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2 years ago

There are 0 bicycles in each bicycle rack if they are a 8

Mathematics

2 answers:

notsponge [240]2 years ago

6 0

0 because there are no bicycles in the bycicle rack

Bezzdna [24]2 years ago

4 0

Yes there is no bikes in the rack hope I helped :)

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A compound experiment consists of drawing at random twice without replacing from a jar with 3 red, 3 white, and 2 black marbles.

Eddi Din [679]

The solution to your problem about the probability of getting first a red and then a black marble is as follows:

(3 red / 8 total) * (2 black / 7 total) = 3/28

Therefore, the probability of getting first a red and then a black marble is 3/28.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

8 0

3 years ago

Determine the number of solutions the system of 1/3y-5=2x and y=3(2x+5)

lorasvet [3.4K]

Answer:

A. (4,1)

Step-by-step explanation:

2x -5y = 3

x -3y = 1

Change the equation to isolate a variable so you can substitute it in the other one.

x = 1+3y so

2(1+3y) -5y = 3

2 + 6y -5y =3

2 + y = 3

y=1

Back to the easy equation we isolated

x = 1+3(1)

x= 4

(4,1) is a point where x=4 and y=1

hendikeps2 and 22 more users found this answer helpful

THANKS

5.0

(9 votes)

3 0

2 years ago

1. Write an equivalent ratio by multiplying: 1 - 4

ahrayia [7]

Answer:

2 4

- _

8 16

Step-by-step explanation:

1 2

- X -

4 2

1. 4

- X -

4. 4

7 0

2 years ago

Rewrite the subtraction number sentence as an addition number sentence. 5- (-2)

GrogVix [38]

Answer:

5 + 2

Step-by-step explanation:

We have to rewrite the given statement in addition form.

The integers have property of:

Negative(-) Negative(-) = Positive(+)

Positive(+) Positive(+) = Positive(+)

Positive(+) Negative(-) = Negative(-)

Negative(-) Positive(+) = Negative(-)

The given statement is:

5-(-2)

Since we have two negative together, it is converted into a positive.

Thus, the given statement can be written in positive form as

5 + 2

4 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of

marusya05 [52]

The Lagrangian is

$L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)$

It has critical points where the first order derivatives vanish:

$L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}$

$L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y$

$L_\lambda=x^2+y^2-4=0$

From the first two equations we get

$-\dfrac1{2x}=-\dfrac4y\implies y=8x$

Then

$x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}$

At these critical points, we have

$f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125$ (maximum)

$f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125$ (minimum)

5 0

3 years ago