Mathematics, 12.12.2020 15:50 untouchedyannaa

heights of adults women have a mean of 63.6 in and a standard deviation of 2.5 indies chebyshevs theorem say about the percentage of women with heights between 58.6 and 68.6in

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Answer from: kyleereeves2007

Chebyshev's Theorem (Tchebysheff's Theorem) suggests that this percentage would be greater than $75\%$ .

Step-by-step explanation:

Let be a random variable with a mean of $\mu$ and a (finite) variance of $\sigma$ . By Chebyshev's Theorem, for any constant where k 0 :

$\displaystyle P\left(\left|Y -\mu\right| \le k\, \sigma\right) \ge 1 - \frac{1}{k^2}$ .

In this question, let denote height (in inches.) Accordingly, $\mu = 63.6$ whereas $\sigma = 2.5$ .

Note, that the interval of interest $(58.6,\, 68.6)$ is centered at $\mu = 63.6$ . Besides, the difference between $\mu$ and either endpoints of this interval is equal to 5.0 , which is the same as $2\, \sigma$ .

Rewrite the interval $(58.6,\, 68.6)$ as $(63.6 - 2\times 2.5,\, 63.6 + 2 \times 2.5)$ .

For the height of one such individual, $Y \in (58.6,\, 68.6)$ is equivalent to $\left|Y - 63.6\right| < 2 \times 2.5$ . With $\mu = 63.6$ whereas $\sigma = 2.5$ , that expression is equivalent to $\left|Y - \mu \right| < 2\, \sigma$ .

The percentage of individuals with a height in the interval $(58.6,\, 68.6)$ would be equal to the probability $P(\left|Y - \mu \right| < 2\, \sigma)$ .

Compare this expression to Chebyshev's Theorem. Notice that k = 2 . Therefore:

$\displaystyle P\left(\left|Y -\mu\right| \le 2\, \sigma\right) \ge 1 - \frac{1}{2^2} = 0.75$ .

In other words, at least $75\%$ of the heights would be between 58.6 inches and 68.6 inches.

Answer from: Quest

the system of inequity is

$y\leq -x+4$

$3y> x$

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

the solid line passing through the points (0,4) and (4,0).

$y-4=\frac{0-4}{4-0}(x-0)$

$y=-x+4$

the related equation of solid line is $y=-x+4$ .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

$y-0=\frac{0-1}{3-0}(x-0)$

$3y=x$

the related equation of dashed line is $3y=x$

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

$1=-0+4$

$1=4$

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be $\leq$ . the required inequality is

$y\leq -x+4$

similarly, check the second related equation.

$3(1)=0$

it is a dashed line, so this statement is true if the sign of inequality is > . the required inequality is

$3y> x$

15 points write a system of inequalities to represent the shaded portion of the graph

Answer from: Quest

sorry i got nothing

step-by-step explanation:

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