The ARL (average run length) after the shift is approximately 222.22.
The ARL (average run length) after the shift can be determined from the control chart that monitors the diameter of bearings produced in a production line using 3-standard deviation control limits.
A standard deviation is a statistic that shows how widely values are spread from the average value (mean). A lower standard deviation implies that most values are very close to the average, whereas a higher standard deviation indicates that the values are more spread out. It is used to measure the amount of variation or dispersion of a set of values. The square root of the variance is the standard deviation.
ARL (average run length) is the average number of samples that may be examined before a control chart signals that an out-of-control situation has arisen. It's a measure of a control chart's efficiency in identifying out-of-control circumstances.
Let's solve the given problem: Mean (μ) = 1.6 cm, Standard deviation (σ) = 0.3 mm, Sample size (n) = 9
The sample mean is shifted to 1.65 cm after operating for a month.
The shift is = 1.65 - 1.6 = 0.05 cm = 0.5 mm.The new mean (μ') = 1.65 cm = 16.5 mm.The new standard deviation (σ') remains the same, which is 0.3 mm.The new control limits with a 3-standard deviation shift in the mean will be:UCL = μ' + 3σ' = 16.5 + 3(0.3) = 17.4 mmLCL = μ' - 3σ' = 16.5 - 3(0.3) = 15.6 mmThe width of the control limits is: WL = UCL - LCL = 17.4 - 15.6 = 1.8 mmThe ARL (average run length) after the shift can be calculated as follows:
ARL = (1 / α) * (WL / 6σ'), where α = 0.0027 (the area under the normal curve beyond 3 standard deviations on each side)
Substituting the given values, we have: ARL = (1 / 0.0027) * (1.8 / (6 * 0.3)) = 222.22.
Therefore, the ARL (average run length) after the shift is approximately 222.22.
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HELP Me PLEASE I'M BEGGING. I GOT TO SEND IT TONIGHT
Answer:
1. -34
Step-by-step explanation:
Duncan has cups of sand. He divides the sand equally into 4 containers. He uses all the sand in 1 container to make pieces of sand art. It takes cups of sand to complete each piece of sand art. How many pieces of sand art does Duncan make? A. 1 2/3 B. 2 C. 4 D. 5 5/9
Answer:
4
Step-by-step explanation:
From the above question:
1/4 cup of sand = 1 sand art
1 cup of sand = x
Cross Multiply
x × 1/4 cup of sand = 1 cup of sand × 1 sand art
x = 1 cup of sand × 1 sand art/1/4 cup of sand
x = 1 ÷ 1/4
x = 1 × 4
x = 4 pieces sand art
Therefore, Duncan can make 4 pieces of sand art.
Find the value of x
Answer:
dkeekek
Step-by-step explanation:
kddeejwkwnqnnwwnwwnwnwnwwnw
Answer:
use pythagoras theorem for solving this
value of x is 12 cm
there are 12 socks in flora drawer 9 are red and 2 are blue and 1 is green she take out one sock without looking at the color. What is the numerical probability of flora picking out a blue sock?
The numerical probability of Flora picking out a blue sock is 1 out of 6, or approximately 0.1667, or 16.67%.
To calculate the numerical probability of Flora picking out a blue sock, we need to consider the total number of socks and the number of blue socks in the drawer.
Given:
Total number of socks = 12
Number of red socks = 9
Number of blue socks = 2
Number of green socks = 1
The probability of Flora picking a blue sock can be calculated as the ratio of the number of blue socks to the total number of socks:
Probability of picking a blue sock = Number of blue socks / Total number of socks
Probability of picking a blue sock = 2 / 12
Simplifying the fraction, we get:
Probability of picking a blue sock = 1 / 6
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Select all the equations that are true when xis -4.
A)-8 = 2x
B) -12 = x.-3
C) -12 = x+x+x
D = -1
Ex+4 = -8
F = -16
Answer:
A,C and D are the correct options.
The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 35.5 for a sample of size 767 and standard deviation 15.2. Estimate how much the drug will lower a typical patients systolic blood pressure (using a 90% confidence leve). Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
_____<µ<______
The tri-linear inequality that estimates how much the drug will lower a typical patient's systolic blood pressure using a 90% confidence level is 34.5 < µ < 36.5.
Given that the sample mean of a blood-pressure drug is 35.5 for a sample size of 767 and standard deviation 15.2, to estimate how much the drug will lower a typical patient's systolic blood pressure, we use the following formula of a confidence interval:
Confidence interval = sample mean ± margin of error,
where the margin of error = z(α/2) * (σ/√n),
σ = 15.2, the standard deviation
n = 767, sample size
α = 0.10, level of significance
z(α/2) = 1.645 (from a standard normal distribution table)
Plugging in the values,
Margin of error = 1.645 * (15.2 / √767)≈ 1.02
Confidence interval = 35.5 ± 1.02≈ 34.5 < µ < 36.5
Therefore, the blood-pressure drug will lower a typical patient's systolic blood pressure within the range of 34.5 and 36.5.
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What is the quotient of 905.8 and 0.2?
Answer: 4529
Step-by-step explanation:
Hope it helps Have a good Day
Just divide it
Help please will give brainlist!!!
MODELING REAL LIFE The equation y=2x + 3 represents the cost y(in dollars) of mailing a package that weighs x pounds.
a. Use a graph to estimate how much costs to mail the package.
b. Use the equation to find exactly how much it costs to mail the package.
It costs $ to mail the package.
Answer:
I can't read the weight of the package due to the image quality, could you type it out please.
Find the area of the circle. Round your answer to the nearest hundredth. Use 3.14 or 22/7 for π (pi)
Answer:
3.14
Step-by-step explanation:
1x1xpi
1x3.14=3.14
Find the value of x.
Answer:
14
Step-by-step explanation:
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snsn
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The arm span and foot length were both measured (in centimeters) for each of 20 students in a biology class. The computer output displays the regression analysis.
Which of the following is the best interpretation of the coefficient of determination r2?
About 37% of the variation in arm span is accounted for by the linear relationship formed with the foot length.
About 65% of the variation in foot length is accounted for by the linear relationship formed with the arm span.
About 63% of the variation in arm span is accounted for by the linear relationship formed with the foot length.
About 63% of the variation in foot length is accounted for by the linear relationship formed with the arm span.
Answer:
The guy above me is completely wrong hahaha.
The correct answer should have a 63%.
It's probably D. The terminology is a little confusing.
The equation of the output looks like:
-7.611 + .186(x) = y
x --> Arm span
y --> Foot span
The linear relationship is formed with the armspan.
Answer: It’s D
Step-by-step explanation:
I took the test
The National Teacher Association survey asked primary school teachers about the size of their classes. Nineteen percent responded that their class size was larger than 30. Suppose 760 teachers are randomly selected, find the probability that more than 22% of them say their class sizes are larger than 30.
The probability for more than 22% of the given data say their class sizes are larger than 30 is equal to 0.0864, or 8.64%.
To find the probability that more than 22% of the randomly selected teachers say their class sizes are larger than 30,
Use the binomial distribution.
Let us denote the probability of a teacher saying their class size is larger than 30 as p.
19% of the teachers responded with a class size larger than 30, we can estimate p as 0.19.
Now, calculate the probability using the binomial distribution.
find the probability of having more than 22% of the 760 teachers .
which is equivalent to more than 0.22 × 760 = 167 teachers saying their class sizes are larger than 30.
P(X > 167) = 1 - P(X ≤167)
Using the binomial distribution formula,
P(X ≤167) = [tex]\sum_{i=0}^{167}[/tex] [C(760, i) × [tex]p^i[/tex] × [tex](1-p)^{(760-i)[/tex]]
where C(n, r) represents the combination 'n choose r' the number of ways to choose r items from a set of n.
Using a statistical calculator, the probability P(X ≤ 167) is determined to be approximately 0.9136.
This implies,
The probability of having more than 22% of the randomly selected teachers say their class sizes are larger than 30 is,
P(X > 167)
= 1 - P(X ≤ 167)
≈ 1 - 0.9136
≈ 0.0864
Therefore, the probability for the given condition is approximately 0.0864, or 8.64%.
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In general, how many variables are there in an experiment? O a. Many including independent and dependent variables n O b. One independent variable and one dependent variable O c. None because experiments are controlled for the best results O d. Moisture and temperature are the only variables
In an experiment, there are many variables including independent and dependent variables. Therefore, the correct option is a.
Many including independent and dependent variables.
Variables are any feature, amount, or state that can be quantified or measured in any way. In a study, the term variable refers to any feature that can be changed or manipulated.
Variables in an Experiment. In an experiment, an independent variable is a variable that is changed or manipulated by the experimenter, and a dependent variable is a variable that is measured in response to the independent variable.
An independent variable is a variable that is changed or manipulated in an experiment by the researcher. The independent variable is the one that the researcher controls in order to examine its effect on the dependent variable.
The dependent variable is the variable that is measured in response to changes in the independent variable. This is the variable that the experimenter is interested in studying and is affected by the independent variable.
An experiment is a scientific study in which the researcher manipulates one or more independent variables in order to observe the effect on the dependent variable.
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Convert 456,300,000 to scientific notation.
Answer:
I believe it would be 4.563 x 10^8
Twelve identical computers are to be distributed to an elementary, middle, high school and community college. How many ways are there to distribute the 12 computers to the four schools, if we assume that some schools might end up with none, but all 12 must be given out
There are 91 ways to distribute the 12 identical computers among the elementary, middle, high school, and community college, ensuring that all 12 computers are given out.
To find the number of ways to distribute 12 identical computers among four schools (elementary, middle, high school, and community college) while ensuring that all 12 computers are given out, we can use the concept of stars and bars or the balls and urns method.
Let's represent the distribution using stars and bars. We have 12 identical stars (representing the computers) and 3 identical bars (representing the separators between the four schools). The bars divide the stars into four groups, each representing the number of computers given to each school.
We need to determine the number of ways to arrange the 12 stars and 3 bars. This can be calculated using the formula:
Number of ways = (n + k - 1) choose (k - 1) where n is the number of stars (12) and k is the number of bars (3).
Using this formula, the number of ways to distribute the 12 computers among the four schools is:
Number of ways = (12 + 3 - 1) choose (3 - 1)
= 14Choosee 2
= 91
Therefore, there are 91 ways to distribute the 12 identical computers among the elementary, middle, high school, and community colleges, ensuring that all 12 computers are given out.
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HURRY PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
Answer:
Option 3 Or number 5
Step-by-step explanation:
Answer:
8
Step-by-step explanation:
Eight people are at least 169.5 cm tall because the frequency of people that height is five. The frequency of people taller than 169.5 (because it said 'at least') is three. 5 + 3 = 8
Ben opened a savings
account with $75,
Every week he added
$20 more. Write on
equation to model this
situation.
Answer:
The equation would be y = 75 + 20x
PLZ HELPPPPPPP AND EXPLAIN BC I HAVE NO CLUE HOW TO DO THIS
A.) 500
B.) 490
C.) 21
D.) 390
i had this but your numbers aren't the same i thought i could help you..
Rosie bought a ring in the USA she paid 345 US dollars work out in pounds the amount rosie paid for the ring
Answer:
Step-by-step explanation:
Find the common difference of the arithmetic sequence 13, 10, 7
Answer: -3
Step-by-step explanation: DeltaMath
Which expression is equivalent to
[tex](3 {u}^{3} {v}^{4}) ^{2} [/tex]
A.
[tex]3 {u}^{5} {v}^{6} [/tex]
B.
[tex]3 {u}^{6} {v}^{8} [/tex]
C.
[tex]9 {u}^{6} {v}^{8} [/tex]
D.
[tex]9 {u}^{5} {v}^{6} [/tex]
Answer:
C
Step-by-step explanation:
When 3 is squared, it becomes 9. That means that both A and B are incorrect.
The powers on u and v are doubled not added to become u^6 and v^8 so that makes C the only possible answer.
10 x 10 = ? for easy points
Answer:
[tex]\huge\mathfrak{Heya\:Mate}[/tex]
[tex]\huge{\boxed{\bold{ANSWER}}}[/tex]
[tex]10 \times 10 \\ = 10^{2} \\ = 100[/tex]
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The depth of a river changes after a heavy rainstorm, Its depth, in feet, is modeled as a function of time, in hours. Consider this graph of the function. Enter the average rate of change for the depth of the river, measured as feet per hour, between hour 9 and hour 18. Round your answer to the nearest tenth
Answer:
The average rate of change for the depth of the river measured as feet per hour is approximately 0.3 feet/hour
Step-by-step explanation:
The depth of the river in feet with time is given by the function with the attached
From the graph, we have;
The depth of the river at hour t = 9 is f(9) = 18 feet
The depth of the river at hour t = 18 is f(18) = 21 feet
The average rate of change, A(x), for the depth of the river measured as feet per hour is given as follows;
[tex]A(X) = \dfrac{f(b) - f(a)}{b - a}[/tex]
Therefore, for the river, we have;
[tex]A(X) = \dfrac{f(18) - f(9)}{18 - 9} = \dfrac{21 - 18}{18 -9} = \dfrac{3}{9} =\dfrac{1}{3}[/tex]
The average rate of change for the depth of the river measured as feet per hour A(X) = 1/3 feet/hour
By rounding the answer to the nearest tenth, we have;
A(X) = 0.3 feet/hour.
what is the area of a trapezoid
Answer:
28 in² = 28
Step-by-step explanation:
[tex]a = \frac{a + b}{2}h[/tex]
[tex]a = 6[/tex]
[tex]b = 8[/tex]
[tex]h = 4[/tex]
[tex]a = \frac{6 + 8}{2}4[/tex]
[tex] \frac{14}{2} 4[/tex]
[tex]7 \times 4[/tex]
[tex] = 28[/tex]
Answer = 28 _
If a population of 10,000 increases by 5% every year, how large will the population be in 5 years?
______________________________
Answer:
10.500
Step-by-step explanation:
step by step explenation
HELP ME!!!!!!!!!!!!!!!!!!!!!!
Answer:
H
Step-by-step explanation:
If they are similar that means that they are racially the same so just divide 60 by 84 to find the rate and then multiply the rate by 210 to get your answer
Calculate the relative error of approximation On methods. y(t) = 8e²-8 (t+1) y (1) for all of three
The relative error is found by approximating y(1) relative to the exact value 8e²-16.
The relative error of approximation for the given method can be calculated using the formula:
Relative error = |(approximated value - exact value) / exact value| * 100%
In this case, we have the function y(t) = 8e²-8 (t+1) and need to approximate the value of y(1) using three different methods. To calculate the relative error for each method, we substitute t = 1 into the function and compare the approximated value with the exact value.
In the first paragraph,
To calculate the relative error of approximation for the given methods in estimating y(t), we substitute t = 1 into the function y(t) = 8e²-8 (t+1). By comparing the approximated values with the exact value, we can determine the relative error for each method.
In the second paragraph:
Let's evaluate the function at t = 1:
y(1) = 8e²-8 (1+1) = 8e²-8(2) = 8e²-16
Now, let's consider the three methods for approximating y(1) and calculate their respective relative errors:
Method 1: Approximated value = 8e²-8 (1+1) = 8e²-16
Relative error = |(8e²-16 - 8e²-16) / 8e²-16| * 100% = 0%
Method 2: Approximated value = 8e²-8 (1+1) - 8 = 8e²-24
Relative error = |(8e²-24 - 8e²-16) / 8e²-16| * 100% = |8e²-24 - 8e²-16| / |8e²-16| * 100%
Method 3: Approximated value = 8e²-8 (1+1) + 8 = 8e²-8
Relative error = |(8e²-8 - 8e²-16) / 8e²-16| * 100% = |8e²-8 - 8e²-16| / |8e²-16| * 100%
By calculating the relative error using the above formulas, we can determine the accuracy of each method in approximating y(1) relative to the exact value 8e²-16.
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A random sample of n1 = 201 people who live in a city were selected and 73 identified as a "dog person." A random sample of n2 = 91 people who live in a rural area were selected and 56 identified as a "dog person." Find the 99% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person."
The 99% confidence interval for the difference in approximately (-0.409123, -0.095277).
Calculating the 99% confidence intervalTo obtain the confidence interval for the difference in the proportions, we use the formula:
Confidence Interval = (p₁ - p₂) ± Z × √((p₁ × (1 - p₁) / n₁) + (p₂ × (1 - p₂) / n₂))
Where:
p₁ and p₂ are the proportionsn₁ and n₂ are the sample sizes of the city and rural areas respectively.Z = Z-score level (99% confidence level means Z = 2.576).Given the parameters:
p₁ = 73 / 201 = 0.3632
p₂ = 56 / 91 = 0.6154
n₁ = 201
n₂ = 91
Z = 2.576
Plugging in the values:
Confidence Interval = (0.3632 - 0.6154) ± 2.576 × √((0.3632 × (1 - 0.3632) / 201) + (0.6154 × (1 - 0.6154) / 91))
Confidence Interval = -0.2522 ± 2.576 × √((0.3632 × 0.6368 / 201) + (0.6154 × 0.3846 / 91))
Confidence Interval = -0.2522 ± 2.576 × √(0.003712)
Confidence Interval = -0.2522 ± 2.576 × 0.060851
Confidence Interval = -0.2522 ± 0.156923
Confidence Interval = (-0.409123, -0.095277)
Therefore, the 99% confidence interval is approximately (-0.409123, -0.095277).
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There is 20 million m3 of water in a lake at the beginning of a month. Rainfall in this month is a random variable with an average of 1 million mº and a standard deviation of 0.5 million mº. The monthly water flow entering the lake is also a random variable, with an average of 8 million mº and a standard deviation of 2 million mº. Average monthly evaporation is 3 million m3 and standard deviation is 1 million mº. 10 million m’ of water will be drawn from the lake this month. a Calculate the mean and standard deviation of the water volume in the lake at the end of the month. b Assuming that all random variables in the problem are normally distributed, calculate the probability that the end-of-month volume will remain greater than 18 million m3.
The probability that the end-of-month volume will remain greater than 18 million m³ is approximately 0.1922.
a)The mean water volume in the lake at the end of the month can be calculated using the formula given below:
Mean water volume = Starting water volume + Total rainfall + Total flow - Total evaporation - Water drawn from the lake
Given:
Starting water volume = 20 million m³
Total rainfall = random variable with mean = 1 million m³ and standard deviation = 0.5 million m³
Total flow = random variable with mean = 8 million m³ and standard deviation = 2 million m³
Total evaporation = 3 million m³
Water drawn from the lake = 10 million m³
Now, let's calculate the mean water volume at the end of the month.
Mean water volume = Starting water volume + Total rainfall + Total flow - Total evaporation - Water drawn from the lake= 20 + 1 + 8 - 3 - 10= 16 million m³
Therefore, the mean water volume at the end of the month is 16 million m³.
The standard deviation of the water volume in the lake at the end of the month can be calculated using the formula given below:
σ = √{σr² + σf² + σe²}
σr = standard deviation of rainfall = 0.5 million m³
σf = standard deviation of flow = 2 million m³
σe = standard deviation of evaporation = 1 million m³σ = √{σr² + σf² + σe²}σ = √{0.5² + 2² + 1²}= √{5.25}≈ 2.29 million m³
Therefore, the standard deviation of the water volume in the lake at the end of the month is approximately 2.29 million m³.b)Given that all the random variables in the problem are normally distributed, we can find the probability that the end-of-month volume will remain greater than 18 million m³ using the z-score formula.
z = (x - μ) / σ
Where,
z = z-scorex = 18 μ = 16σ = 2.29
Now, let's calculate the z-score.
z = (x - μ) / σ= (18 - 16) / 2.29= 0.87
Using the z-table, we can find that the probability of z being less than 0.87 is 0.8078.
Therefore, the probability of the end-of-month volume being greater than 18 million m³ is:
1 - 0.8078 = 0.1922 (rounded to 4 decimal places)
Hence, the probability that the end-of-month volume will remain greater than 18 million m³ is approximately 0.1922.
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Mr. Herman's class is selling candy for a school fundraiser. The class has a goal of raising $500 by selling C
boxes of candy. For every box they sell, they make $2.75.
Write an equation that the students could solve to figure out how many boxes of candy they need to sell.
Answer:
2.75 x (c) = 500
Step-by-step explanation:
i think this is it