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2 years ago

Suppose that X is a random variable with mean and variance both equal to 20. What can be said about P{0 ≤ X ≤ 40}.

Mathematics

1 answer:

Dennis_Churaev [7]2 years ago

5 0

Answer:

$P(0 \leq X \40) = \frac{19}{20}$

Step-by-step explanation:

We know that:

$P(|X - 20| < 20) + P(|X - 20| \geq 20) = 1$

$P(0 \leq X \leq 40) = P(0-20 \leq X-20 \leq 40-20) = P(-20 \leq X \leq 20) = P(|X - 20| \leq 20)$

$P(|X - 20| \leq 20) = 1 - P(|X - 20| > 20) = 1 - \frac{20}{20^{2}} = \frac{19}{20}$

$P(0 \leq X \40) = \frac{19}{20}$

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If Angle 6 is congruent to angle 10 and Angle 5 is congruent to angle 7, which describes all the lines that must be parallel? Li

Nina [5.8K]

Answer:

Third option: Lines "r" and "s" and lines "t" and "u" must be parallel.

Step-by-step explanation:

The missing figure is attached.

You need to remember that:

1- A Transversal is defined as a line that intersects two or more lines.

2- When a transversal cut two parallel lines, several angles are formed, which are grouped in pairs. Some of them are:

a. Vertical angles: are those pairs of angles that share the same vertex and are opposite to each other. These angles are congruent.

b. Corresponding angles: are those pairs of non-adjacent angles located on the same side of the transversal and outside the parallel lines. They are congruent.

In this case, you can identify in the figure that:

$\angle 6$ and $\angle 10$ are Corresponding angles.

$\angle5$ and $\angle 7$ are Vertical angles.

Therefore, based on the explained before, you can conclude that lines "r" and "s" and lines "t" and "u", must be parallel.

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2 years ago

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2 years ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .