Answer : The confidence interval is [9.18, 10.82].
Explanation :
Given:Sample mean, x = 10
Sample standard deviation, s = 2
Sample size, n = 11
Significance level = 0.10
We can find the standard error of the mean, SE using the below formula:
SE = s/√n where, s is the sample standard deviation, and n is the sample size.
Substituting the values,SE = 2/√11 SE ≈ 0.6
Using the t-distribution table, with 10 degrees of freedom at a 0.10 significance level, we can find the t-value.
t = 1.372 Margin of error (ME) can be calculated using the formula,ME = t × SE
Substituting the values,ME = 1.372 × 0.6 ME ≈ 0.82
Confidence interval (CI) can be calculated using the formula,CI = (x - ME, x + ME)
Substituting the values,CI = (10 - 0.82, 10 + 0.82)CI ≈ (9.18, 10.82)
Therefore, the confidence interval is [9.18, 10.82].
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n simple linear regression, r 2 is the _____.
a. coefficient of determination
b. coefficient of correlation
c. estimated regression equation
d. sum of the squared residuals
The coefficient of determination is often used to evaluate the usefulness of regression models.
In simple linear regression, r2 is the coefficient of determination. In statistics, a measure of the proportion of the variance in one variable that can be explained by another variable is referred to as the coefficient of determination (R2 or r2).
The coefficient of determination, often known as the squared correlation coefficient, is a numerical value that indicates how well one variable can be predicted from another using a linear equation (regression).The coefficient of determination is always between 0 and 1, with a value of 1 indicating that 100% of the variability in one variable is due to the linear relationship between the two variables in question.
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answer all of fhem please
Mr. Potatohead Mr. Potatohead is attempting to cross a river flowing at 10m/s from a point 40m away from a treacherous waterfall. If he starts swimming across at a speed of 1.2m/s and at an angle = 40
Mr. Potatohead will be carried downstream by 10 × 43.5 = 435 meters approximately.
Given, Velocity of water (vw) = 10 m/s Velocity of Mr. Potatohead (vp) = 1.2 m/s
Distance between Mr. Potatohead and the waterfall (d) = 40 m Angle (θ) = 40
The velocity of Mr. Potatohead with respect to ground can be calculated by using the Pythagorean theorem.
Using this theorem we can find the horizontal and vertical components of the velocity of Mr. Potatohead with respect to ground.
vp = (vpx2 + vpy2)1/2 ......(1)
The horizontal and vertical components of the velocity of Mr. Potatohead with respect to ground are given as,
vpx = vp cos θ
vpy = vp sin θ
On substituting these values in equation (1),
vp = [vp2 cos2θ + vp2 sin2θ]1/2
vp = vp [cos2θ + sin2θ] 1/2
vp = vp
Therefore, the velocity of Mr. Potatohead with respect to the ground is 1.2 m/s.
Since Mr. Potatohead is swimming at an angle of 40°, the horizontal component of his velocity with respect to the ground is,
vpx = vp cos θ
vpx = 1.2 cos 40°
vpx = 0.92 m/s
As per the question, Mr. Potatohead is attempting to cross a river flowing at 10 m/s from a point 40 m away from a treacherous waterfall.
To find how far Mr. Potatohead is carried downstream, we can use the equation, d = vw t,
Where, d = distance carried downstream vw = velocity of water = 10 m/sand t is the time taken by Mr. Potatohead to cross the river.
The time taken by Mr. Potatohead to cross the river can be calculated as, t = d / vpx
Substituting the values of d and vpx in the above equation,
we get t = 40 / 0.92t
≈ 43.5 seconds
Therefore, Mr. Potatohead will be carried downstream by 10 × 43.5 = 435 meters approximately.
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find all solutions of the equation cos x sin x − 2 cos x = 0 . the answer is a b k π where k is any integer and 0 < a < π ,
Therefore, the only solutions within the given interval are the values of x for which cos(x) = 0, namely [tex]x = (2k + 1)\pi/2,[/tex] where k is any integer, and 0 < a < π.
To find all solutions of the equation cos(x)sin(x) - 2cos(x) = 0, we can factor out the common term cos(x) from the left-hand side:
cos(x)(sin(x) - 2) = 0
Now, we have two possibilities for the equation to be satisfied:
cos(x) = 0In this case, x can take values of the form x = (2k + 1)π/2, where k is any integer.
sin(x) - 2 = 0 Solving this equation for sin(x), we get sin(x) = 2. However, there are no solutions to this equation within the interval 0 < a < π, as the range of sin(x) is -1 to 1.
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Use geometry to evaluate the following integral. ∫1 6 f(x)dx, where f(x)={2x 6−2x if 1≤x≤ if 2
To evaluate the integral ∫[1 to 6] f(x) dx, where f(x) = {2x if 1 ≤ x ≤ 2, 6 - 2x if 2 < x ≤ 6}, we need to split the integral into two parts based on the given piecewise function and evaluate each part separately.
How can we evaluate the integral of the given piecewise function ∫[1 to 6] f(x) dx using geometry?Since the function f(x) is defined differently for different intervals, we split the integral into two parts: ∫[1 to 2] f(x) dx and ∫[2 to 6] f(x) dx.
For the first part, ∫[1 to 2] f(x) dx, the function f(x) = 2x. We can interpret this as the area under the line y = 2x from x = 1 to x = 2. The area of this triangle is equal to the integral, which we can calculate as (1/2) * base * height = (1/2) * (2 - 1) * (2 * 2) = 2.
For the second part, ∫[2 to 6] f(x) dx, the function f(x) = 6 - 2x. This represents the area under the line y = 6 - 2x from x = 2 to x = 6. Again, this forms a triangle, and its area is given by (1/2) * base * height = (1/2) * (6 - 2) * (2 * 2) = 8.
Adding the areas from the two parts, we get the total integral ∫[1 to 6] f(x) dx = 2 + 8 = 10.
Therefore, by interpreting the given piecewise function geometrically and calculating the areas of the corresponding shapes, we find that the value of the integral is 10.
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characterize the likely shape of a histogram of the distribution of scores on a midterm exam in a graduate statistics course.
The shape of a histogram of the distribution of scores on a midterm exam in a graduate statistics course is likely to be bell-shaped, symmetrical, and normally distributed. The bell curve, or the normal distribution, is a common pattern that emerges in many natural and social phenomena, including test scores.
The mean, median, and mode coincide in a normal distribution, making the data symmetrical on both sides of the central peak.In a graduate statistics course, it is reasonable to assume that students have a good understanding of the subject matter, and as a result, their scores will be evenly distributed around the average, with a few outliers at both ends of the spectrum.The histogram of the distribution of scores will have an approximately normal curve that is bell-shaped, with most of the scores falling in the middle of the range and fewer scores falling at the extremes.
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