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3 years ago

15

The table below shows the number of marbles of different colors in a bag: Color of Marbles Number of Marbles Pink 5 Blue 2 Green

3 Ursula draws a marble from the bag randomly without looking. She then draws another marble from the bag without replacing the first one. Which expression shows the probability of drawing green marbles in both the trials? plz help!!!!

2 answers:

JulijaS [17]3 years ago

7 0

$|\Omega|=10\cdot9=90\\
|A|=3\cdot2=6\\\\
P(A)=\dfrac{6}{90}=\dfrac{1}{15}\approx6,7\%$

Airida [17]3 years ago

6 0

$\frac{3}{10}$ x $\frac{2}{9}$ = $\frac{1}{15}$

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2(n+5)=-2 step by step plz

Elodia [21]

Answer:

n=-6

Step-by-step explanation:

2(n+5)=-2

distribute

2(n+5)

2xn 2x5

2n+10

2n+10=-2

subtract 10 from both sides

10-10=0

-2-10=-12

2n=-12

divide each side by 2

2n/2=n

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6 0

3 years ago

If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

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4 years ago

For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th

timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

$(a,b)=\left \{ x\in\Re : a$

2. Functions

1) $\Re \rightarrow \Re \\ f(x)=x$

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2) $\Re\geq0\rightarrow\Re , f(x)=x+1\\$

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3) $f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)$

The cosine function is not injective, bijective nor surjective.

4) $f:\Re\rightarrow\Re \:f(x)=ex$

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5) Quite unclear format

$6) \:f:\Re\rightarrow(0,\infty), f(x) =ex$

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

$7) f:\Re\geq 0 \rightarrow \Re\geq0, f(x) =x^{4}$

Not injective nor surjective therefore not bijective too.

$8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}$

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Injective (one to one), Surjective, and Bijective.

$10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.$

Surjective.

$11.f:R\geq 0[-1,1], f(x) = 0\\$

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

7 0

3 years ago

What is 0.75×2 please show your work need a answer today

attashe74 [19]

1.50 that should be it and i dont know how to show my work on a computer SORRY.

8 0

3 years ago

Read 2 more answers

The formula a= bf-v1 is used to calculate acceleration as the change in velocity over a period of time. Solve the formula for th

cricket20 [7]

Answer:

The formula for the final velocity, bf in terms of initial velocity, v1, acceleration, a, and time is given as

$bf=v1+a\times t$

Step-by-step explanation:

Given:

$a=\dfrac{bf-v1}{t}$

To Find:

bf = ?

Solution:

Acceleration:

Acceleration is defined as the ratio between the change in velocity of an object and the time it takes for the object to change the velocity.

In formula, this is written as:

$a=\dfrac{v_{f}-v_{i}}{t}$

Where,

'a' is the Acceleration.

'vf' is the final velocity .

'vi' is the initial velocity.

't' is the time it takes for the object to accelerate from vf to vi

Here it is given as

$a=\dfrac{bf-v1}{t}$

So,

Where,

'a' is the Acceleration

'bf' is the final velocity .

'v1' is the initial velocity.

't' is the time it takes for the object to accelerate from bf to v1.

On Solving the above equation 'bf' we get

$a=\dfrac{bf-v1}{t}\\\\a\times t=bf-v1\\\\bf=v1+a\times t$ ....As required

7 0

3 years ago