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3 years ago

4. At Eagle Rock High School, the probability that a student takes theatre and choir is 0.052.

Mathematics

1 answer:

aleksklad [387]3 years ago

6 0

Answer:

Step-by-step explanation:

Conditional probability is:

P(A given B) = P(A and B) / P(B)

Here, P(A and B) = 0.052 and P(B) = 0.17:

P(A given B) = 0.052 / 0.17

P(A given B) = 0.306

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3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

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2 years ago

Fiona invested $1000 at 7% compounded continuously. At the same time, Maria invested $1100 at 7% compounded daily. How long will

Natali [406]

9514 1404 393

Answer:

14,201 years

Step-by-step explanation:

The two compound interest formulas are ...

A = P·e^(rt) . . . . . continuous compounding at rate r for t years

A = P·(1 +r/365)^(365t) . . . . . daily compounding at rate r for t years

We went the amounts to be equal:

1000·e^(0.07t) = 1100·(1+0.07/365)^(365t)

Dividing by 1000(1 +0.07/365)^(365t), we have ...

((e^0.07)/(1+0.07/365)^365)^t = 1.1

The base of the exponential on the left is ...

( e^0.07)/(1+0.07/365)^365 ≈ 1.00000671149321522

Taking logs, we have ...

t×ln(1.00000671149321522) = ln(1.1)

t = ln(1.1)/ln(1.00000671149321522) ≈ 0.09531018/(6.7114704·10^-6)

t ≈ 14,201.09 . . . . . years

It will take about 14,201 years for the investments to be equal.

_____

<em>Additional comment</em>

The investment value at that time will be about $5.269·10^434. (That's a larger number than <em>anything</em> countable in the known universe, including energy quanta.)

These calculations are beyond the ability of many calculators, so might need to be carefully rewritten if the calculator only keeps 10 significant digits, or only manages exponents less than 100.

This shows that daily compounding is very close in effect to continuous compounding. It would take almost 150 years to make a difference of 0.1% in value.

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2 years ago