Can someone really help me!

Answer:

27.5 minutes

Step-by-step explanation:

Using,

R/R' = 2ᵃ/ᵇ------------------ Equation 1

From the equation,

R = mass of Element X before radioactive decay, R' = mass of element X after radioactive decay, a = Time taken, b = half life.

Given: R = 870 grams, R' = 154 grams, b = 11 minutes.

Substitute these values into equation 1

870/154 = 2ᵃ/¹¹

(870/154)¹¹ = 2ᵃ

Solve for a

2ᵃ = (5.649)¹¹

2ᵃ = 187061.26

Taking the logarithm of both side,

Log2ᵃ = Log(187061.26)

⇒ a = log(187061.26)/log2

a = 8.272/0.301

a = 27.5 minutes

Answer: 56840

A bit of long multiplication

Answer:

56840

Step-by-step explanation:

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:

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:

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.

Answer:

diameter 33

radius 66

area 360

Answer:

$436

Step-by-step explanation:

Answer:

5x+2(3x+8)

5x+6x+16

11x+16

11x=-16

X=16/11

Answer:

one cup of coffee costs $2, while one bottle of water costs $1

Step-by-step explanation:

We can represent this using two equations:

5x + 2y = 12

4x + 7y = 15

Your answer for the value of x and y should be

x = 2

y = 1

Therefore, one cup of coffee costs $2, while one bottle of water costs $1.

Answer:

a twelth= 1/12

Step-by-step explanation:

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:

I cannot see the picture

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 ☝️

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:

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:

It will be A

Step-by-step explanation:

6453

- 135

--------

6318 dollars needed for Sarah's food.

Hope this helps!

Answer:

A

I’m pretty sure it’s A

Answer:

A, sure

mark me as brainliest plz

Answer:

a= 126 degrees

b= 54 degrees

r= 54 degrees

s= 126 degrees

Step-by-step explanation:

i think it depends on if you have enough money to buy the books you need so C would be the answer

Answer:

687.549 deg

Step-by-step explanation:

12 rad = 687.549 deg

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 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 words to fill each gap, respectively, are Portal of entry, Vector, Reservoir, Fomite, Incubation period, Portal of exit, Vector.

Portal of entry: It refers to the site or pathway through which an infectious agent enters the body of a susceptible host, allowing the agent to establish an infection.

Vector: A vector is a living organism, usually an insect or animal, that transmits disease-causing pathogens from one host to another. It serves as an intermediate carrier of the infectious agent.

Reservoir: Reservoir refers to a population or environment where a pathogen lives and multiplies, serving as a source of infection. The pathogen depends on the reservoir for its survival and can be transmitted from the reservoir to susceptible individuals.

Fomite: A fomite is an inanimate object or surface that becomes contaminated with disease-causing agents, such as bacteria or viruses. These agents can survive on fomites and potentially transmit infection if they come into contact with a susceptible individual.

Incubation period: The incubation period is the interval between exposure to a pathogen and the appearance of the first signs and symptoms of the disease. It represents the time required for the pathogen to replicate and cause noticeable illness in the infected individual.

Portal of exit: It refers to the site or pathway through which an infectious agent leaves the body of an infected individual, allowing it to be transmitted to other hosts. Examples of portals of exit include respiratory secretions, feces, blood, or skin lesions.

Vector: As mentioned earlier, a vector is a living insect or animal that plays a role in transmitting disease. It can acquire the pathogen from an infected host and transmit it to a susceptible individual, potentially causing disease in the process.

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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:

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]