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ioda
3 years ago
6

I NEED HELP ASAP! : An initial population of 175 quail increases at an annual rate of 22%. Write an exponential function to mode

l the quail population. What will the approximate population be after 5 years?
Mathematics
1 answer:
goblinko [34]3 years ago
7 0

Answer:

About 473 quails.

P=175(1.22)^t

Step-by-step explanation:

Exponential equation with P_0 as the initial amount and rate of r is:

P=P_0(1+r)^t

P=175(1+.22)^t

P=175(1.22)^t

After 5 years the population will be:

P=175(1.22)^5

P \approx 472.9739

So the population will be about 473 quails.

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Twin​ brothers, Billy and​ Bobby, can mow their​ grandparent's lawn together in 53 minutes. Billy could mow the lawn by himself
podryga [215]
5353 minutes = 89.22 hours
Maybe the figures should be 53 and 20??

Let's let Billy = A
The formula to use for work problems is
Time = (A*B) / (A+B)

53 = ( (A +20)* A) / (A +20 + A) )
53 = ( A^2 + 20A ) / (2A + 20)
A^2 + 20A -106A -1060
A^2 -86A -1060
A = 96.935
Bobby Mows in 116.935 minutes

3 0
3 years ago
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]

The left and right endpoints of the i-th subinterval, respectively, are

\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8

r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8

for 1\le i\le4, and the respective midpoints are

m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8

  • Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}

  • Midpoint rule

We approximate the area for each subinterval by

M_i=f(m_i)(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}

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We first interpolate the integrand over each subinterval by a quadratic polynomial p_i(x), where

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

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Then the integral is approximately

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

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3 years ago
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elixir [45]
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