statistical software/tool to find these critical values.
Assuming [tex]\( \chi^2_{\alpha/2} \)[/tex] is 19.675 and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] is 2.201, plug these values into the formula:
[tex]\[ \left( \sqrt{\frac{(12-1) \times 4.75^2}{19.675}}, \sqrt{\frac{(12-1) \times 4.75^2}{2.201}} \right) \][/tex]
Calculating this expression gives approximately (2.9, 6.9). Here option B is correct.
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Use the pbinom() function in R to show the cumulative probability of getting 0, 1, 2, 3, or 4 heads when you flip the coin 4 times (this is the same as finding the probability than the value is less than or equal to 0, 1, 2, 3, or 4.)
A. probability of getting no more than 0 heads: 0.0625
B. probability of getting no more than 1 head: 0.3125
C. probability of getting no more than 2 heads: 0.6875
D. probability of getting no more than 3 heads: 0.9375
E. probability of getting no more than 4 heads: 1.0000 4.
Answer:
Half half for both since all three are fair coins.
Step-by-step explanation:
Final answer:
The cumulative probability of getting a certain number of heads in 4 coin flips can be calculated with the pbinom() function in R, which uses the binomial distribution, suitable for independent Bernoulli trials with two outcomes and a constant success probability.
Explanation:
Binomial Probability Distribution in R
To calculate the cumulative probability of getting a certain number of heads in multiple coin flips using binomial distribution, the pbinom() function in R can be used. We are looking at a scenario where a coin is flipped 4 times, and we want to find the cumulative probability of getting 0, 1, 2, 3, or 4 heads.
The binomial distribution is appropriate here because coin flipping is a Bernoulli trial: there are two possible outcomes (heads or tails), the probability of success (head in this case) is constant, and each flip is independent of others.
The cumulative probability can be calculated as follows:
Probability of no more than 2 heads: pbinom(2, 4, 0.5)
Probability of no more than 4 heads (which is certain): pbinom(4, 4, 0.5) equals to 1
You will get results that match options A through E provided in the question.
At time t=0 sec, a tank contains 15 oz of salt dissolved in 50 gallons of water. Then brine containing 88oz of salt per gallon of brine is allowed to enter the tank at a rate of 55 gal/min and the mixed solution is drained from the tank at the same rate.a. How much salt is in the tank at an arbitrary time t? b. How much salt is in the tank after 25 min?
Answer:
a) [tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]
b) [tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]
Step-by-step explanation:
Part a
Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.
For this case we know that the rate of change can be expressed on this way:
[tex]Rate change = In-Out[/tex]
And we can name the rate of change as [tex]\frac{ds}{dt}=rate change[/tex]
And our variable s would represent the amount of salt for any time t.
We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.
For the concentration out we can assume that is [tex]\frac{s}{50gal}[/tex]
And now we can find the expression for the amount of salt after time t like this:
[tex]\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}[/tex]
And we have this differential equation:
[tex]\frac{dS}{dt} +\frac{1}{10} s = 40[/tex]
With the initial conditions y(0)=15 oz
As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:
[tex]\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}[/tex]
And then in order to solve the differential equation we need to multiply with the integrating factor like this:
[tex]e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt[/tex]
[tex]e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C[/tex]
Now we can divide both sides by [tex] e^{\frac{1}{25} t} [/tex] and we got:
[tex]s(t) =400 + C e^{-\frac{1}{10} t}[/tex]
Now we can apply the initial condition in order to solve for the constant C like this:
[tex]15 = 400+C[/tex]
[tex]C=-385[/tex]
And then our function would be given by:
[tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]
Part b
For this case we just need to replace t =25 and see what we got for the value of the concentration:
[tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]
a) The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex].
b) There are 4400 ounces of salt in the tank after 25 minutes.
How to model a dissolution process in a tankIn this question we must model the salt concentration in the tank ([tex]c(t)[/tex]), in ounces per gallon, as a function of time ([tex]t[/tex]), in minutes, in which salt is dissolved into the tank due to a constant inflow rate ([tex]\dot V[/tex]), in gallons. Likewise, an equal outflow rate exists with resulting concentration.
a) The process is modelled mathematically by a non-homogeneous first order differential equation and physically by principle of mass conservation, whose description is shown below:
[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V_{T}}\cdot c(t) = \frac{\dot V}{V_{T}}\cdot c_{in}[/tex] (1)
Where:
[tex]V_{T}[/tex] - Tank volume, in gallons[tex]c_{in}[/tex] - Inflow salt concentration, in ounces per gallonThe solution of this differential equation is:
[tex]c(t) = c_{in} + \left(\frac{m_{T}}{V_{T}}-c_{in} \right)\cdot e^{-\frac{\dot V}{V_{T}}\cdot t }[/tex] (2)
Where [tex]m_{T}[/tex] is the initial salt mass of the tank, in ounces.
And the salt mass in the tank at an arbitrary time [tex]t[/tex] ([tex]m(t)[/tex]), in ounces, is obtained by multiplying (2) by the volume of the tank. That is to say:
[tex]m(t) = c(t)\cdot V_{T}[/tex] (3)
By replacing [tex]c(t)[/tex] in (3) by (2), we have the following expression:
[tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex] (4)
The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]. [tex]\blacksquare[/tex]
b) If we know that [tex]V_{T} = 50\,gal[/tex], [tex]\dot V = 55\,\frac{gal}{min}[/tex], [tex]c_{in} = 88\,\frac{oz}{gal}[/tex], [tex]m_{T} = 15\,oz[/tex] and [tex]t = 25\,min[/tex], then the quantity of salt is:
[tex]m(25) = (50)\cdot (88)+[15-(50)\cdot (88)]\cdot e^{-\left(\frac{55}{50} \right)\cdot (25)}[/tex]
[tex]m(25) = 4400\,oz[/tex]
There are 4400 ounces of salt in the tank after 25 minutes. [tex]\blacksquare[/tex]
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A tank contains 50 kg of salt and 2000 L of water. A solution of a concentration 0.0125 kg of salt per liter enters a tank at the rate 7 L/min. The solution is mixed and drains from the tank at the same rate.
(a) Find the amount of salt in the tank after 3.5 hours.
(amount in kg)
(b) Find the concentration of salt in the solution in the tank as time approaches infinity.
(concentration in kg/L)
a) There is a mass of 25 kilograms of salt in the tank after 3.5 hours.
b) The concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute.
How to model salt concentration in a tank
Salt concentration (c(t)), in kilograms per liter, is usually modelled by first order non-homogeneous linear differential equation, whose form is derived from principle of mass conservation and described below:
[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V} \cdot c(t) = \frac{\dot V\cdot c_{o}}{V}[/tex] (1)
Where:
[tex]\dot V[/tex] - Volume flow, in liters per minute. V - Volume of the tank, in liters. [tex]c_{o}[/tex] - Inflow concentration, in liters per minute.The solution of this differential equation is:
[tex]c(t) = \left(c_{i}-c_{o}\right)\cdot e^{-\frac{\dot V}{V}\cdot t }+c_{o}[/tex] (2)
Where:
[tex]c_{i} = \frac{m_{s}}{V}[/tex] (3)
Where [tex]m_{s}[/tex] is the initial salt mass in the tank, in kilograms and t is the time, in minutes.
a) The amount of salt is found by multiplying the volume occupied by the water and the salt concentration in the tank. If we know that [tex]m_{s} = 50\,kg[/tex], [tex]V = 2000\,L[/tex], [tex]\dot V = 7\,\frac{L}{min}[/tex], [tex]c_{o} = 0.0125\,\frac{kg}{L}[/tex] and [tex]t = 3.5\,h[/tex], then the amount of salt in the tank:
[tex]c_{i} = \frac{50\,kg}{2000\,L}[/tex]
[tex]c_{i} = 0.025\,\frac{kg}{L}[/tex]
[tex]c(12600) = (0.025-0.0125)\cdot e^{-\frac{7}{2000}\cdot (12600)}+0.0125[/tex]
[tex]c(12600) = 0.0125[/tex]
And the salt mass in the tank is:
[tex]m = \left(0.0125\,\frac{kg}{L} \right)\cdot \left(2000\,L\right)[/tex]
[tex]m = 25\,kg[/tex]
There is a mass of 25 kilograms of salt in the tank after 3.5 hours. [tex]\blacksquare[/tex]
b) For [tex]t \to +\infty[/tex], [tex]c(t) \to c_{o}[/tex]. Therefore, the concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute. [tex]\blacksquare[/tex]
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a) The amount of salt in the tank after 3.5hr is 25Kg
b) The concentration of the salt as time approaches infinity is 0.125Kg/min.
How to calculate the values?Let y represents the total amount of salt at time(t)
dy/dt = rate of solution in - rate of solution out of the tank
But, concentration = amount of salt at time t/ volume of water at time t
= y/ 2000
dy/dt = 0.0875 - 7y/2000dy/dt
dy/dt = (175 - 7y/2000)dy/dt
separating variables
(1/7y-175)dy = - 1/200dtIn7y - 175
Integration both sides
y = 175 + e^(–1/2000t) + k/7
when the beginning y = 50, t = 0, k = 175
substituting the values
At t = 3.5 hrs
y = 25kg
When t approaches to infinity
Concentration is 0.125kg/min
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A triangle is measured and two adjacent sides are found to be 3 inches and 4 inches long, with an included angle of π/4. The possible errors in measurement are 1/18 inch for the sides and 0.05 radian for the angle. Approximate the maximum possible error in the computation of the area. (Round your answer to two decimal places.)
The maximum possible error in the area is approximately 0.18 square inches.
Calculate the nominal area:
Using the given measurements, the nominal area is (1/2) * 3 * 4 * sin(π/4) ≈ 3 square inches.
Find the partial derivatives of the area formula:
A = (1/2)ab sin(C), where a and b are the sides and C is the included angle.
∂A/∂a = (1/2)b sin(C)
∂A/∂b = (1/2)a sin(C)
∂A/∂C = (1/2)ab cos(C)
Estimate the maximum possible errors for each variable:
Δa ≈ 1/18 inches
Δb ≈ 1/18 inches
ΔC ≈ 0.05 radians
Approximate the maximum error using differentials:
ΔA ≈ |∂A/∂a|Δa + |∂A/∂b|Δb + |∂A/∂C|ΔC
ΔA ≈ (1/2)(4)(sin(π/4))(1/18) + (1/2)(3)(sin(π/4))(1/18) + (1/2)(3)(4)(cos(π/4))(0.05)
ΔA ≈ 0.18 square inches
Round the answer:
The maximum possible error in the area is approximately 0.18 square inches.
In some year, a candy shop produced 100 boxes of candy per working day in January. In each month following this, the shop produced 25 more boxes of candy per working day in addition to the previous month.
How much boxes did the candy shop produce on each working day in October?
Answer: 325 boxes of candy will be produced on each working day in October
Step-by-step explanation:
The initial number of boxes of candy produced is 100. In each month following this, the shop produced 25 more boxes of candy per working day in addition to the previous month. This means that the number of boxes produced each month is increasing in arithmetic progression. The formula for the nth term of an arithmetic sequence is expressed as
Tn = a + (n-1)d
Where
a is the first term of the sequence
n is the number of terms
d is the common difference.
From the given information,
a = 100
d = 25
n = number of months from January to October = 10
Tn = 100 + (10 - 1)25
Tn = 100 + 9×25
Tn = 100 + 225 = 325
Which polyhedron is convex?
Answer:
The fourth one
Step-by-step explanation:
Because Convex polygons can only be shapes like hexagons and squares and stuff like that. They cant be curved or zig zagged.
Answer:
The fourth one
Step-by-step explanation:
Convex polygons can be hexagons and stuff like that.
Can the following points be on the graph of the equation x-y = 0? Explain
Mr. Kim needs to buy 3 plane tickets for him, his wife & their 2 children at $259 for each person. How much will he spend on all of their plane tickets?
Answer: [tex]\$3108[/tex]
Step-by-step explanation:
Mr. Kim, his wife and two children are 4 persons. Now, he needs to buy 3 plane tickets for each one, which means he has to buy 12 plane tickets:
[tex](3 tickets)(4)=12 tickets[/tex]
On the other hand, we know each ticket costs [tex]\$259[/tex] per person. So, if we have 12 tickets to buy, we will have to multiply [tex]\$259[/tex] by 12:
[tex](12)(\$259)=\$3108[/tex] This is the amount Mr. Kim will spend in the plane tickets
A 1kg projectile is launched from a platform 2m above ground northwards with initial speed of 300m/s and an angle of elevation of π 4 above the horizon. If the wind applies a force of 3N to the east, find the position function of the object.
Answer:
the position equation of the projectile are
[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]
[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]
in x and y direction
Step-by-step explanation:
let the mass of the projectile be m, initial velocity be u
the wind applies a force of 3 newton in east direction.
therefore acceleration due to the force in east direction =[tex]\frac{force}{mass}[/tex]
= [tex]\frac{3}{m} =\frac{3}{1}[/tex]
acceleration due to gravity is in south direction = g
let east be x direction and north be y direction.
therefore acceleration in x direction = 3[tex]\frac{m}{s^{2} }[/tex] and in y direction = -g[tex]\frac{m}{s^{2} }[/tex]
writing equation of motion in x and y direction:
[tex]x = u_{x} t + \frac{1}{2} at^{2}[/tex]
[tex]y = u_{y} t + \frac{1}{2} at^{2}[/tex]
[tex]u_{x}[/tex]= ucos45 = [tex]\frac{300}{\sqrt{2} }[/tex]
[tex]u_{y}[/tex]= usin45= [tex]\frac{300}{\sqrt{2} }[/tex]
therefore
[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]
[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]
here 2 is added as the projectile already 2 meter above the ground