Paola wants to measure the following dependent variable: happiness. How could you measure happiness in a way:
a) physiological?
b) observation?
c) self-report? Search for a scale that already exists.
What is the scale called? :
APA citation:_____

Answer:

what is this a different answer orrrr?

Answer:

The area is approximately [tex]588cm^2[/tex]

Step-by-step explanation:

Given

See attachment for figure

Required

The area of the shaded region

The shaded region is as follows:

First, calculate the area of the major segment using:

[tex]Area = \frac{\theta}{360} * \pi r^2[/tex]

Where

[tex]r = 14[/tex]

[tex]\theta = 360 - 70 =290[/tex]

So, we have:

[tex]A_1 = \frac{290}{360} * 3.14 * 14^2[/tex]

[tex]A_1 = 495.7711[/tex]

Next, the area of the triangle using:

[tex]Area = \frac{1}{2}ab \sin C[/tex]

Where

[tex]a=b=r = 14[/tex]

[tex]C = 70^\circ[/tex]

So, we have:

[tex]A_2 = \frac{1}{2} * 14 * 14 * sin(70)[/tex]

[tex]A_2 = 92.0899[/tex]

So, the area of the shaded region is:

[tex]Area = A_1 + A_2[/tex]

[tex]Area = 495.7711 + 92.0899[/tex]

[tex]Area = 587.8610[/tex]

[tex]Area \approx 588[/tex]

Answer:

7!

Step-by-step explanation:

If you add all of those numbers together it would be 49!

Then you divide that number how many numbers there are.

There are 7 numbers and 49/7 =7!

The expression (1 - cos(x))^2 * tangent(x/2) is equivalent to (1 - cos^2(x))(sin(x)).

We can simplify the expression by using the trigonometric identity: cos^2(x) + sin^2(x) = 1. Rearranging this identity, we have sin^2(x) = 1 - cos^2(x).

Substituting this identity into the expression, we get (1 - cos^2(x))(sin(x)).

Expanding the expression further, we have sin(x) - cos^2(x)sin(x).

Therefore, the expression (1 - cos(x))^2 * tangent(x/2) is equivalent to (1 - cos^2(x))(sin(x)).

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Answer:

It will be A

Step-by-step explanation:

6453

- 135

--------

6318 dollars needed for Sarah's food.

Hope this helps!

Answer:

1. 1/6

2. 1/2

Step-by-step explanation:

A dice has 6 sides. So the sample space for this question =

[1,2,3,4,5,6]

1. From the sample space we can see that 6 occurs only once in a dice

So probability of a 6

= 1/6

2. Number of odds = 3 (1,3,5)

Probability Leah rolls an odd number

= Odd/total

= 3/6

= 1/2

Answer:

a=[tex]\sqrt{28.32}[/tex]≈5.3217 cm

Step-by-step explanation:

Use the Pythagorean Theorem

a^2+4.7^2=7.1^2

a^2+22.09=50.41

a^2=28.32

a=[tex]\sqrt{28.32}[/tex]

≈5.3217

Answer:

5.3

Step-by-step explanation:

Check the attached image for explanation if you are interested, as I did not feel like spending 20 minutes re-formatting the equations to work on Brainly.

Also, I did not do this

Answer:

a twelth= 1/12

Step-by-step explanation:

Step-by-step explanation:

Dang, that's crazy! Thanks for letting us know! People be warned!

Answer:

FR

Step-by-step explanation:

don’t interact just report ☝️

Answer:

Step-by-step explanation:

Rule for the dilation of a point about the origin is,

(x, y) → (kx, ky)

Here, k = scale factor

Dilating points D, E and F about the origin by a scale factor 'k' = [tex]\frac{1}{3}[/tex]

D(-9, -6) → D'(-3, -2)

E(-6, -3) → E'(-2, -1)

F(0, -9) → F'(0, -3)

Now we can plot these points on graph.

Option (b) is the correct answer. Minimum usual value: 54.65

Maximum usual value: 79.92.

The given values are n = 267 and p = 0.239. The minimum usual value and the maximum usual value are to be calculated. We use the formula of the mean and the standard deviation for this purpose:

Mean = µ = np = 267 × 0.239 = 63.93Standard Deviation = σ = sqrt (npq) = sqrt [(267 × 0.239 × (1 - 0.239)] = 5.01The minimum usual value is obtained when the z-value is -2, and the maximum usual value is obtained when the z-value is +2. We use the z-score formula: z = (x - µ) / σwhere µ = 63.93 and σ = 5.01(a) When the z-value is -2, x = µ - 2σ = 63.93 - 2(5.01) = 53.91(b) When the z-value is +2, x = µ + 2σ = 63.93 + 2(5.01) = 73.95

Therefore, the minimum usual value is 53.91, and the maximum usual value is 73.95 (rounded to the nearest hundredth).

Thus, option (b) is the correct answer. Minimum usual value: 54.65Maximum usual value: 79.92.

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The given values are: n=267, p=0.239

We need to find the minimum usual value and the maximum usual value using these values of n and p.

Let X be a random variable with a binomial distribution with parameters n and p.

The mean of the binomial distribution is:μ = np

The standard deviation of the binomial distribution is:σ = sqrt(npq)where q = 1-p

Let X be a binomial distribution with parameters n = [tex]267 and p = 0.239μ = np = 267 × 0.239 = 63.813σ = sqrt(npq) = sqrt(267 × 0.239 × 0.761) = 6.788[/tex]

The minimum usual value is given by:[tex]μ - 2σ = 63.813 - 2 × 6.788 = 50.236[/tex]

The maximum usual value is given by:[tex]μ + 2σ = 63.813 + 2 × 6.788 = 77.39[/tex]

Thus, the minimum usual value is 50.24 and the maximum usual value is 77.39(rounded to the nearest hundredth).

Therefore, the answer is:Minimum usual value: 50.24, Maximum usual value: 77.39

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Answer:

5.71428571429

5 7/10

Step-by-step explanation:

Hope this helps and have a great day!!!!

Answer:

40/7

Step-by-step explanation:

Answer:

H0​:μ=440 g

Ha​:μ does not equal 440 g​

Step-by-step explanation:

kahn

The 90% confidence interval for the mean amount of pretzels per bag (in grams) in this production run for this case is  [441.31, 458.69] approximately.

If the sample size is given to be n < 30, then for finding the confidence interval for mean of population from this small sample, we use t-statistic.

Then, we get the confidence interval in between the limits

[tex]\overline{x} \pm t_{\alpha/2}\times \dfrac{s}{\sqrt{n}}[/tex]

where [tex]t_{\alpha/2}[/tex] is the critical value of 't' that can be found online or from tabulated values of critical value for specific level of significance and degree of freedom n - 1.

For this  case, we're provided;

The critical value of t  at level of significance 0.1 iand at degree of freedom 10-1=9 is:

Thus, the confidence interval in between the limits

[tex]450 \pm 1.833 \times \dfrac{15}{\sqrt{10}}[/tex]

or

[tex]450 \pm 8.69[/tex] approximately or 441.31 to 458.69 or we write it as: [441.31, 458.69] approximately.

Thus, the 90% confidence interval for the mean amount of pretzels per bag (in grams) in this production run for this case is  [441.31, 458.69] approximately.

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Answer:

687.549 deg

Step-by-step explanation:

12 rad = 687.549 deg

Answer: 56840

A bit of long multiplication

Answer:

56840

Step-by-step explanation:

Answer:

$436

Step-by-step explanation:

Answer:

5x+2(3x+8)

5x+6x+16

11x+16

11x=-16

X=16/11

The ordered pair (x, y) that is a solution to the equation -x + 5y = 2 is (0, 2/5).

To find the coordinates of the other endpoint D given that the midpoint of CD is M(2, -1) and one endpoint is C(-3, -3), we can use the midpoint formula:

Midpoint formula:

The coordinates of the midpoint between two points (x₁, y₁) and (x₂, y₂) are given by ((x₁ + x₂) / 2, (y₁ + y₂) / 2).

Using the given information, we can substitute the known values into the midpoint formula and solve for the coordinates of D:

M(2, -1) = ((-3 + x₂) / 2, (-3 + y₂) / 2)

Simplifying the equation:

2 = (-3 + x₂) / 2

-1 = (-3 + y₂) / 2

To solve for x₂:

4 = -3 + x₂

x₂ = -3 + 4

x₂ = 1

To solve for y₂:

-2 = -3 + y₂

y₂ = -3 - 2

y₂ = -5

Therefore, the coordinates of the other endpoint D are D(1, -5).

To find an ordered pair (x, y) that is a solution to the equation -x + 5y = 2, we can choose any value for either x or y and solve for the other variable. Let's choose x = 0:

-0 + 5y = 2

5y = 2

y = 2/5

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Answer:

diameter 33

radius 66

area 360

Answer:

Step-by-step explanation:

10^7 = 1 and 7 zeros after it.

=> 10000000

Answer:

a= 126 degrees

b= 54 degrees

r= 54 degrees

s= 126 degrees

Step-by-step explanation:

The approximate area of the lake where the resort has ownership responsibility is 711.82 meters².

Area of sector = (central angle / 360°) * π * radius²

We know that the central angle is 50°, the radius is 130 meters, and π is approximately equal to 3.14.

Plugging these values into the formula, we get:

Area of sector = (50° / 360°) * 3.14 * 130²

Area of sector = 25/72 * 3.14 * 16900

Area of sector = 15750/22

Area of sector = 711.82 meters²

Therefore, the approximate area of the lake where the resort has ownership responsibility is 711.82 meters².

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Answer:

(21x-25)

Step-by-step explanation:

We need to find an equivalent expression for the following.

5x-3+4(4x-6)+2

We can solve it as follows:

5x-3+4(4x-6)+2 = 5x-3+16x-24+2

= 5x+16x-3-24+2

= 21x-25

So, the equivalent expression is equal to (21x-25).

The probability that the sample proportion of private banks will be greater than 65% is 0.1548.

Given that 60% of banks in Switzerland are private organizations.

If a sample of 469 banks is selected, we need to find the probability that the sample proportion of private banks will be greater than 65%.

Let p be the proportion of private banks in a sample of 469 banks.

Then

q = 1 - p

= 1 - 0.6

= 0.4 (as the percentage of private banks is given to be 60%)

Sample size, n = 469

We are to find the probability that the sample proportion of private banks will be greater than 65%.

That is, we need to find P(p > 0.65).

We use the normal distribution for finding probabilities associated with proportions.

We can use the formula as follows:

z = (p - P) / √[PQ/n],

where P is the population proportion (given as 0.6),

Q = 1 - P

= 0.4, and

n = 469

Putting all the given values in the above formula, we get;

z = (0.65 - 0.6) / √[0.6 × 0.4 / 469]

z = 1.0182

P(p > 0.65) = P(z > 1.0182)

Using the standard normal table, we get; P(z > 1.0182) = 0.1548

Therefore, the probability that the sample proportion of private banks will be greater than 65% is 0.1548 (rounded to four decimal places).

Hence, the answer is 0.1548.

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The probability density function of Y = exp(U) is given by:

                   f(y) = { 1/y, 2 ≤ y ≤ e³ ; 0, elsewhere }.

Given that: U follows the Uniform distribution U ~ U[2, 3]. We have to find the probability density function of Y = exp(U).

The formula used: The probability density function of a random variable X, is denoted by f(x), is the derivative of the cumulative distribution function (cdf), denoted by F(x). We have F(x) = P(X ≤ x).

The probability density function of the uniform distribution U(a,b), is given by

f(x)=1/(b-a), where a ≤ x ≤ b.

Here, U[2,3]So, a = 2, b = 3

Let's find the probability density function of Y = exp(U).

So, for finding the probability density function of Y = exp(U), first, we need to find the cumulative distribution function F(y) of Y. Let's do that.

So, F(y) = P(Y ≤ y) = P(exp(U) ≤ y) = P(U ≤ ln y)

We have, Y = exp(U), which is a one-to-one function of U and increasing in U. Hence, we can use the one-to-one transformation formula. Hence, the probability density function of Y, f(y) = f(u) / |dy/du|.f(u) = 1/ (3-2) = 1

Here, dy/du = d/dy [exp(u)] = exp(u) = Y

Therefore, f(y) = 1/Y, for 2 ≤ u ≤ 3.

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Suppose that U follows the Uniform distribution U ~ U[2, 3].

Find the probability density function of Y = exp(U).

Let fU(u) be the pdf of U.Since U has a Uniform distribution over the interval [2, 3], its pdf is given by:

fU(u) = {1 / (3 - 2)} = 1, 2 ≤ u ≤ 3, and 0 elsewhere.

Now we need to find the pdf of Y = exp(U).

Let FY(y) be the cdf of Y.

Then we can write:

FY(y) = P(Y ≤ y) = P(exp(U) ≤ y) = P(U ≤ ln(y)), for y > 0.

Since U is continuous and its pdf is given by fU(u), we have:

[tex]FY(y) = ∫_{2}^{ln(y)} fU(u)du = ∫_{2}^{ln(y)} 1du = ln(y) - 2, 2 ≤ ln(y) ≤ 3, and 0 elsewhere.[/tex]

We can differentiate FY(y) to find the pdf of Y:

fy(y) = d/dy FY(y) = (1 / y) fY(ln(y)) = (1 / y), 2 ≤ y ≤ e3, and 0 elsewhere.

Therefore, the probability density function (pdf) of Y = exp(U) is given by:

fy(y) = (1 / y), 2 ≤ y ≤ e3, and 0 elsewhere.

In general, if U is a continuous random variable with pdf fU(u) and Y = g(U) is a monotonic transformation of U, then the pdf of Y can be found using the formula:

[tex]fy(y) = fU(g^{-1}(y)) / |dg^{-1}(y) / dy|,[/tex]

where g^{-1}(y) is the inverse function of g(y) and |dg^{-1}(y) / dy|

is the absolute value of the derivative of g^{-1}(y) with respect to y.

The probability density function (pdf) of the random variable

Y = exp(U)

where U is distributed uniformly over the interval [2, 3] can be found as follows:

Let f_U(u) be the pdf of U.

Since U has a Uniform distribution over the interval [2, 3], its pdf is given by:

f_U(u) = {1 / (3 - 2)} = 1, 2 ≤ u ≤ 3, and 0 elsewhere.

Now we need to find the pdf of Y = exp(U).

Let F_Y(y) be the cdf of Y.

Then we can write:

[tex]F_Y(y) = P(Y ≤ y) = P(exp(U) ≤ y) = P(U ≤ ln(y)), for y > 0.[/tex]

Since U is continuous and its pdf is given by f_U(u), we have:

[tex]F_Y(y) = ∫_{2}^{ln(y)} f_U(u)du = ∫_{2}^{ln(y)} 1du = ln(y) - 2, 2 ≤ ln(y) ≤ 3, and 0 elsewhere.[/tex]

We can differentiate F_Y(y) to find the pdf of Y:

[tex]fy(y) = d/dy F_Y(y) = (1 / y) f_Y(ln(y)) = (1 / y), 2 ≤ y ≤ e^3, and 0 elsewhere.[/tex]

Therefore, the probability density function (pdf) of Y = exp(U) is given by:

fy(y) = (1 / y), 2 ≤ y ≤ e^3, and 0 elsewhere.

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Answer:

I cannot see the picture

Step-by-step explanation: