Which of the following expressions is equivalent

A dealer bought 3 gross buttons at $.38 a dozen. He sold the buttons at 5 cents each. How much was his profit per gross?(12 doze

777dan777 [17]

Answer:

$2.64

Step-by-step explanation:

Selling them at 5 cents each ($0.05), he could sell 1 dozen buttons for

12 * $0.05 = $0.60

As he bought them at $0.38 per dozen, the profit per dozen would be

$0.60 - $0.38 = $0.22

As 12 dozen is 1 gross, the profit per gross would be

12 * $0.22 = $2.64

7 0

3 years ago

Read 2 more answers

Consider a sequence whose first three

snow_tiger [21]

Answer:

Below, depends if 27 is term number 1 or term number 0. Answered for both cases.

Step-by-step explanation:

The most common sequences are arithmetic and geometric, so lets check those first.

Arithmetic first since its the easiest.

to go from 27 to 21 we subtract 6, if we subtract 6 from 21 again we get to 15, which is what we need, so it is indeed arithmetic.

Explicit formula is basically of the form of y=mx+b with an arithmetic sequence. the m is the common difference and b is the first term minus the common difference. so lets fill those in. y = -6x + 33

Then it usually has n as the x and y f(n) so we'll just put those in

f(n) = -6n + 33

This si as long as the first term is labeled as term number 1 and not term number 0. if you have 27 as term 0 instead just make 33 back to 27, so f(n) = -6n + 27

Let me know if this doesn't make sense.

5 0

3 years ago

Which statement about the figure below is true?

LiRa [457]

B

they have limited (3) common points where the planes meet.

4 0

3 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

a.

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

3 years ago

if a committee of 4 is to be chosen at random from the class of 11 students what is the probability of any particular commuter b

Darina [25.2K]

i think the answer is 4/11 because the 4 people has a 1% chance out of the 11 people

7 0

3 years ago