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3 years ago

How do you find the median I lowkey forgot

Mathematics

1 answer:

QveST [7]3 years ago

3 0

You arrange the numbers from lowest to highest,And you cover the numbers untill your left ith the number in the middle.

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For each hour he babysit,Anderson earns 1$ more than half of carey's hourly rate.Anderson earns 6$ per hour. wich equation can b

Lena [83]

Answer:

The expression to find Carey's hourly rate is: $c = 2*a - 2$ . Carey's hourly rate is $10.

Step-by-step explanation:

Anderson receives 1 more than half of carey's hourly rate. If we call Anderson's rate by "a" and Carey's by "c", we can express this phrase in the following equation:

$a = \frac{c}{2} + 1$

We want to find the Carey's rate, therefore we need to isolate the "c" variable.

$\frac{c}{2} + 1 = a\\\frac{c}{2} = a - 1\\c = 2*a - 2$

Since Anderson earns $6, then we can find Carey's rate:

$c = 2*6 - 2 = 12 - 2 = 10$

6 0

3 years ago

The width of a rectangle is 1m less than half of its length, and the perimeter is 46m. Find the dimensions.

VLD [36.1K]

The width is 15.
The length is 31.

6 0

4 years ago

Twice a number is more than the sum of three times that number and 5

saw5 [17]

Answer:

31 = x

Step-by-step explanation:

First multiply out

2x + 10 = 3x - 21

Now collect terms

10 + 21 = 3x - 2x

Simplify

31 = x

5 0

3 years ago

Find the slope ! Please answer this !!!

Aleksandr-060686 [28]

Answer:

m = 4

Step-by-step explanation:

The m in y= mx+b is the slope.

7 0

4 years ago

Read 2 more answers

The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial

qaws [65]

Answer:

Step-by-step explanation:

Let P be the population of the community

So the population of a community increase at a rate proportional to the number of people present at a time

That is

$\frac{dp}{dt} \propto p\\\\\frac{dp}{dt} =kp\\\\ [k \texttt {is constant}]\\\\\frac{dp}{dt} -kp =0$

Solve this equation we get

$p(t)=p_0e^{kt}---(1)$

where p is the present population

p₀ is the initial population

If the initial population as doubled in 5 years

that is time t = 5 years

We get

$2p_o=p_oe^{5k}\\\\e^{5k}=2$

Apply In on both side to get

$Ine^{5k}=In2\\\\5k=In2\\\\k=\frac{In2}{5} \\\\\therefore k=\frac{In2}{5}$

Substitute $k=\frac{In2}{5}$ in $p(t)=p_oe^{kt}$ to get

$\large \boxed {p(t)=p_oe^{\frac{In2}{5}t }}$

Given that population of a community is 9000 at 3 years

substitute t = 3 in ${p(t)=p_oe^{\frac{In2}{5}t }}$

$p(3)=p_oe^{3 (\frac{In2}{5}) }\\\\9000=p_oe^{3 (\frac{In2}{5}) }\\\\p_o=\frac{9000}{e^{3(\frac{In2}{5} )}} \\\\=5937.8$

<h3>Therefore, the initial population is 5937.8</h3>

7 0

3 years ago