PLEASE ANSWERRRRRRRRR

You are creating bracelets and then selling them to your friends at a 85% markup rate. If it costs you $1.80 to create the brace

posledela

Answer:

you will add $1.53 to every bracelet you make

Step-by-step explanation:

figure out what 85 percent of 1.80 is and then add that to 1.80 for the total price

85% x 1.8 = 1.53 (markup ammount)

1.80 + 1.53 = 3.33 (total cost)

5 0

3 years ago

What is 1 + 1 i need help.

Anastasy [175]

Answer:

2

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Twelve percent of the population is left handed. Approximate the probability that there are at least 20 left-handers in a school

DochEvi [55]

Answer:

Step-by-step explanation:

We would assume a binomial distribution for the handedness of the population. Let x be a random variable representing the type of handedness in the population. The probability of success, p is that a randomly chosen person is left handed only. Then probability of failure is that a chosen person is not left handed only(right handed only or both).

p = 12/100 = 0.12

number of success, x = 20

n = 200

the probability that there are at least 20 left-handers is expressed as P(x ≥ 20)

From the binomial probability calculator,

P(x ≥ 20) = 0.84

8 0

3 years ago

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}}$