Ascending order= 6, 8, 12, 12, 12, 16

No of observations (n) = 6
Median = Mean of the values of (n/2) and (n/2+1)
= Mean values of 3rd and 4th values
= 12+12/2
= 12
Therefore, median = 12

Mode = Most number of observations= 12

4 0

3 years ago

Read 2 more answers

Another math question <br>if you answer I will give you a cookie

lina2011 [118]

Answer:

Step-by-step explanation:

Start from left one first

12x & 70

Area as Sum= 12x+70

Area as Product = 12(x+5)

Right side

X & 4

Area as Sum= 5X+20

Area as Product = 5(X+4)

8 0

2 years ago

Find the point of intersection of each pair of lines using substitution

lyudmila [28]

Answer:

ok ok ok ok ok ok o kom ko ko ko ko ko komk ko ko ko k kokok ok ok ok ok ok ok ok ok o-k ok ok

Step-by-step explanation:

3 0

2 years ago

The rate of change of revenue (in dollars per calculator) from the sale of x calculators is R′(x)=(x+1)ln(x+1)R ′ (x)=(x+1)ln(x+

jekas [21]

Answer:

Total revenue = $\frac{169}{2}\ln(13)-\frac{169}{4}$ dollars

Step-by-step explanation:

$R'(x)=(x+1)\ln(x+1)$

Integrate with respect to $x$ .

$R(x)=$ ∫ $(x+1)\ln(x+1)$ dx

Take $x+1=u$

Differentiate with respect to $x$

$dx=du$

So,

$R(x)=$ ∫ $u\ln(u)\,du$

Use integration by parts: ∫f g dx = f∫ g dx-∫(∫g dx)f' dx

Therefore,

$R(x)=$ $\frac{u^2}{2}\ln(u)$ $-$ ∫ $\frac{u^2}{2}\frac{1}{u}\,du$

= $\frac{u^2}{2}\ln(u)$ $-$ ∫ $\frac{u}{2}\,du$

Put $u=x+1$

$R(x)= \frac{(x+1)^2}{2}\ln(x+1)-\frac{(x+1)^2}{4}$

To find total revenue from the sale of the first 12 calculators, put $x=12$

$R(x)= \frac{(12+1)^2}{2}\ln(12+1)-\frac{(12+1)^2}{4}\\\\= \frac{(13)^2}{2}\ln(13)-\frac{(13)^2}{4}\\\\=\frac{169}{2}\ln(13)-\frac{169}{4}$

Total revenue = $\frac{169}{2}\ln(13)-\frac{169}{4}$ dollars

4 0

2 years ago