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3 years ago

20 points! plus brainliest.

2 answers:

4 0

The possibility of getting 2 or more marbles are infinite, but to put it simply. Yes, it is definitely possible

They are both independent. Because they are both separate scenarios

This probably didn’t help, but If it did I am glad （╹◡╹）

hoa [83]3 years ago

4 0

You have to say hello

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Is negative 1 4/5 greater than negative 1 7/10

Yakvenalex [24]

Answer:

Step-by-step explanation:

1 4/5 = 1 8/5 if you make a common denominator

3 0

3 years ago

Given the following constants for a standard form quadratic equation, determine the nature of its roots.

anyanavicka [17]

$ax^2+bx+c=0$
$3x^2+3x+2=0$
$D=b^2-4ac=3^2-4*3*2=9-24=-15$
$D$ , so there are no real solutions.

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3 years ago

Marie can read 22 pages in 30 minutes. How long would it take to read a 100 page book?

True [87]

136 minutes or 2 hours, these are both rounded to the nearest whole number

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3 years ago

Read 2 more answers

Plz help asap i have limited time i will give brainliest

Naddik [55]

Answer:

358

Step-by-step explanation:

In 1995, the boys were 50 %

Take the number of people at the convention in 1995 ( 716)

and multiply by 50%

716 * 50%

716 * .50

358

6 0

3 years ago

Read 2 more answers

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

3 years ago