Interpret the coefficient of determination. Choose the correct answer below. A. The coefficient of determination measures the percent of variation not explained by the multiple regression model. B. The coefficient of determination measures the percent of variation explained by the multiple regression model. C. The coefficient of determination measures the expected error of the predicted sales given a specific total square footage and number of shopping centers. D. The coefficient of determination measures the average predicted sales for the multiple regression model.

Answer:

2:3 is equal to 8:12

Step-by-step explanation:

2:3

To get the first number to 8

8/2 = 4

Multiply by all terms 4

2*3 : 3*4

8:12

2:3 is equal to 8:12

8:12 = 8/12

= 2/3

= 2:3

Therefore 2:3 and 8:12 are equalent to each other.

Answered by Gauthmath must click thanks and mark brainliest

Answer:

The third choice down

Step-by-step explanation:

|-2x| = 4

There are two solutions, one positive and one negative

-2x = 4  and -2x = -4

Divide by -2

-2x/-2 = 4/-2    -2x/-2 = -4/-2

x = -2   and x = 2

Step-by-step explanation:

f(x)=log(x)

     =d(log(x)/dx)

=>y=1/x

Answer:

Step-by-step explanation:

In order to find out how long it takes for the rocket to hit the ground, we only need set that position equation equal to 0 (that's how high something is off the ground when it is sitting ON the ground) and factor to solve for t:

[tex]0=-4.9t^2+112t+395[/tex]

Factor that however you are factoring in class to get

t = -3.1 seconds and t = 25.9 seconds.

Since time can NEVER be negative, it takes the rocket approximately 26 seconds to hit the ground.

Answer:

1:3

Step-by-step explanation:

27 : 81

Divide each side by 27

27/27 : 81/27

1:3

Answer:

354°15'38.99''

Step-by-step explanation:

The answer is A and D

good luck

Answer:

[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]

Step-by-step explanation:

Given

[tex]\sqrt{128x^8y^3}[/tex] --- the complete expression

Required

The equivalent expression

We have:

[tex]\sqrt{128x^8y^3}[/tex]

Expand

[tex]\sqrt{128x^8y^3} = \sqrt{128* x^8 * y^3}[/tex]

Further expand

[tex]\sqrt{128x^8y^3} = \sqrt{64 * 2* x^8 * y^2 * y}[/tex]

Rewrite as:

[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2* 2 * y}[/tex]

Split

[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2} * \sqrt{2 * y}[/tex]

Express as:

[tex]\sqrt{128x^8y^3} = (64 * x^8 * y^2)^\frac{1}{2} * \sqrt{2y}[/tex]

Remove bracket

[tex]\sqrt{128x^8y^3} = (64)^\frac{1}{2} * (x^8)^\frac{1}{2} * (y^2)^\frac{1}{2} * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 * x^\frac{8}{2} * y^\frac{2}{2} * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 * x^4 * y * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]

Answer:

9/25

Step-by-step explanation:

3 novels , 1 bio , 1 poetry = 5 books

P( novel) = novels / books

               = 3/5

Book is returned

3 novels , 1 bio , 1 poetry = 5 books

P( novel) = novels / books

               = 3/5

P(novel, return, novel) = 3/5 * 3/5 = 9/25

9514 1404 393

Answer:

Step-by-step explanation:

For continuous compounding the "rule of 69" applies. That is the doubling time can be found from ...

  t = 69.3147/r . . . . where r is the interest rate in percent.

Here, r=2, so ...

  t = 69.3147/2 = 34.6574 . . . years

That's 34 years and 240 days.

Answer:

(x-1)^2 + (y+4)^2 = 4

Step-by-step explanation:

The equation for a circle is given by

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x-1)^2 + (y- -4)^2 = 2^2

(x-1)^2 + (y+4)^2 = 4

Answer:

34

Step-by-step explanation:

let the missing value is x

(25+x) /2 = 29.5

25+x = 29.5(2)

25+x = 59

x = 59-25

x = 34

Step-by-step explanation:

before Tax

Coast = 888

in 2ND Item = $21

• 888/21

= $42.28

Answer:

8 units^2

Step-by-step explanation:

The area of a tringle is 1/2 bh. The base, LK, measures 4 while the height is also 4(you can get these values by counting the squares). This means the area is:

1/2 * (4)(4) = 1/2 * 16 = 8 units^2

The Total mileage is "400" and the further solution can be defined as follows:

Let t become the time he spent commuting on the first day of his vacation.

It is then calculated as [tex]t + 2[/tex].

[tex]\to 40\times(t+2) = 60(t) + 20 \\\\\to 40t+80 = 60t + 20 \\\\\to 80-20 = 60t + 40t \\\\\to 60 = 20t \\\\\to t=\frac{60}{20} \\\\\to t=\frac{6}{2} \\\\\to t= 3\\\\[/tex]

It traveled [tex]40\times (3 + 2) + 20 = 40\times 5 + 20 = 200+20=220[/tex] miles on its first day of operation.

The car traveled [tex]180\ miles[/tex] on the second day, which was [tex]60 \ miles \times 3[/tex].

So,

Total mileage= first day traveled + second day traveled [tex]= 220+ 180= 400 \miles[/tex]

Learn more:

Total distance traveled: brainly.com/question/20670144

Answer:

"A"

Step-by-step explanation:

a+b >c

a+c>b

b+c>a

~~~~~~~~~~~~

A. T,T,T

B. T,T,F

C. T,F,T

Answer:

0.2725

0.0431

Step-by-step explanation:

The distribution here is a poisson distribution :

λ = 1.3

The poisson distribution :

p(x) = [(e^-λ * λ^x)] ÷ x!

Expected probability of finding male with 0 accident ; x = 0

p(0) = [(e^-1.3 * 1.3^0)] ÷ 0!

p(0) = [0.2725317 * 1] ÷ 1

p(0) = 0.2725317

= 0.2725

2.)

P(x ≥ 4) = 1 - P(x < 4)

P(x < 4) = p(x = 0) + p(x. = 1) + p(x = 2) + p(x = 3)

p(x = 0) =  p(0) = [(e^-1.3 * 1.3^0)] ÷ 0! = 0.2725

p(x = 1) = p(1) = [(e^-1.3 * 1.3^1)] ÷ 1! = 0.35429

p(x = 2) = p(2) = [(e^-1.3 * 1.3^2)] ÷ 2! = 0.23029 p(x = 3) = p(3) = [(e^-1.3 * 1.3^3)] ÷ 0! = 0.09979

P(x < 4) = 0.2725 + 0.35429 + 0.23029 + 0.09979 = 0.95687

P(x ≥ 4) = 1 - 0.95687 = 0.0431

Answer:

m=[tex]7\sqrt3[/tex]

n=7

Step-by-step explanation:

Hi there!

We are given a right triangle (notice the 90°) angle, the measure of one of the acute angles as 60°, and the measure of the hypotenuse (the side OPPOSITE from the 90 degree angle) as 14

We need to find the lengths of m and n

Firstly, let's find the measure of the other acute angle

The acute angles in a right triangle are complementary, meaning they add up to 90 degrees

Let's make the measure of the unknown acute angle x

So x+60°=90°

Subtract 60 from both sides

x=30°

So the measure of the other acute angle is 30 degrees

This makes the right triangle a special kind of right triangle, a 30°-60°-90°  triangle

In a 30°-60°-90° triangle, if the length of the hypotenuse is a, then the length of the leg (the side that makes up the right angle) opposite from the 30 degree angle is [tex]\frac{a}{2}[/tex], and the leg opposite from the 60 degree angle is [tex]\frac{a\sqrt3}{2}[/tex]

In this case, a=14, n=[tex]\frac{a}{2}[/tex], and m=[tex]\frac{a\sqrt3}{2}[/tex]

Now substitute the value of a into the formulas to find n and m to find the lengths of those sides

So that means that n=[tex]\frac{14}{2}[/tex], which is equal to 7

And m=[tex]\frac{14\sqrt3}{2}[/tex], which simplified, is equal to [tex]7\sqrt3[/tex]

Hope this helps!

Answer:

y = -x + 5

Step-by-step explanation:

The point is in quadrant 2, so the line must pass through points that look like (a, 0) and (0, a) where a is a positive number.  The slope of such a line is -1.

If (x, y) is a point on the line, then the slope between points (x, y) and (-5, 10) is 1, and you can write

[tex]\frac{y-10}{x-(-5)}=-1\\y-10 = -1(x+5)\\y-10=-x-5\\y=-x+5[/tex]

Answer:

rt3/2

Step-by-step explanation:

first off cosine is the x coordinate

now if you do't want to use a calculator, you can use use the unit circle.

360 - 330 = 30 (360 degrees is a whole circle)

a 30 60 90 triangle is made, then use the law for 30 60 90 triangles:

if the shortest leg is x, the other leg is x*rt3 and the hypotenuse is 2x.

Answer:

D

Step-by-step explanation:

cos 330 = cos (360-330)

= cos 30

= √3 /2

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

Answer:

rectangle

Step-by-step explanation:

Perimeter of 20 feet

rectangle (square is technically a rectangle):

sides 5 and 5

5*5 = 25ft²

Circle:

20/(2π) = 3.18309...

3.1809...²π = 31.831ft²

Max area of rectangle (i.e. square) has a smaller area than a circle.

0.08 cm/min

Step-by-step explanation:

Given:

[tex]\dfrac{dV}{dt}=5\:\text{cm}^3\text{/min}[/tex]

Find [tex]\frac{dr}{dt}[/tex] when diameter D = 20 cm.

We know that the volume of a sphere is given by

[tex]V = \dfrac{4\pi}{3}r^3[/tex]

Taking the time derivative of V, we get

[tex]\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi\left(\dfrac{D}{2}\right)^2\dfrac{dr}{dt} = \pi D^2\dfrac{dr}{dt}[/tex]

Solving for [tex]\frac{dr}{dt}[/tex], we get

[tex]\dfrac{dr}{dt} = \left(\dfrac{1}{\pi D^2}\right)\dfrac{dV}{dt} = \dfrac{1}{\pi(20\:\text{cm}^2)}(5\:\text{cm}^3\text{/min})[/tex]

[tex]\:\:\:\:\:\:\:= 0.08\:\text{cm/min}[/tex]

It looks like the limit you want to find is

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4}[/tex]

One way to compute this limit relies only on the definition of the constant e and some basic properties of limits. In particular,

[tex]e = \displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x[/tex]

The idea is to recast the given limit to make it resemble this definition. The definition contains a fraction with x as its denominator. If we expand the fraction in the given limand, we have a denominator of x - 1. So we rewrite everything in terms of x - 1 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\dfrac{x-1+5}{x-1}\right)^{x-1+5} \\\\ = \left(1+\dfrac5{x-1}\right)^{x-1+5} \\\\ =\left(1+\dfrac5{x-1}\right)^{x-1} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Now in the first term of this product, we substitute y = (x - 1)/5 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(1+\dfrac1y\right)^{5y} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Then use a property of exponentiation to write this as

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\left(1+\dfrac1y\right)^y\right)^5 \times \left(1+\dfrac5{x-1}\right)^5[/tex]

In terms of end behavior, (x - 1)/5 and x behave the same way because they both approach ∞ at a proportional rate, so we can essentially y with x. Then by applying some limit properties, we have

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty} \left(\left(1+\dfrac1x\right)^x\right)^5 \times \left(1+\dfrac5{x-1}\right)^5 \\\\ = \lim_{x\to\infty}\left(\left(1+\dfrac1x\right)^x\right)^5 \times \lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)^5 \\\\ =\left(\lim_{x\to\infty}\left(1+\dfrac1x\right)^x\right)^5 \times \left(\lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)\right)^5[/tex]

By definition, the first limit is e and the second limit is 1, so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = e^5\times1^5 = \boxed{e^5}[/tex]

You can also use L'Hopital's rule to compute it. Evaluating the limit "directly" at infinity results in the indeterminate form [tex]1^\infty[/tex].

Rewrite

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \exp\left((x+4)\ln\dfrac{x+4}{x-1}\right)[/tex]

so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty}\exp\left((x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ = \exp\left(\lim_{x\to\infty}(x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ =\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right)[/tex]

and now evaluating "directly" at infinity gives the indeterminate form 0/0, making the limit ready for L'Hopital's rule.

We have

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\ln\dfrac{x+4}{x-1}\right] = -\dfrac5{(x-1)^2}\times\dfrac{1}{\frac{x+4}{x-1}} = -\dfrac5{(x-1)(x+4)}[/tex]

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{x+4}\right]=-\dfrac1{(x+4)^2}[/tex]

and so

[tex]\displaystyle \exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right) = \exp\left(\lim_{x\to\infty}\frac{-\dfrac5{(x-1)(x+4)}}{-\dfrac1{(x+4)^2}}\right) \\\\ = \exp\left(5\lim_{x\to\infty}\frac{x+4}{x-1}\right) \\\\ = \exp(5) = \boxed{e^5}[/tex]

Answer:

The number of hits would follow a binomial distribution with [tex]n =10,\!000[/tex] and [tex]p \approx 4.59 \times 10^{-6}[/tex].

The probability of finding [tex]0[/tex] hits is approximately [tex]0.955[/tex] (or equivalently, approximately [tex]95.5\%[/tex].)

The mean of the number of hits is approximately [tex]0.0459[/tex]. The variance of the number of hits is approximately [tex]0.0459\![/tex] (not the same number as the mean.)

Step-by-step explanation:

There are [tex](26 + 10)^{6} \approx 2.18 \times 10^{9}[/tex] possible passwords in this set. (Approximately two billion possible passwords.)

Each one of the [tex]10^{9}[/tex] randomly-selected passwords would have an approximately [tex]\displaystyle \frac{10,\!000}{2.18 \times 10^{9}}[/tex] chance of matching one of the users' password.

Denote that probability as [tex]p[/tex]:

[tex]p := \displaystyle \frac{10,\!000}{2.18 \times 10^{9}} \approx 4.59 \times 10^{-6}[/tex].

For any one of the [tex]10^{9}[/tex] randomly-selected passwords, let [tex]1[/tex] denote a hit and [tex]0[/tex] denote no hits. Using that notation, whether a selected password hits would follow a bernoulli distribution with [tex]p \approx 4.59 \times 10^{-6}[/tex] as the likelihood of success.

Sum these [tex]0[/tex]'s and [tex]1[/tex]'s over the set of the [tex]10^{9}[/tex] randomly-selected passwords, and the result would represent the total number of hits.

Assume that these [tex]10^{9}[/tex] randomly-selected passwords are sampled independently with repetition. Whether each selected password hits would be independent from one another.

Hence, the total number of hits would follow a binomial distribution with [tex]n = 10^{9}[/tex] trials (a billion trials) and [tex]p \approx 4.59 \times 10^{-6}[/tex] as the chance of success on any given trial.

The probability of getting no hit would be:

[tex](1 - p)^{n} \approx 7 \times 10^{-1996} \approx 0[/tex].

(Since [tex](1 - p)[/tex] is between [tex]0[/tex] and [tex]1[/tex], the value of [tex](1 - p)^{n}[/tex] would approach [tex]0\![/tex] as the value of [tex]n[/tex] approaches infinity.)

The mean of this binomial distribution would be:[tex]n\cdot p \approx (10^{9}) \times (4.59 \times 10^{-6}) \approx 0.0459[/tex].

The variance of this binomial distribution would be:

[tex]\begin{aligned}& n \cdot p \cdot (1 - p)\\ & \approx(10^{9}) \times (4.59 \times 10^{-6}) \times (1- 4.59 \times 10^{-6})\\ &\approx 4.59 \times 10^{-6}\end{aligned}[/tex].

Answer:

The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.  

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Siberian Huskies:

Sample of 47, mean of 5.2 minutes, standard deviation of 1.4. So

[tex]\mu_1 = 5.2[/tex]

[tex]s_1 = \frac{1.4}{\sqrt{47}} = 0.2042[/tex]

Others:

Sample of 39, mean of 5.5 minutes, standard deviation of 1.1. So

[tex]\mu_2 = 5.5[/tex]

[tex]s_2 = \frac{1.1}{\sqrt{39}} = 0.1761[/tex]

Distribution of the difference:

[tex]\mu = \mu_1 - \mu_2 = 5.2 - 5.5 = -0.3[/tex]

[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.2042^2+0.1761^2} = 0.2692[/tex]

Confidence interval:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = zs[/tex]

In which s is the standard error. So

[tex]M = 1.96(0.2692) = 0.5276[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is -0.3 - 0.5276 = -0.8276.

The upper end of the interval is the sample mean added to M. So it is -0.3 + 0.5276 = 0.2276

The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).

Answer:

2025 cm

Step-by-step explanation:

Given the length of pieces - 45 cm, 75 cm and 81 cm

To find the length of the rope we have to find the L.C.M. of 45, 75 and 81 :

3  |  45, 75, 81

    | ________________      

3  |  15, 25, 27

    |________________

3  |    5, 25, 9

    |________________

3  |    5, 25, 3

    |________________

5  |    5, 25, 1

    |________________

5  |     1, 5, 1

    |________________

    |      1, 1, 1

L.C.M. = 3 × 3 × 3 × 3 × 5 × 5

= 2025 cm

So, the least length of the rope should be 2025 cm which can be cut into a whole number of pieces of length 45 cm, 75 cm and 81 cm.

Answer: C

Step-by-step explanation:

Answer:

m=1/2

y-8=1/2(x-1)

y-8=1/2x-1/2

multiply through by 2

2y-16=x-1

2y-16+1-x=0

2y-15-x=0

2y-x-15=0

Answer:

chocolate chips are $2.00 per pound.

nd walnuts must be $3.50 per pound.

Step-by-step explanation:

Let x be the price of walnuts and y the price of chocolate chips.

2x + 5y = 17 (i)

8x + 3y = 34 (ii)

Multiply (i) by 4 to get

8x + 20y = 68

Subtract (ii) to get

 17y = 34

Dividing by 17, we see that chocolate chips are $2.00 per pound.

Substituting y=2 in (i) or (ii), walnuts must be $3.50 per pound.