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stepan [7]
3 years ago
7

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

dent. (a) If you call 11 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654). (b) If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654). (c) If you call 26 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.
Mathematics
1 answer:
e-lub [12.9K]3 years ago
3 0

Answer with Step-by-step explanation:

Since the given event is binary we can use Bernoulli's probability to sove the problem

Thus for an event 'E' with probability of success 'p' the probability that the event occurs 'r' times in 'n' trails is given by

P(E)=\frac{n!}{(n-r)!\cdot r!}\cdot p^{r}\cdot (1-p)^{n-r}

Part a)

For part a n = 11 , r =9, p = 0.75

Applying values we get

P(E)=\frac{11!}{(11-9)!\cdot 9!}\cdot (0.75)^{9}\cdot (1-0.75)^{11-9}\\\\\therefore P(E)=0.2581

Part b)

For part b n = 20 , r = 16 , p=0.75

Applying values we get

P(E)=\frac{20!}{(20-16)!\cdot 16!}\cdot (0.75)^{16}\cdot (1-0.75)^{20-16}\\\\\therefore P(E)=0.1896

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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)

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\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

It so happens that the integral of p_i(x) reduces nicely to the form you're probably more familiar with,

S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))

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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a homeowner wants to carpet 2 rooms that each measure 12 feet by 15 feet. how many square yards of carpet should be ordered?
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12 x 15 = 180

180 x 2 = 360

360 square yards of carpet should be ordered.
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