Answer:
50/100, simplified to 1/2
Step-by-step explanation:
50% means half of 1.
1/2=.5 which is half of 1.
1=100%
.5=50%
1/2=50%
Answer:
[tex] \displaystyle \frac{1}{2} [/tex]
Step-by-step explanation:
we are given a parcentage
we want to convert it into fraction
remember that,
[tex] \displaystyle \% = \frac{1}{100} [/tex]
therefore
substitute:
[tex] \displaystyle 50 \times \frac{1}{100} [/tex]
reduce fraction:
[tex] \displaystyle \cancel{50} \times \frac{1}{ \cancel{100} \: ^{2} }[/tex]
[tex] \displaystyle 1 \times \frac{1}{2} [/tex]
simplify multiplication:
[tex] \displaystyle \frac{1}{2} [/tex]
hence,
[tex] \displaystyle 50\% = \frac{1}{2} [/tex]
Pls someone help me
Answer:
Step-by-step explanation:
455
The height, h metres, of a soccer ball kicked directly upward can be modelled by the equation h(t) = -4.9t2 + 13.1t + 1, where t is the time, in seconds, after the ball was kicked. a) How high is the ball after 2 s? 15 b) After how many seconds does the ball reach a height of 0.5 m? /3 21. Determine the inverse function for f(x)=(x - 2), state the domain and range for the function and its inverse. Write each step. 14 22. A farmer has 600m of fencing to enclose a rectangular area and divide 14 into three sections as shown. a) write a equation to express the total area enclosed as a function of the length x. b) Determine the domain and the range of this area function.
a) The height of the ball after 2 seconds is 7.6 metres.
b) The given function is f(x) = (x - 2).
a) The inverse function of f(x) is g(x) = x + 2.
b) A farmer has 600 metres of fencing to enclose a rectangular area and divide it into 14 sections.
a) The total area enclosed can be expressed as a function of the length of the rectangle x as A(x) = 300x - x².
b) The domain of the area function is (0, 300) and its range is [0, ∞).
a) h(t) = -4.9t² + 13.1t + 1 When the time, t = 2 seconds;
the height of the ball, h(t) =h(2) = -4.9(2)² + 13.1(2) + 1= -4.9(4) + 26.2 + 1= -19.6 + 27.2= 7.6 metres
Therefore, the height of the ball after 2 seconds is 7.6 metres.
b) h(t) = -4.9t² + 13.1t + 1When the height of the ball, h(t) = 0.5 metres; we can write the equation as:
0.5 = -4.9t² + 13.1t + 1
Rearranging,4.9t² - 13.1t + 0.5 - 1 = 0i.e. 4.9t² - 13.1t - 0.5 = 0
Solving using the quadratic formula, t = [-(-13.1) ± √(13.1² - 4(4.9)(-0.5))]/(2(4.9))= [13.1 ± √(171.61)]/9.8≈ 0.138 and t ≈ 2.23
Thus, the ball reaches a height of 0.5 metres twice, at approximately 0.14 seconds and 2.23 seconds.
The given function is f(x) = (x - 2).
a) We can find the inverse of a function by interchanging x and y and then solving for y.
Here, f(x) = (x - 2) so we can write it as x = (y - 2)
Interchanging x and y: x = (y - 2)
Solving for y, we get: y = x + 2
Hence, the inverse function of f(x) is g(x) = x + 2.
b) Domain and range of the function and its inverse
The domain of f(x) is all real numbers. (i.e. -∞ < x < ∞)
The range of f(x) is all real numbers. (i.e. -∞ < f(x) < ∞)
The domain of g(x) is all real numbers. (i.e. -∞ < x < ∞)
The range of g(x) is all real numbers. (i.e. -∞ < g(x) < ∞)
The steps of the solution are given below:
Therefore, the inverse function of f(x) is g(x) = x + 2.
The domain of f(x) is all real numbers, and its range is all real numbers.
Similarly, the domain and range of g(x) is all real numbers.
A farmer has 600 metres of fencing to enclose a rectangular area and divide it into 14 sections.
Let x be the length and y be the width of the rectangle.
a) Write an equation to express the total area enclosed as a function of the length x.
Perimeter of the rectangle = Total fencing = 600 metres2(x + y) = 600i.e. x + y = 300
Solving for y, we get:
y = 300 - x
Thus, the area of the rectangle is A(x) = xy = x(300 - x) = 300x - x²
Therefore, the total area enclosed can be expressed as a function of the length of the rectangle x as A(x) = 300x - x².
b) Determine the domain and range of the area function.
The domain of the function is the set of all possible values of x, which in this case is the set of all positive real numbers. (i.e. 0 < x < 300)
The range of the function is the set of all possible values of A(x), which in this case is the set of all non-negative real numbers. (i.e. 0 ≤ A(x) < ∞)
Therefore, the domain of the area function is (0, 300) and its range is [0, ∞).
The steps of the solution are given below:
Therefore, the total area enclosed can be expressed as a function of the length of the rectangle x as A(x) = 300x - x². The domain of the area function is (0, 300) and its range is [0, ∞).
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The assembly time for a product is uniformly distributed between 6 to 10 minutes. The probability of assembling the product in 8 minutes or less is a. 0.25 b. 0.75 c. 0.5 d. 1.5
The assembly time for a product is uniformly distributed between 6 to 10 minutes.
The probability of assembling the product in 8 minutes or less is 0.5 (option c).
Solution: Given, the assembly time for a product is uniformly distributed between 6 to 10 minutes. The range is a = 6 to b = 10.The probability of assembling the product in 8 minutes or less is to be determined.
Let's calculate the probability using the formula: P(x < or = 8) = (x - a) / (b - a)Here, a = 6, b = 10, and x = 8.P(x < or = 8) = (8 - 6) / (10 - 6) = 2 / 4 = 0.5Therefore, the probability of assembling the product in 8 minutes or less is 0.5. So, the correct option is (c) 0.5.
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Let be a fixed vector in and vector be a solution to where Q is a m*n matrix.
Prove every solution to the equation is in the form?
Given a fixed vector b and a vector x is a solution to Qx = b, it is required to prove that every solution to the equation is in the form x = xh + xp where xh is a particular solution to Qx = b and xp is a solution to the equation Qxp = 0.
Let xh be a particular solution to Qx = b, so that Qxh = b.
Now consider the homogeneous equation Qx = 0.
This is an m × n system of homogeneous linear equations in the n unknowns x1, x2, ..., xn, whose coefficient matrix is Q.
Since xh is a solution to the equation Qx = b, it follows that the equation Q(x - xh) = Qx - Qxh = b - b = 0.
This means that x - xh is a solution to the homogeneous equation Qx = 0.
Now any solution to Qx = b is of the form x = xh + xp, where xp is any solution to the homogeneous equation Qxp = 0.
Thus, every solution to the equation is in the form x = xh + xp, as required.
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1. Select all of the situations below that model this fraction. *
Captionless Image
A. 243 people were at the park. 20 people were playing volleyball. How many people were not playing volleyball?
B. Marcy's bookshelf had 20 shelves. She had 243 books. How many books would go on each shelf if she put the same number of books on each shelf?
C. Josh made $243 in 20 days. How much did he earn each day?
D. Sally and Len bicycled 243 miles in 20 days. How many miles did they bicycle each day?
Answer:
A. 223 were not playing volleyball
B. 12 books on each shelf with 3 books left over..
C. 12.15$ a day
D. 12.15 miles per day.
Step-by-step explanation:
A. Subtract 20 from 243.
B. Since books cant be divided into fractions, you take 243 divided by 20, which is 12.15. However, we have to say 12 with 3 leftover because books isn't money, a distance measure, etc. and cannot be cut into pieces.
C. 243 / 20 is 12.15
D. 243 / 20 is 12.15
helpppppppp will b marked brainliest!!!!!!!!!! Which system of equations is represented in the graph?
a. y= -2 x-2y=6
b. y= -2 x+2y=6
c. y= -2 2x-y= 3
d. y= -2 2x+y= -3
Answer: d. y= -2 2x+y= -3
Step-by-step explanation: Hope this help :D
Let A = {a,b,c}, B = {1, 2, 3, 4), and C = {w, x, y, z), and let R= {(2, 1), (6, 3), (6, 4), (C, 3)} and S = {(1, y), (1, 2), (2, w), (3, z)}. What is the composition relation ( RS) of R with S?
The pair (6, 3) in R, (6, 4) in R, and (C, 3) in R couldn't be matched with any pair in S, so they are not included in the composition relation RS.
To find the composition relation RS of R with S, we need to combine the ordered pairs in R and S in a way that the second element of each pair in R matches the first element of the pair in S.
Let's perform the composition:
R = {(2, 1), (6, 3), (6, 4), (C, 3)}
S = {(1, y), (1, 2), (2, w), (3, z)}
To form RS, we match the second element of each pair in R with the first element of each pair in S:
(2, 1) in R matches with (1, y) in S, so we combine them to form (2, y).
(2, 1) in R matches with (1, 2) in S, so we combine them to form (2, 2).
(6, 3) in R doesn't match with any pair in S.
(6, 4) in R doesn't match with any pair in S.
(C, 3) in R doesn't match with any pair in S.
Therefore, the composition relation RS is: RS = {(2, y), (2, 2)}
The pair (6, 3) in R, (6, 4) in R, and (C, 3) in R couldn't be matched with any pair in S, so they are not included in the composition relation RS.
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i need help!!!! i have to identify the area and i forgot how to do it and it was due 2 days ago!
Answer: I think B is the answer.
Step-by-step explanation:
Answer:not sure but
Step-by-step explanation: you can multiply each box, so lets say we have 20 times 7cm you multiply that and you get your answer, so put that a side
and once you multiply in each box add the numbers up and that should get your answer and make sure to add the "cm" at the end
The greatest snowfall in a 24- hour period was 76 inches. Which of these is the same as 76 inches?
A. 6 1/3 feet
B. 2.5 yards
C. 7.6 feet
D. 2 1/3 yards
Answer:
the answer is A. 6 1/3 feet
Step-by-step explanation:
There are 12 inches in a foot. 12 x6=72 r of 4
there are 12 inches in a foot 12÷4=3. 1/3 of a foot
6 + 1/3= 6 1/3
Guided Practice
Find the square.
(c + 1)2
A.
c2 + 1
B.
c2 + c + 1
C.
c2 + 2c + 1
Answer:
C
Step-by-step explanation:
(c + 1)(c+1) = c² + c + c + 1 = c² + 2c + 1
which equation has no real solutions?
2x²+2x+15=0
2x²+5x-3=0
x²+7x+2=0
x²-4x+2=0
Answer:
A
Step-by-step explanation:
x=
−b±√b2−4ac
2a
x=
−(2)±√(2)2−4(2)(15)
2(2)
x=
−2±√−116
4
and there is really no solution
help plsss. marking brainliest
Answer:
1. 3
2. 1
Step-by-step explanation:
Answer:
1.3. now mark brainlieet plsssssssssss like you said you would
Here is an equation and all the steps Clare and Lin took to solve it. Are both of their solutions correct? Explain your reasoning.
Answer:
Both solutions are correct
Step-by-step explanation:
Given
Expression:
[tex]14x - 2x +3 = 3(5x + 9)[/tex]
See attachment for Clare and Lins' solution
Required
Determine if they are correct or not
[tex]14x - 2x +3 = 3(5x + 9)[/tex]
Open bracket
[tex]14x - 2x + 3 = 15x + 27[/tex]
Collect like terms
[tex]14x - 2x -15x = -3 + 27[/tex]
[tex]-3x = 24[/tex]
Divide by -3
[tex]x = \frac{24}{-3}[/tex]
[tex]x = -8[/tex]
Both solutions are correct
Let f and g be functions defined on R" and c a real number. Consider the following two problems, Problem 1: max f(x) and Problem 2: max f(x) subject to g(x) = c. 1. Any solution of problem 1 is also a solution of problem 2. True or false? 2. If Problem 1 does not have a solution, then Problem 2 does not have a solution. True or false? 3. Problem 2 is equivalent to min - f(x) subject to g(x) = c. True or false? 4. In Problem 2, quasi-convexity of f is a sufficient condition for a point satisfying the first-order conditions to be a global minimum. True or false? 5. Consider the function f(x,y) = 5x - 17y. f is a) quasi-concave b) quasi-convex c) quasi-concave and quasi-convex d) no correct answer
True. Any solution of Problem 1 (max f(x)) is also a solution of Problem 2 (max f(x) subject to g(x) = c).
True. If Problem 1 does not have a solution, then Problem 2 does not have a solution.
True. Problem 2 (max f(x) subject to g(x) = c) is equivalent to min -f(x) subject to g(x) = c.
False. In Problem 2, the quasi-convexity of f is not a sufficient condition for a point satisfying the first-order conditions to be a global minimum.
The function f(x,y) = 5x - 17y is quasi-concave.
Any solution that maximizes f(x) will also satisfy the constraint g(x) = c. Therefore, any solution of Problem 1 is also a solution of Problem 2.
If Problem 1 does not have a solution, it means that there is no maximum value for f(x). In such a case, Problem 2 cannot have a solution since there is no maximum value to subject to the constraint g(x) = c.
Problem 2 can be reformulated as finding the minimum of -f(x) subject to the constraint g(x) = c. This is because maximizing f(x) is equivalent to minimizing -f(x) since the maximum of a function is the same as the minimum of its negative.
False. Quasi-convexity of f is not a sufficient condition for a point satisfying the first-order conditions to be a global minimum in Problem 2. Quasi-convexity guarantees that local minima are also global minima, but it does not ensure that the point satisfying the first-order conditions is a global minimum.
The function f(x,y) = 5x - 17y is quasi-concave. A function is quasi-concave if the upper contour sets, which are defined by f(x,y) ≥ k for some constant k, are convex. In this case, the upper contour sets of f(x,y) = 5x - 17y are convex, satisfying the definition of quasi-concavity.
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which relation is also a function
Please help me
Find the surface area
If you can explain to that would be great if not that’s fine
4 m
12 m
18 m
Answer
672 meters²
Step-by-step explanation:
2×(18×12 + 18×4 + 12×4) = 672 meters²
hope this helps :))
Evaluate ∫ x ds, where C is a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6) b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)
To evaluate the integral ∫ x ds, we need to parameterize the given curves and compute the arc length integral. In part (a), we evaluate the integral for the straight line segment from (0, 0) to (12, 6). In part (b), we evaluate the integral for the parabolic curve from (0, 0) to (2, 12).
(a) For the straight line segment x = t, y = t/2 from (0, 0) to (12, 6), we can parameterize the curve as follows: x = t, y = t/2. The differential arc length element ds is given by ds = √(dx² + dy²). Substituting the parameterizations, we have ds = √(dt² + (dt/2)²) = √(5/4 dt²). Thus, the integral becomes ∫ x ds = ∫ t √(5/4 dt²) = ∫ t (√5/2) dt. Integrating with respect to t from 0 to 12, we get (√5/2) ∫ t dt = (√5/2) (t²/2) evaluated from 0 to 12. Evaluating this expression, we find that the integral is equal to (√5/2) (144/2) = 36√5.
(b) For the parabolic curve x = t, y = 3t² from (0, 0) to (2, 12), we can parameterize the curve as before: x = t, y = 3t². The differential arc length element ds is given by ds = √(dx² + dy²). Substituting the parameterizations, we have ds = √(dt² + (6t dt)²) = √(1 + 36t²) dt. Thus, the integral becomes ∫ x ds = ∫ t √(1 + 36t²)dt. Integrating with respect to t from 0 to 2, we can use techniques like substitution or numerical methods to evaluate the integral and obtain the result.
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PLEASE HELP WILL MARK BRAINLIEST
Answer:
36
Step-by-step explanation:
There are 30 red marbles and the ratio of black to yellow is 4/5, so 4/5=orange/green=orange/30.
30 is 6 times 5, so black=6 times 4=24. There are 24 black marbles.
Total of black and yellow=24+30=54.
40% of the marbles are green so 60% are black/yellow. Also 60% is 54 marbles.
60% is 60/100=3/5 so 3/5 of the total is 54 marbles, therefore 1/5 is 54/3=18 and 2/5 is the number of red marbles=2 times 18=36 red marbles.
to be honest i can't think of anything original for the visual thing, but i think the pi chart from the person above me would work well
Answer: 36
Step-by-step explanation: 2/5 of marbles are red, and 4:5 ratio is multiplied by 6 because there are 30 yellow marbles and so 3/5 of the marbles are 54 and 2/5 are 36. For visual model draw a pi chart with 40% red, 33% yellow, and 27% black.
Which rectangular equation represents the parametric equations x =t Superscript one-half and y = 4t? y = 4x2, for x ≥ 0 y = one-fourth x squared, for x greater-than-or-equal-to 0 y = 16x2, for x ≥ 0 y = StartFraction 1 Over 16 EndFraction x squared, for x greater-than-or-equal-to 0
Answer:
Answer is Option A
Step-by-step explanation:
the things people do for points smh :/
The rectangular equation which represents the parametric equations; x = t^(¹/2) and y = 4t is; y = 4x2, for x ≥ 0.
Rectangular Equation from Parametric equationsFrom the task content, it follows that the parametric equations given are;
x = t^(¹/2) and y = 4tHence, it follows that; t = x² and y= 4t
Ultimately, upon substitution of x² for t; the resulting rectangular equation is; y = 4t².
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Find the distance between the points (–7,–9) and (–2,4).
Answer:
13.93
Step-by-step explanation:
see attached for explanation
Which equation represents the line that passes through the point (3, –1) and is perpendicular to the graph of x + 4y = 10?
Answer:
y=6x+7
Step-by-step explanation:
What is 40x + 4y = 24
Answer:
sorry but your question can't be solved any further because it doesn't have any like termsand you can't add40x and4y together it's impossible unless you multiply
What is the equation in point-slope form of the line passing through (0,5) and (-2, 11)?
Oy-5=-3(x + 2)
Oy-5= 3(x + 2)
Oy - 11 = -3(x - 2)
Oy - 11 = -3(x + 2)
Answer: y-11 = -3(x+2)
The second quartile for the numbers:
231,423,521,139,347,400,345 is
A 231
B 347
C 330,
D 423
Which of the following measures of variability is dependent on every value in a Set of data?
A Range
B. Standard deviation
C A and B
D. Neither A nor B
Which one of these statistics is unaffected by Outliers?
A Mean
B. Interquartile range
C Standard deviation
D. Range
The second quartile for the numbers 231, 423, 521, 139, 347, 400, 345 is 347. The median value is the second quartile
. So, when the numbers are arranged in ascending order, 347 is in the middle.
The following measures of variability is dependent on every value in a Set of data:
Standard deviation is the measure of variability that is dependent on every value in a Set of data.
Standard deviation is a measure that is used to quantify the amount of variation or dispersion of a set of data values.
Among these statistics, the Interquartile range is unaffected by outliers. The IQR is the distance between the first quartile (Q1) and the third quartile (Q3), which represents the middle 50% of the data.
The interquartile range (IQR) is unaffected by outliers because it only uses the values of the data points located at the 25th and 75th percentile of the data set to measure variability.
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predictive modelling and lifetime value modelling are the same
True or False
The given statement, "predictive modelling and lifetime value modelling are the same" is false. What is Predictive Modeling?
Predictive modeling is a technique for forecasting the probability of a certain event taking place in the future. It entails using current and past data to forecast the future events. In predictive modeling, you use known outcomes of historical data to determine whether specific patterns are likely to recur in the future. What is Lifetime Value Modeling? Lifetime Value Modeling is a method of forecasting the long-term earnings and profit of a business. It's a strategy for calculating the cumulative amount of profit generated by a customer over the life of their relationship with a company. Lifetime Value Modeling is used to decide the most effective ways to engage with consumers, such as personalized deals or special promotions, to maximize their lifetime value to the company by encouraging them to purchase more often and spend more during each transaction.
Predictive modeling and lifetime value modeling are distinct concepts that serve different purposes. Predictive modeling is used to forecast the future occurrence of specific events, whereas lifetime value modeling is used to calculate the long-term value of a customer to a company. So, the given statement is false.
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The fourth-grade students are taking a field trip and need to rent minivans. Each minivan will hold 8 people. There are 135 people
going on the trip. How many people will not be able to go if they only rent 16 minivans?
A)6 people
B)7 people
C)8 people
D)9 people
HELP ASAP ILL GIVE BRAINLIEST
Answer:
B. 7 people
Step-by-step explanation:
If you multiply 8x16 you get: 128. Then you subtract 135 from 128 and get: 7. Therefore, 7 people will not be able to go if they only rent 16 minivans.
1) According to one study, brain weights of men are normally distributed with a mean of 1.10 kg and a standard deviation of 0.14 kg. Use the data to answer questions (a) through (e).
a. Determine the sampling distribution of the sample mean for samples of size 3.
b. Determine the sampling distribution of the sample mean for samples of size 12.
d. Determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg.
e. Determine the percentage of all samples of twelve men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg.
_________________________________________
2) According to a study, brain weights of men in country A are normally distributed with mean 1.60 kg and standard deviation 0.12 kg. Apply the 68.26-95.44-99.74 rule to fill in the blanks.
68.26% of men in country A have brain weights between ___ kg and __kg
_____________________________________________
a) Sample distribution follows normal distribution with mean( μ) = 1.10 kg,
and standard deviation σ = 0.081
b) Sample distribution follows normal distribution with mean( μ) = 1.10 kg,
and standard deviation σ = 0.04
d) The percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg is 79.77%.
e) The percentage of all samples of twelve men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg is 99.3%.
2) 68.26% of men in country A have brain weights between 1.48 kg and 1.72 kg.
Solution:
Population standard deviation is the measure of how spread out the population data is. It measures the difference of the individual items from the mean. A standard deviation is a statistic that measures the dispersion of a dataset relative to its mean. It is calculated as the square root of variance by determining the variation between each data point relative to the mean.
1)
Given mean = 1.10 kg, standard deviation = 0.14 kg
a) To find the sampling distribution of the sample mean for samples of size 3.
Standard error of mean = σ/√n
= 0.14/√3
=0.081
Sample distribution follows normal distribution with mean( μ) = 1.10 kg,
and standard deviation σ = 0.081
b) To find the sampling distribution of the sample mean for samples of size 12.
Standard error of mean = σ/√n
= 0.14/√12
= 0.04
Sample distribution follows normal distribution with mean( μ) = 1.10 kg,
and
standard deviation σ = 0.04
d) Determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg.
Sample distribution follows normal distribution with mean( μ) = 1.10 kg,
and
standard deviation σ = 0.081
Z = (x - μ) / σZ
= (1.1 + 0.1 - 1.1) / 0.081
= 1.23
Z = (1.1 - 0.1 - 1.1) / 0.081
= -1.23
P ( -1.23 < Z < 1.23) = 0.7977
The percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg is 79.77%.
e) Determine the percentage of all samples of twelve men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg.
Sample distribution follows normal distribution with mean( μ )= 1.10 kg,
and
standard deviation σ = 0.04
Z = (x - μ) / σ
Z = (1.1 + 0.1 - 1.1) / 0.04
= 2.5
Z = (1.1 - 0.1 - 1.1) / 0.04 = -2.5
P ( -2.5 < Z < 2.5) = 0.993
The percentage of all samples of twelve men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.10 kg is 99.3%.
2)
Given mean = 1.60 kg,
standard deviation = 0.12 kg
68.26% of men in country A have brain weights between μ - σ and μ + σ
68.26% of men in country A have brain weights between 1.48 kg and 1.72 kg.
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The graph of the function f(x) = ax^2 + bx + c has vertex at (0, 2) and passes through the point
(1, 8). Find a, b and c
Answer:
Step-by-step explanation:
You need to use vertex form of a quadratic to solve this.
Consider the vertex to be [tex](h,k)[/tex]
Another way of representing a quadratic is in "vertex form":
[tex]f(x) = a(x-h)^2+k[/tex]
Now all you have to do is solve for a. You know that the vertex is [tex](0,2)[/tex] and you have know the point of [tex](1,8)[/tex]. Now, all you have to do is plug in these values and solve for a.
[tex]8 = a(1-0)^2+2\\8=a(1)^2+2\\8=a+2\\a=6[/tex]
Now you know the equation is [tex]f(x) = 6(x-0)^2+2[/tex] , but you need it in quadratic form. All you have to do is solve is distribute the 6:
[tex]6x^2+2[/tex]
You get:
a = 6
b = 0
c = 2
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On average, a banana will last 6.2 days from the time it is purchased in the store to the time it is too rotten to eat. Is the mean time to spoil less if the banana is hung from the ceiling? The data show results of an experiment with 16 bananas that are hung from the ceiling. Assume that that distribution of the population is normal.
3.9, 4.9,5.1, 3.9, 4, 5.8, 7, 5, 3.6, 4.3, 4.4, 6, 6.8, 6.7, 7.1, 5.2
What can be concluded at the the α = 0.05 level of significance level of significance?
Using a one-sample t-test, we cannot conclude that the mean time to spoil is significantly different when bananas are hung from the ceiling.
One sample t-test3.9, 4.9, 5.1, 3.9, 4, 5.8, 7, 5, 3.6, 4.3, 4.4, 6, 6.8, 6.7, 7.1, 5.2
We can calculate the sample mean and sample standard deviation:
Sample mean (x) = (3.9 + 4.9 + 5.1 + 3.9 + 4 + 5.8 + 7 + 5 + 3.6 + 4.3 + 4.4 + 6 + 6.8 + 6.7 + 7.1 + 5.2) / 16 = 5.3
Sample standard deviation (s) = √[(Σ(xi - x)²) / (n - 1)] = √[(Σ( - 5.3)²) / 15] ≈ 1.273
We will perform a one-sample t-test using the null hypothesis (H0) that the mean time to spoil is equal to 6.2 days, and the alternative hypothesis (H1) that the mean time to spoil is less than 6.2 days.
The test statistic is calculated as:
t = (x - μ) / (s / √n)
Where μ is the hypothesized mean (6.2), s is the sample standard deviation (1.273), and n is the sample size (16).
Plugging in the values:
t = (5.3 - 6.2) / (1.273 / √16) ≈ -0.887
To determine the critical t-value for a one-tailed test at α = 0.05 level of significance with 15 degrees of freedom (n - 1), we refer to the t-distribution table or use statistical software. The critical t-value is approximately -1.753.
Since the test statistic (-0.887) does not exceed the critical t-value (-1.753), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the mean time to spoil is less when bananas are hung from the ceiling compared to the average time of 6.2 days, at the α = 0.05 level of significance.
Therefore, we cannot conclude that the mean time to spoil is significantly different when bananas are hung from the ceiling.
Learn more on normal distribution: https://brainly.com/question/15414435
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