The method of Lagrange multipliers assumes that the extreme values exist, but that is not always the case. Show that the problem of finding the minimum value of f(x, y) = x^2 + y^2 subject to the constraint can be solved using Lagrange multipliers, but does not have a maximum value with that constraint.
Answers
Answer:
Incomplete question check attachment for complete question
Step-by-step explanation:
Given the function,
F(x, y)=x²+y²
The La Grange is theorem
Solve the following system of equations.
∇f(x, y)= λ∇g(x, y)
g(x, y)=k
Fx=λgx
Fy=λgy
Fz=λgz
Plug in all solutions, (x,y), from the first step into f(x, y) and identify the minimum and maximum values, provided they exist and
∇g≠0 at the point.
The constant, λ, is called the Lagrange Multiplier.
F(x, y)=x²+y²
∇f= 2x i + 2y j
So, given the constraint is xy=1.
g(x, y)= xy-1=0
∇g= y i + x j
gx= y. And gy=x
So, here is the system of equations that we need to solve.
Fx=λgx; 2x=λy. Equation 1
Fy=λgy; 2y=λx. Equation 2
xy=1
Solving this
x=λy/2. From equation 1, now substitute this into equation 2
2y=λ(λy/2)
2y=λ²y/2
2y-λ²y/2 =0
y(2-λ²/2)=0
Then, y=0. Or (2-λ²/2)=0
-λ²/2=-2
λ²=4
Then, λ= ±2
So substitute λ=±2 into equation 2
2y=2x
Then, y=x
So inserting this into the constraint g will give
xy=1. Since y=x
x²=1
Therefore,
x=√1
x=±1
Also y=x
Then, y=±1
Therefore, there are four points that solve the system above.
(1,1) (-1,-1) (1,-1) and (-1,1)
The first two points (1,1) (-1,-1) shows the minimum points because they show xy=1
The other points does not give xy=1
They give xy=-1.
Now,
F(x, y)=x²+y²
F(1,1)=1²+1²
F(1,1)=2
F(-1,-1)= (-1) ²+(-1)²
F(-1,-1)=1+1
F(-1,-1)=2
Then
F(1,1)= F(-1,-1)=2 is the minimum point
This gives the same four points as we found using Lagrange multipliers above.
Related Questions
Suppose that a random sample of size 36 is to be selected from a population with mean 43 and standard deviation 6. What is the approximate probability that X will be more than 0.5 away from the population mean?
Answers
Answer:
61.70% approximate probability that X will be more than 0.5 away from the population mean
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are 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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 36, \sigma = 6, n = 36, s = \frac{6}{\sqrt{36}} = 1[/tex]
What is the approximate probability that X will be more than 0.5 away from the population mean?
This is the probability that X is lower than 36-0.5 = 35.5 or higher than 36 + 0.5 = 36.5.
Lower than 35.5
Pvalue of Z when X = 35.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{35.5 - 36}{1}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085.
30.85% probability that X is lower than 35.5.
Higher than 36.5
1 subtracted by the pvalue of Z when X = 36.5. SO
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{36.5 - 36}{1}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a pvalue of 0.6915.
1 - 0.6915 = 0.3085
30.85% probability that X is higher than 36.5
Lower than 35.5 or higher than 36.5
2*30.85 = 61.70
61.70% approximate probability that X will be more than 0.5 away from the population mean
To determine the probability that a sample mean is more than 0.5 away from the population mean, we calculate the z-score for the sample mean being 0.5 above or below the population mean, look up the cumulative probability for this z-score, double it to account for both tails, and subtract from 1.
To find the probability that a sample mean is more than 0.5 away from the population mean, we can use the concept of a sampling distribution. The central limit theorem (CLT) tells us that for a sample of size 36 (which is sufficiently large), the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the original population distribution.
Given the population mean (μ) is 43 and the population standard deviation (σ) is 6, the standard error of the mean (SEM) can be calculated as σ divided by the square root of the sample size (n), which in this case is 6/√36 = 1. The z-score for a value A that is 0.5 away from the mean (either 43.5 or 42.5) can be calculated using the formula (A-μ)/SEM. Thus, the z-score is (42.5-43)/1 = -0.5 or (43.5-43)/1 = 0.5.
Using the standard normal distribution table or a calculator for the cumulative distribution function, we can find the probability of a z-score being less than -0.5 or greater than 0.5. Since the distribution is symmetric, we can simply look up the probability of z < -0.5, then double it (to account for both tails of the distribution) and subtract from 1 to get the probability that the sample mean is more than 0.5 away from 43.
The length and width of a rectangle are measured as 45 cm and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.
Answers
Answer:
dA(r) = 6.9 cm²
Step-by-step explanation:
Area of a rectangle is
A(r) = L * W (1)
Where L stands for length and W for width.
Taking differentials on both sides f equation (1)
dA(r) = W*dL + L*dW
As error in measurments are at most 0,1 cm ( in each side), then the maximum error in calculating the area is
dA(r) = 24* ( 0,1) (cm²) + 45*(0,1)(cm²) ⇒ dA(r) = 2.4 + 4.5 (cm²)
dA(r) = 6.9 cm²
A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are signal problems, the amount of time in minutes that you have to wait at the platform is described by the pdf probability density function with signal problems = pT|SP(t) = .1e^(−.1t). But when there is good service, the amount of time you have to wait at the platform is probability density function with good service = pT|Good(t) = .3e^(−.3t) You arrive at the subway platform and you do not know if the train has signal problems or running with good service, so there is a 30% chance the train is having signal problems. (a) After 1 minute of waiting on the platform, you decide to re-calculate the probability that the train is having signal problems based on the fact that your wait will be at least 1 minute long. What is that new probability? (b) After 5 minutes of waiting, still no train. You re-calculate again. What is the new probability? (c) After 10 minutes of waiting, still no train. You re-calculate again. What is the new probability?
Answers
Answer:
Part a: The probability is 0.3436
Part b: The probability is 0.5381
Part c: The probability is 0.7600
Step-by-step explanation:
As per the given data
[tex]p_{T | SP} = e^{0.1}\\p_{T | Good} = e^{0.3}\\[/tex]
(a)
Probability that train is delayed by more than 1 minute = P(T > 1) = P(SP) * P(T > 1 | SP) + P(Good) * P(T > 1 | Good)
[tex]= 0.3 e^{-.1 \times 1 }+ 0.7 * e^{-.3 \times 1}\\ = 0.3 e^{-.1} + 0.7 e^{-.3}[/tex]
Probability that after 1 minute of waiting, probability that train has signal problems = P(SP | T > 1)
= P(T > 1 | SP) * P(SP) / P(T > 1) (By Bayes theorem)
[tex]= \dfrac{0.3 e^{-.1 }}{ 0.3 e^{-.1 }+ 0.7 e^{-.3 }}\\\\= \dfrac{0.2714512}{0.790024}\\\\= 0.3435987[/tex]
(b)
Probability that train is delayed by more than 5 minutes = P(T > 5) = P(SP) * P(T > 5 | SP) + P(Good) * P(T > 5 | Good)
[tex]= 0.3 e^{-.1 \times 5 }+ 0.7 * e^{-.3 \times 5}\\ = 0.3 e^{-.5} + 0.7 e^{-1.5}[/tex]
Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 5)
= P(T > 5 | SP) * P(SP) / P(T > 5) (By Bayes theorem)
[tex]= \dfrac{0.3 e^{-.5 }}{ 0.3 e^{-.5 }+ 0.7 e^{-1.5 }}\\\\= \dfrac{0.1819592}{0.3381503}\\\\= 0.5381015[/tex]
(c)
Probability that train is delayed by more than 10 minutes = P(T > 10) = P(SP) * P(T > 10 | SP) + P(Good) * P(T > 10 | Good)
[tex]= 0.3 e^{-.1 \times 10 }+ 0.7 * e^{-.3 \times 10}\\ = 0.3 e^{-1.0} + 0.7 e^{-3.0}[/tex]
Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 10)
= P(T > 10 | SP) * P(SP) / P(T > 10) (By Bayes theorem)
[tex]= \dfrac{0.3 e^{-1.0 }}{ 0.3 e^{-1.0 }+ 0.7 e^{-3.0 }}\\\\= \dfrac{0.1103638}{0.1452148}\\\\= 0.7600038[/tex]
(a) The probabilities recalculated based on the waiting time are: 21.4% after 1 minute,
(b) 54.0% after 5 minutes, and
(c) 75.9% after 10 minutes.
These calculations use Bayes' theorem and the exponential distribution functions provided.
The question pertains to conditional probability and involves exponential distributions.
Let’s solve this step-by-step:
Part (a)
Given:
pT|SP(t) = [tex]0.1e^(-0.1t)[/tex]
pT|Good(t) = [tex]0.3e^(-0.3t)[/tex]
P(Signal Problems) = [tex]0.30[/tex]
P(Good Service) = [tex]0.70[/tex]
We need to find P(Signal Problems | T ≥ 1). Using Bayes’ theorem:
P(T ≥ 1 | Signal Problems) = [tex]\int_{1}^{\infty} 0.1 \, dx[/tex] [tex]e^(-0.1t)[/tex] [tex]dt =[/tex] [tex]e^(-0.1)[/tex]P(T ≥ 1 | Good Service) = [tex]\int_{1}^{\infty} 0.3 \, dx[/tex] [tex]e^(-0.3t)[/tex] [tex]dt =[/tex] [tex]e^(-0.3)[/tex]Using the law of total probability:
P(T ≥ 1) = P(T ≥ 1 | Signal Problems)P(Signal Problems) + P(T ≥ 1 | Good Service)P(Good Service)
= [tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70[/tex]
Now applying Bayes’ theorem:
P(Signal Problems | T ≥ 1) = [tex][[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70][/tex]
[tex]\approx [0.9048 * 0.30] / [0.9048 * 0.30 + 0.7408 * 0.70][/tex]
[tex]\approx 0.214[/tex]
Part (b)
Similarly for T ≥ 5, the probabilities become:
P(T ≥ 5 | Signal Problems) = [tex]e^(-0.5)[/tex]P(T ≥ 5 | Good Service) = [tex]e^(-1.5)[/tex]Using the law of total probability:
P(T ≥ 5) = [tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex][tex]* 0.70[/tex]
Applying Bayes’ theorem:
P(Signal Problems | T ≥ 5) = [tex][[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex] [tex]* 0.70][/tex]
≈ [tex][0.6065 * 0.30] / [0.6065 * 0.30 + 0.2231 * 0.70][/tex]
≈ [tex]0.540[/tex]
Part (c)
For T ≥ 10:
P(T ≥ 10 | Signal Problems) = [tex]e^{-1}[/tex]P(T ≥ 10 | Good Service) = [tex]e^{-3}[/tex]Using the law of total probability:
P(T ≥ 10) = [tex]e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70[/tex]
Applying Bayes’ theorem:
P(Signal Problems | T ≥ 10) = [tex]\frac{e^{-1} \cdot 0.30}{e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70}[/tex]
≈ [tex][0.3679 * 0.30] / [0.3679 * 0.30 + 0.0498 * 0.70][/tex]
≈ [tex]0.759[/tex]
The least-squares regression model y =−3.4+5.2x and correlation coefficient r=−0.66 were calculated for a set of bivariate data with variables x and y . What is closest to the proportion of the variation in y that cannot be explained by the explanatory variable?
Answers
Answer:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case the value of r = -0.66
Now we can calculate the determination coeffcient:
[tex] r^2 = (-0.66)^2 = 0.4356[/tex]
And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable
And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable
Step-by-step explanation:
For this case we need to calculate the slope with the following formula:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
[tex]\bar x= \frac{\sum x_i}{n}[/tex]
[tex]\bar y= \frac{\sum y_i}{n}[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x[/tex]
And the model obtained for this case is:
[tex] y = -3.4 +5.2 x[/tex]
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case the value of r = -0.66
Now we can calculate the determination coeffcient:
[tex] r^2 = (-0.66)^2 = 0.4356[/tex]
And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable
And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable
The Coefficient of determination gives the proportion of variation which can be explained by the regression line.
The proportion of variation that cannot be explained by the given regression line is 56.44%
From the question, the correlation Coefficient, R = - 0.66
The Coefficient of determination, R² can be calculated from the R value given. R² = - 0.66² = 0.4356This means that :
Percentage of variation that can be explained = (0.4356) × 100% = 43.56%The percentage of variation that cannot be explained = 100% - 43.56% = 56.44%Therefore, proportion of the variation in y that cannot be explained is 56.44%
Learn more :https://brainly.com/question/18405415
Consider a value to be significantly low if its z score less than or equal to minus 2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 21.6 and the standard deviation was 5.4. Identify the test scores that are significantly low or significantly high. What test scores are significantly low?
Answers
Answer:
The scores that are less than or equal to 10.8 are considered significantly low.
The scores that are greater than or equal to 32.4 are considered significantly high.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 21.6
Standard Deviation, σ = 5.4
We are given that the distribution of score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Significantly low score:
[tex]z \leq -2\\z = \displaystyle\frac{x-21.6}{5.4} \leq -2\\\\\displaystyle\frac{x-21.6}{5.4} \leq -2\\\\x\leq -2(5.4) + 21.6\\\Rightarrow x \leq 10.8[/tex]
Thus, scores that are less than or equal to 10.8 are considered significantly low.
Significantly high score:
[tex]z \geq 2\\z = \displaystyle\frac{x-21.6}{5.4} \geq 2\\\\\displaystyle\frac{x-21.6}{5.4} \geq 2\\\\x\geq 2(5.4) + 21.6\\\Rightarrow x \geq 32.4[/tex]
Thus, scores that are greater than or equal to 32.4 are considered significantly high.
This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 100 customer orders to fill. Each order requires one component part that is purchased froma supplier. However, typically, 2% of the components are identifiedas defective, and the components can be assumed to beindependent.a)If the manufacturer stocks 100 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?b) If the manufacturer stocks 102 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?c) If the manufacturer stocks 105 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?
Answers
Answer:
(a) 0.1326
(b) 0.2732
(c) 0.0410
Step-by-step explanation:
Let X = number of defective components.
The probability of X is, P (X) = p = 0.02.
The random variable X follows a Binomial distribution with parameters n and p. The probability mass function of a Binomial distribution is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3,...[/tex]
(a)
Compute the probability that the 100 orders can be filled without reordering components as follows:
n = 100
[tex]P(X=0)={100\choose 0}0.02^{0}(1-0.02)^{100-0}=1\times1\times0.13262=0.1326[/tex]
Thus, the probability that the 100 orders can be filled without reordering components is 0.1326.
(b)
Compute the probability that out of 102 orders 2 orders needs reordering as follows:
n = 102
[tex]P(X=2)={102\choose 2}0.02^{2}(1-0.02)^{102-2}=5151\times0.0004\times0.13262=0.2732[/tex]
Thus, the probability that out of 102 orders 2 orders needs reordering is 0.2732.
(c)
Compute the probability that out of 105 orders 2 orders needs reordering as follows:
n = 105
[tex]P(X=5)={105\choose 5}0.02^{5}(1-0.02)^{105-5}=96560646\times0.0000000032\times0.13262=0.0410[/tex]
Thus, the probability that out of 105 orders 5 orders needs reordering is 0.0410.
A researcher would like to evaluate the claim that large doses of vitamin C can help prevent the common cold. One group of participants is given a large dose of the vitamin (500 mg per day), and a second group is given a placebo (sugar pill). The researcher records the number of colds each individual experiences during the 3-month winter season. a. Identify the dependent variable for this study.b. Is the dependent variable discrete or continuous?c. What scale of measurement (nominal, ordinal, or interval/ratio) is used to measure the dependent variable?d. What is the independent variable?e. What research method is being used (experimental or correlational?
Answers
Answer:
(a) the dependent variable here are the Participants.
(b) the dependent variable is discrete.
(c) The scale measurement of measurement used is interval/ratio to measure the dependent variable.
(d) the independent variable is Vitamin C
(e) The research method being used is experimental.
Step-by-step explanation:
The dependent variable (sometimes known as the responding variable) is what is being studied and measured in the experiment.
Examples of continuous dependent variable may include costs, profits and sales.But some dependent variables are discrete – that is, they take on a relatively small number of integer values.
Interval scale and ratio scale are the two variable measurement scales where they define the attributes of the variables quantitatively.A ratio scale is a measurement scale which has more or less all the properties of an interval scale. Ratio data on this scale has measurable intervals.
Experimental research is a study that strictly adheres to a scientific research design. It includes a hypothesis, a variable that can be manipulated by the researcher, and variables that can be measured, calculated and compared.
Answer: 1. The dependent variable is the common cold.
2. It is a discrete variable.
3. The interval/ratio scale.
4. The independent value is the large doses of Vitamin C administered.
5. The research method used is Experimental.
Step-by-step explanation:
1. The dependent variable is the factor that we are trying to understand. In this case, it is the common cold and how it is affected by large doses of vitamins.
2. A discrete value is computed by counting. So the dependent variable - the common cold was computed by counting the number of occurrences.
3. The Interval/ratio scale is used because both the order of measurement and the differences between them are observed.
4. The Independent variable is the control in the experiment which can be compared to changes in the dependent variable. The large doses of vitamin C affects the common cold.
5. Correlational research observes patterns in variables that occur naturally while Experimental research introduces a change and monitors its effect. The research is Experimental because a change in the form of the placebo is introduced and observed.
2. A canoe requires 8 hours of fabrication. A row boat requires 5 hours of fabrication. The fabrication department has at most 110 hours to labor each week. Write the equation and solve the problems.
Answers
Answer:
Let x is the number of the cannon and y is the number of the row boat.Thus the equation will become
8x+5y=110
Answer: 8C+5R≤110
Step-by-step explanation: Let hours of fabrication of canoes = C
Hours of fabrication of rowboat = R
If a canoe requires 8 hours of fabrication. And a row boat requires 5 hours of fabrication. Then
8C + 5R
The fabrication department has at most 110 hours to labor each week.
Therefore
8C+5R≤110
At most means less than or equal to.
Since the ratios for sides of triangles are the same, when the angles are the same, ? have been developed which show these ratios for every angle encountered in a triangle.
Answers
Answer:
Step-by-step explanation:
Mathematically, if two triangles are similar, then all the three of their angles are congruent to each other and their corresponding sides are in the same proportion. This means that the ratio of their corresponding sides are equal to each other.
Answer: A trigonometric table will provide all the angles in respect to the ratios of the sides.
Step-by-step explanation:
A trigonometric table would provide all of the ratios of the sides in respect to the angles. Sin, cos, and tan are parts of trigonometric table.
Trigonometry table, tabulated values for some or all of the six trigonometric functions for various angular values.
Since the ratios for sides of triangles are the same, when the angles are the same, trigonometric tables have been developed which show these ratios for every angle.
The given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. y = c1 + c2 cos(x) + c3 sin(x), (−[infinity], [infinity]); y''' + y' = 0, y(π) = 0, y'(π) = 9, y''(π) = −1
Answers
To solve the initial-value problem for the differential equation y''' + y' = 0 with the general solution y = c1 + c2 cos(x) + c3 sin(x), we differentiate y to find y' and y'', apply the initial conditions at x = π, and solve for c1, c2, and c3. The specific solution is y = 1 + cos(x) - 9 sin(x).
We're given the general solution of the third-order differential equation y''' + y' = 0 as y = c1 + c2 cos(x) + c3 sin(x), and we need to find specific constants c1, c2, and c3 that satisfy the initial conditions y(π) = 0, y'(π) = 9, and y''(π) = -1.
First, differentiate y = c1 + c2 cos(x) + c3 sin(x) to find y' and y''.
Substitute x = π into the resulting equations to apply the initial conditions.
Solve the system of equations to find the values of c1, c2, and c3.
Let's carry out these steps:
Differentiation gives us:
y' = -c2 sin(x) + c3 cos(x)
y'' = -c2 cos(x) - c3 sin(x)
Applying initial conditions at x = π:
y(π) = c1 - c2 = 0
y'(π) = -c3 = 9
y''(π) = -c2 = -1
Solve the system:
c1 = c2
c2 = 1
c3 = -9
Therefore, the specific solution to the given initial-value problem is y = 1 + cos(x) - 9 sin(x).
what is the recursive formula for this geometric sequence? -4,-24,-144,-864,...
Answers
[tex]\bf -4~~,~~\stackrel{-4\cdot 6}{-24}~~,~~\stackrel{-24\cdot 6}{-144}~~,~~\stackrel{-144\cdot 6}{-864}~\hfill \impliedby \begin{array}{llll} \textit{we're multiplying by 6 the previous}\\ \textit{term(n-1) to get the current one} \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_n = a_{n-1}(6)\qquad for~~a_1=-4~~,~~n>2\qquad \impliedby \textit{recursive formula}[/tex]
In this question, f : R → R is a differentiable function. We will use linear algebra try to find a good linear approximation to f near a point x = a. Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.
Answers
Answer:
See step by step explanations to get answer.
Step-by-step explanation:
Given that:
Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.
A 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall. What is the 5-day growth factor for the height of the sunflower
Answers
Answer:
The growth factor is approximately 0.11 or 11%.
Step-by-step explanation:
We have been given that a 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall.
We will use exponential growth formula to solve our given problem.
[tex]y=a\cdot (1+r)^x[/tex], where,
y = Final value,
a = Initial value,
r = Growth rate in decimal form,
x = Time.
Upon substituting initial value [tex]a=8[/tex], [tex]x=5[/tex] and [tex]y=13.4805[/tex], we will get:
[tex]13.4805=8\cdot(1+r)^5[/tex]
[tex]8\cdot(1+r)^5=13.4805[/tex]
[tex]\frac{8\cdot(1+r)^5}{8}=\frac{13.4805}{8}[/tex]
[tex](1+r)^5=\frac{13.4805}{8}[/tex]
Now, we will take 5th root of both sides of equation as:
[tex]\sqrt[5]{(1+r)^5} =\sqrt[5]{\frac{13.4805}{8}}[/tex]
[tex]1+r =\sqrt[5]{1.6850625}[/tex]
[tex]1+r =1.1100005724234515[/tex]
[tex]1-1+r =1.1100005724234515-1[/tex]
[tex]r=0.1100005724234515[/tex]
[tex]r\approx 0.11[/tex]
Therefore, the growth factor is approximately 0.11 or 11%.
The 5-day growth factor of the sunflower, which grows exponentially, is calculated by dividing the final height (13.4805 inches) by the initial height (8 inches) giving a growth factor of approximately 1.6850625.
Explanation:The 5-day growth factor of the sunflower that increases exponentially can be calculated by dividing the final height by the initial height. The formula is:
5-day growth factor = final height / initial height
To calculate the 5-day growth factor, we substitute the heights into the formula. We have:
5-day growth factor = 13.4805 inches (final height) / 8 inches (initial height)
So, 5-day growth factor = 1.6850625. This means that the height of the sunflower is multiplied by approximately 1.685 each day for 5 days, which accounts for the exponential growth.
Learn more about growth factor here:https://brainly.com/question/32122796
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There are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class project. How divisions are possible?
Answers
Answer:
27720
Step-by-step explanation:
Given that there are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class project.
From 12 students 3 students for group I can be selected in 12C3 ways.
Now from remaining 9, 4 students can be selected for II group in 9C4 ways
The remaining 5 have to be placed in III group.
Hence possible divisions for grouping the 12 students in the class into three groups
= 12C3 *9C4
= [tex]=220*126=27720[/tex]
Final answer:
There are 27,720 possible divisions of the 12 students into three groups of 3, 4, and 5 members.
Explanation:
To determine the number of possible divisions, we need to find the number of ways to choose 3 students from 12 for the first group, then 4 students from the remaining 9 for the second group, and finally 5 students from the remaining 5 for the third group.
Using the combination formula, the number of ways can be calculated as:
Number of ways = C(12, 3) * C(9, 4) * C(5, 5)
C(12, 3) = 12! / (3! * (12-3)!) = 220
C(9, 4) = 9! / (4! * (9-4)!) = 126
C(5, 5) = 5! / (5! * (5-5)!) = 1
Therefore, the number of possible divisions is 220 * 126 * 1 = 27,720.
Below are the jersey numbers of 11 players randomly selected from a football team. Find the range, variance, and standard deviation for the given sample data. What do the results tell us? 59 19 30 75 42 49 57 81 11 87 91
Answers
Answer:
Range = The difference between the highest and lowest which is
The mean is the sum of all values divided by the number of values
= 59+19+30+75+42+49+57+81+11+87+91 divided by n (which is 11)
601÷11
= 54.6
Variance = sum of squared deviations from the mean divided by n-1
=679.65
Standard Deviation = This is the square root of variance which gives us;
26.07
One method used to distinguish between Granitic (G) and Basaltic (B) rocks is to examine a portion of the infrared spectrum of the sun’s energy reflected from the rock surface. Let R1, R2, and R3 denote measures of spectrum intensities at three different wave lengths; typically, for granite R1 < R2 < R3, whereas for basalt R3 < R1 < R2. When measurements are made remotely (using aircraft), various orderings of the Ri′s may arise whether the rock is basalt of granite. Flights over regions of know composition have yielded the following information: Reading R_1 P(basalt|R1 < R2 < R3). If measurements yielded R1 < R2 < R3, would you classify the rock as granite? b. If measurements yielded R1 < R3 < R2, how would you classify the rock? Answer the same question forR3
Answers
Answer:
Step-by-step explanation:
so from the question i have here, i will be giving a step by step analysis of the question.
(a). Here we are showing a relationship, i.e.
P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║ R₁ ∠ R₂ ∠ R₃)
from the LHS;
P(Granite ║ R₁ ∠ R₂ ∠ R₃) = P(Granite ║ R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)
= P(Granite)P(R₁ ∠ R₂ ∠ R₃ ║ Granite) / [ P(Granite R₁ ∠ R₂ ∠ R₃) + P(Basaltic R₁ ∠ R₂ ∠ R₃) ]
= 0.25 × 0.6 / [(0.25×0.6)+(0.75×0.1)] = 0.667
from the RHS;
P (Basalt ║ R₁ ∠ R₂ ∠ R₃) = P(Basalt R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)
= P(Basalt)P(R₁ ∠ R₂ ∠ R₃ ║ Basalt) / [ P(Basalt R₁ ∠ R₂ ∠ R₃) + P(Granite R₁ ∠ R₂ ∠ R₃) ]
= 0.75 × 0.1 / [(0.25 × 0.6)+(0.75 × 0.1)] = 0.333
Therefore from this we can infer that;
P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║ R₁ ∠ R₂ ∠ R₃)
(b). here we are asked to classify the rocks considering the measurement yield.
Measurement yielded R1 < R3 < R2
P(Granite ║ R₁ ∠ R₃ ∠ R₂) = P(Granite R₁ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)
= P(Granite)P(R₁ ∠ R₃ ∠ R₂ ║ Granite) / [ P(Granite R₁ ∠ R₃ ∠ R₂) + P(Basaltic R₁ ∠ R₃ ∠ R₂) ]
= 0.25 × 0.25 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.294
also for RHS;
P (Basalt ║ R₁ ∠ R₃ ∠ R₂) = P(Basalt ║ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)
= P(Basalt)P(R₁ ∠ R₃ ∠ R₂ ║ Basalt) / [ P(Basalt R₁ ∠ R₃∠ R₂) + P(Granite R₁ ∠ R₃ ∠ R₂) ]
= 0.75 × 0.2 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.706
from this we can infer that;
P(Granite ║ R₁ ∠ R₃ ∠ R₂) ∠ P (Basalt ║ R₁ ∠ R₃ ∠ R₂)
Also considering measurements yielded R₃ ∠ R₁ ∠ R₂
P(Granite ║ R₃ ∠ R₁ ∠ R₂) = P(Granite R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)
= P(Granite)P(R₃ ∠ R₁ ∠ R₂ ║ Granite) / [ P(Granite R₃ ∠ R₁ ∠ R₂) + P(Basaltic R₃ ∠ R₁ ∠ R₂) ]
= 0.25 × 0.15 / [(0.25×0.15)+(0.75×0.7)] = 0.067
from the RHS;
P (Basalt ║ R₃ ∠ R₁ ∠ R₂) = P(Basalt R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)
= P(Basalt)P(R₃ ∠ R₁ ∠ R₂ ║ Basalt) / [ P(Basalt R₃ ∠ R₁ ∠ R₂) + P(Granite R₃ ∠ R₁ ∠ R₂) ]
1 - 0.067 = 0.933
from this we can infer that;
P(Granite ║ R₃ ∠ R₁ ∠ R₂) ∠ P (Basalt ║ R₃ ∠ R₁ ∠ R₂)
cheers i hope this helps
Final answer:
Measurements yielding R1 < R2 < R3 suggest a granitic composition, while the pattern R1 < R3 < R2 leans granitic, and R3 < R1 < R2 lean towards basaltic. The rock type can be further supported by the appearance of the rock, with granitic rocks being coarse and light-colored due to feldspar and quartz, while basaltic rocks are fine-grained and dark with ferromagnesian minerals. If granite contains basaltic inclusions, the granite is younger.
Explanation:
In analyzing rock compositions from infrared spectrum measurements, if measurements yield R1 < R2 < R3, based on typical patterns, the rock would more likely be classified as granitic because, for granite, the intensity measurements usually increase in that order. Conversely, when measurements produce R1 < R3 < R2, it does not fit neatly into the categories provided for basalt or granite, but it leans closer to a granitic composition given that R3 is not the least among the three, which is a characteristic of basaltic rocks. Similarly, if R3 presents a lower intensity compared to R1 and R2, and you have to classify without further information, it might suggest a basaltic nature following the typical pattern for basalt (R3 < R1 < R2), although it is important to note that real-world applications may require additional contextual information to accurately determine the rock type.
When studying rock samples, you should also consider differences in mineral size (coarse-textured like granite versus fine-textured like basalt), mineral content (dark, mafic minerals vs light, felsic minerals), and elemental content of the igneous rock types. For instance, granitic rock contains feldspar and quartz, which is reflected in its coarse and light-colored appearance. In contrast, basaltic rock, which is fine-grained and dark-colored, includes ferromagnesian minerals and feldspars. Additionally, if a granitic rock has inclusions of basalt (xenoliths), this implies that the granitic rock would typically be younger than the basalt since the inclusions must have been present already for the granite to incorporate them during its formation process.
This Excel file Undergrad Survey shows the data resulting from a survey of 50 undergraduate students at Clemson University. Majors of students in the survey are accounting (A), economics and finance (EF), management (M), marketing (MR), computer information systems (IS), other (O), and undecided (UN). "Number of affiliations" is the number of social networking sites at which the student is registered; "Spending" is the amount spent on textbooks for the current semester. The other variables are self-explanatory.
We will assume that this sample is a representative sample of all Clemson undergraduates. Use Excel or statcrunch to make a histogram of GPA to verify that the distribution of GPA can be approximated by the N(3.12, 0.4) normal model.
Question 1. The School of Business at Clemson has created a rigorous new International Business Studies major to better prepare their students for the global marketplace. A GPA of 3.69 or higher is required for a Clemson undergraduate to change his/her major to International Business Studies. What is the probability that a randomly selected Clemson undergraduate has a GPA of at least 3.69? (Use 4 decimal places in your answer).
Question 2. To attract high-quality current Clemson undergraduates into the new International Business Studies major, scholarships in International Business Studies will be offered to a Clemson undergraduate if his/her GPA is at or above the 95.54th percentile. What is the minimum GPA required to meet this criterion?
The GPA Values:
2.38
2.42
2.45
2.50
2.60
2.61
2.65
2.67
2.74
2.75
2.75
2.76
2.80
2.87
2.88
2.91
2.92
2.93
2.94
3.00
3.02
3.09
3.10
3.11
3.13
3.14
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.26
3.28
3.33
3.34
3.43
3.44
3.48
3.50
3.55
3.62
3.62
3.63
3.71
3.72
3.77
3.85
4.00
Answers
Answer:
Question 1:
Here total number of students GPA is given = 50
Number of people who have GPA more than 3.69 = 5
Therefore,
Pr(That a student has GPA more than 3.69) = 5/50 = 0.1
Here X ~ NORMAL (3.12, 0.4)
so Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4)
Z = (3.69 - 3.12)/ 0.4 = 1.425
Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4) = 1 - Pr(Z < 1.425) = 1 - 0.9229 = 0.0771
Question 2
Here GPA would be above 95.54th percentile
so as per Z table relative to that percentile is = 1.70
so Z = (X - 3.12)/ 0.4 = 1.70
X = 3.12 + 0.4 * 1.70 = 3.80
so any person with GPA above or equal to 3.80 is eligible for that.
A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. Calculate how far the car travels in the time it takes to stop. Round your answer to one decimal place.
Answers
Answer:
80 feet
Step-by-step explanation:
Given:
Initial speed of the car ([tex]v_0[/tex]) = 40 ft/sec
Deceleration of the car ([tex]\frac{dv}{dt}[/tex]) = -10 ft/sec²
Final speed of the car ([tex]v_x[/tex]) = 0 ft/sec
Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.
Now, deceleration means rate of decrease of velocity.
So, [tex]\frac{dv}{dt}=-10\ ft/sec^2[/tex]
Negative sign means the velocity is decreasing with time.
Now, [tex]\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})[/tex] using chain rule of differentiation. Therefore,
[tex]\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx[/tex]
Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,
[tex]\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft[/tex]
Therefore, the car travels a distance of 80 feet before stopping.
he purpose of an x-bar chart is to determine whether there has been a: change in the AOQ. change in the number of defects in a sample. change in the percent defective in a sample. change in the dispersion of the process output. change in the central tendency of the process output.
Answers
Answer:
E. Change in the central tendency of the process output.
Step-by-step explanation:
The x-bar chart is a control chart for the central tendency of the process output. Therefore it's purpose is to determine whether there has been a change in the central tendency of the process output.
Answer:
The change in the central tendency of the process output.
Step-by-step explanation:
Lets examine each of the listed options
The change in the percent defective in a sample is associated with the change in average outgoing quality (AOQ).
The change in the dispersion of the process output is associated with R-chart since the function of R-chart is to detect changes in the dispersion.
The change in the number of defects in a sample is associated with C-chart since the function of a C-chart is to show the number of flaws per unit in a sample.
The change in the central tendency of the process output is associated with X-bar chart since the function of X-bar chart is to check whether the values are within the appropriate limits (central tendency) of the process output or not.
Therefore, the purpose of an X-bar chart is to determine whether there has been a change in the central tendency of the process output.
A factorial experiment was designed to test for any significant differences in the time needed to perform English to foreign language translations with two computerized language translators. Because the type of language translated was also considered a significant factor, translations were made with both systems for three different languages: Spanish, French, and German. Use the following data for translation time in hours.
Language
Spanish French German
System 1 9 12 13
13 16 17
System 2 9 15 19
13 17 25
Test for any significant differences due to language translator system (Factor A), type of language (Factor B), and interaction. Use = .05.
Answers
Answer:
See the attached pictures for answer.
Step-by-step explanation:
See the attached pictures for explanation.
Answer:
Step-by-step explanation:
We have a factorial experiment with Factors A and B, levels a=2, b=3, and n=2 observations.
It's necessary to apply a Two-Factor with replication ANOVA by using Excel or calculating by hand.
1. Sum of Squares could be calculated with the following formulas:
[tex]SS_{T}=\sum\limits^a_{i=1}\sum\limits^b_{j=1}\sum\limits^n_{k=1} {(y_{ijk}-y_{...} ^{2} )}\\SS_{A}=bn\sum\limits^a_{i=1} {(y_{i..}^{2}) -Ny_{...} ^{2} }\\SS_{B}=an\sum\limits^b_{j=1} {(y_{.j.}^{2}) -Ny_{...} ^{2} }\\SS_{AB}=n\sum\limits^a_{i=1}\sum\limits^b_{j=1} {(y_{ij.}^{2}) -Ny_{...} ^{2} -SC_{A} -SC_{B}-SC_{AB}}[/tex]
2. Calculate Degrees of Freedom
Factor A: [tex]a-1[/tex]
Factor B: [tex]b-1[/tex]
Interaction: [tex](a-1)(b-1)[/tex]
Within: [tex]ab(n-1)[/tex]
Total: [tex]abn-1[/tex]
3. Mean Square:
Factor A: [tex]MS_{A}=\frac{SS_{A}}{a-1}[/tex]
Factor B: [tex]MS_{B}=\frac{SS_{B}}{b-1}[/tex]
Interaction: [tex]MS_{AB}=\frac{SS_{AB}}{(a-1)(b-1)}[/tex]
Within:[tex]MS_{E}=\frac{SS_{E}}{ab(n-1)}[/tex]
4. Estimate F and P-value
[tex]F_{A}=\frac{SS_{A} }{SS_{E}}[/tex]
[tex]F_{B}=\frac{SS_{B} }{SS_{E}}[/tex]
[tex]F_{AB}=\frac{SS_{AB} }{SS_{E}}[/tex]
Open the attachment to see the results.
5. Analysis of P-values
Factor A: P-value (0.1280 )>(0.05)
Factor B: P-value (0.0315 )<(0.05)
Interaction: P-value (0.2963 )<(0.05)
P-value of Factor B is less than 0.05, then type of language is insignifficant.
P-value of Factor A and interaction are greater than 0.05, then the translator system and the interaction of both factors are signifficant.
A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor decides to randomly sample 100 students by using the registrar's database of students. The histogram below shows the distribution of the number of credits taken by these students. Sample statistics for this distribution are also provided.Min Q1 Median Mean SD Q3 Max
8 13 14 13.65 1.91 15 18
(a) Based on this data, would you accept or reject the hypothesis that the usual load is 13 credits?
(b) How unlikely is it that a student at this college takes 16 or more credits?
Answers
Answer:
(a) The usual load is not 13 credits.
(b) The probability that a a student at this college takes 16 or more credits is 0.1093.
Step-by-step explanation:
According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.
The information provided is:
[tex]Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18[/tex]
The sample size is, n = 100.
The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.
So,
[tex]\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191[/tex]
(a)
The null hypothesis is:
H₀: The usual load is 13 credits, i.e. μ = 13.
Assume that the significance level of the test is, α = 0.05.
Construct a (1 - α) % confidence interval for population mean to check the claim.
The (1 - α) % confidence interval for population mean is given by:
[tex]CI=\bar x\pm z_{\alpha/2}\times SE[/tex]
For 5% level of significance the two tailed critical value of z is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Construct the 95% confidence interval as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)[/tex]
As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.
Thus, it can be concluded that the usual load is not 13 credits.
(b)
Compute the probability that a a student at this college takes 16 or more credits as follows:
[tex]P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z<1.23)\\=1-0.8907\\=0.1093[/tex]
Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.
Answer:
(a) We reject the null hypothesis that usual load is 13 credits.
(b) Probability that student at this college takes 16 or more credits = 0.10935
Step-by-step explanation:
We are given that the histogram below shows the distribution of the number of credits taken by these students;
Min Q1 Median Mean SD Q3 Max
8 13 14 13.65 1.91 15 18
Also, the counselor decides to randomly sample 100 students by using the registrar's database of students, i.e., n = 100.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 13 {means that the usual load is 13 credits}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 13 {means that the usual load is not 13 credits}
The test statistics we will use here is ;
T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, Xbar = sample mean = 13.65
s = sample standard deviation = 1.91
n = sample size = 100
So, test statistics = [tex]\frac{13.65 - 13}{\frac{1.91}{\sqrt{100} } }[/tex] ~ [tex]t_9_9[/tex]
= 3.403
Now, since significance level is not given to us so we assume it to be 5%.
At 5% significance level, the t tables gives critical value of 1.987 at 99 degree of freedom. Since our test statistics is more than the critical value which means our test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that the usual load is not 13 credits.
(b) Let X = credits of students
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
So, probability that student at this college takes 16 or more credits = P(X [tex]\geq[/tex] 16)
P(X [tex]\geq[/tex] 16) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{16-13.65}{1.91}[/tex] ) = P(Z [tex]\geq[/tex] 1.23) = 1 - P(Z < 1.23)
= 1 - 0.89065 = 0.10935 or 11%
Therefore, probability that student at this college takes 16 or more credits is 0.10935.
Ask Your Teacher What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United States gave the following information about a prehistoric extended family group of 86 members on what is now a Native American reservation. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint. (Round your answers to one decimal place.)
Answers
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
Step-by-step explanation:
We assume the following dataset
Age range (years) 1-10 11-20 21-30 >31
Number of individuals 30 18 23 10
Solution to the problem
We can solve the problem creating the following table:
Class Midpoint(xi) fi xi*fi xi^2 *fi
1-10 5.5 30 165 907.5
11-20 15.5 18 279 4324.5
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
___________________________________
Total 81 1385.5 32790.25
The midpoint is calculated as the average between the lower and the upper interval values.
We can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study27 has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is 0.183. The standard error for this estimate is 0.016.
a.Give notation for the quantity being estimated, and define any parameters used.
b.Give notation for the quantity that gives the best estimate, and give its value.
c.Give a 95% confidence interval for the quantity being estimated. Interpret the interval in context.
Answers
Answer:
a) [tex]\mu[/tex]
b) [tex]\bar{x}[/tex]
c) (0.152, 0.214)
Step-by-step explanation:
We are given the following in the question:
Sample mean = 0.183 ppm
Standard error = 0.016
a) quantity being estimated
We have to estimate the population mean.
Notation for population mean:
[tex]\mu[/tex]
b) Best estimate for population mean is the sample mean
Notation for sample mean:
[tex]\bar{x}[/tex]
The point estimate for population mean is
[tex]\mu = \bar{x} = 0.183[/tex]
c) 95% confidence interval
[tex]\bar{x}\pm z_{critical}(\text{Standard Error})[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get,
[tex]0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)[/tex]
Thus, the 95% confidence interval is:
(0.152, 0.214)
Interpretation:
After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.
The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The best estimate for this quantity is 0.183 ppm. The 95% confidence interval for this estimate is (0.15164, 0.21436) ppm.
Explanation:a. The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The parameter used is the standard error, which measures the variability of the estimate.
b. The notation for the quantity that gives the best estimate is the mean level of the toxin ergovaline after treatment with moose drool, denoted as μ.
c. To calculate a 95% confidence interval, we need to determine the margin of error. Since the standard error is given as 0.016, the margin of error is 1.96 times the standard error, which is approximately 0.03136. The 95% confidence interval is then calculated by subtracting and adding the margin of error from the mean level of the toxin ergovaline after treatment with moose drool, resulting in an interval of (0.15164, 0.21436). This means that we are 95% confident that the true mean level of the toxin ergovaline after treatment with moose drool is between 0.15164 and 0.21436 ppm.
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According to a July 31, 2013, posting on cnn subsequent to the death of a child who bit into a peanut, a 2010 study in the journal Pediatrics found that 8% of children younger than 18 in the United States have at least one food allergy. Among those with food allergies, about 39% had a history of severe reaction. a. If a child younger than 18 is randomly selected, what is the probability that he or she has at least one food allergy and a history of severe reaction? b. It was also reported that 30% of those with an allergy in fact are allergic to multiple foods. If a child younger than 18 is randomly selected, what is the probability that he or she is allergic to multiple foods?
Answers
Answer:
a. 0.0312 or 3.12%
b. 0.024 or 2.40%
Step-by-step explanation:
a. The probability that randomly selected child has at least one food allergy and a history of severe reaction is determined by the probability of having a food allergy (8%) multiplied by the probability of having a history of severe reaction (39%):
[tex]P = 0.08*0.39\\P=0.0312 = 3.12\%[/tex]
b. The probability that randomly selected child is allergic to multiple foods is determined by the probability of having a food allergy (8%) multiplied by the probability of being allergic to multiple foods (30%):
[tex]P = 0.08*0.30\\P=0.024 = 2.40\%[/tex]
I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and provide statistical explanations. Please help me and answer these before 9: 15 in the morning EST on March 3rd, 2020.
Answers
Answer:
question 1[tex]n^{2} - 20n -96 = 0[/tex]
use product and sum method
product = -96
sum = -20
numbers needed = ( -24 , 4)
n - 24 = 0
n + 4 = 0
hence n = 24 and n = -4
Question 2
[tex]x^{2} + 12 x = 48[/tex]
in the form [tex]ax^{2} +bx +c = 0[/tex]
= [tex]x^{2} +12x - 48[/tex]
make use of the formula :
[tex]\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]
replace values to make 2 equations :
1.[tex]\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = 3.17
2.[tex]\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = -15.2
hence x = 3.17 and x = -15.2
Question 3
[tex]x^{2} -14x+40=0[/tex]
use product and sum method
product = 40
sum = -14
numbers needed = (-10 , -4)
x - 10 = 0
x - 4 = 0
hence x = 10 and x = 4
Question 4
[tex]5b^{2} -20b-18 = 7[/tex]
in the form [tex]ax^{2} +bx +c = 0[/tex]
this becomes [tex]5b^{2} -20b-18-7[/tex]
= [tex]5b^{2} -20b-25[/tex]
can simplify by 5
= [tex]b^{2} -4b-5 =0\\[/tex]
use product and sum method
product = -5
sum = -4
numbers needed (-5 , 1)
b-5 = 0
b + 1 = 0
hence b = 5 and b = -1
Answer:
Step-by-step explanation:
1) n² - 20n - 96 = 0
n² - 20n + (- 20/2)² = 96 + (- 20/2)²
(n - 10)² = 96 + 100
(n - 10)² = 196
Taking square root of both sides
n - 10 = √196 = 14
n = 14 + 10
n = 24
2) x² + 12x = 48
x² + 12x + (12/2)² = 48 + (12/2)²
(x + 6)² = 48 + 36 = 84
Taking square root of both sides,
x + 6 = 9.2
x = 9.2 - 6
x = 3.2
3) x² - 14x + 40 = 0
x² - 14x = - 40
x² - 14x + (- 14/2)² = - 40 + (- 14/2)²
(x - 7)² = - 40 + 49 = 9
Taking square root of both sides,
x - 7 = 3
x = 3 + 7
x = 10
4) 5b² - 20b - 18 = 7
5b² - 20b = 7 + 18
5b² - 20b = 25
Dividing both sides by 5, it becomes
b² - 4b = 5
b² - 4b + (-4/2)² = 5 + (-4/2)²
(b - 2)² = 5 + 4 = 9
Taking square root of both sides
b - 2 = 3
b = 3 + 2
b = 5
Military radar and missile detection systems are designed to warn a country of an enemy attack. A reliability question is whether a detection system will be able to identify an attack and issue a warning. Assume that a particular detection system has a 0.80 probability of detecting a missile attack. Use the binomial probability distribution to answer the following questions. (a) What is the probability that a single detection system will detect an attack
Answers
Answer:
0.8 is the probability that a single detection system will detect a missile attack.
Step-by-step explanation:
We are given the following information:
We treat detection a missile attack as a success.
P(detecting a missile attack) = 80% = 0.8
Then the number of missile attack follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 1
We have to evaluate:
[tex]P(x = 1)\\= \binom{1}{1}(0.8)^1(1-0.8)^0\\= 0.8[/tex]
0.8 is the probability that a single detection system will detect a missile attack.
Answer:
(a) Probability that a single detection system will detect an attack is 0.80
Step-by-step explanation:
We are given that a reliability question is whether a detection system will be able to identify an attack and issue a warning. Assuming that a particular detection system has a 0.80 probability of detecting a missile attack.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials(samples) taken = 1 detection system
r = number of success
p = probability of success which in our question is probability of
detecting a missile attack, i.e., 80%
LET X = a particular detection system
Also, it is given that a single detection system is taken,
So, it means X ~ [tex]Binom(n=1,p= 0.80)[/tex]
Now, Probability that a single detection system will detect an attack is given by = P(X = 1)
P(X = 1) = [tex]\binom{1}{1}0.8^{1} (1-0.8)^{1-1}[/tex]
= [tex]1 \times 0.8 \times 1[/tex] = 0.80 .
An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 6 cm, and the height is 12 cm. Find the rate at which the water level is rising when the water level is 5 cm.
Answers
Answer:
The volume of a pyramid is 1/3 Sh, where S is the area of the base and h is the height. Since the area of base (Square) is , S = [tex]s^{2}[/tex], where s is the side of the base. So the volume is
V = 1/3 [tex]s^{2}[/tex]h.
Further solution is on paper (Pictures attached)
5)
Solve the equation for x, in the simplest form.
2/3x = 2/5
A) x = 2/5
B) x = 3/5
C) x = 1/2
D) x = 5/6
Answers
Answer:
B x = 3/5
Step-by-step explanation:
2/3 x = 2/5
Multiply each side by 15 to get rid of the fractions
15(2/3 x )= 2/5*15
10x = 6
Divide each side by 10
10x/10 =6/10
x = 6/10
We can simplify this fraction. Divide each side by 2
x = 3/5
Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt.(a) What population is under consideration in the data set?(b) What parameter is being estimated?(c) What is the point estimate for the parameter?(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?(e) Compute the value from part (d) for this context.(f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?(g) Suppose the true population value was found to be 40%. If we use this proportion to recompute the value in part (e) using p = 0:4 instead of ^p, does the resulting value change much?
Answers
Step-by-step explanation:
(a)
The population under study are the adults of United States.
(b)
A parameter the population characteristic that is under study.
In this case the researcher is interested in the proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt.
So the parameter is the population proportion of US adults who say this.
(c)
A point estimate is a numerical value that is the best guesstimate of the parameter. It is computed using the sample values.
For example, sample mean is the point estimate of population mean.
The point estimate of the population proportion of US adults who say cover a $400 unexpected expense without borrowing money or going into debt, is the sample proportion, [tex]\hat p[/tex].
[tex]\hat p=\frac{322}{765}=0.421[/tex]
(d)
The uncertainty of the point estimate can be measured by the standard error.
The standard error tells us how closer the sample statistic is to the parameter value.
[tex]SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
(e)
The standard error is:
[tex]SE_{\hat p}=\sqrt{\frac{0.421(1-0.421)}{765}} =0.018[/tex]
(f)
The sample proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt, is approximately 42.1%.
As the sample size is quite large this value can be used to estimate the population proportion.
If the proportion is believed to be 50% then she will be surprised because the estimated percentage is quite less than 50%.
(g)
Compute the standard error using p = 0.40 as follows:
[tex]SE=\sqrt{\frac{ p(1- p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{765}}=0.0177\approx0.018[/tex]
The standard error does not changes much.
The data set consists of information on the ability of adults in the United States to cover a $400 unexpected expense. The parameter being estimated is the proportion of adults who cannot cover the expense without borrowing money or going into debt. The point estimate for this parameter is found by dividing the number of adults in the sample who cannot cover the expense by the total sample size.
Explanation:(a) The population under consideration in the data set is all adults in the United States.
(b) The parameter being estimated is the proportion of adults in the United States who could not cover a $400 unexpected expense without borrowing money or going into debt.
(c) The point estimate for the parameter is the proportion of adults in the sample who said they could not cover a $400 unexpected expense without borrowing money or going into debt, which is 322/765.
(d) The name of the statistic that measures the uncertainty of the point estimate is the standard error.
(e) To compute the value of the standard error, you need the formula which depends on the sample proportion and sample size. Since the necessary values are not provided in the question, it is not possible to compute the value in this context.
(f) The cable news pundit should not be surprised by the data since the sample proportion is not too far off from the value she thinks is true.
(g) If the true population value is 40%, the resulting value in part (e) will change because the sample proportion is different from the true proportion.
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Define a random variable x = number of cups of coffee consumed on an average day. Let x = 4 represent four or more cups. Round your answers to four decimal places.
Answers
Answer:
E (X) = 6.4
Step-by-step explanation:
SOLUTION:
A random variable x = number of cups of coffee consumed on an average day.
∴Let x = 4 represent four or more cups. Round your answers to four decimal places.
X Probability (X)
0 0.1
1 0.15
2 0.3
3 0.75
4 0. 25
5 0.21
∴ E (X) = Ux(Mean)
0x.0.1 + 1 x.15 + 2 x 0.3 + 3 x 0.75 + 4 x 0.25 + 5 x 0.21 = 6.4