I will mark as the brainiest if you get this correct<br> thx so much xxx

How do you find the distance between -2 and 3 on a number line

Masteriza [31]

If you draw the line and mark -2 and 3 you find that they are 5 apart. By calculation, subtract the lowest from the highest: 3 - -2 = 3+2 = 5

6 0

3 years ago

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What is the answer ?

Softa [21]

Answer:

353

Step-by-step explanation:

355-42=x-40

I just need to subtract 2 from 355 to get x because I subtracted 2 from 42 to get 40.

355-2=353=x

8 0

3 years ago

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2x + y = -2

cupoosta [38]

Answer is A
if x+y=5
y= -x+5
2x + (-x+5) =-2
2x -x +5 = -2
x+5 = -2
x=-7

5 0

3 years ago

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A box contains 12 balls numbered 1 through 12. Two balls are drawn in succession without replacement. If the second ball has the

Nikitich [7]

<h2>Answer with explanation:</h2>

A box contains 12 balls numbered 1 through 12.

Also the ball are drawn without replacement.

a)

The probability that the first ball had a smaller number on it.

i.e. the number on the first ball could be: {1,2,3}

Hence, the probability that the first ball had a smaller number on it is:

$\dfrac{3}{12}=\dfrac{1}{4}=0.25$

b)

The probability that the first ball has a even number on it is:

There are total 6 even numbers {2,4,6,8,10,12}

but 4 can't be considered as it comes on the second draw, so we are left with just 5 balls with even number.

Hence, the probability is:

$\dfrac{5}{12}=0.4166$

8 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.

8 0

3 years ago

I will mark as the brainiest if you get this correct thx so much xxx