The answer to this problem involves applying integrals, norms, and concepts of angles between vectors to the functions f(x) and g(x). The INNER PRODUCT is the integral of the products of the two functions, the norms are the square roots of the inner products of the functions with themselves, and the angle between the functions is calculated using the dot product and norms.
Explanation:To find the inner product 〈f,g〉, the norms ∥f∥ and ∥g∥, and the angle αf,g between the functions f(x)=−10x2−6 and g(x)=−9x−4, we'll apply concepts from vector calculus. The inner product (also known as the dot product) is the integral from 0 to 1 of the products of the two functions. The norm of a function is the square root of the inner product of the function with itself. The angle between two vectors in a Vector Space, in this case the space of continuous functions C0[0,1], is given by cos(α) = 〈f,g〉/( ∥f∥∙ ∥g∥). Integrating and solving these equations will give us the desired values.
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These are the corrected values for the inner product, norms, and angle between [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]in the vector space [tex]\(C^0[0,1]\).[/tex]
1. The inner product \(\langle f,g \rangle\) is given by the integral [tex]\(\int_{1}^{0} f(x)g(x) \, dx\).[/tex] We will compute this integral with[tex]\(f(x) = -10x^2 - 6\) and \(g(x) = -9x - 4\).[/tex]
[tex]\[ \langle f,g \rangle = \int_{1}^{0} (-10x^2 - 6)(-9x - 4) \, dx \][/tex]
Expanding the integrand, we get:
[tex]\[ \int_{1}^{0} (90x^3 + 40x^2 + 54x + 24) \, dx \][/tex]
Integrating term by term, we have:
[tex]\[ \left[ \frac{90}{4}x^4 + \frac{40}{3}x^3 + \frac{54}{2}x^2 + 24x \right]_{1}^{0} \][/tex]
[tex]\[ = \left( \frac{90}{4} \cdot 0^4 + \frac{40}{3} \cdot 0^3 + \frac{54}{2} \cdot 0^2 + 24 \cdot 0 \right) - \left( \frac{90}{4} \cdot 1^4 + \frac{40}{3} \cdot 1^3 + \frac{54}{2} \cdot 1^2 + 24 \cdot 1 \right) \][/tex]
[tex]\[ = - \left( \frac{90}{4} + \frac{40}{3} + \frac{54}{2} + 24 \right) \][/tex]
[tex]\[ = - \left( 22.5 + 13.333 + 27 + 24 \right) \][/tex]
[tex]\[ = -87.833 \][/tex]
So, [tex]\(\langle f,g \rangle = -87.833\).[/tex]
2. The norm of f is given by [tex]\(\|f\|[/tex] = [tex]\sqrt{\langle f,f \rangle}\)[/tex], where:
[tex]\[ \langle f,f \rangle = \int_{1}^{0} (-10x^2 - 6)^2 \, dx \][/tex]
[tex]\[ = \int_{1}^{0} (100x^4 + 120x^2 + 36) \, dx \][/tex]
[tex]\[ = \left[ \frac{100}{5}x^5 + \frac{120}{3}x^3 + 36x \right]_{1}^{0} \][/tex]
[tex]\[ = - \left( \frac{100}{5} + \frac{120}{3} + 36 \right) \][/tex]
[tex]\[ = - \left( 20 + 40 + 36 \right) \][/tex]
[tex]\[ = -96 \][/tex]
[tex]\[ \|f\| = \sqrt{96} \][/tex]
3. Similarly, the norm of g is given by [tex]\(\|g\|[/tex] = [tex]\sqrt{\langle g,g \rangle}\)[/tex], where:
[tex]\[ \langle g,g \rangle = \int_{1}^{0} (-9x - 4)^2 \, dx \][/tex]
[tex]\[ = \int_{1}^{0} (81x^2 + 72x + 16) \, dx \][/tex]
[tex]\[ = \left[ \frac{81}{3}x^3 + 36x^2 + 16x \right]_{1}^{0} \][/tex]
[tex]\[ = - \left( \frac{81}{3} + 36 + 16 \right) \][/tex]
[tex]\[ = - \left( 27 + 36 + 16 \right) \][/tex]
[tex]\[ = -79 \][/tex]
[tex]\[ \|g\| = \sqrt{79} \][/tex]
4. The angle \[tex](\alpha_{f,g}\)[/tex] between [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] is found using the formula:
[tex]\[ \cos(\alpha_{f,g}) = \frac{\langle f,g \rangle}{\|f\| \cdot \|g\|} \][/tex]
[tex]\[ \alpha_{f,g} = \cos^{-1}\left(\frac{-87.833}{\sqrt{96} \cdot \sqrt{79}}\right) \][/tex]
[tex]\[ \langle f,g \rangle = -87.833, \quad \|f\| = \sqrt{96}, \quad \|g\| = \sqrt{79}, \quad \alpha_{f,g} = \cos^{-1}\left(\frac{-87.833}{\sqrt{96} \cdot \sqrt{79}}\right) \][/tex]
The correct calculations for the inner product and norms should be as follows:
[tex]\[ \langle f,g \rangle = \int_{0}^{1} (90x^3 + 40x^2 + 54x + 24) \, dx \][/tex]
[tex]\[ \|f\| = \sqrt{\int_{0}^{1} (100x^4 + 120x^2 + 36) \, dx} \][/tex]
[tex]\[ \|g\| = \sqrt{\int_{0}^{1} (81x^2 + 72x + 16) \, dx} \][/tex]
The angle calculation remains the same. Let's correct the integrals:
[tex]\[ \langle f,g \rangle = \left[ \frac{90}{4}x^4 + \frac{40}{3}x^3 + \frac{54}{2}x^2 + 24x \right]_{0}^{1} \][/tex]
[tex]\[ = \frac{90}{4} + \frac{40}{3} + \frac{54}{2} + 24 \][/tex]
[tex]\[ = \frac{90}{4} + \frac{40}{3} + \frac{54}{2} + 24 \][/tex]
[tex]\[ = 87.833 \][/tex]
So,[tex]\(\langle f,g \rangle = 87.833\).[/tex]
[tex]\[ \|f\| = \sqrt{\left[ \frac{100}{5}x^5 + \frac{120}{3}x^3 + 36x \right]_{0}^{1}} \][/tex]
[tex]\[ = \sqrt{20 + 40 + 36} \][/tex]
[tex]\[ = \sqrt{96} \][/tex]
[tex]\[ \|g\| = \sqrt{\left[ \frac{81}{3}x^3 + 36x^2 + 16x \right]_{0}^{1}} \][/tex]
[tex]\[ = \sqrt{27 + 36 + 16} \][/tex]
[tex]\[ = \sqrt{79} \][/tex]
The corrected values are:
[tex]\[ \langle f,g \rangle = 87.833, \quad \|f\| = \sqrt{96}, \quad \|g\| = \sqrt{79} \][/tex]
The angle calculation remains the same:
[tex]\[ \alpha_{f,g} = \cos^{-1}\left(\frac{87.833}{\sqrt{96} \cdot \sqrt{79}}\right) \][/tex]
The following information was obtained from matched samples. Individual Method 1 Method 2 1 7 5 2 5 9 3 6 8 4 7 7 5 5 6 Refer to Exhibit 10-5. If the null hypothesis is tested at the 5% level, the null hypothesis _____.
a. should be rejected
b. should be revised
c. should not be rejected
d. None of these answers are correct
Answer:
[tex]p_v =2*P(t_{(8)}<-1.195)=0.266[/tex]
c. should not be rejected
Step-by-step explanation:
1) Data given and notation
The data given is:
Method 1 : 7,5,6,7,5
Method 2: 5,9,8,7,6
[tex]\bar X_{1}=6[/tex] represent the mean for the method 1
[tex]\bar X_{2}=7[/tex] represent the mean for the method 2
[tex]s_{1}=1[/tex] represent the sample standard deviation for the method 1
[tex]s_{2}=1.58[/tex] represent the sample standard deviation for the method 2
[tex]n_{1}=5[/tex] sample size selected for the Consultant A
[tex]n_{2}=5[/tex] sample size selected for the Consultant B
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean for the difference is equal to 0, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}- \mu_{2}=0[/tex]
Alternative hypothesis:[tex]\mu_{1} - \mu_{2}\neq 0[/tex]
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{6-7}{\sqrt{\frac{1^2}{5}+\frac{1.58^2}{5}}}=-1.195[/tex] (1)
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{1}+n_{2}-2=5+5-2=8[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(8)}<-1.195)=0.266[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. So the best option would be:
c. should not be rejected
Suppose y varies directly with x if y=27 when x=3 what is y when x=5
The value of y when x = 5 is 45
Solution:Given that y varies directly with x. This can be written mathematically as:
y ∝ x
[tex]y = k \times x[/tex] ---- eqn 1
Where "k" is the constant of propotionality
Given y = 27 when x = 3. Substitute these values in eqn 1
[tex]27 = k \times 3\\\\k = \frac{27}{3} = 9[/tex]
k = 9Substitute k = 9 in eqn 1
y = 9x ----- eqn 2
To find: y = ? and x = 5
Substitute x = 5 in eqn 2
[tex]y = 9 \times 5 = 45[/tex]
Thus the value of y when x = 5 is 45
For a field trip the school bought 39 sandwiches for 4.35 each and 42 bags of chips for 15 cent each how much did the school spend in all
Answer:
Step-by-step explanation:
For a field trip, the school bought 39 sandwiches and 42 bags of chips. Each sandwich cost 4.35. That means the cost of 39 sandwiches would be
39 × 4.35 = 169.65
Each bag of chip costs 15 cents. Converting 15 cents to dollars, it becomes
15/100 =$ 0.15
Therefore, the cost of 42 bags of chips would be
42 × 0.15 = 6.3
Total cost of 39 sandwiches and 42 bags of chips would be
169.65 + 6.3 = 175.95
A car has 200 joules of gravitational potential energy. How far is the height of the car weighs 300 Newton’s?
Final answer:
To find the height of the car, we can rearrange the equation for potential energy and solve for h. Using the given values of 200 J for potential energy and 300 N for weight, the height of the car is approximately 0.683 meters.
Explanation:
To calculate the height of a car that weighs 300 Newtons and has 200 joules of gravitational potential energy, we can use the formula for potential energy, PE = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height. Since we know the potential energy and the weight (force) of the car, we can rearrange the formula to solve for h. Here's how:
Start with the equation for potential energy: PE = mghRearrange the equation to solve for h: h = PE / (mg)Plug in the given values: h = 200 J / (300 N * 9.81 m/s²)Solve the equation to find the height: h ≈ 0.683 metersTherefore, the height of the car is approximately 0.683 meters.
The time to fly between New York City and Chicago is uniformly distributed with a minimum of 120 minutes and a maximum of 150 minutes. What is the distribution's standard deviation? Select one: a. 8.66 minutes b. 75 minutes c. 135 minutes d. 270 minutes
Answer:
a. 8.66 minutes
Step-by-step explanation:
Since the flight times are uniformly distributed, the standard deviation can be calculated as follows:
[tex]\sigma = \frac{b-a}{\sqrt{12}}[/tex]
Where 'b' is the maximum flight time (150 minutes) and 'a' is the minimum flight time (120 minutes):
[tex]\sigma = \frac{150-120}{\sqrt{12}}\\\sigma = 8.66\ minutes[/tex]
The distribution's standard deviation is 8.66 minutes.
What is the difference between the votes for red and the votes for blue and yellow combined?
Red 11
Blue 8
Green 10
Yellow 3
Answer:
0
Step-by-step explanation:
8+3=11.
11-11=0
So the answer is zero.
Hope that helps!
The marginal cost of manufacturing an item when x thousand items are produced is dC/dx= 4x^3 - 6x + 5 dolars/item.Find the cost function C(x) if C(0)=550
Answer:
The cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].
Step-by-step explanation:
It is given that the marginal cost of manufacturing an item when x thousand items are produced is
[tex]\frac{dC}{dx}=4x^3-6x+5[/tex]
We need to find the cost function.
Multiply both sides by dx.
[tex]dC=(4x^3-6x+5)dx[/tex]
Integrate both sides to find the cost function.
[tex]\int dC=\int (4x^3-6x+5)dx[/tex]
[tex]\int dC=4\int x^3dx-6\int xdx+5\int 1 dx[/tex]
[tex]C(x)=4(\frac{x^4}{4})-6(\frac{x^2}{2})+5x+C[/tex]
where, C(x) is const function and C is a constant.
[tex]C(x)=x^4-3x^2+5x+C[/tex]
It is given that C(0)=550. Substitute x=0 in the above function.
[tex]C(0)=(0)^4-3(0)^2+5(0)+C[/tex]
[tex]C(0)=C[/tex]
[tex]550=C[/tex]
Therefore, the cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].
A chef needs vinegar with an 8% acidity level to make her special sauerbraten. She only has two types of vinegar, one with an acidity level of 5% and the other with 12% acidity level. How many ounces of the 12% vinegar should she add to 40 ounces of the 5% vinegar to obtain the needed vinegar with an 8% acidity level
Answer:
She should add 30 ounces of the 12% vinegar she should add to 40 ounces of the 5% vinegar to obtain the needed vinegar with an 8% acidity level.
Step-by-step explanation:
Let X be ounces of the 12% vinegar she should add to 40 ounces of the 5% vinegar to obtain the needed vinegar with an 8% acidity level.
We have the equation
[tex]\frac{(X*0.12)+(40*0.05}{X+40} =0.08[/tex] when 8% acidity level is reached.
Solving the equation we have
X*0.12+2=0.08(X+40)=0.08X+3.2
That is 0.12X-0.08X=3.2-2=1.2
Finally 0.04X=1.2 and
X=30
Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1 PM and 7 PM. Indepen- dently of when the plumber arrives, the time it takes to fix the broken faucet is exponentially distributed with mean 30 minutes.
1. Find the expectation and variance of the time at which the plumber completes the project?
Final answer:
The expectation of the time at which the plumber completes the project is 4:30 PM, and the variance is 901.33 [tex]minutes^2[/tex]
Explanation:
To find the expectation and variance of the time at which the plumber completes the project, we need to use the properties of the uniform and exponential distributions.
1. The arrival time of the plumber is uniformly distributed between 1 PM and 7 PM, which means it has a continuous uniform distribution. The expectation of a continuous uniform distribution is calculated as the average of the lower and upper bounds, so the expectation of the arrival time is (1 + 7) / 2 = 4 PM. The variance of a continuous uniform distribution is calculated as [tex](upper bound - lower bound)^2 / 12,[/tex] so the variance of the arrival time is [tex](7 - 1)^2 / 12 = 1.33.[/tex]
2. The time it takes to fix the broken faucet is exponentially distributed with a mean of 30 minutes, which means it has an exponential distribution. The expectation of an exponential distribution is equal to its mean, so the expectation of the fixing time is 30 minutes. The variance of an exponential distribution is equal to the square of its mean, so the variance of the fixing time is [tex]30^2 = 900 minutes^2[/tex].
Since the arrival time and fixing time are independent, the total time at which the plumber completes the project is the sum of the arrival time and fixing time. The expectation of the total time is the sum of the expectations, which is 4 PM + 30 minutes = 4:30 PM. The variance of the total time is the sum of the variances, which is 1.33 + 900 = 901.33 [tex]minutes^2.[/tex]
When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population correction factor:Image for When obtaining a confidence interval for a population mean in the case of a finite population of size N and aFind the 95% confidence interval for the mean of 200 weights if a sample of 32 of those weights yields a mean of 150.6 lb and a standard deviation of 24.4 lb.a. 143.9 lb < µ < 157.3 lbb. 141.4 lb < µ < 159.8 lbc. 142.8 lb < µ < 158.4 lbd. 142.1 lb < µ < 159.1 lb
Answer:
95% Confidence interval for the mean
[tex]142.8 \leq\mu\leq158.4[/tex]
Step-by-step explanation:
We have to calculate a 95% confidence interval for the mean of a finite population.
The error is multiplied by the following finite population correction factor:
[tex]cf=\sqrt{\frac{N-n}{N-1} }[/tex]
The standard deviation can be estimated as
[tex]\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963[/tex]
The 95% confidence interval has a z value of 1.96, so it becomes:
[tex]M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4[/tex]
A financial advisor believes that the proportion of investors who are risk-averse (that is, try to avoid risk in their investment decisions) is at least 0.6. A survey of 33 investors found that 21 of them were risk-averse. Formulate a one-sample hypothesis test for a proportion to test this belief.
Based on the given data, cannot conclude that the financial advisor's belief is accurate. Since the test statistic (Z-score) of 0.4483 is less than the critical value of 1.645, fail to reject the null hypothesis
To formulate a one-sample hypothesis test for a proportion, we can use the following steps:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (H1).
The null hypothesis assumes that the proportion of risk-averse investors is equal to or less than 0.6. The alternative hypothesis states that the proportion of risk-averse investors is greater than 0.6.H0: p ≤ 0.6 (proportion of risk-averse investors is less than or equal to 0.6)H1: p > 0.6 (proportion of risk-averse investors is greater than 0.6)Step 2: Determine the significance level (α) for the hypothesis test.
The significance level represents the probability of rejecting the null hypothesis when it is true.
Let's assume a significance level of α = 0.05, which is a commonly used value.
Step 3: Collect the data and calculate the test statistic.
Given the data from the survey: out of 33 investors, 21 were risk-averse. To calculate the test statistic, we need to find the sample proportion ([tex]\bar{p}[/tex]) and the standard error (SE).Sample proportion ([tex]\bar{p}[/tex]) = Number of risk-averse investors / Total number of investors[tex]\bar{p}= 21 / 33[/tex]
On dividing gives:
[tex]\bar{p}\approx 0.6364[/tex]
Standard Error (SE) [tex]= \sqrt{(\bar{p} * (1 -\bar{p})) / n}[/tex]
plugging given data gives:
[tex]= \sqrt{(0.6364 * (1 - 0.6364)) / 33}[/tex]
On simplifying gives:
≈ 0.0811
Step 4: Determine the critical value or the p-value.
Since we are testing the proportion of risk-averse investors is greater than 0.6, this is a one-tailed test. Either use the critical value approach or calculate the p-value. Let's use the critical value approach.To determine the critical value, to use the Z-table . For a significance level of α = 0.05 (one-tailed),
the critical value corresponds to a z-score of approximately 1.645.
Step 5: Make a decision and interpret the results.
Calculate the test statistic (Z-score):
[tex]Z = (\bar{p} - p) / SE[/tex]
Plugging the given values gives:
[tex]= (0.6364 - 0.6) / 0.0811[/tex]
On simplifying gives:
[tex]\approx 0.4483[/tex]
Since the test statistic (Z-score) of 0.4483 is less than the critical value of 1.645, fail to reject the null hypothesis. There is not enough evidence to support the claim that the proportion of risk-averse investors is greater than 0.6 at a significance level of 0.05.
Therefore, based on the given data, we cannot conclude that the financial advisor's belief is accurate.
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In a large Introductory Statistics lecture hall, the professor reports that 55% of the students enrolled have never taken a Calculus course, 32% have taken only one semester of Calculus, and the rest have taken two or more semesters of Calculus. The professor randomly assigns students to groups of three to work on a project for the course. What is the probability that the first groupmate you meet has studied
a) two or more semesters of Calculus?
b) some Calculus?
c) no more than one semester of Calculus?
Answer:
a) There is a 13% probability that a student has taken 2 or more semesters of Calculus.
b) 45% probability that a student has taken some calculus.
c) 87% probability that a student has taken no more than one semester of calculus.
Step-by-step explanation:
We have these following probabilities:
A 55% that a student hast never taken a Calculus course.
A 32% probability that a student has taken one semester of a Calculus course.
A 100-(55+32) = 13% probability that a student has taken 2 or more semesters of Calculus.
a) two or more semesters of Calculus?
There is a 13% probability that a student has taken 2 or more semesters of Calculus.
b) some Calculus?
At least one semester.
So there is a 32+13 = 45% probability that a student has taken some calculus.
c) no more than one semester of Calculus?
At most one semester.
So 55+32 = 87% probability that a student has taken no more than one semester of calculus.
The correct probabilities for the given scenarios are:
a) The probability that the first groupmate you meet has studied two or more semesters of Calculus is [tex]$\boxed{\frac{13}{100}}$[/tex].
b) The probability that the first groupmate you meet has studied some Calculus (which includes those who have taken only one semester and those who have taken two or more semesters) is [tex]$\boxed{\frac{45}{100}}$[/tex].
c) The probability that the first groupmate you meet has studied no more than one semester of Calculus is [tex]$\boxed{\frac{32}{100}}$[/tex].
To calculate these probabilities, we use the information provided by the professor:
- 55% of the students have never taken a Calculus course.
- 32% have taken only one semester of Calculus.
- The rest, which is 100% - (55% + 32%) = 13%, have taken two or more semesters of Calculus.
Now, let's calculate the probabilities for each part of the question:
a) The probability that the first groupmate has studied two or more semesters of Calculus is the percentage of students who have taken two or more semesters, which is 13%. In probability terms, this is [tex]$\frac{13}{100}$[/tex]
b) The probability that the first groupmate has studied some Calculus is the sum of the percentages of students who have taken at least one semester of Calculus. This includes both those who have taken only one semester (32%) and those who have taken two or more semesters (13%). Therefore, the probability is 32% + 13% = 45%. In probability terms, this is [tex]$\frac{45}{100}$[/tex].
c) The probability that the first groupmate has studied no more than one semester of Calculus is the sum of the percentages of students who have never taken Calculus and those who have taken only one semester. This is 55% + 32% = 87%. However, since we are looking for the probability of no more than one semester, we exclude those who have taken two or more semesters, so we subtract the 13% from the 87%, giving us 87% - 13% = 74%. But this calculation is incorrect because we have double-counted the 32% of students who have taken only one semester. The correct calculation is to consider only the students who have never taken Calculus (55%) and those who have taken only one semester (32%), which gives us 55% + 32% - 0% (since we do not need to subtract any percentage this time) = 87%. In probability terms, this is [tex]$\frac{32}{100}$[/tex], as we are only considering those who have taken no Calculus or just one semester.
The times that a cashier spends processing individual customer’s order are independent random variables with mean 2.5 minutes and standard deviation 2 minutes. What is the approximate probability that it will take more than 4 hours to process the orders of 100 people?
Answer:
0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 2.5 minutes
Standard Deviation, σ = 2 minutes
Since the sample size is large, by central limit theorem, the distribution of sample means is approximately normal.
[tex]P(\sum x_{i} > 4)\\P(\sum x_i > 4\times 60\text{ minutes})\\\\P(\displaystyle\frac{1}{100}\sum x_i > \frac{4\times 60}{100}\text{ minutes})\\\\P(\bar{x} > 2.4)[/tex]
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
P(it will take more than 4 hours to process the orders of 100 people)
P(x > 2.4)
[tex]P( x > 2.4) = P( z > \displaystyle\frac{2.4-2.5}{\frac{2}{\sqrt{100}}}) = P(z > -0.5)[/tex]
Calculation the value from standard normal z table, we have, [tex]P(x > 2.4) = 1 - 0.3085 = 0.6915= 69.15\%[/tex]
0.6915 is the probability that it will take more than 4 hours to process the orders of 100 people.
Final answer:
Using the Central Limit Theorem, the mean and standard deviation for 100 customers were calculated. The z-score was computed to find the probability from the standard normal distribution. The probability of taking more than 240 minutes to serve 100 customers is less than 0.5.
Explanation:
To solve this problem, we need to use the Central Limit Theorem which states that the sum or the average of a large number of independent and identically distributed random variables will be approximately normally distributed, irrespective of the original distribution of the variables.
Step by Step Solution:
First, we calculate the mean total time to process orders for 100 people. Mean total time = mean time per customer * number of customers = 2.5 minutes * 100 = 250 minutes.
Next, we calculate the standard deviation of the total time for 100 customers. Since the customer service times are independent, we can use the formula for the standard deviation of the sum of independent random variables: Standard deviation of the total time = standard deviation per customer * sqrt(number of customers) = 2 minutes * sqrt(100) = 20 minutes.
To find out if it will take more than 4 hours (which is 240 minutes), we find the z-score: z = (Total minutes to consider - Mean total time) / Standard deviation of total time = (240 - 250) / 20 = -0.5.
The probability of service taking more than 4 hours is the same as the probability of the sum being greater than 240 minutes. We use the z-score to find this probability from the standard normal distribution table or using a calculator with normal distribution functions.
Since a z-score of -0.5 corresponds to a probability higher than 0.5 (due to the symmetry of the normal distribution), the probability of taking less than 240 minutes is above 0.5. Therefore, the probability of taking more than 240 minutes is less than 0.5.
A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line.
The machine that dispenses dressing is working properly when 8 ounces are dispensed. The standard deviation of the process is 0.15 ounces.
A sample of 48 bottles is selected periodically, and the filling line is stopped if there is evidence that the mean amount dispensed is different from 8 ounces.
Suppose that the mean amount dispensed in a particular sample of 48 bottles is 7.983 ounces.
Calculate the P-Value.
Answer:
P-value = 0.4324
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 8 ounces
Sample mean, [tex]\bar{x}[/tex] = 7.983 ounces
Sample size, n = 48
Population standard deviation, σ = 0.15 ounces
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 8\text{ ounces}\\H_A: \mu \neq 8\text{ ounces}[/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{7.983 - 8}{\frac{0.15}{\sqrt{48}} } = -0.7851[/tex]
Now, we calculate the p-value with the help of standard normal z table.
P-value = 0.4324
Final answer:
To calculate the P-Value, we need to perform a hypothesis test. The test statistic is calculated by subtracting the hypothesized mean from the sample mean and dividing by the standard error. The P-Value is the probability of observing a test statistic as extreme as the calculated value or more extreme under the null hypothesis.
Explanation:
To calculate the P-Value, we need to perform a hypothesis test. The null hypothesis, H0, is that the mean amount dispensed is equal to 8 ounces, while the alternative hypothesis, Ha, is that the mean amount dispensed is different from 8 ounces.
Using the given information, we can calculate the test statistic, which is the standardized value of the sample mean. First, we calculate the standard error of the sample mean, which is the standard deviation divided by the square root of the sample size. In this case, the standard error is 0.15 / sqrt(48) = 0.021.
The test statistic is then calculated by subtracting the hypothesized mean (8 ounces) from the sample mean (7.983 ounces) and dividing by the standard error. So, the test statistic is (7.983 - 8) / 0.021 = -0.619.
To calculate the P-Value, we need to find the probability of observing a test statistic as extreme as -0.619 (or more extreme) under the null hypothesis. We can use a standard normal distribution table or a calculator to find the corresponding area under the curve. From the table or calculator, we find that the area to the left of -0.619 is 0.269.
Since the alternative hypothesis is two-sided (the mean could be either greater or less than 8 ounces), we double the area to get the P-Value. So, the P-Value is 2 * 0.269 = 0.538.
How do I solve this (see attachment)
Answer: above
Step-by-step explanation:
ok so angle is 30 and we have adjacent and we have to figure out opposite so adjacent and opposite, we need tan so
let x be height of triangle
tan 30= x/81
x= (tan 30) *81
x= about 48ft
48 + 4 is 52 which is more than 50
The equation of a circle is (x + 6)^2 + (y - 4)^2 = 16. The point (-6, 8) is on the circle.
What is the equation of the line that is tangent to the circle at (-6, 8)?
Answer:
y = 8 is the equation of tangent.
Step-by-step explanation:
The equation of the tangent to the circle at (-6,8) is of the form:
y = mx + c
where m is the slope of the tangent and c is the y-intercept.
The point (-6,8) lies on the circle and the tangent line as well.
Hence (-6,8) satisfies the line equation:
8 = m(-6) + c ⇒ c-6m = 8 -------------1
We know that slope of two perpendicular lines are related as:
[tex]m_{1}\times m_{2}=-1[/tex]
At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.
Hence :
[tex]m_{normal} \times m_{tangent}=-1[/tex]
We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
[tex]=\frac{8-4}{-6+6}[/tex]
= ∞
Slope of the normal is infinity and hence slope of tangent is -1/∞ = 0
Hence m=0
Putting m=0 in equation 1 we get:
c = 8
The equation of tangent line at (-6,8) is:
y = 8
Answer:
y = 8
Step-by-step explanation:
The equation of the tangent to the circle at (-6,8) is of the form:
y = mx + c
where m is the slope of the tangent and c is the y-intercept.
The point (-6,8) lies on the circle and the tangent line as well.
Hence (-6,8) satisfies the line equation:
8 = m(-6) + c ⇒ c-6m = 8 -------------1
The electricity rates charged by Monroe Utilities in the summer months are as follows: Base Charge = $ 8 $8 First 800 800 kWh or less at $ 0.05 $0.05/kWh Over 800 800 kWh at $ 0.08 $0.08/kWh The base is a fixed monthly charge, independent of the kWh (kilowatt-hours) used during the month. a . a. Find an expression for the cost function C ( x ) C(x) for usage of or under 800 800 kWh.
Answer:
[tex]C(x) = \$8 + \$0.05*x[/tex]
Step-by-step explanation:
Let 'x' be the usage in kWh. When usage is at or under 800 kWh, the cost function C(x) is given by the base charge of $8 added to the rate of $0.05/kWh multiplied by the consumption 'x', in kWh.
Therefore, for 0 ≤ x ≤ 800, the cost function is:
[tex]C(x) = \$8 + \$0.05*x[/tex]
Do SAT coaching classes work? Do they help students to improve their test scores? Four students were selected randomly from all of the students that completed an SAT coaching class. For each student, we recorded their first SAT score (before the coaching class) and their second SAT score (after the coaching class).
Student
1
2
3
4
First SAT score
920
830
960
910
Second SAT score
1010
800
1000
980
To analyze these data, we should use
A.
the one-sample t test., one answer Yahoo,
B.
the matched pairs t test.
C.
the two-sample t test.
D.) Any of the above are valid. It just needs to be a t since ? is unknown.
Answer:
B. the matched pairs t test.
Step-by-step explanation:
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.
Let put some notation
x=first test value , y = second test value
x: 920, 830, 960, 910
y: 1010, 800, 1000, 980
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x \leq 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x >0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: 90, -30, 40, 70
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{170}{4}=42.5[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =52.520[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{42.5 -0}{\frac{52.520}{\sqrt{4}}}=1.618[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=4-1=3[/tex]
Now we can calculate the p value, since we have a right tailed test the p value is given by:
[tex]p_v =P(t_{(3)}>1.618) =0.1024[/tex]
So the p value is higher than any significance level assumed (0.05), so then we can conclude that we reject the null hypothesis. So we can conclude that the difference between the after and the initial score it's not significantly higher at 5% of significance.
Solve the following system of equations by the substitution method. 3x = y + 6 x = 2y - 1
Answer:
[tex]x=\frac{13}{5}, y= \frac{9}{5}[/tex]
Step-by-step explanation:
First Equation:
3x = y + 6
Or,
y = 3x - 6
2nd Equation:
x = 2y - 1
Lets substitute the expression for "y" from 1st equation in 2nd and solve for x:
x = 2y - 1
x = 2(3x - 6) - 1
x = 6x - 12 - 1
x = 6x - 13
5x = 13
x = 13/5
Now we can substitute this value of x into bold 1st equation and get y:
y = 3x - 6
y = 3(13/5) - 6
y = 9/5
This is the solution
Suppose a package delivery company purchased 14 trucks at the same time. Five trucks were purchased from manufacturer A, four from manufacturer B, and five from manufacturer C. The cost of maintaining each truck was recorded. The company used ANOVA to test if the mean maintenance costs of the trucks from each manufacturer were equal. To apply the F test, how many degrees of freedom must be in the denominator?
Answer:
14
Step-by-step explanation:
Given that a package delivery company purchased 14 trucks at the same time. Five trucks were purchased from manufacturer A, four from manufacturer B, and five from manufacturer C. The cost of maintaining each truck was recorded.
No of companies = 3
No of items in total = 14
Total df = 17-1=16
degrees of freedom between groups = 3-1 =2
Hence degrees of freedom for denominator
=[tex]16-2 =14[/tex]
Consider two populations for which μ1 = 31, σ1 = 3, μ2 = 29, and σ2 = 2. Suppose that two independent random samples of sizes n1 = 4 and n2 = 53 are selected. Describe the approximate sampling distribution of x1 − x2 (center, spread, and shape).a. What is the mean of the distribution?b. What is the standard deviation of the distribution?
Answer:
Part (a): The mean of the distribution is 2.
Part (b): The standard deviation of the distribution is 1.525
Step-by-step explanation:
Consider the provided information.
Part (a) What is the mean of the distribution?
The mean of the sampling distribution of [tex](\bar x_1 -\bar x_2)[/tex] is (µ₁ − µ₂).
It is given that μ₁ = 31, σ₁ = 3, μ₂ = 29, and σ₂ = 2.
Mean of distribution = (x₁ - x₂)
Mean of distribution = (31 - 29) = 2
Hence, the mean of the distribution is 2.
Part (b) What is the standard deviation of the distribution?
If the two samples are independent, the standard deviation of the sampling distribution is: [tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}\right)}}[/tex]
Substitute the respective values in the above formula.
[tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{3^2}{4}+\dfrac{2^2}{53}\right)}}[/tex]
[tex]\sigma_{(x_1-x_2)}=\sqrt{\left(\dfrac{9}{4}+\dfrac{4}{53}\right)}}[/tex]
[tex]\sigma_{(x_1-x_2)}\approx1.525[/tex]
Hence, the standard deviation of the distribution is 1.525
The approximate sampling distribution of x1 - x2 has a mean equal to μ1 - μ2 and a standard deviation equal to √((σ1^2/n1) + (σ2^2/n2)).
Explanation:A sampling distribution is the probability distribution of a statistic, like the mean or proportion, calculated from multiple samples of the same size drawn from a population. It provides insights into the variability of the statistic and is fundamental in statistical inference.
The approximate sampling distribution of x1 - x2 has a mean equal to the difference of the population means, μ1 - μ2. The standard deviation of the distribution is equal to the square root of the sum of the variances of the two populations divided by their respective sample sizes, √((σ1^2/n1) + (σ2^2/n2)).
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The output of an economic system Q, subject to two inputs, such as labor L and capital K, is often modeled by the Cobb-Douglas production function Q=cL^a K^b. When a+b=1, the case is called constant returns to scale. Suppose Q=12,200, a = 1/6, b= 5/6, and c=42. Find the rate of change of capital with respect to labor, dK/dL.
Answer:
Step-by-step explanation:
Given
Economic system Q is given by
[tex]Q=cL^aK^b[/tex]
also [tex]a+b=1[/tex]
if [tex]Q=12,200[/tex]
[tex]a=\frac{1}{6}[/tex]
[tex]b=\frac{5}{6}[/tex]
[tex]c=42[/tex]
substitute these values
[tex]12,200=42\times (L)^{\frac{1}{6}}K^{\frac{5}{6}}[/tex]
[tex](L)^{\frac{1}{6}}K^{\frac{5}{6}}=\frac{12,200}{42}[/tex]
[tex]K^{\frac{5}{6}}=\frac{12,200}{42(L)^{\frac{1}{6}}}[/tex]
[tex]K=(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^{5}}[/tex]
differentiate w.r.t to L to get [tex]\frac{dK}{dL}[/tex]
[tex]\frac{dK}{dL}=(\frac{12,200}{42})^{\frac{6}{5}}\times (-5)\times L^{-6}[/tex]
[tex]\frac{dK}{dL}=-5(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^6}[/tex]
[tex]\frac{dK}{dL}=-\frac{4515.466}{L^6}[/tex]
The department of public safety has an old memo stating that the number of accidents per week at a hazardous intersection varies according to a Normal distribution, with a mean of 2.2 and a standard deviation of 1.4. Department officials implemented a new safety plan, heavier police patrols and new signs, to see if they could reduce the average number of accidents at this intersection. They recorded the number of accidents per week for 52 weeks. They find that the average over that period was two accidents per week. What is the P ‑value for the test of H 0 : μ = 2.2 against H a : μ < 2.2 ?
Answer:
P-value = 0.1515
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 2.2
Sample mean, [tex]\bar{x}[/tex] = 2
Sample size, n = 52
Alpha, α = 0.05
Population standard deviation, σ = 1.4
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu =2.2\\H_A: \mu < 2.2[/tex]
We use one-tailed(left) 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{2 - 2.2}{\frac{1.4}{\sqrt{52}} } = -1.03[/tex]
Now, we calculate the p-value from the normal standard table.
P-value = 0.1515
g A coffee-dispensing machine is supposed to deliver eight ounces of liquid into each paper cup, but a consumer believes that the actual mean amount is less. The critical value for z for a one-tailed test with the tail in the left end is minus1.645 and the obtained value is minus1.87. The appropriate decision is ________.
Answer: Reject the eight- ounces claim.
Step-by-step explanation:
For left tailed test , On a normal curve the rejection area lies on the left side of the critical value.
It means that if the observed z-value is less than the critical value then it will fall into the rejection region other wise not.
As per given ,
Objective : A coffee-dispensing machine is supposed to deliver eight ounces of liquid or less.
Then ,
[tex]H_0: \mu=8\\\\H_a: \mu<8[/tex] , since alternative hypothesis is left-tailed thus the test is an left-tailed test.
the critical value for z for a one-tailed test with the tail in the left end is -1.645 and the obtained value is -1.87.
Clearly , -1.87 < -1.645
⇒ -1.87 falls under rejection region.
⇒ Decision : Reject null hypothesis.
i.e. we reject the eight- ounces claim.
The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 16-inch pizzas. She takes a random sample of 25 pizzas and records their mean and standard deviation as 16.10 inches and 1.8 inches, respectively. She subsequently computes a 95% confidence interval of the mean size of all pizzas as [15.36, 16.84]. However, she finds this interval to be too broad to implement quality control and decides to reestimate the mean based on a bigger sample. Using the standard deviation estimate of 1.8 from her earlier analysis, how large a sample must she take if she wants the margin of error to be under 0.5 inch?
Answer:
n=50
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=16.10[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=1.8 represent the sample standard deviation
n=25 represent the sample size
2) Solution to the problem
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)
We need to find the degrees of freedom given by:
[tex]df=n-1=25-1=24[/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 table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=\pm 1.96[/tex]
Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =0.5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
Replacing into formula (b) we got:
[tex]n=(\frac{1.96(1.8)}{0.5})^2 =49.79 \approx 50[/tex]
So the answer for this case would be n=50 rounded up to the nearest integer
The average life expectancy of tires produced by the Whitney Tire Company has been 40,000 miles. Management believes that due to a new production process, the life expectancy of their tires has increased. In order to test the validity of their belief, the correct set of hypotheses is a. H0: μ > 40,000 Ha: μ ≤ 40,000. b. H0: μ ≥ 40,000 Ha: μ < 40,000. c. H0: μ < 40,000 Ha: μ ≥ 40,000. d. H0: μ ≤ 40,000 Ha: μ > 40,000.
Answer:
[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 40,000 miles
Management believes that due to a new production process, the life expectancy of their tires has increased.
We design the null and the alternate hypothesis in the following manner:
[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]
The null hypothesis states before the new production process life expectancy of tires is equal to or less than 40,000 miles, while the alternate hypothesis states that after the new production process life expectancy of tires is greater than 40,000 miles.
Option D)
[tex]H_{0}: \mu \leq 40,000\text{ miles}\\H_A: \mu > 40,000\text{ miles}[/tex]
Final answer:
The correct hypotheses for checking if the average life expectancy of tires has increased are : μ ≤ 40,000 and : μ > 40,000, where μ represents the mean mileage for the population of tires. Option d) is the correct answer.
Explanation:
The correct set of hypotheses for testing whether the average life expectancy of tires produced by Whitney Tire Company has increased due to a new production process is:
μ ≤ 40,000 (null hypothesis)
μ > 40,000 (alternative hypothesis)
This hypothesis testing is set up to check if there is evidence to support the claim that the average life expectancy of the tires is greater than 40,000 miles. Thus, option d. : μ ≤ 40,000 : μ > 40,000 is correct. Here, μ stands for the mean mileage for the population of tires. The null hypothesis always contains an equality claim (≤, =, ≥), and the alternative hypothesis contains the opposite inequality, representing the claim we are trying to find evidence for.
Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying within their school. They select a random sample of 200 students who are asked to complete a survey anonymously. Of these students, 63 report that they have experienced bullying. Use this information to find a 90% z-confidence interval for p, the true proportion of students at this school who have experienced bullying. Give the limits (bounds) of the confidence interval as a proportion, precise to three decimal places.
lower limit =______________
upper limit =___________
Answer:
lower limit =0.261
upper limit =0.369
Step-by-step explanation:
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Description in words of the parameter p
[tex]p[/tex] represent the real population proportion of people that have experienced bullying
X= 63 people in the random sample that have experienced bullying
n=200 is the sample size required
[tex]\hat p=\frac{63}{200}=0.315[/tex] represent the estimated proportion of people that have experienced bullying
[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Numerical estimate for p
In order to estimate a proportion we use this formula:
[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.
[tex]\hat p=\frac{63}{200}=0.315[/tex] represent the estimated proportion of people that they were planning to pursue a graduate degree
Confidence interval
The confidence interval for a proportion is given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.64[/tex]
And replacing into the confidence interval formula we got:
[tex]0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261[/tex]
[tex]0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369[/tex]
And the 90% confidence interval would be given (0.261;0.369).
We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).
lower limit =0.261
upper limit =0.369
The lower limit of the confidence interval is 0.270, and the upper limit is 0.360.
Explanation:To find the 90% z-confidence interval for the true proportion of students who have experienced bullying, we can use the formula:
CI = p ± z * sqrt((p(1-p))/n)
Where:
p is the sample proportion (63/200 = 0.315)
z is the z-value for a 90% confidence interval (z = 1.645)
n is the sample size (200)
Plugging in these values, we get:
CI = 0.315 ± 1.645 * sqrt((0.315(1-0.315))/200)
Simplifying the equation gives us:
CI = 0.315 ± 0.045
So, the lower limit of the confidence interval is 0.315 - 0.045 = 0.270, and the upper limit is 0.315 + 0.045 = 0.360.
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In a recent survey, 80% of the community favored building a police substation in their neighborhood. If 15 citizens are chosen, what is the mean number favoring the substation?
Answer:
The mean number favoring the substation is 12 citizens.
Step-by-step explanation:
For each citizen surveyed, there are only two possible outcomes. EIther they favored building a police substation in their neighborhood, or they opposed. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, with p probability, and X can only have two outcomes.
Has an expected value of:
[tex]E(X) = np[/tex]
In this problem, we have that:
[tex]n = 15, p = 0.8[/tex]
E(X) = 15*0.8 = 12.
The mean number favoring the substation is 12 citizens.
Final answer:
The mean number of citizens out of 15 who favor the construction of a police substation is 12, calculated by multiplying the sample size (15) by the probability of favoring the substation (0.80).
Explanation:
To find the mean number favoring the substation, we'll use the probability of a citizen favoring the substation to calculate the expected value in a sample of 15 citizens. Since 80% of the community favored building a police substation, the mean number favoring the substation can be computed by multiplying the sample size by the probability of favoring the substation.
Mean = Sample Size × Probability of Favoring the Substation
Mean = 15 × 0.80
Mean = 12
Therefore, the mean number of citizens out of 15 who favor the construction of the police substation is 12.
You want to test your newly created Web site, so you have 250 people access it from random locations at random times. Of the people accessing the site, 75 of them experience computer crashes. You want to estimate the proportion of crashes within a margin of error of 4% at a 95% confidence interval. What sample size do you need?
Answer: 505
Step-by-step explanation:
The formula to find the sample size n , if the prior estimate of the population proportion (p) is known:
[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex] , where E= margin of error and z = Critical z-value.
Let p be the population proportion of crashes.
Prior sample size = 250
No. of people experience computer crashes = 75
Prior proportion of crashes [tex]p=\dfrac{75}{250}=0.3[/tex]
E= 0.04
From z-table , the z-value corresponding to 95% confidence interval = z=1.96
Required sample size will be :
[tex]n=0.3(1-0.3)(\dfrac{1.96}{0.04})^2[/tex] (Substitute all the values in the above formula)
[tex]n= (0.21)(49)^2= 0.21\times2401[/tex]
[tex]n= 504.21\approx505[/tex] (Rounded to the next integer.)
∴ Required sample size = 505
A differential equation of the form y prime (t )equalsF(y) is said to be autonomous (the function F depends only on y). The constant function yequals y 0 is an equilibrium solution of the equation provided Upper F (y 0 )equals0 (because then y prime (t )equals0, and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of t. Consider the equation y prime (t )equals3 sine y .
a. Find all equilibrium solutions in the interva
Answer:
y=n*pi, where n is an integer.
Step-by-step explanation:
We have the equation:
[tex]y'(t)=3sin(y)[/tex]
Which is autonomous because the independent variable does not appear explicitly.
Now, we are asked about all the equilibrium solutions for such autonomous equation. Let's remember that the solution equilibrium y=y0 when F(y0)=0.
Then, matching the given equation to zero, we have:
[tex]y'(t)=3sin(y)=0[/tex]
Which is fulfilled when the sine function has a value of zero. The sine function is worth zero in y=n*pi where n is an integer.