<span>Cancel <span>r2</span>
</span><span><span><span>(s4t)</span>/(</span><span>s<span>t3)</span></span></span>
2 <span>Use Quotient Rule: <span><span><span>xa/</span><span>xb</span></span>=<span>x^<span>a−b</span></span></span>
</span><span><span>s^<span>4−1</span></span><span>t^<span>1−3</span></span></span>
3 <span>Simplify <span>4−1</span> to 3
</span><span><span>s^3</span><span>t^<span>1−3</span></span></span>
4 <span>Simplify <span>1−3</span> to <span>−2</span>
</span><span><span>s^3</span><span>t^<span>−2</span></span></span>
5 <span>Use Negative Power Rule:<span><span>x^<span>−a</span></span>=<span>1/<span>xa</span></span></span>
</span><span><span>s^3</span>×<span>1/<span>t2</span></span></span>
6 <span>Simplify
</span><span><span>(s3)/(</span><span>t2)</span></span>
Done so the answer is a. then
6
6
h/2(sum of bases) =75=area
h=3*2
Mark brainliest please
Answer:
The probability that a household has at least one of these appliances is 0.95
Step-by-step explanation:
Percentage of households having radios P(R) = 75% = 0.75
Percentage of households having electric irons P(I) = 65% = 0.65
Percentage of households having electric toasters P(T) = 55% = 0.55
Percentage of household having iron and radio P(I∩R) = 50% = 0.5
Percentage of household having radios and toasters P(R∩T) = 40% = 0.40
Percentage of household having iron and toasters P(I∩T) = 30% = 0.30
Percentage of household having all three P(I∩R∩T) = 20% = 0.20
Probability of households having at least one of the appliance can be calculated using the rule:
P(at least one of the three) = P(R) +P(I) + P(T) - P(I∩R) - P(R∩T) - P(I∩T) + P(I∩R∩T)
P(at least one of the three)=0.75 + 0.65 + 0.55 - 0.50 - 0.40 - 0.30 + 0.20 P(at least one of the three) = 0.95
The probability that a household has at least one of these appliances is 0.95
Answer:
aasha here it is.
Step-by-step explanation:
3x−1)(2x+3)
6x2+7x−3
=6x2+9x−2x−3
=3x(2x+3)−1(2x+3)
=(2x+3)(3x−1)
and hiii
<span>2 x^2 y^6 z^5
5 x^4 y^5 z^3
---------------
10 ^6 ^11 ^8</span>