For this case we must find the product of the following expression:
We apply distributive property term to term:
We add similar terms:
Finally, the expression is as:
Answeer:
Shira's math test included a survey question asking how many hours students spent studying for the test. The scatter plot below
shows the relationship between how many hours students spent studying and their score on the test. A line was fit to the data to model the relationship. Which of these linear equations best describes the given model? Choose 1 answer: Choose 1 answer: (Choice A) A \hat y=x+45 y ^ =x+45y, with, hat, on top, equals, x, plus, 45 (Choice B) B \hat y=10x+45 y ^ =10x+45y, with, hat, on top, equals, 10, x, plus, 45 (Choice C) C \hat y=-10x+45 y ^ =−10x+45y, with, hat, on top, equals, minus, 10, x, plus, 45 Based on this equation, estimate the score for a student that spent 555 hours studying.
1 answer:
Answer:
Part 1) y=10x+45
Part 2) The score is 95
Step-by-step explanation:
Part 1) Linear equation that best describes the given model
Let
x - number of hours students spent studying
y - their score on the test
Looking at the line that was fit to the data to model the relationship
The slope is positive
The y-intercept is the point (0,45)
For x=1, y=55 ----> point (1,55)
Find the slope
The formula to calculate the slope between two points is equal to
substitute the points (0,45) and (1,55)
Find the equation of the line in slope intercept form
we have
substitute
y=10x+45
Part 2)
Estimate the score for a student that spent 5 hours studying.
For x=5 hours
substitute in the linear equation and solve for y
y=10(5)+45=95
y = 95
You might be interested in
Answer:
a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544
b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States
c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390
ii. There are no statistical significant evidence that the prep course helped
d. i. The 95% confidence interval for the change in average test scores is 3.47467 < μ₁ - μ₂ < 14.52533
ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course
iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience
Step-by-step explanation:
The mean of the standardized test = 1,000
The number of students test to which the test is administered = 453 students
The mean score of the sample of students, = 1013
The standard deviation of the sample, s = 108
a. The 95% confidence interval is given as follows;
At 95% confidence level, z = 1.96, therefore, we have;
Therefore, we have;
1,022.94559 < μ < 1,003.0544
b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States
c. The parameters of the students taking the test are;
The number of students, n = 503
The number of hours preparation the students are given, t = 3 hours
The average test score of the student, = 1019
The number of test scores of the student, s = 95
At 95% confidence level, z = 1.96, therefore, we have;
The confidence interval, C.I., for the difference in mean is given as follows;
Therefore, we have;
Which gives;
-18.955390 < μ₁ - μ₂ < 6.955390
ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped
d. The given parameters are;
The number of students taking the test = The original 453 students
The average change in the test scores, = 9 points
The standard deviation of the change, Δs = 60 points
Therefore, we have;
C.I. = + 1.96 × Δs/√n
∴ C.I. = 9 ± 1.96 × 60/√(453)
i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533
ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course
iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience