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2 years ago

I need help please help me! I don’t understand

Mathematics

1 answer:

Lunna [17]2 years ago

3 0

Step-by-step explanation:

Calculate the cost per ounce of each can by dividing the cost of the can by the size of the can.

For example, $0.89 / 10 oz = $0.089/oz.

$\left[\begin{array}{ccc}Size&Cost&Cost\ per\ ounce\\10 oz&\$0.89&\$0.089/oz\\15 oz&\$1.29&\$0.086/oz\\18 oz&\$2.26&\$0.126/oz\\32 oz&\$3.39&\$0.106/oz\end{array}\right]$

The 15oz can has the lowest cost per ounce.

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State how many significant figures 0.03810 has?

guapka [62]

Answer:

Step-by-step explanation:

the figures are

0,0,3,8,1,0

we repeat 0

so we have 4 figures,

0,3,8,1

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2 years ago

Algebra 1 pls ty! Brainilest if correct!

Elodia [21]

The answer would be 3

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2 years ago

If the point (-2, 5) is reflected across the y-axis, its image will be at (-2, -5). true false

Ivan

False, the image will be (+2; 5)

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2 years ago

What is the answer for 2/5/6

qwelly [4]

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3 years ago

Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)

ioda

(1) Looks like the joint density is

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0$

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

$\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1$

$\implies\boxed{c=\dfrac1{32}}$

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

$P(X>2,Y$

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0$

In the second case, we instead first find the marginal density of Y:

$f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0$

Then use the marginal density to compute the conditional density of X given Y:

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y$

6 0

3 years ago