Need help, my teacher is not great at explaining.

How many ways can eight different people stand in a circle?

grin007 [14]

Eight people can stand in a cirlce in 6 different arrangements .

3 0

3 years ago

Ai Lun, a management trainee at a large New York based bank, is trying to estimate the real rate of return expected by investors

salantis [7]

Answer:

option B

Step-by-step explanation:

given,

3-month T-bill currently yields = 3%

customer price have been rising = 2% rate

Ai Lun's estimate of the real rate = ?

real rate return

= $\dfrac{1 + risk\ free\ rate}{1 +inflation\ rate} - 1$

= $\dfrac{1 + 0.03}{1 +0.02} - 1$

= $\dfrac{1.03}{1 .02} - 1$

= $1.0098 - 1$

= 0.0098

= 0.98 % ≅ 1 %

hence, the correct answer is option B

6 0

3 years ago

Chas and Tyler have been collecting video games. Chas

marshall27 [118]

Answer:

40

Step-by-step explanation:

m+25=65

65-25=40

m=40

4 0

3 years ago

Over the first five years of owning her car, Gina drove about 12,200 miles the first year, 16,211 miles the second year, 12,050

weeeeeb [17]

(12,200 + 16,211 + 12,050 + 11,350 + 13,325) / 5 = 65136/5 =
13027.2 <== this is the mean (average)

11,350 , 12,050 , 12,200 , 13,325, 16,211
median (middle number) = 12,200

there is no mode...a mode is a number that appears most often...all the numbers appear once in this data.

4 0

4 years ago

Read 2 more answers

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago