Answer:
The relation is antisymmetric and transitive
Step-by-step explanation:
Let a,b,c be elements of the set of all people.
1) Let a be a person who is 20 years old. aRa means that this person is younger than themselves, which it's false because 20<20 is false. Then R is not reflexive.
2) Let a be a person who is 20 years old and b a person who is 30 years old. Then a is younger than b, that is, aRb.
However, it is not true that b is younger than a, as 30<20 is false, therefore bRa is false and R is not symmetric.
3) Suppose that aRb, so that a is younger than b. Then, b is not younger than a. If n denotes the age of a and m denotes the age of b, we have that n<m which implies that m<n is false. Then bRa is false, thus R is antisymmetric.
4) Suppose that aRb and bRc. Let n,m,p denote the ages of a,b,c respectively. Then n<m and m<p (a is younger than b and b is younger than c), and by transitivity of the ordering of numbers, n<p, that is, a is younger than c. Thus aRc, and R is transitive.
A researcher wishes to determine the average number of vehicles are registered to a typical Houston residence. In order to do this, he sends a survey to 250 randomly selected residences asking for them to indicate the number of registered and return the survey. Identify the population.
Answer: Houston residences
Step-by-step explanation:
A population is the group of members comes under the same criteria by the researcher's point of view.Here , The objective of the researcher is to determine the average number of vehicles are registered to a typical Houston residence.
Clearly , the population is this situation is "Houston residences" having vehicles.
Note : 250 randomly selected residences are defining the sample of the entire population of Houston residences which is a subset of population.
Hence, the correct answer is Houston residences.
A continuous random variable may assume:
-any value in an interval or collection of intervals
-only integer values in an interval or collection of intervals
-only fractional values in an interval or collection of intervals
-only the positive integer values in an interval
Answer:
-any value in an interval or collection of intervals
Step-by-step explanation:
Discrete random variables are only integers in the interval.
Continuous random variables are all the values(integers, fractional, negative, positive) in an interval, or a collection of intervals.
The correct answer is:
-any value in an interval or collection of intervals
A continuous random variable can assume any value in an interval or collection of intervals. Unlike discrete random variables, which can only take integer values, continuous variables can take a range of values, like measurements such as height.
Explanation:A continuous random variable is a type of variable in statistics that can assume any value within an interval or collection of intervals. This is different from a discrete random variable which can only take on a finite or countable number of values (often integer values).
For example, if we measure the height of a person, it could be any value within the feasible range, say between 0 meter and 3 meters. This is a continuous variable because there are infinite possibilities between 0 and 3, including measurements like 1.73 meters or 2.45 meters etc. It's not limited to integer values (like 1 m, 2 m, etc.), nor only fractional values, nor just positive integers within an interval.
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The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen's School of Business in Kingston, Ontario requires a minimum LSAT score of 157 for admission. Find the 35th percentile of the LSAT scores. Give your answer accurate to one decimal place. Use the applet. (Example: 124.7) Your Answer:
To find the 35th percentile of LSAT scores, a z-score corresponding to the 35th percentile is needed, which can then be applied to the formula using the provided mean and standard deviation of LSAT scores.
Explanation:To find the 35th percentile of LSAT scores, we need to use a normal distribution with the given mean and standard deviation. We use a z-table or a percentile calculator, looking to find the z-score that corresponds with the 35th percentile. Once the z-score is identified, we can use the mean and standard deviation of the LSAT scores to calculate the actual score corresponding to that percentile.
The process involves the following calculations:
Identify the z-score that corresponds to the 35th percentile using the z-table or percentile calculator.Apply the formula: actual score = mean + (z-score * standard deviation).Unfortunately, without the z-score or the use of applets, as suggested in the question, we cannot provide the exact LSAT score that corresponds to the 35th percentile.
The probability density function f(x) of a random variable X that has a uniform distribution between a and b is:
-(b + a)/2
-(a − b)/2
-1/b − 1/a
-None of these choices.
The probability density function (pdf) is;
f(x)=[tex]\frac{1}{b - a}[/tex] for a ≤ x ≤ b.
while the mean is given as [tex]\frac{a + b}{2}[/tex]
And the standard deviation given as [tex]\sqrt{\frac{(b - a)^{2} }{12} }[/tex]
Answer: -None of these choices.
PDF = 1/(b-a)
Step-by-step explanation:
The probability density function f(x) of a random variable X that has a uniform distribution between a and b is given by:
PDF = 1/(b-a) for X€[a,b]
otherwise zero.
Solve the following equation. log Subscript 2 Baseline (3 x plus 7 )equals 5 The solution set is StartSet nothing EndSet . (Simplify your answer.)
Answer: [tex]x=\dfrac{25}{3}[/tex]
Step-by-step explanation:
The given equation : [tex]\log_2(3x+7)=5[/tex]
Using logarithmic property : [tex]\log_a N=M\to N=a^M[/tex]
The given equation will be equivalent to [tex](3x+7)=2^5[/tex]
[tex]\Rightarrow\ 3x+7=32[/tex]
Subtract 7 from both sides , we get
[tex]3x=25[/tex]
Divide both sides by 3 , we get
[tex]x=\dfrac{25}{3}[/tex]
Hence, the solution is [tex]x=\dfrac{25}{3}[/tex]
Robin Inc. feared that the average company loss is running beyond $34,000. It initially conducted a hypothesis test on a sample extracted from its database. The hypothesis was formulated as H0: average company loss ≤ $34,000 H1: average company loss > $34,000. The test resulted in favor of Robin Inc.'s loss not exceeding $34,000. Detailed study of company accounts later revealed that the average company loss had run up to $37,896. Which of the following errors were made during the hypothesis test? Select one: a) Type III error b) Type II error c) Type I error d) Type IV error
Answer
b. Type II error
Step-by-step explanation:
The hypothesis was formulated as
the solution can be seen in the attached document
A paint manufacturer made a modification to a paint to speed up its drying time. Independent simple random samples of 11 cans of type A (the original paint) and 9 cans of type B (the modified paint) were selected and applied to similar surfaces. The drying times, in hours, were recorded.
The summary statistics are as follows.
Type A Type B
x1 = 76.3 hrs x2 = 65.1 hrs
s1 = 4.5 hrs s2 = 5.1 hrs
n1 = 11 n2 = 9
The following 98% confidence interval was obtained for μ1 - μ2, the difference between the mean drying time for paint cans of type A and the mean drying time for paint cans of type B:
4.90 hrs < μ1 - μ2 < 17.50 hrs
What does the confidence interval suggest about the population means?
Answer:
We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X_1 =76.3[/tex] represent the sample mean 1
[tex]\bar X_2 =65.1[/tex] represent the sample mean 2
n1=11 represent the sample 1 size
n2=9 represent the sample 2 size
[tex]s_1 =4.5[/tex] sample standard deviation for sample 1
[tex]s_2 =5.1[/tex] sample standard deviation for sample 2
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
Confidence interval
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =76.3-65.1=11.2[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n_1 +n_2 -1=11+9-2=18[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,18)".And we see that [tex]t_{\alpha/2}=\pm 2.55[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]
And replacing we have:
[tex]SE=\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=2.175[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]11.2-2.55\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=5.65[/tex]
[tex]11.2+2.55\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=16.75[/tex]
So on this case the 98% confidence interval would be given by [tex]5.65 \leq \mu_1 -\mu_2 \leq 16.75[/tex]
But let's assume that the confidence interval given is true 4.90 hrs < μ1 - μ2 < 17.50 hrs
What does the confidence interval suggest about the population means?
We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.
Use trigonometric identities and algebraic methods, as necessary, to solve the following trigonometric equation. Please identify all possible solutions by including all answers in [0,2Ï€) and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution." cos2(5x)=sin2(5x)
Answer:
[tex]\large\boxed{x=\pm\dfrac{3\pi}{20}+\dfrac{2n\pi}{5}\ \vee\ x=\pm\dfrac{\pi}{20}+\dfrac{2n\pi}{5}}[/tex]
Step-by-step explanation:
[tex][tex]\cos^2(5x)=\sin^2(5x)\qquad\text{substitute}\ t=5x\\\\\cos^2t=\sin^2t\qquad\text{use}\ \sin^2\theta+\cos^2\theta=1\to\sin^2\theta=1-\cos^2\theta\\\\\cos^2t=1-\cos^2t\qquad\text{add}\ \cos^2t\ \text{to both sides}\\\\2\cos^2t=1\qquad\text{divide both sides by 2}\\\\\cos^2t=\dfrac{1}{2}\Rightarrow \cos t=\pm\sqrt{\dfrac{1}{2}}\\\\\cos t=\pm\dfrac{\sqrt2}{2}\\\\\cos t=-\dfrac{\sqrt2}{2}\Rightarrow t=\pm\dfrac{3\pi}{4}+2n\pi\\\\\cos t=\dfrac{\sqrt2}{2}\Rightarrow t=\pm\dfrac{\pi}{4}+2n\pi[/tex]
[tex]t=5x\\\\5x=\pm\dfrac{3\pi}{4}+2n\pi\ \vee\ 5x=\pm\dfrac{\pi}{4}+2n\pi\qquad\text{divide both sides by 5}\\\\x=\pm\dfrac{3\pi}{20}+\dfrac{2n\pi}{5}\ \vee\ x=\pm\dfrac{\pi}{20}+\dfrac{2n\pi}{5}[/tex]
The trigonometric equation 2cos^2(x) + 2 = 4 has solutions x = 0 and x = π in the interval [0, 2π). To include all possible solutions, we express them as x = 2nπ and x = π + 2nπ with n as any integer.
To solve the trigonometric equation 2cos^2(x) + 2 = 4, we start by simplifying the equation:
Subtract 2 from both sides to get 2cos^2(x) = 2.Divide both sides by 2 to get cos^2(x) = 1.Take the square root of both sides, considering both positive and negative roots, to get cos(x) = ±1.Find angles x in the interval [0, 2π) where the cosine is ±1. For cos(x) = 1, x = 0. For cos(x) = -1, x = π.This equation has two solutions in the interval [0, 2π): x = 0 and x = π (which are 0 and 3.1416 when rounded to four decimal places). To express the infinite set of solutions that occur at every period of cos(x), we add 2nπ to each solution, where n is any integer, giving the final solutions as x = 2nπ and x = π + 2nπ.
the complete Question is given below:
Use trigonometric identities and algebraic methods, as necessary, to solve the following trigonometric equation. Please identify all possible solutions by including all answers in [0, 2π)
and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."
2cos^2 (x) + 2 = 4
Enter your answer in radians, as an exact answer when possible. Multiple answers should be separated by commas.
What are the possible rational zeros of f(x) = x4 − 4x3 + 9x2 + 5x + 14?
Answer:
±1, ±2, ±7, ±14
Step-by-step explanation:
Your mathematics instructor claims that, over the years, 88% of his students have said that math is their favorite subject. In this year's class, however, only 21 out of 32 students named math as their favorite class. The instructor decides to construct a confidence interval for the true population proportion based on the sample value. What's the correct value for the standard error of pˆ in this case?
Answer: 0.084
Step-by-step explanation:
Formula to find standard error of [tex]\hat{p}[/tex] for finding confidence interval for p:
[tex]SE=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where [tex]\hat{p}[/tex] = sample proportion and n= sample size.
Let p be the population proportion of students named math as their favorite class.
As per given , we have
n= 32
[tex]\hat{p}=\dfrac{21}{32}=0.65625[/tex]
Substitute these values in the formula, we get
[tex]SE=\sqrt{\dfrac{0.65625(1-0.65625)}{32}}\\\\=\sqrt{0.00705}\\\\=0.0839642781187\approx0.084[/tex]
∴ The correct value for the standard error of [tex]\hat{p}[/tex] in this case = 0.084
Final answer:
The standard error of pˆ in this case is 0.0417.
Explanation:
The correct value for the standard error of pˆ in this case is 0.0417.
To calculate the standard error of pˆ, you can use the formula: SE = √((pˆ * q) / n), where pˆ is the sample proportion, q is 1 - pˆ (the proportion of non-favorites), and n is the sample size.
In this case, pˆ = 21/32 = 0.6563, q = 1 - 0.6563 = 0.3437, and n = 32. Plugging these values into the formula gives SE = √((0.6563 * 0.3437) / 32) = 0.0417.
Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). An article reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for m = 7 normal subjects was 1.83 mm, and the sample standard deviation was 0.54; for n = 10 CTS subjects, the sample mean and sample standard deviation were 2.35 and 0.88, respectively. Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of 0.01. (Use μ1 for normal subjects and μ2 for CTS subjects.)
Answer:
[tex]t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507[/tex]
[tex]p_v =P(t_{(15)}>1.507)=0.076[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X_{CTS}=2.35[/tex] represent the mean for the sample CTS
[tex]\bar X_{N}=1.83[/tex] represent the mean for the sample Normal
[tex]s_{CTS}=0.88[/tex] represent the sample standard deviation for the sample of CTS
[tex]s_{N}=0.54[/tex] represent the sample standard deviation for the sample of Normal
[tex]n_{CTS}=10[/tex] sample size selected for the CTS
[tex]n_{N}=7[/tex] sample size selected for the Normal
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{CTS} \leq \mu_{N}[/tex]
Alternative hypothesis:[tex]\mu_{CTS} > \mu_{N}[/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_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}[/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{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{CTS}+n_{N}-2=10+7-2=15[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(15)}>1.507)=0.076[/tex]
We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.
Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) y = x , x = 1, Δx = 1 Δy = dy =
Answer:
Δy = 1
dy = 1
Step-by-step explanation:
Data provided in the question:
dx = Δx
y = x
x = 1,
Δx = 1
Now,
we know,
Δy = f( x + Δx ) - f(x)
also, we have
y = f(x) = x
thus,
f( x + Δx ) = x + Δx
Therefore,
Δy = ( x + Δx ) - x
on substituting the respective values, we get
Δy = ( 1 + 1 ) - 1
or
Δy = 1
and,
dy = f'(x) = [tex]\frac{d(x)}{dx}[/tex]
or
dy = 1
......Help Please......
Answer:
Step-by-step explanation:
the sign here means division
7 divided by the number in the box equals to 8
simply meaning 7 multiplied by 8 will give the number
7 x 8 = 56
If two events are mutually exclusive, what is the probability that one or the other occurs?
In probability, if two events are mutually exclusive (cannot occur at the same time), the probability of either event occurring is calculated by adding the probabilities of each event separately.
Explanation:The concept being referred to in your question is related to probability within the study of mathematics. When two events are mutually exclusive, it means they cannot occur at the same time. For example, when rolling a die, the events of throwing a '6' and a '3' are mutually exclusive; you cannot throw both on a single toss.
In reference to your question, the probability that one event or the other occurs, given that they are mutually exclusive, is calculated by adding the probabilities of each individual event. So if the probability of Event A is P(A) and the probability of Event B is P(B), then the probability that either A or B will occur (P(A U B)) equals P(A) + P(B).
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Listed are 32 ages for Academy Award winning best actors in order from smallest to largest. (Round your answers to the nearest whole number.) 18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77 (a) Find the percentile of 31. th percentile
Answer:
Step-by-step explanation:
The ages for Academy Award winning best actors in order from smallest to largest are
18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
The total number of terms, n is 32
31℅ of 32 = 31/100 × 32 = 0.31 × 32 = 9.92. It is approximately 10
Counting from left to right, the 10th term is 31. This means that the 31st percentile is 31
Find a particular solution to y′′+25y=−40sin(5t). y″+25y=−40sin(5t).
The particular solution to the differential equation is:
[tex]y_p(t)= 16/25tcos(5t) - 8/5tsin(5t)[/tex]
Here, we have,
To find a particular solution to the given differential equation
y′′+25y=−40sin(5t),
we'll assume a particular solution of the form:
[tex]y_p(t)=Asin(5t)+Bcos(5t)[/tex]
where A and B are constants to be determined.
Now, let's find the first and second derivatives of [tex]y_p(t)[/tex] :
[tex]y_p'(t)=5Acos(5t)-5Bsin(5t)\\y_p''(t)=-25Asin(5t)-25Bcos(5t)[/tex]
Now, substitute these derivatives back into the original differential equation:
y′′+25y=(−25Asin(5t)−25Bcos(5t))+25(Asin(5t)+Bcos(5t))
Now, equate the coefficient of sin(5t) and cos(5t) on both sides of the equation:
For sin(5t):
−25A+25A=0⟹0=0
For cos(5t):
−25B+25B=−40⟹0=−40
The above equations show that there is no solution for A and B that satisfy the original equation.
This means that our initial assumption for the particular solution is not appropriate for this case.
To find a particular solution that works, we'll make another assumption. Since the right-hand side of the differential equation is −40sin(5t), we'll try a particular solution in the form:
[tex]y_p(t)=Atcos(5t)+Btsin(5t)[/tex]
where A and B are constants to be determined.
Now, let's find the first and second derivatives of this new
[tex]y_p'(t)=Acos(5t)-5Atsin(5t)+Bsin(5t)+5Btcos(5t)[/tex]
[tex]y_p''(t)=-10Asin(5t)-25Atcos(5t)+25Btsin(5t)-10Bcos(5t)[/tex]
Now, substitute these derivatives back into the original differential equation:
[tex]y''+25y=(-10Asin(5t)-25Atcos(5t)+25Btsin(5t -10Bcos(5t))+25(Atcos(5t)+Btsin(5t))[/tex]
Now, equate the coefficient of sin(5t) and cos(5t) on both sides of the equation:
For sin(5t):
25Bt=−40⟹B=− 8/5
For cos(5t):
−25At−10B=0
⟹A= 16/25
So, the particular solution to the differential equation is:
[tex]y_p(t)= 16/25tcos(5t) - 8/5tsin(5t)[/tex]
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To find a particular solution to the differential equation y″+25y=−40sin(5t), assume a particular solution in the form y(t) = Asin(5t) + Bcos(5t) and solve for the coefficients A and B.
Explanation:To find a particular solution to the given differential equation y″+25y=−40sin(5t), we can assume a particular solution in the form of y(t) = Asin(5t) + Bcos(5t). Differentiating this equation twice and substituting it back into the original differential equation, we can solve for the coefficients A and B. In this case, the particular solution is y(t) = -8sin(5t) - 0.64cos(5t).
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PLEASE PLEASE HELP
Answer:68.3 degrees
Step-by-step explanation:
The diagram of the triangle ABC is shown in the attached photo. We would determine the length of side AB. It is equal to a. We would apply the cosine rule which is expressed as follows
c^2 = a^2 + b^2 - 2abCos C
Looking at the triangle,
b = 75 miles
a = 80 miles.
Angle ACB = 180 - 42 = 138 degrees. Therefore
c^2 = 80^2 + 75^2 - 2 × 80 × 75Cos 138
c^2 = 6400 + 5625 - 12000Cos 138
c^2 = 6400 + 5625 - 12000 × -0.7431
c^2 = 12025 + 8917.2
c = √20942.2 = 144.7
To determine A, we will apply sine rule
a/SinA = b/SinB = c/SinC. Therefore,
80/SinA = 144.7/Sin 138
80Sin 138 = 144.7 SinA
SinA = 53.528/144.7 = 0.3699
A = 21.7 degrees
Therefore, theta = 90 - 21.7
= 68.3 degees
Let A = (0, 0), B = (8, 1), C = (5, −5), P = (0, 3), Q = (7, 7), and R = (1, 10). Prove that angles ABC and P QR have the same size.
Answer:
To prove
∠ABC = ∠PQR
Using euclidean distance, Length of each side can be found as
[tex]AB=\sqrt{65} ,BC=\sqrt{45} ,CA=\sqrt{50} \\PQ=\sqrt{65} ,QR=\sqrt{45} ,RP=\sqrt{50}[/tex]
As can be seen
AB ≅ PQ
BC ≅ QR
CA ≅ RP
As all the sides of ΔABC are equal and congruent to ΔPQR, this Proves that measure of all angles inside both triangles must be equal.
Some sources report that the weights of full-term newborn babies in a certain town have a mean of 7 pounds and a standard deviation of 1.2 pounds and are normally distributed.a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the meanlong dashthat is, between 5.8 and 8.2 pounds, or within one standard deviation of the mean?b. What is the probability that the average of nine babies' weights will be within 1.2 pounds of the mean; will be between 5.8 and 8.2 pounds?c. Explain the difference between (a) and (b).
Answer:
a) [tex]P(5.8<X<8.2)=P(\frac{5.8-7}{1.2}<Z<\frac{8.2-7}{1.2})=P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.683[/tex]
b) [tex]P(5.8<\bar X <8.2) = P(Z<3)-P(Z<-3)=0.999-0.0014=0.9973[/tex]
c) For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.
For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values
Step-by-step explanation:
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]
Let X the random variable that represent the weights of full-term newborn babies of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(7,1.2)[/tex]
a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the meanlong dashthat is, between 5.8 and 8.2 pounds, or within one standard deviation of the mean?
We are interested on this probability
[tex]P(5.8<X<8.2)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(5.8<X<8.2)=P(\frac{5.8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{8.2-\mu}{\sigma})[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(5.8<X<8.2)=P(\frac{5.8-7}{1.2}<Z<\frac{8.2-7}{1.2})=P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.683[/tex]
b. What is the probability that the average of nine babies' weights will be within 1.2 pounds of the mean; will be between 5.8 and 8.2 pounds?
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(7,\frac{1.2}{\sqrt{9}})[/tex]
The z score on this case is given by this formula:
[tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace the values that we have we got:
[tex]z_1=\frac{5.8-7}{\frac{1.2}{\sqrt{9}}}=-3[/tex]
[tex]z_2=\frac{8.2-7}{\frac{1.2}{\sqrt{9}}}=3[/tex]
For this case we can use a table or excel to find the probability required:
[tex]P(5.8<\bar X <8.2) = P(Z<3)-P(Z<-3)=0.999-0.0014=0.9973[/tex]
c. Explain the difference between (a) and (b).
For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.
For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values
A study was made to determine whether more Italians than Americans prefer white champagne to pink champagne at weddings. Of the 300 Italians selected at random, 74 preferred white champagne, and of the 400 Americans selected, 60 preferred white champagne. Can we conclude that a higher proportion of Italians than Americans prefer white champagne at weddings? Use a 0.05 level of significance.
Answer:
[tex]z=\frac{0.247-0.15}{\sqrt{0.191(1-0.191)(\frac{1}{300}+\frac{1}{400})}}=3.23[/tex]
[tex]p_v =P(Z>3.23)=0.000619[/tex]
Comparing the p value with 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 reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at weddings it's significantly higher than the proportion of Americans.
Step-by-step explanation:
1) Data given and notation
[tex]X_{I}=74[/tex] represent the number of Italians that preferred white champagne
[tex]X_{A}=60[/tex] represent the number of Americans that preferred white champagne
[tex]n_{I}=300[/tex] sample of Italians selected
[tex]n_{A}=400[/tex] sample of Americans selected
[tex]p_{I}=\frac{74}{300}=0.247[/tex] represent the proportion of Italians that preferred white champagne
[tex]p_{A}=\frac{60}{400}=0.15[/tex] represent the proportion of Americans that preferred white champagne
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if a higher proportion of Italians than Americans prefer white champagne at weddings, the system of hypothesis would be:
Null hypothesis:[tex]p_{I} - p_{A} \leq 0[/tex]
Alternative hypothesis:[tex]p_{I} - p_{A} > 0[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{I}-p_{A}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{I}}+\frac{1}{n_{A}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{I}+X_{A}}{n_{I}+n_{A}}=\frac{74+60}{300+400}=0.191[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.247-0.15}{\sqrt{0.191(1-0.191)(\frac{1}{300}+\frac{1}{400})}}=3.23[/tex]
4) Statistical decision
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>3.23)=0.000619[/tex]
Comparing the p value with 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 reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at weddings it's significantly higher than the proportion of Americans.
To determine whether a higher proportion of Italians than Americans prefer white champagne at weddings, we need to conduct a hypothesis test. First, state the null and alternative hypotheses. Then, calculate the test statistic and compare it to the critical value at a significance level of 0.05.
Explanation:To determine whether a higher proportion of Italians than Americans prefer white champagne at weddings, we need to conduct a hypothesis test. First, we need to state the null hypothesis (H0) and the alternative hypothesis (Ha). In this case, H0: p1 = p2 (the proportion of Italians who prefer white champagne is equal to the proportion of Americans who prefer white champagne) and Ha: p1 > p2 (the proportion of Italians who prefer white champagne is greater than the proportion of Americans who prefer white champagne).
Next, we calculate the test statistic using the formula: z = (p1 - p2) / √(p*(1-p)*((1/n1) + (1/n2))), where p = (x1 + x2) / (n1 + n2), x1 and x2 are the number of Italians and Americans who prefer white champagne respectively, and n1 and n2 are the sample sizes of Italians and Americans respectively.
We then compare the test statistic to the critical value at a significance level of 0.05. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that a higher proportion of Italians than Americans prefer white champagne at weddings.
The mean consumption of bottled water by a person in the United States is 28.5 gallons per year. You believe that a person consume more than 28.5 gallons in bottled water per year. A random sample of 100 people in the United States has a mean bottled water consumption of 27.8 gallons per year with a standard deviation of 4.1 gallons. At α = 0.10 significance level can you reject the claim?
Answer:
[tex]t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707[/tex]
[tex]p_v =P(t_{99}<-1.707)=0.0455[/tex]
If we compare the p value with a significance level for example [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.
Step-by-step explanation:
Data given and notation
[tex]\bar X=27.8[/tex] represent the mean for the account balances of a credit company
[tex]s=4.1[/tex] represent the population standard deviation for the sample
[tex]n=1000[/tex] sample size
[tex]\mu_o =28.5[/tex] represent the value that we want to test
[tex]\alpha=0.1[/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 determine if the mean for the person consume is more than 28.5 gallons, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 28.5[/tex]
Alternative hypothesis:[tex]\mu < 28.5[/tex]
We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707[/tex]
Calculate the P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=100-1=99[/tex]
Since is a one-side lower test the p value would be:
[tex]p_v =P(t_{99}<-1.707)=0.0455[/tex]
In Excel we can use the following formula to find the p value "=T.DIST(-1.707,99)"
Conclusion
If we compare the p value with a significance level for example [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.
The cash operating expenses of the regional phone companies during the first half of 1994 were distributed about a mean of $29.93 per access line per month, with a standard deviation of $2.65. Company A's operating expenses were $27.00 per access line per month. Assuming a normal distribution of operating expenses, estimate the percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean. (Round your answer to two decimal places.)
Answer:
The percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean is 73%.
Step-by-step explanation:
The Company A's operating expenses were $27.00. This is $2.93 less than the regional mean.
[tex]\Delta E=29.93-27.00=2.93[/tex]
The companies whose operating expenses are closer to the mean are the ones that have expenses $2.93 below or above the mean.
The fraction of companies that are closer to the mean is equal to the proability of having expenses between those two limits:
[tex]z_1=(M-\mu)/\sigma=-2.93/2.65=-1.105\\\\z_2=+1.105[/tex]
[tex]P(|z|\leq1.105)=P(z\leq 1.105)-P(z<-1.105)=0.86542-0.13458=0.73[/tex]
To find the percentage of companies with operating expenses closer to the mean than Company A, calculate the Z score for Company A. Then find the percentile, and double it to account for both sides of the normal distribution.
Explanation:To find out the percentage of regional phone companies that had operating expenses closer to the mean than the ones of Company A, we need to calculate the Z-score for Company A. The
Z-score
is a measure of how many standard deviations an element is from the mean. In this particular case, you can calculate the Z score using the formula:
Z = (X - μ) / σ
, where X is the value from the dataset (in this case, Company A's operating expenses), μ is the mean of the dataset, and σ is the standard deviation of the dataset.
So using the data from the question:
Z = ($27.00 - $29.93) / $2.65 = -1.109. Next, recognize that finding the percentage closer to the mean is the same as finding the percentile of company A's Z-score. You can then use a standard normal distribution table or online Z-score calculator to find the percentile associated with a Z-score of -1.109. If it's a two-tailed test (as normally implied by the word 'closer' in statistical analysis), you can multiply the cumulative probability by 2, as it would be the probability on either side of the distribution.This will give you a percentage that can be understood as the percentage of regional phone companies which had operating expenses closer to the mean than Company A's were to the mean.
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On the Richter Scale, the magnitude R of an earthquake of intensityI is
R = log10 I/Io
where lo is a reference intensity. At 7:23am on March 19, 2003, an earthquake measuring 3.0 on the Richter scale occurred near the town of Nephi. An earthquake of that magnitude is often felt, but rarely causes damage. By comparison, the earthquake that struck San Francisco in 1906 measured 8.25 on the Richter scale. It was ___________times as intensive as the Nephi earthquake.
Answer:
177,827.941
Step-by-step explanation:
The magnitude of an earthquake is given by:
[tex]R=log(\frac{l}{l_0})[/tex]
The intensity of the Nephi earthquake is:
[tex]3.0=log(\frac{l_N}{l_0})\\10^3 =\frac{l}{l_0} \\l_N=1,000*l_0[/tex]
The intensity of the San Francisco earthquake is
[tex]8.25=log(\frac{l_S}{l_0})\\10^{8.25} =\frac{l_S}{l_0} \\l_S=177,827,941*l_0[/tex]
The intensity ratio is:
[tex]r= \frac{l_S}{l_N}=\frac{177,827,941*l_0}{1,000*l_0}\\r= 177,827.941[/tex]
The San Francisco earthquake was 177,827.941 times as intensive as the Nephi earthquake.
When comparing the intensity of the San Francisco earthquake (magnitude 8.25) and the Nephi earthquake (magnitude 3.0) on the Richter Scale, we find that the San Francisco earthquake was approximately 177,827.94 times more intensive than the Nephi earthquake.
Explanation:To compare the intensity of two earthquakes using the Richter Scale, we need to apply the formula:
R = log10 I/Io, where R is the Richter magnitude, I is the intensity of the earthquake, and Io is a reference intensity.
Now, we know that the Richter scale is a base-10 logarithmic scale. This means that each whole number increase on the scale represents a tenfold increase in measured amplitude and roughly 31.6 times more energy release.
The earthquakes in question have magnitudes of 3.0 and 8.25. So, the intensity of the San Francisco earthquake compared to the Nephi earthquake is 10^(8.25-3) = 10^5.25.
Therefore, the San Francisco earthquake was approximately 177,827.94 times more intensive as the Nephi earthquake.
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Compare - 40% of A is equal to $300 and 30% of B is equal to $400
Answer:
B is greater than A.
Step-by-step explanation:
We have been given that 40% of A is equal to $300 and 30% of B is equal to $400. We are asked to compare both quantities.
Let us find A and B using our given information.
[tex]\frac{40}{100}*A=300[/tex]
[tex]0.40A=300[/tex]
[tex]\frac{0.40A}{0.40}=\frac{300}{0.40}[/tex]
[tex]A=750[/tex]
Similarly, we will find B.
[tex]\frac{30}{100}*B=400[/tex]
[tex]0.30*B=400[/tex]
[tex]\frac{0.30B}{0.30}=\frac{400}{0.30}[/tex]
[tex]B=1333.3333[/tex]
We know that smaller percent of big number is greater than bigger percent of a small number.
Therefore, B is greater than A.
Answer:
Step-by-step explanation:
40% Of A = 300
30% of B = 400
40a / 100 = 300, 40a = 3 divide both sides by 3 ,
then a = 13.33 approximately 13
30b = 400 = 400, 30b = 4 divide both sides by 30, then b =7.5
comparing both results where a = 13 , b = 7.5
a is greater than b with difference of 5.5
simplify (3/4 + 4/5i)-(1/2 - 3/10i)
For this case we must simplify the following expression:
[tex](\frac {3} {4} + \frac {4} {5} i) - (\frac {1} {2} - \frac {3} {10} i) =[/tex]
By law of multiplication signs we have to:
[tex]- * + = -\\- * - = +\\\frac {3} {4} + \frac {4} {5} i- \frac {1} {2} + \frac {3} {10} i =[/tex]
We add similar terms:
[tex]\frac {3} {4} - \frac {1} {2} + \frac {4} {5} i + \frac {3} {10} i =\\\frac {3 * 2-4 * 1} {4 * 2} + \frac {4 * 10 + 5 * 3} {10 * 5} i =\\\frac {2} {8} + \frac {55} {50} i =[/tex]
We simplify:
[tex]\frac{1}{4}+\frac{11}{10}i[/tex]
Answer:
[tex]\frac{1}{4}+\frac{11}{10}i[/tex]
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 80% confident that his estimate is within six percentage points of the true populationpercentage?
Complete parts (a) through (c) below.
1) Assume that nothing is known about the percentage of adults who have heard of the brand.
a.n=_________(Round up to the nearest integer.)
2) Assume that a recent survey suggests that about 85% of adults have heard of the brand.
b.n=_________(Round up to the nearest integer.)
3) Given that the required sample size is relatively small, could he simply survey the adults at the nearest college?
Answer:
1) n=114
2) n=59
3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by [tex]\alpha=1-0.80=0.2[/tex] and [tex]\alpha/2 =0.1[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28[/tex]
Part 1
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.06[/tex] or 6% points, and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
Since we don't have a prior estimate of [tex]\het p[/tex] we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.06}{1.28})^2}=113.77[/tex]
And rounded up we have that n=114
Part 2
On this case we have a prior estimate for the population proportion and is [tex]\hat p =0.85[/tex] so replacing the values into equation (b) we got:
[tex]n=\frac{0.85(1-0.85)}{(\frac{0.06}{1.28})^2}=58.027[/tex]
And rounded up we have that n=59
Part 3
On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.
There is a strong correlation between the temperature and the number of skinned knees on playgrounds. Does this tell us that warm weather causes children to trip? Choose the correct answer below. A. Yes. In warm weather, more children will go outside and play. B. No. Warm weather will cause less children to trip and suffer skinned knees. C. No. In warm weather, more children will go outside and play. D. Yes. Warm weather will cause more children to trip and suffer skinned knees.
Answer:
C. No. In warm weather, more children will go outside and play
Step-by-step explanation:
The correct answer is option C: No. In warm weather, more children will go outside and play.
Correlation means that two variables are related, but it does not necessarily mean that one causes the other. In this case, there is a strong correlation between temperature and the number of skinned knees on playgrounds, but it does not tell us that warm weather causes children to trip.
Option A is incorrect because warm weather does not directly cause children to go outside and play. It may be a factor, but it is not the sole reason. Option B is incorrect because warm weather does not cause fewer children to trip and suffer skinned knees. In fact, the correlation suggests that more children are likely to be outside playing in warm weather, which could potentially increase the number of skinned knees. Option D is incorrect because warm weather does not directly cause more children to trip and suffer skinned knees.
The correlation suggests that the increase in skinned knees is likely due to more children being outside and playing, rather than the warm weather itself. To summarize, the strong correlation between temperature and the number of skinned knees on playgrounds indicates that in warm weather, more children will go outside and play. However, it does not tell us that warm weather causes children to trip.
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Conduct a test at the a=0.05 level of significance by determining (a) null and alternative hypothesis, (b) the test statistic, (c) the P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether p1 > p2. The sample data are x1=117 n1=249 x2=141 n2=312
Answer:
Since p > alpha, we accept H0, there is no evidence to prove that p1 is greater than p2
Step-by-step explanation:
Set up hypotheses as
[tex]H_0: p1 = p2\\H_a: p1>p2[/tex]
(Right tailed t test)
Alpha = 0.05
Sample I II Total
X 117 141 248
N 249 312 561
p 0.4700 0.4519 0.4420
p difference = 0.0181
Std dev = [tex]\sqrt{p(1-p)(\frac{1}{n_1} +\frac{1}{n_2} )}[/tex]
=0.0427
z statistic = 0.424
p value = 0.33724
Since p > alpha, we accept H0, there is no evidence to prove that p1 is greater than p2
Final answer:
To test whether p1 is greater than p2 using hypothesis testing, one needs to state the null and alternative hypotheses, calculate the test statistic and p-value, and then decide whether to reject or accept the null hypothesis based on the level of significance and the p-value. Calculating the test statistic and p-value requires specific formulas and is not provided here.
Explanation:
Conducting a Hypothesis Test for Two Population Proportions
A student would like assistance in conducting a hypothesis test for two population proportions. The steps to perform such a test include:
State the null and alternative hypotheses: The null hypothesis (H0) is that there is no difference between the two population proportions (p1 - p2 = 0), while the alternative hypothesis (Ha) posits that there is a difference (p1 > p2).
The random variable (P') represents the difference between the two sample proportions.
Calculate the test statistic: Using the provided sample sizes and successful outcomes, the test statistic is calculated using a formula based on the Z-distribution.
Calculate the p-value: The p-value is determined from the test statistic, indicating the probability of observing such a result if the null hypothesis were true.
At the 5 percent level of significance, compare the p-value to the alpha value of 0.05 to make a decision. If the p-value is less than alpha, reject the null hypothesis.
The Type I error would occur if the null hypothesis is incorrectly rejected when it is actually true.
The Type II error would occur if the null hypothesis is not rejected when it is actually false.
The test statistic and p-value cannot be calculated from the information provided without the appropriate formulas or statistical tools.
Decision Making and Errors
If the alpha level is greater than the p-value, the null hypothesis should be rejected, indicating there is evidence to suggest p1 is greater than p2.
Failure to reject the null hypothesis when it is false constitutes a Type II error.
Assume a device manufacturer tests 100 devices. The first device fails at 100 hours. The last device fails at 200 hours. What is the device MTBF:a. 10,000 Hoursb. 15,000 hours c. 20,000 hours d. 100 hours
Answer:
a. 10,000 Hours
Step-by-step explanation:
MTBF (Mean Time Between Failures) can be calculated using the formula:
[tex]MTBF={(L-F)}*{n}[/tex] where
L is the time at which last device fails (200 hours) F is the time at which first device fails (100 hours) n is the number of devices tested (100)[tex]{MTBF=(200-100)}*{100}=10000[/tex]
Final answer:
The MTBF for a set of devices tested by the manufacturer, with failures evenly distributed from 100 to 200 hours, is 15,000 hours, which is calculated using the sum of an arithmetic series formula.
Explanation:
The student is asking about the mean time between failures (MTBF) for a set of devices. MTBF is a basic measure of reliability for repairable items and represents the average time expected between failures. To calculate MTBF, you divide the total operational time by the number of failures. In this case, the device manufacturer tests 100 devices, with the first failing after 100 hours and the last after 200 hours, which suggests a linear distribution of failures over time.
Assuming a linear and even distribution of failures from 100 to 200 hours for the 100 devices, you would calculate the total operational time before each device failed and then divide by the number of devices. The calculation would be the sum of an arithmetic series:
MTBF = (100 hours + 101 hours + ... + 200 hours) / 100 devices
We can use the formula for the sum of an arithmetic series:
S = n/2 * (a1 + an)
Where:
n is the number of terms in the series (here, 100 devices),
a1 is the first term in the series (100 hours),
an is the last term in the series (200 hours),
Now apply the formula:
MTBF = 100/2 * (100 + 200)
MTBF = 50 * 300
MTBF = 15,000 hours
So, the answer is b. 15,000 hours.
What point on the line y 9x+4 is closest to the origin? Let D be the distance between the two points. What is the objective function in terms of the x-coordinate? (Type an expression.)
Answer:
The objective function in terms of the x-coordinate is [tex]d=\sqrt{82x^2+72x+16}[/tex].
The point closest to the origin is [tex](-\frac{18}{41},\frac{2\sqrt{82}}{41})[/tex].
Step-by-step explanation:
The formula for the distance from point (x, y) to the origin is
[tex]d=\sqrt{x^{2}+y^{2} }[/tex]
So, in our case, [tex]y=9x+4[/tex] and the distance is
[tex]d=\sqrt{x^{2}+(9x+4)^{2} }\\\\d=\sqrt{x^2+81x^2+72x+16} \\\\d=\sqrt{82x^2+72x+16}[/tex]
This is the objective function.
Next, we need to find the derivative of the function
[tex]d=\sqrt{82x^2+72x+16}\\\\\frac{d}{dx}d= \frac{d}{dx}\sqrt{82x^2+72x+16}\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt{u},\:\:u=82x^2+72x+16\\\\\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(82x^2+72x+16\right)\\\\\frac{1}{2\sqrt{u}}\left(164x+72\right)\\\\\mathrm{Substitute\:back}\:u=82x^2+72x+16\\\\\frac{1}{2\sqrt{82x^2+72x+16}}\left(164x+72\right)\\\\\frac{d}{dx}d=\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}[/tex]
Now, we set the derivative function equal to zero to find the critical points
[tex]\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}=0[/tex]
[tex]\frac{f\left(x\right)}{g\left(x\right)}=0\quad \Rightarrow \quad f\left(x\right)=0\\\\2\left(41x+18\right)=0\\\\\frac{2\left(41x+18\right)}{2}=\frac{0}{2}\\\\41x+18=0\\\\x=-\frac{18}{41}[/tex]
We need to check that the value that we found is a minimum point for this we analyze the intervals of increase or decrease (First derivative test).
We can use a sign chart. In a sign chart, we pick a test value at each interval that is bounded by the critical points and check the derivative's sign on that value.
This is the sign chart for our function:
[tex]\left\begin{array}{ccc}\mathrm{Interval}&\mathrm{Test \:x-value}&f'(x)\\(-\infty,-\frac{18}{41} )&-1&-9.02\\(-\frac{18}{41},\infty )&1&9.05\end{array}\right\\[/tex]
d(x) decreases before [tex]x=-\frac{18}{41}[/tex], increases after it, and is defined at [tex]x=-\frac{18}{41}[/tex]. So d(x) has a relative minimum point at [tex]x=-\frac{18}{41}[/tex].
The point closest to the origin is
[tex]d=y=\sqrt{82(-\frac{18}{41})^2+72(-\frac{18}{41})+16}=\frac{2\sqrt{82}}{41}[/tex]
[tex](-\frac{18}{41},\frac{2\sqrt{82}}{41})[/tex]
The objective function in this problem is the x-coordinate of the closest point on the line y = 9x + 4 to the origin. To find this point, we need to find the intersection of the line and the perpendicular line passing through the origin.
Explanation:The objective function in this case is the distance between the closest point on the line y = 9x + 4 and the origin. To find this point, we need to find the intersection of the line and the perpendicular line passing through the origin. The x-coordinate of this closest point will be our objective function.
The slope of the given line is 9, the negative reciprocal of which is -1/9. This means the perpendicular line will have a slope of -1/9 as well. Since the perpendicular line passes through the origin, its equation is given by y = -1/9x.
To find the intersection point, we can set the equations of the two lines equal to each other: 9x + 4 = -1/9x. Solving this equation, we find x = -4/81. Therefore, the objective function in terms of the x-coordinate is -4/81.
Learn more about Objective function here:https://brainly.com/question/34856872
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