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Diano4ka-milaya [45]

3 years ago

11

At a high school movie night, the refreshments stand sells popcorn and soft drinks. Of the 100 students who came to the movie, 6

2 bought popcorn and 47 bought a drink. 38 students bought both popcorn and a drink. What is the probability that a student buys a drink, given that he or she buys popcorn? Express your answer as a percent, rounded to the nearest tenth... best answer wins brainliest!!!

1 answer:

Olenka [21]3 years ago

3 0

Answer:

47% and 62%

Step-by-step explanation:

1. Drink

The probability that a student buys a drink is 0.47

Step-by-step explanation:

The probability that a student buys a drink will be given by;

( the number of students who bought a drink)/(the total number of students)

We are told that;

Of the 100 students who came to the movie, 62 bought popcorn and 47 bought a drink. Therefore, the required probability is;

47/100= 0.47

0.47 = 47%

2. Popcorn

For popcorn probability, it's basically the same.

The probabilty that a student buys popcorn is 0.62

The probability that a student buys popcorn will be given by;

( the number of students who bought popcorn)/(the total number of students)

So therefore,

62/100 = 0.62

0.62 = 62%

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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

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

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

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

Simpson's rule

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,

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

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

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

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Step-by-step explanation:

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Step-by-step explanation:

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