Find the sum of all angles of the quadrilateral if sides are in ratio 2:3:6:9​

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).

Answer:

No coz 2:3 is 2+3=5

1:2 is 1+2=3 so they are not equivalent.

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:

$436

Step-by-step explanation:

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:

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:

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 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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The standard deviation of the fish Zagreus will catch is approximately 0.6833.

Given probability of catching fish by Zagreus in a single run of Hades is as follows: P(0 fish) = 0.10 P(1 fish) = 0.40 P(2 fish) = 0.35 P(3 fish) = 0.15

To calculate the standard deviation of the fish Zagreus will catch, we need to follow these steps:

Find the expected value, µ, of the fish he will catch.

Then, calculate the variance, σ², using the formula:σ² = Σ [(x - µ)² P(x)]

Finally, calculate the standard deviation, σ, which is the square root of the variance.μ = Σ [xP(x)]μ = (0 × 0.10) + (1 × 0.40) + (2 × 0.35) + (3 × 0.15)μ = 0.75

The expected value, µ, of the fish he will catch is 0.75.

To find the variance:σ² = [(0 - 0.75)² × 0.10] + [(1 - 0.75)² × 0.40] + [(2 - 0.75)² × 0.35] + [(3 - 0.75)² × 0.15]σ² = 0.4675

Finally, the standard deviation, σ, is the square root of the variance:σ = √σ²σ = √0.4675σ ≈ 0.6833

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Given: In a single run of hades, zagreus has a 10% chance of catching 0 fish, 40% chance of catching 1 fish, 35% chance of catching 2 fish, and a 15% chance of catching 3 fish. The standard deviation of the fish that Zagreus will catch is 0.49 fish.

The standard deviation of the fish that Zagreus will catch can be calculated using the following formula

σ = sqrt [∑(x-μ)²/N], where σ is the standard deviation, ∑ is the sum of, x is the fish, μ is the mean, and N is the total number of chances.

The mean value of the fish Zagreus is expected to catch is given by:

μ = (0 x 10/100) + (1 x 40/100) + (2 x 35/100) + (3 x 15/100)

μ = 0 + 0.4 + 0.7 + 0.45

μ = 1.55.

Therefore, the mean value of the fish Zagreus will catch is 1.55 fish.

To calculate the standard deviation, we first calculate the deviation of each value from the mean as shown below: Deviation = x - μ

The deviations for each value of fish that Zagreus could catch are: -1.55, -0.55, 0.45, and 1.45.

Now, we can plug in these values into the formula above to calculate the standard deviation as shown below:

σ = sqrt [(-1.55² x 10/100) + (-0.55² x 40/100) + (0.45² x 35/100) + (1.45² x 15/100)]

σ = sqrt [0.24025]

σ = 0.49

Therefore, the standard deviation of the fish that Zagreus will catch is 0.49 fish.

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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:

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: x=3/5

Step-by-step explanation:

The sum of n consecutive integers, where n is an odd positive integer, is divisible by n.

To prove the statement, let's consider a set of n consecutive integers starting from a.

The sum of n consecutive integers can be expressed as:

S = a + (a+1) + (a+2) + ... + (a+n-1)

To find the sum, we can use the formula for the sum of an arithmetic series:

S = (n/2) × (2a + (n-1))

Since n is an odd positive integer, we can represent it as n = 2k + 1, where k is a non-negative integer.

Substituting this value of n into the sum formula, we get:

S = ((2k+1)/2) × (2a + ((2k+1)-1))

Simplifying further:

S = (k+1) × (2a + 2k)

S = 2(k+1)(a + k)

Since k is an integer, (k+1) is also an integer. Therefore, we can rewrite the sum as:

S = 2m(a + k)

Now, we can see that S is divisible by n = 2k + 1, where m = (k+1).

Thus, we have proven that the sum of n consecutive integers, where n is an odd positive integer, is divisible by n.

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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:

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:

a twelth= 1/12

Step-by-step explanation:

Answer:

what is this a different answer orrrr?

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!

Answer:

I cannot see the picture

Step-by-step explanation:

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

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:

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:

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.