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cupoosta [38]
3 years ago
5

Four children randomly line up, single file. What is the probability that they are in height order, with the shortest child in f

ront? All of the children are of different heights.
A 1/4
B 1/256
C 1/16
D 1/24
Mathematics
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer:

the answer is D.  1/24

Step-by-step explanation:

The number of ways for the children to line up is 4!=4×3×2×1=24 -- there are 4 choices for who is first in line, then 3 for who is second, etc. Only one of these lines has the children in the order specified.

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Find the intercepts of the line.y=−5x−17
GaryK [48]

Answer:

Step-by-step explanation:

y = -5x - 17

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y = mx + b

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u can also find the y intercept by subbing in 0 for x and solving for y

u can find the x intercept by subbing in 0 for y and solving for x

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3 years ago
Write each expression as an algebraic​ (nontrigonometric) expression in​ u, u &gt; 0.
max2010maxim [7]

Answer:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

Step-by-step explanation:

We want to write the trignometric expression:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)

Take the secant of both sides:

\displaystyle \sec(\theta)=\frac{u}{10}

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}

By substitutition:

\displaystyle= \sin(2\theta)

Using an double-angle identity:

=2\sin(\theta)\cos(\theta)

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)

Simplify. Therefore:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

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2 years ago
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