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Sphinxa [80]
4 years ago
9

Find the median of the list 12,18,21,25,30,43

Mathematics
2 answers:
BlackZzzverrR [31]4 years ago
4 0
The median is 21 and 23
Anarel [89]4 years ago
3 0

Answer:

23

21+25=46 ÷2 =23 what you do is cancel out the numbers and find the middle

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Does anyone know how to directly ask a tutor a question? It’s like the option to is gone. Answer asap please!
tia_tia [17]

Answer:

Just find the button to ask a question. Search around for it.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is the pattern in the values as the exponents increase?
uranmaximum [27]

Answer:

Option D is correct.

Step-by-step explanation:

We need to find the pattern in the values as the exponents increase.

First value is 1/3

if we multiply 3 with 1/3 i.e. 1/3*3 we get 1

The next value is 1 in the table.

Now, if we multiply 1 with 3 i.e. 1*3 we get 3

The next value is 3 in the table.

Now, if we multiply 3 with 3 i.e 3*3 we get 9

So, the next value is 9 in the table.

So, if we multiply 3 with the previous value we get the next value in the table.

Hence Option D multiply the previous value by 3 is correct.

6 0
3 years ago
8(3n + 5) = -32<br> what does n =
olga_2 [115]

Answer: n = -3

Step-by-step explanation:

8(3n + 5) = -32

  • <em>Divide both sides by 8.</em>

(3n + 5) = -4

  • <em>Subtract 5 by both sides.</em>

3n = -4 - 5

3n = -9

  • <em>Divide both sides by 3.</em>

n = -3

8 0
3 years ago
The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),
pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}

(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}

When t=0, S(t)=15

15=30+Ce^{-k*0}\\C=15-30\\C=-15

When t = 2, S(t)=20

20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}

Therefore:

S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22

8 0
3 years ago
Solve the following equation for x.
makkiz [27]

Answer:

you're answer is d

Step-by-step explanation:

you make the 9 positive and add 9 to -51 and you get -42 then you divide by 6 and get -7

3 0
4 years ago
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