I suppose you just have to simplify this expression.
(2ˣ⁺² - 2ˣ⁺³) / (2ˣ⁺¹ - 2ˣ⁺²)
Divide through every term by the lowest power of 2, which would be <em>x</em> + 1 :
… = (2ˣ⁺²/2ˣ⁺¹ - 2ˣ⁺³/2ˣ⁺¹) / (2ˣ⁺¹/2ˣ⁺¹ - 2ˣ⁺²/2ˣ⁺¹)
Recall that <em>n</em>ª / <em>n</em>ᵇ = <em>n</em>ª⁻ᵇ, so that we have
… = (2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺³⁾ ⁻ ⁽ˣ⁺¹⁾) / (2⁽ˣ⁺¹⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾)
… = (2¹ - 2²) / (2⁰ - 2¹)
… = (2 - 4) / (1 - 2)
… = (-2) / (-1)
… = 2
Another way to get the same result: rewrite every term as a multiple of <em>y</em> = 2ˣ :
… = (2²×2ˣ - 2³×2ˣ) / (2×2ˣ - 2²×2ˣ)
… = (4×2ˣ - 8×2ˣ) / (2×2ˣ - 4×2ˣ)
… = (4<em>y</em> - 8<em>y</em>) / (2<em>y</em> - 4<em>y</em>)
… = (-4<em>y</em>) / (-2<em>y</em>)
… = 2
Answer:
There are 36 possibilities per character (26 lower case letters + 10 numbers), so the total number of permutations are 36^4 = 1,679,616
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
Given that a teacher prepares 26 tiles with 5 vowels numbered 1 and 21 consonants numbered 2.
The probability for drawing vowel =
Prob for consonant =
If number of trials is atleast 30 we can expect reliable results.
Here the results are recorded for 120 times at random.
Since number of trials is large, we can expect a reliable and accurate results representing the actual probability.
This is because more the number of trials, the less would be the margin of error i.edeviationfrom the expected probability would be minimum
Alright, so we'd use the combinations with repetition formula, so we choose from 4 schools to distribute to and distribute 8 blackboards. It's then
( 8+4-1)!/8!(4-1)!=11!/(3!*8!)=165
For at least one blackboard, we first distribute 1 to each school and then have 4 blackboards left, getting (4+4-1)!/4!(4-1)!=7!/(4!*3!)=35
Answer:
8 units
Step-by-step explanation:
Since both are on the same place in terms of the x-axis, this makes it much easier. Just go from (-1,4) and move upwards upon the graph until your at (-1,12). You counted 8 units. *mic drop*
