We start to solve this by looking for the formula that let us calculate an average value, this formula would be:
where Save is the average score and s is the score of each exam, Since we know the value of the score Save and 4 of the exams we can replace them in the formula, like this:
And we can solve for s5, which is the score that we need to find, like this:
Then, Pedro must have a 55 score on the fifth test
Answer:
0.00067969413
Step-by-step explanation:
Move all terms that don't contain K to the right side and solve.
K= Z/3 + 3
Step-by-step explanation:
<h3><u>Given</u><u>:</u><u>-</u></h3>
(√3+√2)/(√3-√2)
<h3><u>To </u><u>find</u><u>:</u><u>-</u></h3>
<u>Rationalised</u><u> form</u><u> </u><u>=</u><u> </u><u>?</u>
<h3><u>Solution</u><u>:</u><u>-</u></h3>
We have,
(√3+√2)/(√3-√2)
The denominator = √3-√2
The Rationalising factor of √3-√2 is √3+√2
On Rationalising the denominator then
=>[(√3+√2)/(√3-√2)]×[(√3+√2)/(√3+√2)]
=>[(√3+√2)(√3+√2)]×[(√3-√2)(√3+√2)]
=>(√3+√2)²/[(√3-√2)(√3+√2)]
=> (√3+√2)²/[(√3)²-(√2)²]
Since (a+b)(a-b) = a²-b²
Where , a = √3 and b = √2
=> (√3+√2)²/(3-2)
=> (√3-√2)²/1
=> (√3+√2)²
=> (√3)²+2(√3)(√2)+(√2)²
Since , (a+b)² = a²+2ab+b²
Where , a = √3 and b = √2
=> 3+2√6+2
=> 5+2√6
<h3><u>Answer:-</u></h3>
The rationalised form of (√3+√2)/(√3-√2) is 3+2√6+2.
<h3>
<u>Used formulae:-</u></h3>
→ (a+b)² = a²+2ab+b²
→ (a-b)² = a²-2ab+b²
→ (a+b)(a-b) = a²-b²
→ The Rationalising factor of √a-√b is √a+√b
Answer:
Data can be defined as a collection of facts or information from which conclusions may be drawn. Example of Data EXAMPLE :The data shown below are Mark's scores on five Math tests conducted in 10 weeks.
Step-by-step explanation:
