After considering all the given data we conclude that the 90% uncertainty interval of the system up is 5 under the condition that a experiment is observed and the following random and systematic uncertainties are determined for the system.
To evaluate the degrees of freedom of the system, we use the formula
DF = N - P
Here,
N = sample size
P = number of parameters or relationships.
For the given case, the degrees of freedom are not given in the problem statement. Then, we can evaluate it using the formula
DF = N - P.
So the degrees of freedom for systematic uncertainties are very large, we can consider that it is equal to infinity.
Therefore,
DF = N - P = N - infinity = N.
So, the degrees of freedom of the system is equal to the sample size
To evaluate the 90% uncertainty interval of the system up (90%), we can apply the formula
x' ± tα/2 × s/√n
Here
x' = sample mean,
tα/2 = t-distribution value for α/2
n = sample size.
The value of tα/2 can be obtained using a t-distribution table
For a 90% confidence interval with 1 degree of freedom, tα/2 = 1.833.
The sample mean is given as 120 mm and the standard deviation can be calculated using the formula s = √(sa² + sb² + sc²) where sa, sb and sc are the standard deviations of random uncertainties in V1, V2 and V3 respectively.
Staging the values given in the problem statement, we get s = √(1.7² + 3.5² + 1.4²) = 4.0 mm.
The sample size is not given in the problem statement but we can assume that it is large enough to use a t-distribution table . Therefore, substituting all values in the formula, we get:
x' ± tα/2
s/√n = 120 ± 1.833 × 4.0/√n
We need to find n such that this expression gives us an interval width of 90%. Therefore,
tα/2 × s/√n = 0.9 × x'
Staging all values in this equation and solving for n, we get:
n = (tα/2 × s / (0.9 × x'))²
Staging all values in this equation, we get:
n = (1.833 × 4.0 / (0.9 × 120))²
≈ 5
Therefore, the sample size required to obtain a 90% uncertainty interval with an interval width of up (90%) is approximately equal to 5.
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33. PROBLEM SOLVING How many revolutions does the smaller gear complete during a single revolution of the larger gear?
The required number of revolution for small gear is 2
The given figure is a circle,
Then,
Radius of big circle = 7
radius of small circle = 3
Since we know that perimeter of circle = 2πr
Therefore,
Perimeter of big circle = 2x7x(22/7)
= 44 square units
Perimeter of small circle = 2x3x(22/7)
= 18.84 square units
Now the umber of revolution for small gear to complete a single revolution of the larger gear = 44/18.84 = 2.33 ≈ 2
Hence, number of revolution = 2.
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A survey asked 700 people for their favorite genre of book. The table shows the data. How many people surveyed responded with a genre other than one of those listed? Find the probabilities for a complete probability model for the responses.
The probabilities for a complete probability model for the responses are approximately as follows:
Adventure: 0.24
Comedy: 0.18
Mystery: 0.15
Romance: 0.17
Other: 0.26
How did we get the values?To find the number of people surveyed who responded with a genre other than the listed ones, subtract the sum of people who chose the listed genres from the total number of people surveyed.
Total number of people surveyed: 700
Number of people who chose the listed genres:
Adventure: 168
Comedy: 126
Mystery: 105
Romance: 119
Sum of people who chose the listed genres: 168 + 126 + 105 + 119 = 518
Number of people who responded with a genre other than the listed ones: 700 - 518 = 182
Therefore, 182 people surveyed responded with a genre other than the listed ones.
To find the probabilities for a complete probability model for the responses, divide the number of people who chose each genre by the total number of people surveyed (700).
Probability of choosing Adventure: 168/700 ≈ 0.24
Probability of choosing Comedy: 126/700 ≈ 0.18
Probability of choosing Mystery: 105/700 ≈ 0.15
Probability of choosing Romance: 119/700 ≈ 0.17
To find the probability of choosing a genre other than the listed ones, divide the number of people who responded with a genre other than the listed ones (182) by the total number of people surveyed (700).
Probability of choosing other: 182/700 ≈ 0.26
Therefore, the probabilities for a complete probability model for the responses are approximately as follows:
Adventure: 0.24
Comedy: 0.18
Mystery: 0.15
Romance: 0.17
Other: 0.26
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lena is going to rent a truck for one day. there are two companies she can choose from, and they have the following prices. company a charges and allows unlimited mileage. company b has an initial fee of and charges an additional for every mile driven. for what mileages will company a charge less than company b? use for the number of miles driven, and solve your inequality for .
Therefore, company A will charge less than company B for any mileage greater than 65 miles.
To help you with your question, I need the specific prices for both companies, as well as any additional information you can provide about the charges.
To solve for the mileages at which company A will charge less than company B, we can set up the following inequality:
A < B
where A is the cost of renting from company A and B is the cost of renting from company B. We can plug in the given information to get:
x ≤ A
x > B = y + 40z
where x is the number of miles driven, y is the initial fee for company B, and z is the additional cost per mile for company B.
Now we can substitute the given values to get:
x ≤ A
x > y + 40z
For company A, the cost is a flat rate with unlimited mileage, so A is just the cost listed for one day of rental. Let's say that cost is $100.
For company B, the initial fee is $50 and the additional cost per mile is $0.25. So:
y = 50
z = 0.25
Now we can plug in these values and solve for x:
x ≤ 100
x > 50 + 40(0.25)
x > 65
Therefore, company A will charge less than company B for any mileage greater than 65 miles.
To help you with your question, I need the specific prices for both companies, as well as any additional information you can provide about the charges.
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Find all the values of x where the tangent line is horizontal for f(x) = x^3 - 4x^2 - 9x
The values of x where the tangent line is horizontal are:
x = 3.52
x =-0.85
How to find the values of x?The tangent line is horizontal when the derivate of f(x) is zero, here we have:
f(x) = x³ - 4x² - 9x
If we differentiate this, we will get:
f'(x) = 3x² - 8x - 9
Now we need to find the zeros:
0 = 3x² - 8x - 9
Using the quadratic formula we will get:
[tex]x = \frac{8 \pm \sqrt{(-8)^2 - 4*3*-9} }{2*3}\\ \\x = \frac{8 \pm 13.1 }{6}[/tex]
The two solutions are:
x+ = (8 + 13.1)/6 = 3.52
x- = (8 - 13.1)/6 = -0.85
At these values the tangent line is horizontal.
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Which statement best describes the difference between medium and format?
O A medium is the way content is shared, while a format is the way information will be processed by the five senses.
O A medium is the way information is delivered, while a format is the way information is organized.
O A medium is the way information is organized, while a format is the way the writer or speaker presents the
information.
O A medium is the way information is designed to be processed by the five senses, while a format helps the reader
understand the facts.
The distinction lies in the fact that medium relates to the delivery or transmission of information, while format relates to the organization or presentation of information.
The statement that best describes the difference between medium and format is:
O A medium is the way information is delivered, while a format is the way information is organized.
A medium refers to the channel or method through which information is transmitted or shared. It could be a newspaper, television, radio, internet, or any other means of communication. The medium determines how the content reaches the audience.
On the other hand, a format pertains to the structure or arrangement of information. It focuses on how the content is organized, presented, or displayed. For example, a format can refer to the layout of a book, the structure of a report, the design of a website, or the style of a presentation.
Therefore, the distinction lies in the fact that medium relates to the delivery or transmission of information, while format relates to the organization or presentation of information.
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3
TIME
54:
34
4
8
9
What is the approximate percent change in a temperature that went down from 120 degrees to 100 degrees?
VX
O The percent change is approximately 17%.
O The percent change is approximately 20%.
O The percent change is approximately 80%.
O The percent change is approximately 120%.
Please helpppppp I have a timer
To find the percent change in temperature, we can use the formula: percent change = [tex](\frac{(new value - old value)}{old value } )X 100[/tex] i.e Percent Change = [tex]\frac{difference in temperature}{original temperature}[/tex] x 100
In this case, the old value is 120 degrees and the new value is 100 degrees. Substituting these values into the formula, we get: percent change = [tex]\frac{(100 - 120)}{120} X 100[/tex]%
percent change = [tex]\frac{-20}{120}[/tex] x 100%
percent change = -0.1667 x 100%
percent change = -16.67%
Since the temperature went down, the percent change is negative. Therefore, the approximate percent change in temperature that went down from 120 degrees to 100 degrees is approximately 16.67%. So, the correct answer is: O The percent change is approximately 17%. (rounded to the nearest whole number).
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please help:
if triangle PRT∼ triangle QRS, find PT
Answer:
C. 40
Step-by-step explanation:
36/(5x+13)=30/(6x-2)
Cross multiply
36·(6x-2)=30·(5x+13)
216x-72=150x+390
216x-150x=390+72
66x=462
x=7
Substituting 7 in for x,
6(7)-2
42-2
40
Find the critical value (or values) for the ttest for each. a. n-12, α-0.01, left-tailed b. n-16, α-0.05, right-tailed C. n-7, α 0.10, two-tailed d. n-11, α-0.025, right-tailed e. n-10, α-0.05, two-tailed
The critical values for the t-test depend on the sample size, significance level, and whether the test is one-tailed or two-tailed. For a left-tailed test with n=12 and α=0.01, the critical value is -2.680,for a right-tailed test with n=16 and α=0.05, the critical value is 1.746.
a. For a left-tailed t-test with n = 12 and α = 0.01, the critical value is -2.718.
b. For a right-tailed t-test with n = 16 and α = 0.05, the critical value is 1.746.
c. For a two-tailed t-test with n = 7 and α = 0.10, the critical values are -1.895 (for the left tail) and 1.895 (for the right tail).
d. For a right-tailed t-test with n = 11 and α = 0.025, the critical value is 2.718.
e. For a two-tailed t-test with n = 10 and α = 0.05, the critical values are -2.306 (for the left tail) and 2.306 (for the right tail).
Note: These critical values were calculated using a t-distribution table or a statistical software.
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Probability Distributions for Discrete Random Variables
Which of the following are discrete random variables?
Select all that apply
1-The number of CDs that a college student owns
2- The number of dogs you own
3- The amount of gas in your car
4- Number of 6s you get when you throw 5 number cubes
5- The number of dog sleds that a competitor uses in an annual sled dog race
The discrete random variables from the given options are: 1, 2, 4, and 5.
The number of CDs that a college student owns: This is a discrete random variable because the number of CDs can only be a whole number. You cannot have a fractional or continuous value for the number of CDs.
The number of dogs you own: This is a discrete random variable because you can only own a whole number of dogs. You cannot own a fractional or continuous number of dogs.
Number of 6s you get when you throw 5 number cubes: This is a discrete random variable because the number of 6s can only be a whole number from 0 to 5. You cannot have a fractional or continuous value for the number of 6s obtained.
The number of dog sleds that a competitor uses in an annual sled dog race: This is a discrete random variable because the number of dog sleds can only be a whole number. You cannot have a fractional or continuous value for the number of dog sleds used.
On the other hand, the following option is not a discrete random variable:
The amount of gas in your car: This is a continuous random variable because the amount of gas can be any non-negative real number. It can have fractional or continuous values, such as 10.5 liters or 20.25 gallons. Option 1,2,3,4 and 5
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16->
7 Determine whether each relation is a function. Explain your reasoning
The relation that describes a function is graph in option B because each of the input values has only one output
What is a functionIn mathematics a function is a relation between a set of inputs (called the domain) and a set of outputs (called the range) where each input is associated with exactly one output.
It is a rule or mapping that assigns a unique output value to each input value.
In the graph, only option B typically shows aa unique output value for all the input values. hence this is the function
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Determine the correct nth term formula for the following sequence. 78. 65. 5,53,40. 5
an=90-12. 5n
an=78-12. 5(n-1)
an=78(12. 5)^n-1
an=78-12. 5n
The correct option for the nth-term formula is:
aₙ = 78 - 12.5(n-1)
What is a sequence?
A sequence is an enumerated collection of objects in which repetitions are allowed. Like a set, it contains members (also called elements, or terms).
We can find the correct nth term formula for the given sequence by analyzing the pattern of the terms.
Starting from the first term, 78, we see that each successive term is obtained by subtracting 12.5 from the previous term.
Therefore, the sequence is a linear sequence with a common difference of -12.5.
The nth term formula for a linear sequence with first term a1 and common difference d is given by:
aₙ = a1 + (n-1)d
Applying this formula to the given sequence, we have:
a₁ = 78 (the first term)
d = -12.5 (the common difference)
Therefore, the nth term formula for the sequence is
aₙ = 78 - 12.5(n-1)
Hence, the correct option for the nth-term formula is:
aₙ = 78 - 12.5(n-1)
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find the values of the trigonometric functions of from the information given. cos() = 8 11 , sin() < 0
Therefore, we have: sin() = -√(57/121) = -3√57/11 , To find the other trigonometric functions, we can use the definitions: tan() = sin()/cos() = (-3√57/11)/(8/11) = -3√57/8
The given information is that cos() = 8/11 and sin() is negative. From this, we can use the Pythagorean identity to solve for sin():
sin²() = 1 - cos²() = 1 - (8/11)² = 1 - 64/121 = 57/121
Since sin() is negative, we know that it must be in the third or fourth quadrant, where the sine function is negative. To determine which quadrant exactly,
we can use the fact that cos() is positive and recall that cosine is also positive in the first quadrant. Since cosine decreases as we move to the right, we know that angle must be in the fourth quadrant, where cosine is positive and sine is negative.
Therefore, we have:
sin() = -√(57/121) = -3√57/11
To find the other trigonometric functions, we can use the definitions:
tan() = sin()/cos() = (-3√57/11)/(8/11) = -3√57/8
csc() = 1/sin() = -11/(3√57)
sec() = 1/cos() = 11/8
cot() = 1/tan() = -8/(3√57)
These values give us a complete description of the trigonometric properties of angle .
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Estimate the endurance limit, Se', (in kpsi) for the following materials. Consider 2024 T4 aluminum. A) Aluminum has an endurance limit of 115 KPI b) Aluminum has an endurance limit of 105 KPI c) Aluminum has an endurance limit of 95 KPI d) Aluminum has no endurance limit
Option d) Aluminum has no endurance limit of 100 KPI is not a valid statement.
2024 T4 aluminum does have an endurance limit which is defined as the stress level below which the material can withstand an infinite number of cycles without failure.
Now the value of the endurance limit depends on various factors such as the material's processing heat treatment, and surface finish, as well as the type of loading and environmental conditions.
In general, the endurance limit of 2024 T4 aluminum ranges from 45 to 85 percent of its ultimate tensile strength (UTS). The UTS of 2024 T4 aluminum is typically around 65-75 kpsi (kilo pounds per square inch).
Using the given options, we can estimate the endurance limit of 2024 T4 aluminum as:
a) Aluminum has an endurance limit of 115 KPI:
This option suggests that the endurance limit of 2024 T4 aluminum is higher than its UTS, which is not possible. Therefore, this option is not valid.
b) Aluminum has an endurance limit of 105 KPI:
Assuming this option to be true, the endurance limit of 2024 T4 aluminum is around 68 kpsi (i.e., 105/1.55). This value is within the typical range of endurance limit for this material.
c) Aluminum has an endurance limit of 95 KPI:
Assuming this option to be true, the endurance limit of 2024 T4 aluminum is around 61 kpsi (i.e., 95/1.55). This value is also within the typical range of endurance limit for this material.
Therefore, based on the given options, it is reasonable to estimate the endurance limit of 2024 T4 aluminum to be around 68-61 kpsi (or 105-95 KPI).
So, Aluminum has no endurance limit of 100 KPI is not a valid statement.
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find the area of the infinite region in the first quadrant between the curve y=e^-x and teh x-axis
Thus, the area of infinite region in the first quadrant between the curve y=e^-x and the x-axis is 1 square unit.
To find the area of the infinite region in the first quadrant between the curve y=e^-x and the x-axis, we need to integrate the function y=e^-x from x=0 to x=∞.
First, let's find the indefinite integral of e^-x:
∫e^-x dx = -e^-x + C
Next, we can use this indefinite integral to find the definite integral from x=0 to x=∞:
∫[0,∞]e^-x dx = lim┬(t→∞)∫[0,t]e^-x dx
= lim┬(t→∞)[-e^-t + e^0]
= lim┬(t→∞)[-e^-t + 1]
= 1
Therefore, the area of the infinite region in the first quadrant between the curve y=e^-x and the x-axis is 1 square unit.
It is important to note that this region is infinite because the curve y=e^-x approaches the x-axis but never actually touches it.
As we integrate from x=0 to x=∞, we are essentially adding up an infinite number of infinitely small rectangles, resulting in an infinitely large area.
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1) Imagine that you want to clean the window of a 1st floor bedroom and you have a 13-meter-long ladder. To reach the window, you place the ladder such that the foot of the ladder is 5 meters away from the wall. Can you tell the height of the window from the ground? (Please show your work for full points.)
The height of the window according to the diagram is 12 m
How to determine the height of the windowThe height of the window is worked using Pythagoras theorem, This is used for a right triangle.
The diagram shows a right triangle of and the parts are compared as follows
hypotenuse = length of ladder
opposite = height of the window and
adjacent = distance of ladder from wall
The equation is written below
(length of ladder)² = (height of the window)² + (distance of ladder from wall)²
(height of the window)² = 13² - 5²
height of the window = √(13² - 5²)
height of the window = 12 m
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Give me two real world questions about angle pairs
In architecture, how can understanding angle pairs help in designing and constructing buildings with stability and strength?
In surveying and navigation, how can angle pairs be used to calculate distances between two points or to determine the direction of a particular location?
Angle pairs refer to two or more angles that are related to each other in some way.
Here are two real-world questions about angle pairs:
In architecture, how can understanding angle pairs help in designing and constructing buildings with stability and strength?
In surveying and navigation, how can angle pairs be used to calculate distances between two points or to determine the direction of a particular location?
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if angle c=(2x+3) and angle d= (2x+1) what does x equal
if angle c=(2x+3) and angle d= (2x+1) Then, we can only say that: 4x + angle e = 176.
We know that the sum of angles in a triangle is always 180 degrees. Therefore, we can write an equation based on the given information:
angle c + angle d + angle e = 180
Substituting the given expressions for angles c and d, we get:
(2x + 3) + (2x + 1) + angle e = 180
Simplifying and combining like terms, we get:
4x + 4 + angle e = 180
Subtracting 4 from both sides, we get:
4x + angle e = 176
We do not have enough information to solve for x or angle e, as we do not know the value of angle e.
Therefore, we can only say that:
4x + angle e = 176.
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suppose that the p.d.f of a random variable x with a continuous distribution is f(x) = 2x 0 < x < 1. find the expectation of 1/x
Thus, the expectation of 1/x for random variable x the given pdf is 2.
To find the expectation of 1/x, we need to compute the expected value E(1/x) for the given probability density function (pdf) f(x) = 2x, where 0 < x < 1.
The expected value E(1/x) is calculated using the following integral:
E(1/x) = ∫[1/x * f(x)] dx, evaluated over the range 0 < x < 1.
Substitute f(x) = 2x into the integral:
E(1/x) = ∫[(1/x) * (2x)] dx, from x = 0 to x = 1.
Simplify the integral:
E(1/x) = ∫2 dx, from x = 0 to x = 1.
Now, integrate and evaluate:
E(1/x) = [2x] from x = 0 to x = 1.
E(1/x) = 2(1) - 2(0) = 2.
So, the expectation of 1/x for the given pdf is 2.
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Solve for x
√3x + 4 = 6
The value of x that satisfies the equation √3x + 4 = 6 is x = 4/3.
To solve the equation √3x + 4 = 6, we'll need to isolate the variable x. Let's go through the steps to find the solution:
Subtract 4 from both sides of the equation:
[tex]\sqrt{3x}[/tex] + 4 - 4 = 6 - 4
[tex]\sqrt{3x }[/tex]= 2
Square both sides of the equation to eliminate the square root:
[tex](\sqrt{3x)^2} = 2^{-2}[/tex]
3x = 4
Divide both sides of the equation by 3 to solve for x:
(3x)/3 = 4/3
x = 4/3
Therefore, the solution to the equation √3x + 4 = 6 is x = 4/3.
By substituting x = 4/3 back into the original equation, we can verify if it is indeed a solution:
√3(4/3) + 4 = 6
2 + 4 = 6
6 = 6
The equation holds true, confirming that x = 4/3 is the correct solution.
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AD and BE are the altitudes of triangle ABC intersecting at point O. AD+BE=35 dm, AO=9 dm, BO=12 dm. Find OE and OD.
The values of the OD and OE are 20 dm and 15 dm respectively.
What is the triangle?
A triangle is a three-sided polygon made up of three line segments that connect at three endpoints, called vertices. The study of triangles is an important part of geometry, and it has applications in various fields such as engineering, architecture, physics, and computer graphics.
Let's use the fact that the product of two segments of the same line through a point is equal:
AOOD = BOOE
We also have the equation:
AD + BE = 35
Using the fact that triangles AOD and BOE are similar (because they share angle AOB), we can write:
OD / OE = AO / BO = 9 / 12 = 3 / 4
We can use the fact that OD = (35 - BE) and OE = (35 - AD) to eliminate AD and BE:
AOOD = BOOE
9(35-BE) = 12(35-AD)
315 - 9BE = 420 - 12AD
AD + BE = 35
Solving these two equations simultaneously, we get:
AD = 15 dm
BE = 20 dm
Substituting into the expressions for OD and OE, we get:
OD = 20 dm
OE = 15 dm
Therefore, OD = 20 dm and OE = 15 dm.
Hence, the values of the OD and OE are 20 dm and 15 dm respectively.
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Suppose that a population grows according to a logistic model with carrying capacity 5900 and k = 0.0013 per year.(a) Write the logistic differential equation for these data.\frac{dP}{dt}\, =\, 0.0013P(1-\frac{P}{5900})
The logistic differential equation for these data is [tex]\frac{dP}{dt}\, =\, 0.0013P(1-\frac{P}{5900})[/tex]
The logistic differential equation is a mathematical model used to describe the growth of a population when there is a limiting factor that affects the growth rate. It is based on the idea that the growth rate of the population decreases as it approaches a maximum capacity or carrying capacity.
The equation is typically written as:
dP/dt = rP(1 - P/K)
where dP/dt is the rate of change of the population over time, P is the population size at any given time, r is the intrinsic growth rate, and K is the carrying capacity.
In the given problem, the carrying capacity is 5900, which means that the population cannot exceed 5900 individuals. The growth rate is given by k = 0.0013 per year. Thus, the logistic differential equation can be written as:
dP/dt = 0.0013P(1 - P/5900)
This equation represents the rate at which the population grows over time, taking into account the limiting factor of the carrying capacity. The solution to this differential equation can be used to predict the population size at any future time, given the initial population size and the growth rate.
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The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4
parts per million (ppm). A consumer health group measured the level of the chemical in a
random sample of tomatoes obtained from one producer. The levels, in ppm, are shown below.
0.31 0.47 0.19 0.72 0.56
0.91 0.29 0.83 0.49 0.28
0.31 0.46 0.25 0.34 0.17
0.58 0.19 0.26 0.47 0.81
Does the data provide sufficient evidence to support the claim that the mean level of the
chemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm?
Use a 0.05 significance level to test the claim that these sample levels come from a population
with a mean greater than 0.4 ppm. Use the P-value method of testing hypotheses Assume that
the standard deviation of levels of the chemical in all such tomatoes is 0.21 ppm.
Choose the correct test statistic and P-value associated with this experiment.
O Test statistic: z-0.95 P-value: 0.1711
O Test statistic: z=-0.95 P-value: 0.1711
O Test statistic: z-0.95 P-value: 0.8289
O Test statistic: z--0.95 P-value: 0.8289
To determine the p-value associated with this test statistic, we would need to consult the t-distribution table or use statistical software.
The given options do not provide the p-value or the correct test statistic.
None of the provided options is the correct test statistic and p-value associated with this experiment.
To determine if the data provides sufficient evidence to support the claim that the mean level of the chemical in tomatoes from this producer is greater than 0.4 ppm, we can perform a one-sample t-test.
Given the sample data, the sample mean, denoted as [tex]\bar X[/tex], can be calculated by taking the average of the observed levels:
[tex]\bar X[/tex] = (0.31 + 0.47 + 0.19 + 0.72 + 0.56 + 0.91 + 0.29 + 0.83 + 0.49 + 0.28 + 0.31 + 0.46 + 0.25 + 0.34 + 0.17 + 0.58 + 0.19 + 0.26 + 0.47 + 0.81) / 20
[tex]\bar X[/tex] ≈ 0.457
The standard deviation, denoted as σ, is given as 0.21 ppm.
Since the sample size (n) is 20, we can use a t-distribution to calculate the test statistic.
The test statistic is given by:
t = ( [tex]\bar X[/tex] - μ) / (σ / √n)
μ is the population mean (0.4 ppm), σ is the population standard deviation (0.21 ppm), and n is the sample size (20).
Substituting the given values:
t = (0.457 - 0.4) / (0.21 / √20)
= 0.057 / (0.21 / 4.472)
≈ 0.057 / 0.094
t ≈ 0.6064
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h=-16t²+36 where t represents the time in seconds after launch. How long is
the ball in the air?
Considering the definition of zeros of a function, the time the ball remains in the air is 1.5 seconds.
Definition of zeros of a functionThe points where a polynomial function crosses the axis of the independent term (x) represent the zeros of the function.
Then, the zeros of a function are those values of x for which the expression is equal to 0, and they correspond to the abscissa of the points where the parabola intersects the x-axis.
Zeros of the function h= -16t² +36Considering the function h= -16t² +36, to calculate the time that the ball remains in the air, I must consider when the height is zero. That is, I must calculate the zeros of the function:
-16t² +36= 0
Solving:
-16t² = -36
t² = (-36)÷ (-16)
t²= 2.25
t=√2.25
t= ±1.5
Since time cannot be negative, the time the ball remains in the air is 1.5 seconds.
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TRUE OR FALSE question 6when gathering data through a survey, companies can save money by surveying 100% of a population
False. Surveying 100% of a population is not always necessary and can be costly.
Instead, companies can use sampling techniques to survey a representative subset of the population, which can be more cost-effective and still provide accurate results. The key is to ensure that the sample is representative of the larger population to avoid biased results. Statistical methods such as margin of error and confidence intervals can be used to estimate the accuracy of the survey results based on the sample size and level of representativeness.
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Angle ABC and angle CBD are complementary. What is the value of x?
Answer:
x = 26
Step-by-step explanation:
complementary angles sum to 90° , that is
∠ ABC + ∠ CBD = 90
2x + 38 = 90 ( subtract 38 from both sides )
2x = 52 ( divide both sides by 2 )
x = 26
which equation can be used to find x, the length of the hypotenuse of the right triangle? a triangle has side lengths 63, 16, x. 16 63
The equation that can be used to find the length of the hypotenuse of a right triangle is the Pythagorean theorem, which states that the sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of the hypotenuse.
In the given triangle with side lengths 63, 16, and x, the length of the hypotenuse is represented by x. Thus, we can apply the Pythagorean theorem to find the value of x.
Using the Pythagorean theorem, we get:
x^2 = 63^2 + 16^2
Simplifying the equation, we get:
x^2 = 3969 + 256
x^2 = 4225
Taking the square root of both sides, we get:
x = 65
Therefore, the length of the hypotenuse of the right triangle is 65.
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Probability Distributions for Discrete Random Variables
Consider the discrete random variable, X = customer satisfaction, shown:
X 1 2 3 4 5
P(x) 0.1 0.2 ? 0.3 0.2
a. What is P(×=3)?
b. What is P(x < 3)?
c. What is P(2<_ X < 5) ?
Probability Distributions for Discrete Random Variables are
a. P(X = 3) = 0.2
b. P(X < 3) = 0.3
c. P(2 < X < 5) = 0.5
To solve the given problems, we need to determine the missing probability value for X = 3. Let's calculate it and then proceed with the rest of the questions.
Given:
X 1 2 3 4 5
P(x) 0.1 0.2 ? 0.3 0.2
To find the missing probability value:
0.1 + 0.2 + P(X = 3) + 0.3 + 0.2 = 1
0.8 + P(X = 3) = 1
P(X = 3) = 1 - 0.8
P(X = 3) = 0.2
Now, we can answer the questions:
a. P(X = 3) = 0.2
b. P(X < 3) = P(X = 1) + P(X = 2)
= 0.1 + 0.2
= 0.3
c. P(2 < X < 5) = P(X = 3) + P(X = 4)
= 0.2 + 0.3
= 0.5
Therefore, Probability Distributions for Discrete Random Variables are
a. P(X = 3) = 0.2
b. P(X < 3) = 0.3
c. P(2 < X < 5) = 0.5
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In kite WXYZ, mzWXY = 104°, and mzVYZ = 49°. Find each measure.
X
1. m2VZY =
2. m/VXW =
3. mzXWZ =
W
Z
Answer:
a) <VZY = (180°- 2×49°)/2 = 41°
b) <VXW = 104°- 41° = 63°
c) <XWZ = 360°- (98°+2×104°) = 54°
Which expression matches the graph? 0 1 2 3 4 5 6 7 8 9 A. x < 7 B. x >= 7 O c. x = 7 D. x > 7 E. x <= 7
The inequality that matches the graph is given as follows:
E. n ≥ 1.
What are the inequality symbols?The four most common inequality symbols, and how to interpret them, are presented as follows:
> x: the amount is greater than x -> the number is to the right of x with an open dot at the number line. On the coordinate plane, these are the points above the dashed line y = x.< x: the amount is less than x. -> the number is to the left of x with an open dot at the number line. On the coordinate plane, these are the points below the dashed line y = x.≥ x: the amount is at least x. -> the number is to the right of x with a closed dot at the number line. On the coordinate plane, these are the points above the continuous line y = x.≤ the amount is at most x. -> the number is to the left of x with a closed dot at the number line. On the coordinate plane, these are the points below the continuous line y = x.The graph in this problem is composed by the values to the right of n = 1, with a closed interval, hence the inequality is given as follows:
n ≥ 1.
Missing InformationThe graph is presented at the end of the answer.
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the weights in grams of a sample of 24 walnuts are shown. if the mean is 20 grams, and the standard deviation is 2.45 grams, do the data appear to be normally distributed? explain.
Yes, it appears that the data are normally distributed. This is because the standard deviation (2.45 grammes) and the mean (20 grammes) both fall within acceptable bounds.
Since the data points are evenly spaced from the mean, the distribution is symmetric and corresponds to a normal distribution. The standard deviation is also not excessive in comparison to the mean, supporting the notion that the data is regularly distributed.
A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed.
In contrast, a high or low standard deviation indicates that the data points are, respectively, above or below the mean. A standard deviation that is close to zero implies that the data points are close to the mean.
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