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3 years ago

Do bank tellers need a community college degree?please help me

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

5 0

Answer:

NO: FALSE

Step-by-step explanation:

becauase I am just that smart.

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What is the answer to 2z+12=30

Romashka [77]

It’s z=9 my works here

5 0

3 years ago

Read 2 more answers

Need answer thanks! only answer if you know please!

Vaselesa [24]

Answer:

a) $k \approx 0.0602\,\frac{1}{min}$ ( $k\approx 6.02\,\frac{\%}{min}$ ), b) $k \approx 0.0648\,\frac{1}{min}$ ( $k \approx 6.48\,\frac{\%}{min}$ ), c) $t_{1/2} \approx 60.274\,days$ , d) $t_{1/2}\approx 4.305\,min$ , e) $k \approx 0.3\,\frac{1}{min}$ ( $k\approx 30\,\frac{\%}{min}$ ), f) $k \approx 0.0120\,\frac{1}{min}$ ( $k \approx 1.20\,\frac{\%}{min}$ ), g) $k \approx 2.876\times 10^{-5}\,\frac{1}{yr}$ ( $k \approx 2.876\times 10^{-3}\,\frac{\%}{yr}$ )

Step-by-step explanation:

All radioactive isotopes decay exponentially, the mass of the isotope as a function of time ( $m(t)$ ), measured in grams, is defined by:

$m(t) = m_{o}\cdot e^{-k\cdot t}$ (1)

Where:

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

$k$ - Decay rate, measured in $\frac{1}{min}$ or $\frac{1}{day}$ or $\frac{1}{yr}$ .

$t$ - Time, measured in minutes, days or years.

We define the decay rate in terms of half-life by using the following expression:

$k = \frac{\ln 2}{t_{1/2}}$ (2)

Where $t_{1/2}$ is the half-life of the isotope, measured in minutes or years.

Now we proceed to determine the missing values:

a) Polonium-200 ( $t_{1/2} = 11.5\,min$ )

$k = \frac{\ln 2}{11.5\,min}$

$k \approx 0.0602\,\frac{1}{min}$

b) Lead-194 ( $t_{1/2} = 10.7\,min$ )

$k = \frac{\ln 2}{10.7\,min}$

$k \approx 0.0648\,\frac{1}{min}$

c) Iodine-125 ( $k = 0.0115\,\frac{1}{day}$ )

$t_{1/2} = \frac{\ln 2}{k}$

$t_{1/2} = \frac{\ln 2}{0.0115\,\frac{1}{day} }$

$t_{1/2} \approx 60.274\,days$

d) Kryption-75 ( $k = 0.161\,\frac{1}{min}$ )

$t_{1/2} = \frac{\ln 2}{k}$

$t_{1/2} = \frac{\ln 2}{0.161\,\frac{1}{min} }$

$t_{1/2}\approx 4.305\,min$

e) Strontium-79 ( $t_{1/2} = 2.3\,min$ )

$k = \frac{\ln 2}{2.3\,min}$

$k \approx 0.3\,\frac{1}{min}$

f) Uranium-229 ( $t_{1/2} = 58\,min$ )

$k = \frac{\ln 2}{58\,min}$

$k \approx 0.0120\,\frac{1}{min}$

g) Plutonium-239 ( $t_{1/2} = 24100\,yr$ )

$k = \frac{\ln 2}{24100\,yr}$

$k \approx 2.876\times 10^{-5}\,\frac{1}{yr}$

6 0

3 years ago

What is the volume of the Sphere below?

marysya [2.9K]

$\bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}\quad 
\begin{cases}
r=radius\\
-----\\
r=3
\end{cases}\implies V=\cfrac{4\pi 3^3}{3}\implies V=36\pi$