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3 years ago

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Math question down below

2 answers:

Anna35 [415]3 years ago

7 0

Eliminate the numbers, one from the start of the list, and one from the end of it, until you get to the middle. If you wind up with two numbers, the median is in between them. (Ex: If you get 33 and 33, the median would be 33.5) If you get one number, then that is your median. Hope this helps!

Alchen [17]3 years ago

4 0

the answer will be 36

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Tu went to the movies. He bought 2 tickets for $9.50 each, popcorn for $5.00 and 2 drinks for $4.25 each. He gave the clerk two

AnnyKZ [126]

Answer:

A

Step-by-step explanation:

A is the only equation that does not account for the fact that Tu bought 2 tickets and 2 drinks.

The rest do: B subtracts the appropriate amounts twice, C multiplies the doubled expenses, and D does the same, just the amount paid is also separated.

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3 years ago

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The GCF (Greatest Common Factor) of 45 and 75. Show your work too. Thanks for your help!

Studentka2010 [4]

$45=3 \cdot 3 \cdot 5 \\
75=3 \cdot 5 \cdot 5$

The GCF is $3\cdot 5 =15$ .

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3 years ago

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It costs $35 per hour to rent a boat at the lake. You also need to pay a $25 fee for safety equipment. You have $200. For how lo

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$200 - $25 = $175

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3 years ago

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The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

Use the distributive property to write an equivalent expression.

Tasya [4]

Answer:

32m + 24n

Step-by-step explanation:

Distribute the 8 to the other terms.

8(4m + 3n)

32m + 24n

6 0

3 years ago