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grin007 [14]
3 years ago
5

Virus A will damage the system with probability 0.3. Independently of it, virus B will damage the system with probability 0.5. I

ndependently of A and B, virus C will damage the system with probability 0.6. What is the probability that the system will be damaged (by at least one of the viruses)?
Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

0.86.

Step-by-step explanation:

Probability (None of the viruses will do any damage) = (1 - 0.3)(1-0.5)(1-0.6)

= 0.14.

Therefore the probability that at least one will do damage = 1 - 0.14 = 0.86.

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PLEASE HELP WILL MARK BRAINLIEST
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Answer:

You may or may not need to include the units.

A = 18x - 18

P = 6x + 6

Graph is attached below. (2, 18)

Step-by-step explanation:

Substitute the information we need, "l" and "w", into the formulas.

l is for length, 6cm.

w is for width, (3x - 3)cm.

Use the formula for area of a rectangle.

A = lw

A = (6)(3x-3)cm²

A = (18x - 18)cm²  or 18x - 18

Use the formula for perimeter of a rectangle.

P = 2(l + w)

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P = 2(3x + 3)cm

P = (6x + 6)cm  or 6x + 6

Linear equations are written in the form y = mx + b, so we do not need to factor or further simplify the formulas.

To graph, first turn the "m" value into a fraction form.

8 -> 8/1

6 -> 6/1

You need two points to graph each line.

For each equation, the first point is on the y-axis at the "b" value. Then use the "m" in the equation to count the number of units up (numerator) and to the right (denominator).

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3 years ago
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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

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3 years ago
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