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3 years ago

What is the approximate value of the median? A 18 B 20 C 25 D 27

Mathematics

2 answers:

alexdok [17]3 years ago

6 0

The answer is 25 from what I remember about box plots I’m not shure

Crazy boy [7]3 years ago

3 0

Answer: from what i remember, the median is that line thats right in the center of the box so 25?

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A ramp 18 ft long rises to a platform. The bottom of the platform is 13 ft from the foot of the ramp. Find x, the angle of eleva

Varvara68 [4.7K]

Answer:

The answer to your question is: Ф = 43.95°

Step-by-step explanation:

Data

length = 18 ft

distance bottom to is 13 ft

x = angle = ?

Formula

cos Ф = opposite side / hypotenuse

Process

cos Ф = 13 ft / 18 ft substitution

cos Ф = 0.72 simplify

Ф = cos⁻¹ (0.72)

Ф = 43.95° result

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3 years ago

What is a factor (ﾉ◕ヮ◕)ﾉ I'm dumbヽ(◕ヮ◕ヽ)

Minchanka [31]

it’s basically just any number that can be multiplied with another to get a new expression

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3 years ago

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the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a

Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - First derivative of mass in time, measured in miligrams per year.

$\tau$ - Time constant, measured in years.

$m$ - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

$\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt$

$\ln m = -\frac{1}{\tau} + C$

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of isotope, measured in miligrams.

$t$ - Time, measured in years.

And time is cleared within the equation:

$t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]$

Then, time constant can be found as a function of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $t_{1/2} = 5730\,yr$ and $\frac{m(t)}{m_{o}} = 0.35$ , then:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.35$

$t \approx 8678.505\,yr$

The wood was cut approximately 8679 years ago.

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3 years ago

Which letter on the number line represents 4.55? F E D C

REY [17]

Answer:

Step-by-step explanation:

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3 years ago

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What is this answer?

sergij07 [2.7K]

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3 years ago