All categories

3 years ago

A marble is selected at random from a jar containing 4 red marbles, 2 yellow marbles, and 3 green marbles.

Mathematics

2 answers:

s344n2d4d5 [400]3 years ago

7 0

Answer:

4 / 9 probability of getting a red

Step-by-step explanation:

How many marbles are there total?

4 red + 2 yellow + 3 green = 9 marbles total

P( red marble) = 4 red / 9 total = 4/9

Sophie [7]3 years ago

5 0

Answer:

Step-by-step explanation:

Red marbles- 4

Yellow marbles-2

Green marbles-3

Total marbles=9

Probability= 4/9

=0.4444

You might be interested in

Does someone mind helping me with this problem? Thank you!

Damm [24]

Using the Pythagorean theorem:

c = sqrt(8^2 + 7^2)

c = sqrt(64 + 49)

c = sqrt(113)

c = 10.6

7 0

2 years ago

Read 2 more answers

(-3x+4)+(8x-5)

VikaD [51]

Answer:

answer is 5x-1

its A

hopes this helps u

4 0

3 years ago

Read 2 more answers

What is the average rate of change of the function f(x)=20(14)xf(x)=20(14)x from x = 1 to x = 2? Enter your answer, as a decimal

Advocard [28]

The average rate of a function f(x) over an interval x = a and x = b is given by:

$Average \ Rate \ of \ Change= \frac{f(b)-f(a)}{b-a}$

Given the function

$f(x)=20(14)^x$

The average rate of change from x = 1 to x = 2 is given by:

$\frac{f(2)-f(1)}{2-1} = \frac{20(14)^2-20(14)^1}{1} \\ \\ =20(196-14)=20(182)=3,640$

6 0

3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

When solving for -1/5(x-25)=7 what is the correct sequence of operations

slamgirl [31]

A sequence of operations that will get you to the solution is ...

... d. multiply each side by-5, add 25 to each side

_____

That is not the only correct sequence of operations. However the other ones listed here are incorrect ways to get to x = -10.

You can do whatever multiplication and addition you like to each side of the equation, but if those operations are poorly chosen, they will make finding the solution more difficult.

4 0

3 years ago