A vowel will be chosen when a letter is randomly

Ответ:

korban23

25.01.2022

Answer (Assuming you are asking for the probability.):

75%

Step-by-step explanation:

A, E, I, O, and U are the common vowels.

The word "Lieu" contains three vowels. (i,w,u)

Total number of preferred events: 3

There are 4 total letters.

Total number of events: 4

$\frac{preferred}{total} =\frac{3}{4} = 0.75$

Convert into a percent: 0.75 * 100 = 75

There is a 75% chance that a vowel will be chosen when a letter is randomly picked from the word LIEU.

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Hfruit

14.05.2021

Given:

The expression is:

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\left(\_\_\_+\dfrac{1}{19}n\right)+\left(-8+\_\_\_\right)$

$=\_\_\_n+\_\_\_$

To find:

The missing values.

Solution:

We have,

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)$

It can be written as

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\left(\dfrac{11}{19}n+\dfrac{1}{19}n\right)+\left(-8+10\right)$

So, the values for the first two boxes are $\dfrac{11}{19}n$ and 10 respectively.

On simplification, we get

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\left(\dfrac{11+1}{19}n\right)+\left(2\right)$

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\dfrac{12}{19}n+2$

So, the values for third and fourth boxes are $\dfrac{12}{19}$ and 2 respectively.

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