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RideAnS [48]
3 years ago
6

For the data below what is the outlier? 46, 39, 61, 38, 50, 65, 60, 45, 64, 55

Mathematics
2 answers:
Lubov Fominskaja [6]3 years ago
8 0
The first 3 choices are not outliers, so the answer is D.
vampirchik [111]3 years ago
4 0
Hello there.

<span>For the data below what is the outlier? 46, 39, 61, 38, 50, 65, 60, 45, 64, 55

</span>D. There are no outliers.
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43.77 as a mixed number in simplest form
valina [46]
First you have to take the whole number (43) and then take the decimal and put it over 100 (77/100).

5 0
2 years ago
What is the common ratio of these numbers written as a fraction <br> 768,480,300,187.5
kolbaska11 [484]

Answer:

Step-by-step explanation:

im pretty sure it is 768/187.5 = about 4

because you are taking the highest number and putting it over the lowest number

4 0
3 years ago
Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot
Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

\frac{dm}{dt} = -\frac{t}{\tau}

Where:

m - Current mass of the isotope, measured in kilograms.

t - Time, measured in days.

\tau - Time constant, measured in days.

The solution of the differential equation is:

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

Where m_{o} is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}

\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }

The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2

The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

The approximate difference in the half-lives of the isotopes is 66 days.

4 0
3 years ago
Read 2 more answers
What is the value of 4 (3) - 17​
yawa3891 [41]

Answer:

4 times 3 is 12 and 12-17 is -5.

Step-by-step explanation:

8 0
2 years ago
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Write 243 as a power of 3
Vlad [161]
Um... Maybe it's 6^3+3^3
5 0
3 years ago
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