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3 years ago

What is the slope of the line segment with endpoints at (2,-5) and (-1, -1)?

Mathematics

1 answer:

Advocard [28]3 years ago

3 0

Answer:

The slope of the line segment is $m=-\frac{4}{3}$

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have the points

(2,-5) and (-1, -1)

substitute in the formula

$m=\frac{-1+5}{-1-2}$

$m=\frac{4}{-3}$

$m=-\frac{4}{3}$

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Calculus 283 Chapter 16

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3 years ago

An urn contains 2 red marbles and 3 blue marbles. 1. One person takes two marbles at random from the urn and does not replace th

Ghella [55]

Answer:

A) The best way to picture this problem is with a probability tree, with two steps.

The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.

The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.

If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.

If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.

B) So a person can have a red marble and a blue marble in two ways:

1) Picking the red first and the blue last

2) Picking the blue first and the red last

C) P(R&B) = 3/5 = 60%

Step-by-step explanation:

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3 years ago

What is the mean for the data shown in the dot plot? 4 5 6 10

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Answer:

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3 years ago

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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

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

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

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

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

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

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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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