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3 years ago

(−19, 12), (−9, 1) Find the slope of the line through each pair of points.

Mathematics

2 answers:

Alchen [17]3 years ago

7 0

Answer:

-11/10

Step-by-step explanation:

Slope = rise/run

m = (12-1) / (-19-(-9))

m = -11/10

steposvetlana [31]3 years ago

6 0

To find the slope(m), use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$ And plug in the two points

(-19, 12) = (x₁, y₁)

(-9, 1) = (x₂, y₂)

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{1-12}{-9-(-19)}$ (two negative signs cancel each other out and become positive)

$m=\frac{1-12}{-9+19}$

$m=\frac{-11}{10}$

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3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago