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3 years ago

7

According to the bar graph above, which salary range is the least frequent?

1 answer:

Elis [28]3 years ago

5 0

9514 1404 393

Answer:

D. $101,000 – $120,000

Step-by-step explanation:

The bar graph is not completely labeled, but in the context of the question it seems safe to assume that the vertical scale can be considered to represent relative frequency.

So, the shortest bar is the one with the lowest frequency. The horizontal scale identifies that as 101-120. If we assume that is salary in thousands of dollars, then Choice D is appropriate.

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What is a system of equations for the following situation? There is a total of 30 questions on a test. Some questions are multip

Alexxx [7]

Answer:

m + s = 30

2m + 6s = 100

Step-by-step explanation:

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3 years ago

30 points!!!Find m A) 5 degrees <br> B) 10 degrees<br> C) 30 degrees <br> D) 40 degrees

Neporo4naja [7]

Answer:

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Step-by-step explanation:

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3 years ago

What is the slope of the line that passed through (3,-4) and (4,9)

Advocard [28]

Step-by-step explanation:

<u>Step 1: Find the slope</u>

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3 years ago

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

Monica has a 24-inch square frame. There are 4 equal sides. She paints 3/4 of one side of the frame blue. What fraction of the f

adelina 88 [10]

18/4 of 24 = 3/16

Hope it helped

8 0

3 years ago

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