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2 years ago

5

I need help!!!!!!!!!!!¡!!!!!!!!!!!¡!!!!!!!!!!¡!!!!!

1 answer:

rusak2 [61]2 years ago

5 0

Can you give brainliest to all of your questions bc it really hard to get brainliest and it be really nice you can choose who i dont really care but just please do it it helps a lot

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Last year Betty's annual salary was $42000. This year she received a raise and now eams $46500. On a weekly basis, how much more

bearhunter [10]

Answer:

About 87$

Step-by-step explanation:

Yu take both salary amounts an divide by 52 since there are 52 weeks in a year and then find the difference between them

3 0

3 years ago

Anyone know the mean.median.and mode of <br> 7,5,0,2,6,4,2

kenny6666 [7]

First put in ascending order.
0, 2, 2, 4, 5, 6, 7
The median is the middle number: 4
The mode is the number that repeats the most often: 2
The mean is the average. Add the numbers and divide by 7.
26/7 = 3.71

7 0

3 years ago

Read 2 more answers

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Tpy6a [65]

Before going to B.two-digit addition

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3 years ago

we bought a dozen of apples. we used half of them to make apple juice. my dad ate two more.how many left?

Aleks [24]

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3 years ago

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