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natulia [17]
3 years ago
6

The melt in your mouth Chocolate factory sells three times as many mints as it sells almond bars. It sells half as many almond b

ars as it does caramels. If it sells 3750 cases of mints each month, how many caramels does it sell in one month.
Mathematics
2 answers:
HACTEHA [7]3 years ago
6 0

Answer:

sells 2500 candies per month

Step-by-step explanation:

m = mints

a = almond bar

c = caramels

m = 3a

1/2 c =  a

m = 3750

3a = m

a = m/3

a = 3750/3

a = 1250

1/2 c = 1250

c = 2 * 1250

c = 2500

Yuri [45]3 years ago
5 0

Answer:

The chocolate factory sells 2500 cases of caramel bars

Step-by-step explanation:

first you would look at where it says mints sells 3 times more than almond therefore the number of mint cases 3750 cases would be divided by 3. It then says caramel sell 2 times more than almond so you would take your new number which would be 1,250 cases of almond and multiply that by 2 and you get 2500 cases of caramel

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Sandeep has twice as much as sonia. together they have rs 150. how much money does sonia have.
Nimfa-mama [501]

It says Sandeep has twice as much as Sonia. Thus, let Sandeep's amount be 2x and Sonia's amount be x.

Now, together they have 150.

Thus, the equation;

x+2x=150

Now let's solve the equation;

3x=150

x=150/3

x=50

Thus, Sonia has 50

6 0
3 years ago
One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P
Darya [45]

Answer:

The differential equation for the model is

\frac{dP}{dt}=kP(1-P)

The model for P is

P(t)=\frac{1}{1-0.99e^{t/447}}

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

\frac{dP}{dt}=kP(1-P)

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

Solving the diferential equation

\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}

P(0)=  \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)=  \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447

Then the model for the population infected at time t is:

P(t)=\frac{1}{1-0.99e^{t/447}}

Now, we can calculate t for P(t)=90,000

P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

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