<img src="https://tex.z-dn.net/?f=12%20%3D%201.35%20%5Csqrt%7Bx%7D%20" id="TexFormula1" title="12 = 1.35 \sqrt{x} " alt="12 = 1.

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satela [25.4K]

3 years ago

35 \sqrt{x} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Artyom0805 [142]3 years ago

4 0

Answer:

(12/1.35)^2 ~= 79.012346

(roughly equal to ~=)

Step-by-step explanation:

divide both sides by 1.35

12/1.35 = √x

square both sides to get rid of √

(12/1.35)^2 ~= 79.012346

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Use conversion factors to find the answer, distance tracked in feet after 12 seconds at 87 miles per hour.

Assoli18 [71]

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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)

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