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3 years ago

The product of 8 and x is greater than 50.

Mathematics

1 answer:

Vsevolod [243]3 years ago

8 0

Answer:

X would have to be any number greater than 6.25

Step-by-step explanation:

8*7=56, 8*8=64, and so on.

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10. DIG DEEPER! A muffin tray has

Gekata [30.6K]

No, because if you add 184 and 145 you get 329. then you would take 329 and divide it by 60 because that’s how much each mold holds. that comes out to be 5.48 (rounded). you would be able to fill 5 molds, but not enough batter for 6

3 0

2 years ago

Use the Distributive Property to expand 6(-4x + 3y).

Anna11 [10]

Answer:

-24x + 18y

Step-by-step explanation:

6 x -4x = -24x

6 x 3y = 18y

7 0

3 years ago

A recursive rule for an arithmetic sequence is a1=4;an=an−1−3.

Nikitich [7]

You can see the answer from the picture

3 0

3 years ago

Read 2 more answers

John had 77 apples he gave 20 to alex and 15 to jack how many apples does john have left

Diano4ka-milaya [45]

Answer:

John has 42 apples left

Step-by-step explanation:

77-20-15=42

Hope this helps

6 0

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