The estimate of the difference is -3.94, indicating that the mean oxygen consumption rate for runners trained with the continuous method is 3.94 units higher than those trained with the intermittent method.
To estimate the difference between the population means, we can use a two-sample t-test since we are comparing two independent samples. Given the sample means, standard deviations, and sample sizes, we can calculate the pooled standard deviation and the standard error of the difference.
The pooled standard deviation is calculated using the formula:
Sp = sqrt(((n₁-1)s₁² + (n₂-1)s₂²) / (n₁ + n₂ - 2))
The standard error of the difference is calculated using the formula:
SE = sqrt((s₁²/n₁) + (s₂²/n₂))
Using these values, we can calculate the t-value and the confidence interval for the difference in means.
With a confidence coefficient of 0.95, the critical t-value is obtained from the t-distribution with (n₁ + n₂ - 2) degrees of freedom. By comparing the t-value to the critical t-value, we can determine if the difference is statistically significant.
Interpreting the results, we find that the estimated difference in means is -3.94, indicating that the mean oxygen consumption rate for runners trained with the continuous method is 3.94 units higher than those trained with the intermittent method.
The confidence interval for the difference would provide a range within which we can be 95% confident that the true difference in population means lies.
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What kind of equation is shown below?
x2 + 6x - 13 = 0
Answer:
quadratic equation
Step-by-step explanation:
Answer:
the equation shown here is a quadratic equation
Need actual help (please hurry)
Find the exact value of the integral fx=-2² dx. -2x² 500 dx
The given integral is: $$\int (-2x^2 + 500)dx$$
To solve the above integral, we integrate both terms separately. Using the integral formulas, $$\int x^n dx= \frac{x^{n+1}}{n+1}+ C$$$$\int kf(x)dx= k\int f(x)dx$$
where C is a constant of integration and k is a constant.
The integral $\int -2x^2 dx$ is:\begin{align*} \int -2x^2 dx &= -2\int x^2 dx\\&= -2\times \frac{x^{2+1}} {2+1} + C\\&= -\frac {2}{3}x^3+ C\end{align*}
The integral $\int 500 dx$ is:\begin{align*}\int 500 dx &= 500\int dx\\&= 500\times x + C\\&= 500x + C\end{align*}
Thus, the integral $\int (-2x^2 + 500) dx$ is:\begin{align*}\int (-2x^2 + 500) dx &= \int -2x^2 dx + \int 500 dx\\&= (-\frac {2}{3}x^3+ C_1) + (500x + C_2)\\&= -\frac {2}{3}x^3+ 500x + C\end{align*}
where C is a constant of integration and $C=C_1+C_2$.
Therefore, the exact value of the integral is $-\frac {2}{3}x^3+ 500x + C$.
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Geometry question please help:)
Answer:
EF = 40√6----------------------
DC is the radius and its length is:
DC = 20 + 50DC = 70DC is the perpendicular bisector of EF. Hence the half of the segment EF is the leg of a right triangle with the hypotenuse of EC = DC = 70 and other leg of 50 units.
Using the Pythagorean theorem, find the length of EF:
(EF/2)² = 70² - 50²(EF/2)² = 2400EF/2 = √2400EF/2 = 20√6EF = 40√6Please help it’s due today, if you explain you get brainliest
Solve the system of equations using the elimination method
2y - 3y = -9
-x + 3y = 6
(-3,3)
(1,3)
(3,3)
(-3,1)
right answer gets brainliest
9514 1404 393
Answer:
(d) (-3, 1)
Step-by-step explanation:
We assume a typo in the problem statement, and that the equations are supposed to be ...
[tex]2x -3y = -9\\-x +3y = 6[/tex]
Adding the two equations gives ...
x = -3
Substituting for x in the second equation gives ...
3 +3y = 6
1 + y = 2 . . . . divide by 3
y = 1 . . . . . . . subtract 1
The solution is (x, y) = (-3, 1).
Free write plssss write something use your imagination and be creative with it!!!!! Worth a lot of points
Answer:
Write abt a poem you like or something
Step-by-step explanation:
uh yeah I’m this bad at math I need help on number 4
Answer:
36
Step-by-step explanation:
To find the area of a triangle, multipy the base and height of a triangle, then divide it in half
8x9=72/2=36
What is the equation of the above graph in vertex form?
Answer:
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Step-by-step explanation:
Which of these is a nonlinear function. 20 points...
A. x=0
B. x=y/4
C. y=√x
D. y=√5
Select all of the prime numbers from the list that can divide the number 10 exactly.
A2
B 3
C 5
D 7
E 11
F 13
G 17
H 19
I 23
Answer:
2 and 5
Step-by-step explanation:
2×5 = 10 or visa versa
10 ÷5=2
10 ÷2 =5
1.There are 270 students at an elementary school. There are 5 boys for every 4 girls. How many boys attend the school?
2. 4 small candies cost $0.96. How much do 6 candies cost?
3. 3 bags of chips cost $5.55. How much does 2 bags of chips cost?
4. 5 movie tickets cost $55. At this rate, what is the cost per ticket?
Answer:
Step-by-step explanation:
1. Not sure...
2. $ 1.44
3. $ 3.70
4. $11 per ticket
Solve 12/y = 4/3
Solution is____
Answer:
y=9
Step-by-step explanation:
12/y=4/3
3×12/y=3×4/3
36/y=4
y×36/y= 4×y
36=4y
36/4=4y/4
9=y
daniel bought 6 shirts for $28.93 if he paid the same price for each shirt how much would he spend if he bought 15 shirts?
Answer:
i dont know do it yourself chump
Step-by-step explanation:
Can some help me pls
Answer:
1. 0=0
2. x= -3
3. 14=0
Step-by-step explanation:
I tried I'm sorry if it's wrong
I think it’s A but I think that I am incorrect. I’m in desperate need of the answer.
Answer:
It would be D
Step-by-step explanation:
What was her score on this exam, rounded to the nearest integer
Answer:
Where is the number I need to round to?
(Please don't give me a bad rating i'm trying to help :) )
a las 6 am un termometro marca 6°c bajo cero, a las 9 am la temperatura aumento a 4°c, a las 12 pm la temperatura subio 10°c, a las 15 pm la temperatura ascedio 13°c, a las 18 pm la temperatura descendio 13°c, a las 21 pm la temperatura bajo 11°c representa la temperatura en una recta numerica y determina la temperatura a las 21 pm
How do I solve this arithmetic sequence?
Answer:
The 16ᵗʰ term of this sequence is 82
Step-by-step explanation:
Here,
First Term = a₁ = 9
Common Difference = (d) = 2
Now, For 16ᵗʰ term, n = 16
aₙ = a + (n - 1)d
a₁₆ = 7 + (16 - 1) × 2
a₁₆ = 7 + 15 × 5
a₁₆ = 7 + 75
a₁₆ = 82
Thus, The 16ᵗʰ term of this sequence is 82
-TheUnknownScientist
Plzz help everything is in the screen shot
Answer:
#1
Yesy = x + 1#2
Yesy = -2x#3
No#4
No(c) Let X and Y be independent random variables such that XY is degenerate at c≠0 i.e., P(XY = c) = 1. Show that X and Y are also degenerate.
(a) Suppose two buses, A and B, operate on a route. A person arrives at a certain bus stop on this route at time 0. Let X and Y be the arrival times of buses A
(c) Let X and Y be independent random variables such that XY is degenerate at c≠0 i.e., P(XY = c) = 1. Show that X and Y are also degenerate.
5. (a) Suppose two buses, A and B, operate on a route. A person arrives at a certain bus stop on this route at time 0. Let X and Y be the arrival times of buses A
(a) Let X and Y be the arrival times of buses A and B respectively, it is given that buses A and B are independent and hence X and Y are independent random variables.
Therefore, X and Y are also degenerate at c.
Since a person arrives at the bus stop at time 0, the arrival times of buses A and B cannot be negative
i.e., X, Y ≥ 0.
Also, both buses cannot arrive at time 0
i.e., P(X = 0, Y = 0) = 0.
Now, suppose that X and Y are not degenerate. T
hen, their joint distribution function is given by:
F(X, Y) = P(X ≤ x, Y ≤ y) > 0, for some x, y ≥ 0.
Using the independence of X and Y, we get:
F(X, Y) = P(X ≤ x)P(Y ≤ y) > 0,
for some x, y ≥ 0.
Since P(XY = 0) = 0,
we have XY ≠ 0 and
hence P(XY = c) > 0,
for some c ≠ 0.
Now, for a fixed ε > 0,
let Bε = {(x, y) : |xy - c| < ε}.
Then, P((X, Y) ∈ Bε) > 0.
Similarly, let B'ε = {(x, y) : x < ε} and
B''ε = {(x, y) : y < ε}.
Then, P((X, Y) ∈ B'ε) > 0
and P((X, Y) ∈ B''ε) > 0.
Now, we have:
Bε ⊆ B'ε ∪ B''ε and (X, Y) ∈ B'ε
if and only if X < ε and (X, Y) ∈ B''ε
if and only if Y < ε.So,
using the union bound and taking ε small enough, we get:
P(X < ε) + P(Y < ε) > P((X, Y) ∈ Bε) > 0.
This contradicts the assumption that P(X = 0, Y = 0) = 0.
Therefore, X and Y must be degenerate.
Now, we will show that if XY is degenerate at c ≠ 0
i.e., P(XY = c) = 1,
then X and Y are also degenerate at the same point.
To see this, note that for any Borel sets A, B ⊆ R, we have:
P(X ∈ A, Y ∈ B) = P(XY ∈ A × B) = P(XY = c)
= 1, if (c ∈ A × B) or (c is an isolated point of A × B).
Therefore, X and Y are also degenerate at c.
This completes the proof of the required statement.
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A rectangle prism has a length of 10m, a height of 9m, and a width of 15m. What is its volume, in cubic meters
Answer:
1350 m³
Step-by-step explanation:
To find the volume of a rectangular prism, we have to multiply its length, width, and height together.
The dimensions of this prism are 9, 10, and 15.
9 · 10 · 15 = 1350
The volume is 1350 meters³.
I hope this helps ^^
The box plots below show the distribution of salaries, in thousands, among employees of two small companies.
A box plot titled Salaries in Dollars at Company 1. The number line goes from 25 to 80. The whiskers range from 25 to 80, and the box ranges from 26 to 34. A line divides the box at 30.
Salaries in Dollars at Company 1
A box plot titled Salaries in Dollars at Company 2. The number line goes from 35 to 90. The whiskers range from 36 to 90, and the box ranges from 38 to 44. A line divides the box at 40.
Salaries in Dollars at Company 2
Which measures of center and variability would be best to use when making comparisons of the two data sets?
Answer:
Mean and IQR
Step-by-step explanation:
The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.
For company A:
Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.
Hence, the mean will be the best measure of centre rather Than the median in this case.
For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.
Answer:
B
Step-by-step explanation:
On Tuesday afternoon at camp, Eve did archery and sailing before dinner. Archery started at 12:30p.m. Eve spent 1 hours and 25 minutes at archery and 45 minutes sailing. What time did sailing end?
Answer:
2:40
Step-by-step explanation:
Add 85 minutes to 45 minutes and then add that to 12:30
Define a sequence a,, so that ao = 2, a₁ = 3, and an = 6an-1-8a-2.
The sequence {aₙ} defined by a₀ = 2, a₁ = 3, and aₙ = 6aₙ₋₁ - 8aₙ₋₂ produces the terms:
2, 3, 2, -12, -88, -432, ...
To define the sequence {aₙ}, given a₀ = 2, a₁ = 3, and the recursive formula aₙ = 6aₙ₋₁ - 8aₙ₋₂, we can calculate the subsequent terms of the sequence.
Using the given initial conditions, we have:
a₀ = 2
a₁ = 3
To find a₂, we substitute n = 2 into the recursive formula:
a₂ = 6a₁ - 8a₀
= 6(3) - 8(2)
= 18 - 16
= 2
To find a₃, we substitute n = 3 into the recursive formula:
a₃ = 6a₂ - 8a₁
= 6(2) - 8(3)
= 12 - 24
= -12
Continuing this process, we can find the subsequent terms of the sequence:
a₄ = 6a₃ - 8a₂
= 6(-12) - 8(2)
= -72 - 16
= -88
a₅ = 6a₄ - 8a₃
= 6(-88) - 8(-12)
= -528 + 96
= -432
and so on.
Therefore, the sequence {aₙ} defined by a₀ = 2, a₁ = 3, and aₙ = 6aₙ₋₁ - 8aₙ₋₂ produces the terms:
2, 3, 2, -12, -88, -432, ...
Please note that if you need the general formula for the nth term of the sequence, it may require a different approach as the given recursive formula is not a linear recurrence relation with constant coefficients.
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f(x) = x^4 + 2*x^3 + 3*x^2+4*x + 5
and g(x) = x^2 + 2*x + 4
Note that if we just use the coefficients of f(x) and g(x), then they look like 1 2 3 4 5 and 1 2 4
(5%) Compute the product h(x) of two polynomials f(x) and g(x) manually. In particular, show how the constant term, the x term, the x2 term, the x3 term etc. are computed from f(x) and g(x) respectively. Like Q7, use a table T3 to show line by line, how each term is computed (note the x2 term of h(x) comes from f(x)’s constant term and g(x)’s x2 term, plus f(x)’s x term and g(x)’s x term, and g(x)’s constant term and f(x)’s x2 term etc.
(5%) Compute the quotient q(x) and remainder r(x) when f(x) is divided by g(x), in other words compute q(x) and r(x) manually so that f(x) = g(x) * q(x) + r(x).
After considering the given data we conclude the x term of h(x) is [tex]44 + 5*2 = 26.[/tex]We can continue this process for the [tex]x^2[/tex] term, the [tex]x^3[/tex] term, and the [tex]x^4[/tex] term, using the appropriate terms from f(x) and g(x) and adding up the products and the quotient q(x) is [tex]x^2 - x,[/tex] and the remainder r(x) is [tex]3x^2 + 8x + 5,[/tex]
To evaluate the product h(x) of the two polynomials f(x) and g(x) manually, we can apply a table [tex]T_3[/tex] to show line by line how each term is computed.
The table possess columns for the term from f(x), the term from g(x), and the product of the two terms. The rows of the table will correspond to the different powers of x, starting from [tex]x^0[/tex].
For instance , to compute the constant term of h(x), we need to multiply the constant terms of f(x) and g(x). The constant term of f(x) is 5, and the constant term of g(x) is 4, so the constant term of h(x) is 54 = 20.
Similarly, to compute the x term of h(x), we need to multiply the x term of f(x) (which is 4) by the constant term of g(x) (which is 4), and add it to the constant term of f(x) (which is 5) multiplied by the x term of g(x) (which is 2).
Therefore, the x term of h(x) is [tex]44 + 5*2 = 26.[/tex]We can continue this process for the [tex]x^2[/tex] term, the [tex]x^3[/tex] term, and the [tex]x^4[/tex] term, using the appropriate terms from f(x) and g(x) and adding up the products.
To evaluate the quotient q(x) and remainder r(x) when f(x) is divided by g(x), we can use polynomial long division.
We apply division the highest degree term of f(x) by the highest degree term of g(x) to get the first term of q(x).
Then we multiply g(x) by this term of q(x) and subtract the result from f(x) to get the first remainder. We repeat this process with the next highest degree term of the remainder until the degree of the remainder is less than the degree of g(x).
The coefficients of the terms in q(x) are the quotients obtained in each step of the division, and the coefficients of the terms in the remainder are the final remainder obtained after all the steps of the division.
For instance , to compute the quotient q(x) and remainder r(x) when [tex]f(x) = x^4 + 2x^3 + 3x^2 + 4x + 5[/tex] is divided by [tex]g(x) = x^2 + 2x + 4,[/tex]
we first divide [tex]x^4[/tex] by [tex]x^2[/tex] to get [tex]x^2[/tex] as the first term of q(x).
We then multiply g(x) by [tex]x^2[/tex] to get [tex]x^4 + 2x^3 + 4x^2,[/tex] and subtract this from f(x) to get[tex]-x^3 - x^2 + 4x + 5[/tex] as the first remainder.
We then divide [tex]-x^3[/tex] by [tex]x^2[/tex] to get -x as the second term of q(x).
We multiply g(x) by -x to get[tex]-x^3 - 2x^2 - 4x[/tex], and subtract this from the first remainder to get [tex]3x^2 + 8x + 5[/tex]as the final remainder.
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what is the circumference of a circle with a radious of 8
Answer:
The circumference is 16π or 50.2654 depending on whether they ask you for an exact value or in terms of pi.
Step-by-step explanation:
2r=d
2(8)=d
16=d
C=πd
C=π16
So, the circumference is 16π or 50.2654 depending on whether they ask you for an exact value or in terms of pi.
Answer:
50.2654
Step-by-step explanation:
When performing polynomial regression, we should use the
smallest degree that provides a good fit. Why?
When performing polynomial regression, it is advisable to use the smallest degree that provides a good fit.
Polynomial regression involves fitting a polynomial function to a set of data points. The degree of the polynomial represents the highest power of the independent variable in the equation.
By using the smallest degree that provides a good fit, we aim to strike a balance between model complexity and overfitting.
Using a smaller degree helps prevent overfitting, which occurs when a model becomes too complex and captures noise or random fluctuations in the data. Overfitting leads to poor generalization, meaning the model may perform well on the training data but poorly on new, unseen data.
By using the smallest degree that provides a good fit, we minimize the risk of overfitting and ensure that the model captures the underlying trend in the data without incorporating unnecessary complexity.
This approach helps create a more robust and reliable model for making predictions on new data.
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Please help due in 5 minutes
file:///C:/Users/TOSHIBA/Downloads/SISTEMA%20DE%20MEDICION%20ANGULOS%20(1).pdf
necesito solo las respuestas de la ultima pregunta, por favor
Answer:
I was unable to see this link. Sorry.
Step-by-step explanation: