Ask question

Ask question

All categories

3 years ago

7

What is the median of the data? 3 8 5 5.5

2 answers:

ladessa [460]3 years ago

6 0

Answer:

Median= middle number

(Example) 1, 4 ,6, 7 ,12

6 Would be the median here

so, 3, 3, 4, 4, 5, 5, 5, 5, 6, 7, 8,

5 Is the middle number and the median

Step-by-step explanation:

PtichkaEL [24]3 years ago

5 0

Answer:

5

Step-by-step explanation:

<u>3, 3, 4, 4</u>,<u> 5,</u> <em>5, </em><u>5, 5, 6, 7, 8</u>

<u />

Hope this helped! Please leave any questions or concerns in the comments byeee <3

You might be interested in

Evaluate the expression when g=8 and h=7. <br> 4g-h

Lelechka [254]

25 because 4(8)-7 = 32-7 = 25

3 0

2 years ago

Read 2 more answers

Write an EQUATION for the nth term of each arithmetic sequence<br><br> -1,-0.5,0,0.5,...

kaheart [24]

Ninth term is 2.5

you are adding 0.5 each time

-1,-0.5,0,0.5,1,1.5,2,2.5

8 0

2 years ago

This equation shows how the total number of necklaces Denise owns is related to the amount of money she spends on additional nec

PtichkaEL [24]

Answer: n=90

please mark brainliest

6 0

2 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

HELP!!!!!! will give brainlist

stealth61 [152]

Answer:

1. 112 cm

2 135 ft

3 53.67

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers