Answer:
Statement A - True.
Statement B - False.
Statement C - True.
Statement D - False.
Step-by-step explanation:
(a) A sequence is convergent if and only if all of its subsequences are convergent - this statement is correct.
(b) A sequence is bounded if and only if all of its subsequences are bounded - this statement is incorrect.
(c) A sequence is monotonic if and only if all of its subsequences are monotonic - this statement is correct.
(d) A sequence is divergent if and only if all of its subsequences are divergent - this statement is incorrect.
Final answer:
In the context of mathematical sequences, statements (a), (b), and (c) regarding convergence, boundedness, and monotonicity are true as all subsequences follow the property of the original sequence. However, statement (d) is false because even divergent sequences can have convergent subsequences.
Explanation:
When analyzing the behavior of sequences and series in mathematics, it's important to understand various characteristics such as convergence, boundedness, monotonicity, and divergence. We can assess the truthfulness of the given statements based on these characteristics.
Assessment of the Statements:
(a) True: A sequence is convergent if and only if all of its subsequences are convergent. This is a fundamental property of convergent sequences, implying that if the original sequence approaches a specific value, every subsequence will also approach that same value.
(b) True: A sequence is bounded if and only if all of its subsequences are bounded. Every subsequence of a bounded sequence also has to be bounded, because it cannot exceed the bounds set by the original sequence.
(c) True: A sequence is monotonic if and only if all of its subsequences are monotonic. Regardless of which elements are chosen to form a subsequence, if the original sequence preserves its direction of progression (either increasing or decreasing), so will the subsequences.
(d) False: A sequence is divergent if and only if all of its subsequences are divergent. Here we have a counterexample: consider the sequence (-1)ⁿ, which diverges. However, it has convergent subsequences, such as the constant subsequences of all 1s or all -1s.
Therefore, statements (a), (b), and (c) are true, while statement (d) is false. Understanding the conditions for convergence, boundedness, and monotonicity is key to studying the behaviors of sequences and leveraging properties like the comparison test for evaluating whether a series converges or diverges.
Enter the equation that describes the line in slope-intercept form.
slope =-4, (5,6) is on the line.
Answer:
y = -4x + 26
Step-by-step explanation:
y = mx + b m: slope -4 b: y intercept
pass (5 , 6)
b = y - mx = 6 - (-4) x5 = 6 + 20 = 26
equation: y = -4x + 26
I don’t understand my homework... and don’t go at me I’m an slow learner I never done this... before... but NOTE I already got the first one...
Answer:
52 weeks
365 days
10 years
100 years
Step-by-step explanation:
Josephine’s father is 5 times as old as Josephine. In 6 years, he will be only three times as old. How old is Josephine now?
Answer:
Josephine is 6 years old.
Step-by-step explanation:
6 * 5 = 30
12 * 3= 36
Josephine is currently 6 years old.
We are given two conditions about the ages of Josephine and her father.
Josephine’s father is 5 times as old as Josephine currently.
In 6 years, he will be only three times as old as she will be at that time.
Let's let 'J' represent Josephine's current age.
Then her father's current age would be 5J. In 6 years, Josephine will be J+6 and her father will be 5J+6.
According to the second condition, at that time her father's age will be three times Josephine's age.
So, we set up the equation 5J+6 = 3(J+6).
To find Josephine's current age, we solve the equation:
5J + 6 = 3J + 18
Subtract 3J from both sides:
2J + 6 = 18
Now subtract 6 from both sides:
2J = 12
And divide both sides by 2:
J = 6
Therefore, Josephine is currently 6 years old.
Sunhee had four plastic shapes a square, a circle, and a pentagon in how many ways can she line up the four shapes of the circle cannot be next to the square
There are 12 valid arrangements for Sunhee to line up the four plastic shapes so that the circle is not next to the square.
In how many ways can Sunhee line up four plastic shapes (a square, a circle, and a pentagon) if the circle cannot be next to the square?
To solve this problem, we can count the total number of ways to arrange the shapes and then subtract the cases where the circle is next to the square.
1. Total ways to arrange the shapes:
There are 4 shapes, so there are 4! = 24 ways to arrange them.2. Cases where the circle is next to the square:Consider the circle and square together as one unit. We now have 3 units (circle and square, pentagon, and an empty spot).The circle and square can be arranged within this unit in 2! = 2 ways.The total number of ways the four shapes can be arranged with the circle next to the square is 2! × 3! = 12 ways.3. Subtract the cases where the circle is next to the square from the total:24 total ways - 12 ways = 12 ways to line up the shapes with the circle not next to the square.A study by a federal agency concludes that polygraph tests given to truthful persons have probability 0.2 of suggesting that the person is deceptive. A firm asks 12 job applicants about thefts from previous employers, using a polygraph to assess their truthfulness. Suppose that all 12 answer truthfully. Let X be the number of applicants who are classified deceptive. a) Describe the probability mass function of X. b) What is the probability that the polygraph says at least 1 is deceptive? c) What is the mean number among 12 truthful persons who will be classified as deceptive? What is the standard deviation of this number? d) What is the probability that the number classified deceptive is less than the mean?
Answer:
b) 0.9313
c) mean = 2.4
standard deviation= 1.3856
d) 0.5583
Step-by-step explanation:
Given:
p = 0.2
n = 12
a) X= number of applicants classified as deceptive.
Probability mass function of X will be:
[tex] P(X=x) = \left(\begin{array}{c}12\\x\end{array}\left) (0.2)^x(1-0.2)^1^2^-^x,x=0, 1, 2, .....,12 [/tex]
b) Probability that the polygraph says at least 1 is deceptive:
[tex] P(X≥1) = 1 -P(X=0) = 1 -\left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0[/tex]
= 1 - 0.0687
= 0.9313
c) The mean number among 12 truthful persons who will be classified as deceptive:
E(X) = n•p
= 12 * 0.2
= 24
Standard deviation:
[tex]s.d = \sqrt{12*0.2*(1-0.2)}[/tex]
= 1.3856
d) Probability that the number classified deceptive is less than the mean:
[tex] P(X<2.4) = P(X≤2) = E^2_x_=_0 \left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0 [/tex]
= 0.5583
Two dice: Find the probability of rolling an odd number on the first die and an even number on the second die.
Answer:
1/4; 25%
Step-by-step explanation:
Both events happen with probability 1/2: there are 3 even numbers and 3 odd numbers in a die.
Since the two events are also independent (the outcome of the first die doesn't affect the outcome of the second), we have to multiply those probability.
So, you roll an odd number on the first die and an even number on the second die with probability
[tex]\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}[/tex]
In a one-way ANOVA, if the computed F statistic is greater than the critical F value you may Question 1 options: reject H0 since there is evidence that not all the means are different. not reject H0 since there is no evidence of a difference in the means. not reject H0 because a mistake has been made. reject H0 since there is evidence all the means differ.
Answer:
And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:
reject H0 since there is evidence all the means differ.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
Solution to the problem
The hypothesis for this case are:
Null hypothesis: [tex]\mu_{A}=\mu_{B}=....=\mu_{k}[/tex]
Alternative hypothesis: Not all the means are equal [tex]\mu_{i}\neq \mu_{j}, i,j=A,B,...,k[/tex]
If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:
reject H0 since there is evidence all the means differ.
In a one-way ANOVA, if the computed F statistic is greater than the critical F value you may eject H0 since there is evidence all the means differ.
What is F-test?F test is a method used in statistics to determine which models best fits the population from which the sample is derived.
The formula for calculating the F-test statistic = explained variance / unexplained variance
The null hypothesis usually states that the means are the means are the same while the alternative hypothesis states that the means are not the same.
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A store sells televisions for $360 and video cassette recorders for $270. At the beginning of the week its entire stock is worth $56,430. During the week it sells three quarters of the televisions and one third of the video cassette recorders for a total of $32,310. How many televisions and video cassette recorders did it have in its stock at the beginning of the week
Answer:
The number of Television at the beginning was 90
The number of video cassette recorders at the beginning was 89
Step-by-step explanation:
Selling Price of 1 Television =$360.
Selling Price of 1 video cassette recorders for $270.
Let the number of Television at the beginning=x
Let the number of video cassette recorders at the beginning =y
Opening Stock =$56,430.
Therefore:
360x+270y=$56,430.It sells three quarters of the televisions and one third of the video cassette recorders for a total of $32,310.
[tex]\frac{3}{4}(360)x+\frac{1}{3}(270)y= \$32,310[/tex]
270x+90y=32310
We then solve the two equations to obtain x and y.
360x+270y=$56,430. (Multiply by 270)270x+90y=32310 (Multiply by 360)97200x+72900y=15236100
97200x+32400y=11631600
Subtract
40500y=3604500
y=89
Substitute y=89 into 270x+90y=32310 to obtain x
270x+90(89)=32310
270x=32310-8010=24300
x=90
Therefore:
The number of Television at the beginning was 90
The number of video cassette recorders at the beginning was 89
Answer:
90 televisions and 89 video cassette recorders
Step-by-step explanation:
The unit cost of television = $360
The unit cost of video cassette recorder = $270
Let "T" represent the number of televisions and "R" represent the number of recorders, so that we can make representations using equations from the statements.
At the beginning of the week, Total Stock is worth $56,430, where
Total Stock = Total cost of televisions + Total cost of recorders
Total Cost = Unit Cost × Number of items
$56,430 = 360T + 270R This is the first equation
Next, During the week, Number of Sales = [tex]\frac{3}{4}[/tex] T + [tex]\frac{1}{3}[/tex] R
Total Sales Price = 360 ([tex]\frac{3}{4}[/tex] T) + 270 ([tex]\frac{1}{3}[/tex] R)
$32,310 = 270T + 90R This is the second equation
Solving both equations simultaneously, let us use elimination method which involves equating one of the two terms in both equations. Let us multiply the second equation by 3. This doesn't affect the equation, since we are doing it to all the terms in it.
56,430 = 360T + 270R
32,310 = 270T + 90R × 3
So, we have;
56,430 = 360T + 270R
96,930 = 810T + 270R
Subtracting both equations, we have;
96,930 - 56,430 = 810T - 360T
40,500 = 450T
T = [tex]\frac{40,500}{450}[/tex] = 90
Since we now have the number of televisions, we can get the number of recorders by putting 90 in any (say, the second) equation.
32,310 = 270 (90) + 90R
32,310 = 24,300 + 90R
32,310 - 24,300 = 90R
8010 = 90R
R = [tex]\frac{8010}{90}[/tex] = 89
At the beginning of the week, the store had 90 televisions and 89 video cassette recorders
A manufacturer sells video games with the following cost and revenue functions (in dollars), where x is the number of games sold. Determine the interval(s) on which the profit function is increasing. Upper C (x )equals 0.17 x squared minus 0.00016 x cubed Upper R (x )equals 0.362 x squared minus 0.0002 x cubed
Answer:
Therefore profit function is increasing on the interval (0,3200)
Step-by-step explanation:
The cost function of the manufacturer C(x) is given as:
[TeX]C(x)= 0.17x^{2}-0.00016x^{3}[/TeX]
The Revenue function is also given as:
[TeX]R(x)= 0.362x^{2}-0.0002x^{3}[/TeX]
Profit=Revenue-Cost
Therefore:
P(x)=R(x)-C(x)
[TeX]= 0.362x^{2}-0.0002x^{3}-[0.17x^{2}-0.00016x^{3}][/TeX]
[TeX]= 0.362x^{2}-0.0002x^{3}-0.17x^{2}+0.00016x^{3}[/TeX]
The Profit Function, [TeX]P(x)= 0.192x^{2}-0.00004x^{3}[/TeX]
To determine the point where the function is increasing, we take the derivative and examine it's critical points.
The derivative of the profit function is:
[TeX]P^{'}(x)= 0.384x-0.00012x^{2}[/TeX]
Set the derivative equal to zero.
[TeX]0.384x-0.00012x^{2}=0[/TeX]
[TeX]x(0.384x-0.00012x)=0[/TeX]
x=0 or [TeX]0.384-0.00012x=0[/TeX]
x=0 or [TeX]0.384=0.00012x[/TeX]
x=0 or x=3200
Now let's choose 2000 and 4000 as test points.
[TeX]P^{'}(2000)= 0.384(2000)-0.00012(2000)^{2}=288[/TeX]
[TeX]P^{'}(4000)= 0.384(4000)-0.00012(4000)^{2}=-384[/TeX]
Therefore profit function is increasing on the interval (0,3200)
Answer:
The is profit function is increasing on (0, 3200).
Step-by-step explanation:
Given
C(x) = 0.17x² - 0.00016x³
R(x) = 0.362x² - 0.0002x³
The profit function is given by
P(x) = R(x) - C(x)
P(x) = (0.362x² - 0.0002x³) - (0.17x² - 0.00016x³)
P(x) = 0.362x² - 0.0002x³ - 0.17x² + 0.00016x³
P(x) = 0.192x² - 0.00004x³
The derivative of the profit function is given by
P'(x) = 0.384x - 0.00012x²
Determine the critical number of P(x) to get interval where the profit function is increasing,
Set P'(x) = 0
0.384x - 0.00012x² = 0
x(0.384 - 0.00012x) = 0
x = 0 or 0.384 - 0.00012x = 0
0.384 - 0.00012x = 0
x = 0.384/0.00012 = 3200
Therefore the is profit function is increasing on (0, 3200)
If n - 2 = 6 -2(7 + n), then n = ?
Salim has 5 boxes of paint jars. Each box has the same number of paint jars. His teacher gives him 6 more paint jars. Now he has 41 paint jars. How many paint jars were in each box
Answer:
7 jars per box
Step-by-step explanation:
He now has 41 jars. How many did he have before his teacher gave him 6?
Well, 41-6=35. He had 35 jars before his teacher gave him more.
It says there was an equal amount of jars in each box. There are 5 boxes. Divide the total amount of jars (35) by the amount of boxes to find out how many jars are in each box.
35 jars / 5 boxes = 7 jars / box
covert 4.5 yards to inches what is the answer ?
Answer:
162 inches
Step-by-step explanation:
To get the answer you would have to know how many inches are a yard. The answer is 36. So you would have to multiply 4.5 by 36.
Answer:
Conversions :
1 ft = 12 in.
3 ft = 1 yard
Step-by-step explanation:
[tex]4.5 yards (\frac{3 ft}{1 yard} ) ( \frac{12 in.}{1 ft} ) = 162[/tex]
MODELING REAL LIFE The total height of the Statue of Liberty and its pedestal is $153$ feet more than the height of the statue. What is the height of the statue?
A picture shows the Statue of Liberty on a pedestal. The total height of the Statue of Liberty and its pedestal is labeled “305 feet”.
Answer: 152
Step-by-step explanation:
305 - 153 = 152
The height of the Statue of Liberty itself, without the pedestal, is 152 feet. This is found by subtracting the pedestal height (153 feet) from the total height (305 feet).
Explanation:The student is asked to determine the height of the Statue of Liberty excluding its pedestal. Given that the total height of the Statue of Liberty including its pedestal is labeled as 305 feet, and the total height is 153 feet more than the height of the statue alone, we can set up the following equation to solve for the height of the statue (let's call it S):
S + 153 = 305
To find the height of the Statue of Liberty without the pedestal, we subtract 153 from both sides of the equation:
S = 305 - 153
S = 152
Therefore, the height of the Statue of Liberty itself is 152 feet.
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the specification for a plastic handle calls for a length of 6.0 inches +- .2 inches. The standard deviation of the process is estimated to be 0.05 Inches. what are the upper and lower specification limits for this product.
Answer:
a)
USL = 6.2 inches
LSL = 5.8 inches
b) Cp = 1.33
Cpk = 0.67
c)
Yes it meets all specifications
Step-by-step explanation:
The specification for a plastic handle calls for a length of 6.0 inches ± .2 inches. The standard deviation of the process is estimated to be 0.05 inches. What are the upper and lower specification limits for this product? The process is known to operate at a mean thickness of 6.1 inches. What is the Cp and Cpk for this process? Is this process capable of producing the desired part?
Given that:
Mean (μ) = 6.1 inches, Standard deviation (σ) = 0.05 inches and the length of the plastic handle is 6.0 inches ± .2
a) Since the length of the plastic handle is 6.0 inches ± .2 = (6 - 0.2, 6 + 0.2)
The Upper specification limits (USL) = 6 inches + 0.2 inches = 6.2 inches
The lower specification limits (LSL) = 6 inches - 0.2 inches = 5.8 inches
b) The Cp is given by the formula:
[tex]Cp=\frac{(USL-LSL)}{6\sigma} =\frac{(6.2-5.8)}{6*0.05} =1.33[/tex]
The Cpk is given by the formula:
c)
The upper specification limit lies about 3 standard deviations from the centerline, and the lower specification limit is further away, so practically all units will meet specifications
[tex]Cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma})=min(\frac{6.2-6.1}{3*0.05},\frac{6.1-5.8}{3*0.05})=min(0.67,2)=0.67[/tex]
Final answer:
The upper specification limit for the plastic handle is 6.2 inches, and the lower specification limit is 5.8 inches, with these limits defining the acceptable range for the handle length.
Explanation:
The specification for a plastic handle is given as a length of 6.0 inches with a tolerance of ± 0.2 inches. This means the upper specification limit (USL) and the lower specification limit (LSL) are defined by adding and subtracting the tolerance to the target length respectively. The process standard deviation is 0.05 inches, but this does not affect the USL and LSL directly; it's a measure of the process variation.
The USL and LSL are calculated as follows:
USL = Target Length + Tolerance = 6.0 inches + 0.2 inches = 6.2 inchesLSL = Target Length - Tolerance = 6.0 inches - 0.2 inches = 5.8 inchesThese limits are the range within which the plastic handle lengths should fall according to the given specifications.
Which diagram represents a cylinder with a base area equal to 50Pi square meters?
1. A cylinder with height of 50 meters and volume = 250 pi meters cubed.
2. A cylinder with height 25 meters and volume = 250 pi meters cubed.
3. A cylinder with height 5 meters and volume = 250 pi meters cubed.
Answer: C
Step-by-step explanation:
Answer:The answer Is The Third One On The Right.
Step-by-step explanation:
A project is graded on a scale of 1 to 5. If the random variable, X, is the project grade, what is the mean of the probability
distribution below?
Answer:
the mean is 3
answer choice D
Step-by-step explanation:
60/20
Hence, the correct answer is 1/5
What is a random variable?A random variable is a numerical valued variable on a defined sample space of an experiment with expressions such as X
How to solve?probability distribution of random variable X is given by,
X 1 2 3 4 5
P(X) 1/5 1/5 1/5 1/5 1/5
mean of probability distribution = [tex]\frac{\sum{P(X)}}{5}[/tex]
= 1/5
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In this assignment, your team is managing a software development project with a total project budget of $178,500. Total work effort is 1,536 hours and the timeline for completion 24 weeks. At the end of week 12, the plan was to have completed 55% of the project scope. However, actual progress was calculated at 650 worth of hours completed on project activities and actual cost (in hours) for these activities is 780. With your team, please answer the following questions. For each response, show the pertinent formula(s) as it applies and "show your work" in addition to providing a narrative statement.
a. How much money was supposed to have been spent at the end of week 12?b. Will the project finish on time and within the given budget?
Answer:
im lost good luck
Step-by-step explanation:
Patty is making a poster for science class. She spends. 50 minutes on each. Of 2 days. She completes 1/3 of the poster the first day and another 1/3 the next day what fraction of the poster has she completed so far?
Answer:
She has completed 2/3 of the poster
Explanation:
1/3 of the poster the first day
+
1/3 of the poster the next day
1/3 + 1/3 = 2/3
Parallelogram ABCD is dilated to form parallelogram EFGH. Side AB is proportional to side EF. What corresponding side is proportional to segment AD? Type the answer in the box below. (2 points)
The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased form a nationally known manufacture is actually 1 gallon. You know from the manufacturer’s specifications that the standard deviation of the amount of pant is 0.02 gallon. You select a random sample of 50 cans, and the mean amount of paint per 1-gallon cans is 0.995 gallon.
a. Is there evidence that the mean amount is different from 1.0 gallon (use α = 0.01)?
b. Compute the p-value and interpret the meaning
c. Construct a 99% confidence interval estimate of the population mean amount of paint.
d. Compare the results of (a) and (c). What conclusions do you reach?
Answer:
a) There is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.
b) The p-value obtained = 0.076727 > significance level (0.01), hence, we fail to reject the null hypothesis and conclude that there is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.
That is, the mean amount of paint per 1-gallon cans is not significantly different from 1 gallon.
c) The 99% confidence for the population mean amount of paint per 1-gallon cans is
(0.988, 0.999) in gallons.
d) The result of the 99% confidence interval does not agree with the result of the hypothesis testing performed in (a) because the right amount of paint in 1-gallon cans, 1 gallon, does not lie within this confidence interval obtained.
Step-by-step explanation:
a) This would be answered after solving part (b)
b) For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and is always about the absence of significant difference between two proportions being compared. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis takes the other side of the hypothesis; that there is indeed a significant difference between two proportions being compared. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, the null hypothesis is that there is no significant difference in the mean amount of paint per 1-gallon cans and 1 gallon. That is, the mean amount of paint per 1-gallon cans should be 1 gallon.
And the alternative hypothesis is that there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon. That is, the mean amount of paint per 1-gallon cans is not 1 gallon.
Mathematically,
The null hypothesis is
H₀: μ₀ = 1 gallon
The alternative hypothesis is
Hₐ: μ₀ ≠ 1 gallon
To do this test, we will use the z-distribution because we have information on the population standard deviation.
So, we compute the z-test statistic
z = (x - μ)/σₓ
x = the sample mean = 0.995 gallons
μ₀ = what the amount of paint should be; that is 1 gallon
σₓ = standard error = (σ/√n)
σ = standard deviation = 0.02 gallon
n = sample size = 50
σₓ = (0.02/√50) = 0.0028284271 = 0.00283 gallons.
z = (0.995 - 1) ÷ 0.00283
z = -1.77
checking the tables for the p-value of this z-statistic
p-value (for z = -1.77, at 0.01 significance level, with a two tailed condition) = 0.076727
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.01
p-value = 0.076727
0.076727 > 0.01
Hence,
p-value > significance level
So, we fail to reject the null hypothesis and conclude that there is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.
That is, the mean amount of paint per 1-gallon cans is not significantly different from 1 gallon.
c) To compute the 99% confidence interval for population mean amount of paint per 1-gallon paint cans.
Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample Mean) ± (Margin of error)
Sample Mean = 0.995 gallons
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error)
Critical value at 99% confidence interval is obtained from the z-tables because we have information on the population standard deviation.
Critical value = 2.58 (as obtained from the z-tables)
Standard error = σₓ = 0.00283 (already calculated in b)
99% Confidence Interval = (Sample Mean) ± [(Critical value) × (standard Error)]
CI = 0.995 ± (2.58 × 0.00283)
CI = 0.995 ± 0.0073014
99% CI = (0.9876986, 0.9993014)
99% Confidence interval = (0.988, 0.999) in gallons.
d) The result of the 99% confidence interval does not agree with the result of the hypothesis testing performed in (a) because the right amount of paint in 1-gallon cans, 1 gallon, does not lie within this confidence interval obtained.
Hope this Helps!!!
By calculating a z-value and comparing it to the critical value at alpha = 0.01, evidence can be determined. The p-value, as extreme as the calculated one assuming the null hypothesis is true, can be used to interpret the findings. Additionally, a 99% confidence interval estimate can be constructed to provide a range of values that is 99% confident in containing the true population mean amount of paint.
Explanation:a. To determine whether the mean amount of paint contained in 1-gallon cans is different from 1.0 gallon, we can perform a hypothesis test. The null hypothesis (H0) is that the mean amount is 1.0 gallon, and the alternative hypothesis (Ha) is that the mean amount is different from 1.0 gallon. We can perform a z-test using the formula Z = (sample mean - population mean) / (standard deviation / sqrt(sample size)). In this case, the sample mean is 0.995 gallon, the population mean is 1.0 gallon, the standard deviation is 0.02 gallon, and the sample size is 50. By calculating the z-value, we can compare it to the critical value at alpha = 0.01 to determine whether there is evidence to reject the null hypothesis.
b. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, we can calculate the p-value using the standard normal distribution table or a calculator. If the p-value is less than the significance level (alpha = 0.01), we reject the null hypothesis. The interpretation of the p-value is that there is strong evidence to suggest that the mean amount of paint is different from 1.0 gallon.
c. To construct a 99% confidence interval estimate of the population mean amount of paint, we can use the formula CI = sample mean ± (z-score * (standard deviation / sqrt(sample size))). In this case, the z-score for a 99% confidence level is approximately 2.61. Plugging in the values, we can calculate the confidence interval, which gives us a range of values that we are 99% confident contains the true population mean amount of paint.
d. By comparing the results of (a) and (c), we can draw conclusions about whether the mean amount of paint is different from 1.0 gallon. If the null hypothesis is rejected in (a) and the 99% confidence interval in (c) does not include 1.0 gallon, then we can conclude that there is evidence to suggest that the mean amount is different from 1.0 gallon.
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Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.542.54 and a standard deviation of 0.420.42. Using the empirical rule, what percentage of the students have grade point averages that are between 1.281.28 and 3.83.8?
Answer:
[tex] P(1.28< X< 3.8) [/tex]
And we can use the z score formula to calculate how many deviations we are within the mean
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And if we use this formula we got:
[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]
[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]
And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values
Step-by-step explanation:
Previous concepts
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
Let X the random variable who represent the grade point averages of undergraduate students.
From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=2.54, Sd(X)=0.42[/tex]
So we can assume [tex]\mu=2.54 , \sigma=0.42[/tex]
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
For this case we want to find this probability:
[tex] P(1.28< X< 3.8) [/tex]
And we can use the z score formula to calculate how many deviations we are within the mean
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And if we use this formula we got:
[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]
[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]
And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values
Approximately 99.7 percent of the students have grade point averages between 1.28 and 3.8 according to the empirical rule.
Explanation:The empirical rule states that approximately 68 percent of the data lies within one standard deviation of the mean, 95 percent lies within two standard deviations, and more than 99 percent lies within three standard deviations. In this case, the mean grade point average is 2.54 and the standard deviation is 0.42. To find the percentage of students with grade point averages between 1.28 and 3.8, we need to find the z-scores for both values and calculate the area under the curve between those z-scores.
First, we find the z-score for 1.28 using the formula: z = (x - µ) / σ = (1.28 - 2.54) / 0.42 = -3.0476. Then, we find the z-score for 3.8 using the same formula: z = (3.8 - 2.54) / 0.42 = 3.0476. Using a standard normal distribution table or a calculator, we can find the area between these two z-scores, which is approximately 99.7 percent.
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To transfer into a particular technical department, a company requires an employee to pass a screening test. A maximum of 3 attempts are allowed at 6-month intervals between trials. From past records it is found that 40% pass on the first trial; of those that fail the first trial and take the test a second time, 60% pass; and of those that fail on the second trial and take the test a third time, 20% pass. For an employee wishing to transfer:
(A) What is the probability of passing the test on the first or second try?
(B) What is the probability of failing on the first 2 trials and passing on the third?
(C) What is the probability of failing on all 3 attempts?
Answer:
a) 0.760
b) 0.048
c) 0.192
Step-by-step explanation:
The step by step solution is attached as an image.
A) The probability of passing the test on the first or second try is 0.760.
That is he pass in the first trial or second trial.
(B) The probability of failing on the first 2 trials and passing on the third is 0.048.
That is the employee fail the first the trial and pass the third trial.
(C) The probability of failing on all 3 attempts is 0.192.
That is the employee fail all the three trial.
The probability of passing on the first or second try is 76%, the probability of failing the first 2 trials and passing on the third is 4.8%, and the probability of failing all 3 attempts is 19.2%.
Explanation:This problem relates to the field of probability. Let's break it down.
For part A, the probability of passing on the first or second try is the sum of the probability of passing on the first try and the product of the probability of failing on the first try and passing on the second. This is calculated as 0.4 + (0.6*0.6) = 0.76 or 76%.
For part B, the probability of failing the first 2 trials and passing on the third is calculated by multiplying the probability of failing the first trial, failing the second, and passing the third: (0.6*0.4*0.2) = 0.048 or 4.8%.
For part C, the probability of failing all 3 attempts is equal to the product of the probability of failing each attempt: (0.6*0.4*0.8) = 0.192 or 19.2%.
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Create a cylinder with a height of 14 cm and a radius of 10 cm.
Create a cylinder with a height of 14 cm and a radius of 10 cm. What can be concluded about the cylinder’s volume when the radius is halved?
The volume is One-fourth of the original.
The volume is One-third of the original.
The volume is One-half of the original.
The volume is twice the original.
Answer:
a on edge2021
Step-by-step explanation:
The volume is One-fourth of the original.
What is a cylinder?"It is a three dimensional structure having two parallel bases joined by a curved surface, at a fixed distance."
The formula of the volume of a cylinder:V = π × r² × h, where 'r' is the radius of the circular base and 'h' is the height of the cylinder.
In the given question,
the radius of the cylinder is 10 cm, and the height is 14 cm.
⇒ r = 10 cm, h = 14 cm
The volume of the cylinder would be,
[tex]V_1=\pi\times 10^2\times 14\\\\V_1=\frac{22}{7}\times 100 \times 14\\\\V_1=4400~~cu.~cm.[/tex]
If the radius is halved, the radius becomes 5 cm.
The new volume of a cylinder would be,
[tex]V_2=\pi \times 5^2\times 14\\\\V_2=\frac{22}{7}\times 25 \times 14\\\\V_2=1100~~cu.~cm.[/tex]
The ratio between the original and the new volume of the cylinder is:
[tex]\Rightarrow \frac{1100}{4400}=\frac{1}{4}\\\\\Rightarrow V_2=\frac{1}{4}V_2[/tex]
This means, the volume is One-fourth of the original.
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The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 indicate that the manufacturer’s claim is overstated? Choose the appropriate hypothesis. a. H1: μ < 530. Reject H1 if tcalc > –1.753 b. H0: μ < 530. Reject H0 if tcalc > –1.753 c. H1: μ ≥ 530. Reject H1 if tcalc < –1.753 d. H0: μ ≥ 530. Reject H0 if tcalc < –1.753 a b c d (a-2) State the conclusion. a. tcalc = –1.6. There is not enough evidence to reject the manufacturer’s claim. b. tcalc = –1.6. There is significant evidence to reject the manufacturer’s claim. a b
Answer:
(a) H1: μ < 530. Reject H1 if tcalc > –1.753
(b) t calc = –1.6. There is not enough evidence to reject the manufacturer’s claim.
Step-by-step explanation:
We are given that the manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour.
A sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 is given.
Let [tex]\mu[/tex] = average bags an airport baggage scanning machine can handle
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 530 bags {means that an airport baggage scanning machine can handle an average of more than or equal to 530 bags per hour}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 530 bags {means that an airport baggage scanning machine can handle an average of less than 530 bags per hour}
The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean = 510
s = sample standard deviation = 50
n = sample of hours = 16
So, test statistics = [tex]\frac{510-530}{\frac{50}{\sqrt{16} } }[/tex] ~ [tex]t_1_5[/tex]
= -1.60
The value of t test statistics is -1.60.
Now, at 0.05 significance level the t table gives critical value of -1.753 at 15 degree of freedom for left-tailed test. Since our test statistics is more than the critical values of t as -1.60 > -1.753, so we have insufficient evidence to reject our null hypothesis as it will not in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that an airport baggage scanning machine can handle an average of more than or equal to 530 bags per hour.
W is less than or equal to 9 and greater than -7
Answer:
(-7, 9]
Step-by-step explanation:
Assuming you want an answer in interval notation:
W is less than or equal to 9 >> [tex]w \leq 9[/tex]
W is greater than- 7 >> [tex]-7 < w[/tex]
Combine the two >> [tex]-7 < w \leq 9[/tex]
So in interval notation, (-7, 9]
In order to estimate the mean 30-year fixed mortgage rate for a home loan in the United States, a random sample of 26 recent loans is taken. The average calculated from this sample is 7.20%. It can be assumed that 30-year fixed mortgage rates are normally distributed with a standard deviation of 0.7%. Compute 95% and 99% confidence intervals for the population mean 30-year fixed mortgage rate.
Answer:
The 95% CI is (6.93% , 7.47%)
The 99% CI is (6.85% , 7.55%)
Step-by-step explanation:
We have to estimate two confidence intervals (95% and 99%) for the population mean 30-year fixed mortgage rate.
We know that the population standard deviation is 0.7%.
The sample mean is 7.2%. The sample size is n=26.
The z-score for a 95% CI is z=1.96 and for a 99% CI is z=2.58.
The margin of error for a 95% CI is
[tex]E=z\cdot \sigma/\sqrt{n}=1.96*0.7/\sqrt{26}=1.372/5.099=0.27[/tex]
Then, the upper and lower bounds are:
[tex]LL=\bar x-z\cdot\sigma/\sqrt{n}=7.2-0.27=6.93\\\\ UL=\bar x+z\cdot\sigma/\sqrt{n} =7.2+0.27=7.47[/tex]
Then, the 95% CI is
[tex]6.93\leq x\leq 7.47[/tex]
The margin of error for a 99% CI is
[tex]E=z\cdot \sigma/\sqrt{n}=2.58*0.7/\sqrt{26}=1.806/5.099=0.35[/tex]
Then, the upper and lower bounds are:
[tex]LL=\bar x-z\cdot\sigma/\sqrt{n}=7.2-0.35=6.85\\\\ UL=\bar x+z\cdot\sigma/\sqrt{n} =7.2+0.35=7.55[/tex]
Then, the 99% CI is
[tex]6.85\leq x\leq 7.55[/tex]
equation of the line through (-10,3) and (-8 -8)
Answer:
(-8-3)/(-8+10)= -11/2
y - 3 = -11/2(x+10)
y-3=-11/2x-55
y=-11/2x-52
Step-by-step explanation:
According to a survey, 62% of murders committed last year were cleared by arrest or exceptional means. Fifty murde committed last year are randomly selected, and the number cleared by arrest or exceptional means is recorded. When technology is used, use the Tech Help button for further assistance.
a. Find the probability that exactly 41 of the murders were cleared.
b. Find the probability that between 36 and 38 of the murders, inclusive, were cleared.
c. Would it be unusual if fewer than 19 of the murders were cleared? Why or why not?
a.The probability that exactly 41 of the murders were cleared is ____. (Round to four decimal places as needed.)
Answer:
a) The probability that exactly 41 of the murders were cleared is 0.0013
b) The probability that between 36 and 38 of the murders, inclusive, were cleared is 0.0809.
c) Yes, it would be unusual
Step-by-step explanation:
Let p=62% considered as the probability of having a commited that is cleared by arres or exceptional means. We assume that choosing each of the 50 commited is independent of each choose. Then, let X be the number of cleared. In this case, X is distributed as a binomial random variable. Recall that, in this case,
[tex] P(X=k) = \binom{50}{k} p^{k}(1-p)^{50-k}[/tex] for[tex]0\leq k \leq 50[/tex], with p=0.62
a) We have that
[tex] P(X=40) = \binom{50}{40} p^{40}(1-p)^{50-40} =0.001273487 [/tex]
b) We are asked for the following
[tex]P(36\leq X \leq 38) = P(X=36)+P(X=37)+P(X=38) = 0.080888936
[/tex] (The specific calculation is omitted.
c) We will check for the following probability [tex]P(X\leq 19)[/tex]
[tex]P(X\leq 19 ) = \sum_{k=0}^{19} P(X=k) = 0.000499222 [/tex]
Given that the probability of this event is really close to 0, it would be unusual if less than 19 murders are cleared.
The question deals with the application of binomial probability distribution in a real life situation involving crime investigation. The probability values for a certain range or exact number of cleared murders can be calculated by using the binomial probability formula. It would be statistically unusual for fewer than 19 murders to be cleared given a 62% clearance rate.
Explanation:This question can be approached using the binomial distribution, where the number of successes in a sequence of n independent experiments (in this case, the number of murders being cleared) follows a binomial distribution.
a. The probability that exactly 41 of the murders were cleared can be found by calculating the binomial probability. This can be done by using the formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)). In this case, n=50 (number of trials/murders), k=41 (number of successes/murders cleared), and p=0.62 (probability of success/clearing a murder). You need to substitute these values into the formula and calculate the value.
b. Finding the probability that between 36 and 38 murders were cleared involves the same process, but you need to calculate for k=36, 37, and 38, and then add the results together to get the total probability.
c. If fewer than 19 of the murders were cleared, it would be statistically unusual considering the 62% clearance rate according to the survey. The reasoning being that, given a 62% probability, the expectation would be significantly higher.
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Multiply: (-3/10)(-2/9)
Answer:
1/15
Step-by-step explanation:
512 = (m + 7) ^ 3/2
It’s algebra!!!
Answer:
m = 57
Step-by-step explanation:
If we assume your equation is supposed to be ...
512 = (m +7)^(3/2)
we can raise both sides to the 2/3 power to get ...
512^(2/3) = m +7
64 = m +7
57 = m . . . . . . subtract 7