Answers
The analysis of the data to obtain the best fit line equations that model the data, indicates;
5. a. Quadratic
b. y = -0.1179·x² + 2.1124·x + 4.215
c. Please find the completed chart showing the predicted values and the in the residuals in the following section.
6. a. A linear model may not be appropriate for the data in the residual plot
b. 4
c. 5
4. a. The exponential model of the function is; y = 100·e^(-0.124·t)
b. The weight after 8 weeks is about 37.1 grams
c. The sample will be 4 grams in about 26 weeks
5. a. The equation is; y = 22.703·x + 2.5733
b. The predicted y-value at x = 10 is; y = 229.60
c. The first time is about x ≈ 3.54
What is the best fit line?The best fit line is a line that is drawn through a set of data points in such a way that it minimizes the sum of squared errors of the data.
5. The best fit model can be obtained by plotting a scatter plot of the graph, which indicates that the best fit line resembles the shape of a parabola.
Therefore;
a. The best fit equation is; Quadratic
b. The best fit equation, obtained using technology is; y = -0.1179·x² + 2.1124·x + 4.215
The square of the correlation coefficient is; R² = 0.9584
The chart can be filled with the best fit equation as follows;
[tex]\begin{tabular}{ | c | c | c | c | c | }\cline{1-4}Distance (foot) & Height (foot) & Predicted Value &Residual \\ \cline{1-4}0 & 4 & 4.215 & -0.215 \\\cline{1-4}2 & 8.4 & 8.16588 & 0.23412 \\\cline{1-4}6 & 12.1 & 13.23804 & -1.13804 \\\cline{1-4}9 & 14.2 & 14.56626 & -0.36626\\\cline{1-4}12 & 13.2 & 13.77228 & -0.57228 \\\cline{1-4}13 & 10.5 & 13.03602 & -2.53602 \\\cline{1-4}15 & 9.8 & 10.8561 & -1.0561 \\\cline{1-4}\end{tabular}[/tex]
The predicted value are obtained by plugging in the x-value into the best fit equation and the residuals is the difference between the actual value and the predicted value.
6. a. A residual plot can be used to as assessment with regards to meeting the assumptions of a linear regression model.
A residual plot that is randomly scattered about zero indicates that a linear model is appropriate for the data.
The data points in the residual plot are not expressed as being randomly scattered around zero.
The pattern that exists in the residual plot indicates that the values are positive for x-values that are either low or high, and the middle x-values have a negative residuals. Therefr;
The pattern indicates that a linear regression model may not be appropriate for the datab. The number of positive residuals = 4
c. The number of negative residuals = 5
4. The general form of the exponential model of a function, y = A × e^(-k·t) can be used to find the function for the data as follows;
A = The initial amount = The value at 0 = 100
y = The amount of radioactive material at a given time t
e = Euler's number = 2.71828
The datapoints in the table indicates;
88.3 = 100 × e^(-k × 1)
k = -㏑(0.883) ≈ 0.124
The exponential model is therefore; y = 100 × e^(-0.124·t)
b. The weight of the sample after 8 weeks can be obtained by plugging in t = 8 in the exponential function for the weight of the radioactive substance as follows;
y = 100 × e^(-0.124 × 8) ≈ 37.1
The weight of the sample after 8 weeks is about 37.1 grams
c. When the sample is 4 grams we get;
y = 4
4 = 100 × e^(-0.124 × t)
e^(-0.124 × t) = 4/100 = 1/25
-0.124 × t = ln(1/25) = -ln(25)
t = ln(25)/0.124 ≈ 26
t ≈ 26 weeks
Therefore; The weight will be 4 grams after approximately 26 weeks
5. A linear regression can be performed using MS Excel to obtain the equation that models the data as follows;
y = 22.703·x + 2.5733
The square of the regression coefficient is; R² = 0.9995
a. The most appropriate equation to model the data in the table is; y = 22.703·x + 2.5733
b. The y-value when x = 10 can be predicted by plugging in x = 10 into the model equation for the data as follows;
y = 22.703 * 10 + 2.5733 ≈ 229.60
Therefore;
The predicted y-value when x is 10 is 229.60 (rounded to the nearest tenth)c. The first time the y-value is 83, can be found by setting y = 83 in the equation and solve for x as follows;
83 = 22.703·x + 2.5733
22.703·x = (83 - 2.5733)
x = (83 - 2.5733)/22.703 ≈ 3.54
Therefore, the first time the y-value is 83 is when x is approximately 3.54
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Related Questions
the 98.4% confidence interval for snapdragons grown in compost is (20.91, 38.43). what is the margin of error of this confidence interval?
Answers
The margin of error of the 98.4% confidence interval for snapdragons is 3.71.
The midpoint of the range is calculated by adding the upper and lower bounds and then dividing by two. So, the sample mean is `(20.91 + 38.43) / 2 = 29.67`.
The margin of error is calculated by multiplying the critical value of z* (1.96 for a 98.4% confidence level) by the standard error of the mean. The formula for calculating the margin of error is:
`Margin of error z*(standard deviation/√n`).
The formula is `range/4 = 1.96 * standard deviation/√n`.Now, solve for the standard deviation:
`standard deviation = (range/4) * √n / 1.96`
Substituting the values: `(38.43 - 20.91)/4 = 1.96 * standard deviation/√n`
Simplifying the equation: `4.26 = (1.96*standard deviation)/√n`
Squaring both sides: `4.26^2 = 3.8416 = (1.96^2 * standard deviation^2)/n`
Substituting the value of the standard deviation: `3.8416 = (1.96^2 * ((38.43 - 20.91)/4)^2) / n`
Solving for n: `n = ((1.96^2 * ((38.43 - 20.91)/4)^2) / 3.8416) = 31.54`
Now that we know the sample size, we can calculate the standard error of the mean:
`standard error = standard deviation/√n = ((38.43 - 20.91)/4)/√31.54 = 1.89`.
The margin of error is `1.96 * 1.89 = 3.71`.
The 98.4% confidence interval for snapdragons grown in compost is (20.91, 38.43). The margin of error of this confidence interval is 3.71.
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what is the probability that a customer purchases biography book given that they purchase cooking and bobvilla books? round your answer to two decimal places.
Answers
The probability that a customer purchases a biography book given that they purchase cooking and BobVilla books is 0.08.
To calculate this probability, we need to consider the following components:
1. The total number of customers purchasing cooking and BobVilla books: This is the denominator of our equation, and it represents the total number of customers who purchased the two books.
2. The number of customers purchasing the biography book: This is the numerator of our equation, and it represents the number of customers who purchased the biography book.
3. The probability that a customer purchases a biography book given that they purchase cooking and BobVilla books: This is the fraction of customers who purchased the biography book over the total number of customers who purchased the two books.
To calculate the probability that a customer purchases a biography book given that they purchase cooking and BobVilla books, we need to divide the numerator (the number of customers purchasing the biography book) by the denominator (the total number of customers purchasing the two books).
This probability can be expressed as a decimal, which is 0.08. This value can also be rounded to two decimal places, which is 0.08.
In conclusion, the probability that a customer purchases a biography book given that they purchase cooking and BobVilla books is 0.08.
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mental ability
hardest queston for grade 7
you are god if you did and explained properly
i will mark you as brainliest
optinons are-:
173
153
182
142
Answers
Answer:
153
Step-by-step explanation:
The relationship in the row is below
4³ +2³ +1³ =64 + 8 +1 = 72
1³ + 2³ +6³= 1 + 8 + 216
3³ + 1³ + 5³ = 27 +1 +125 = 153
All numbers below the first level are raised to power 3 and added together
after a (not very successful) trick or treating round, candice has 12 tootsie rolls and 10 twizzlers in her pillow case. her mother asks her to share the loot with her three younger brothers. (a) how many different ways can she do this?
Answers
Using the stars and bars technique, Candice can distribute her 24 pieces of candy among her four siblings in 2,925 different ways. If she must give each sibling at least one of each type of candy, there are 67,200 ways to distribute the candy among the four siblings.
(A) To solve this problem, we can use the technique of stars and bars. We have a total of 24 pieces of candy to share among four children. We can represent this using 24 stars, with 3 bars to separate the stars into four groups, one for each child. For example, the following arrangement represents giving 6 pieces of candy to the first child, 10 pieces to the second child, 3 pieces to the third child, and 5 pieces to the fourth child:
*****|**********|***|****
The number of ways to arrange the stars and bars is equal to the number of ways to choose 3 positions out of the 27 possible positions for the stars and bars. Therefore, the number of different ways that Candice can share her candy with her three younger brothers is:
C(27, 3) = 27! / (3! * 24!) = 2925
(B) Now, we need to ensure that each child receives at least one Tootsie roll and one Twizzler. We can give each child one of each candy to start, and then distribute the remaining 13 Tootsie rolls and 7 Twizzlers using the stars and bars technique. We have 13 Tootsie rolls and 7 Twizzlers to distribute among four children, which can be represented using 13 stars and 3 bars for the Tootsie rolls, and 7 stars and 3 bars for the Twizzlers. The number of ways to arrange the stars and bars for each type of candy is:
C(16, 3) = 560 for the Tootsie rolls
C(10, 3) = 120 for the Twizzlers
To find the total number of ways to distribute the candy, we can multiply the number of ways for each type of candy:
560 * 120 = 67200
Therefore, there are 67,200 different ways for Candice to share her candy with her three younger brothers after her mother asks her to give at least one of each type of candies to each of her brothers.
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Complete question:
After a (not very successful) trick or treating round, Candice has 15 Tootsie rolls and 9 Twizzlers in her pillow case. Her mother asks her to share some of the loot with her three younger brothers.
(A) How many different ways can she do this?
(B) How many different ways can she do this after her Mother asks her to give at least one of each type of candies to each of her brothers?
in a congressional district, 55% of the registered voters are democrats. which of the following is equivalent to the probability of getting less than 50% democrats in a random sample of size 100?
A. P( z< 50 — 55/ 100 )
B. P( z< 50 — 55/ √55(45)/100)
C. P( z< 55 — 5 / √55(45)/100)
D. P( z< 50 — 55/√100(55) (45))
Answers
The correct answer to the question, "Which of the following is equivalent to the probability of getting less than 50% democrats in a random sample of size 100?" is: B. P( z < 50 — 55/ √55(45)/100).
To find the probability, we first calculate the z-score using the formula:
z = (x - μ) / σ
where x is the value (50%), μ is the mean (55%), and σ is the standard deviation.
The standard deviation can be calculated as:
σ = √(np(1-p))
where n is the sample size (100) and p is the proportion of democrats (0.55).
Now, plug in the values into the z-score formula:
z = (50 - 55) / √(100 * 0.55 * 0.45)
The probability is then found as P(z < z-score), which is represented by the option B.
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What percentage of people would exed to score higher than a 2.5, but lower than 3.5? The mean: X=3.00 The SDis= + 0.500 18% 999 o 50% 03%
Answers
Therefore, approximately 68.26% of people are expected to score higher than 2.5 but lower than 3.5.
Based on the information provided, the mean (X) is 3.00 and the standard deviation (SD) is 0.50. To find the percentage of people expected to score higher than 2.5 but lower than 3.5, we will use the standard normal distribution (z-score) table.
First, we need to calculate the z-scores for both 2.5 and 3.5:
z1 =[tex] (2.5 - 3.00) / 0.50 = -1.0[/tex]
z2 = [tex](3.5 - 3.00) / 0.50 = 1.0[/tex]
Now, we can use the standard normal distribution table to find the probability of the z-scores. For z1 = -1.0, the probability is 0.1587 (15.87%). For z2 = 1.0, the probability is 0.8413 (84.13%).
To find the percentage of people expected to score between 2.5 and 3.5, subtract the probability of z1 from the probability of z2:
Percentage = [tex](0.8413 - 0.1587) x 100 = 68.26%[/tex]
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