School A : 750 + 200 = 950 per semester
School B : if u get a 10% discount, then ur actually paying 90%
0.90(1000) = 900 per semester
The best deal would be school B...u would save $ 50 per semester by choosing B over A
Find 0% of 132: <u>0</u>
Find 100% of 132: <u>132</u>
Find 50% of 132: <u>66</u>
Find 25% of 132: <u>33</u>
Find 75% of 132: <u>99</u>
1) Marnie pays $0.06 per kilowatt-hour of electricity.
2)Marnie budgeted $72 for her electricity.
3)You are supposed to find the greatest number of kilowatt-hours Marnie can use and stay within her budget of $72.
Answer: Third quartile =$21.58
Step-by-step explanation:
Let x be value of a stock .
For uniform distribution,
probability density function = 
Let a be the stock value such that P(x<a) =75% or 0.75
![\Rightarrow\ \int ^{a}_{10.82}f(x)\ dx=0.75\\\\\Rightarrow\ \int ^{a}_{10.82}\dfrac{100}{1435}\ dx=0.75\\\\\Rightarrow\dfrac{100}{1435}[x]^{a}_{10.82}=0.75\\\\\Rightarrow \dfrac{100}{1435}(a-10.82)=0.75\\\\\Rightarrow a-10.82=0.75\times\dfrac{1435}{100}\\\\\Rightarrow a-10.82=10.7625\\\\\Rightarrow a= 10.7625+10.82= $$21.5825\approx $$21.58](https://tex.z-dn.net/?f=%5CRightarrow%5C%20%5Cint%20%5E%7Ba%7D_%7B10.82%7Df%28x%29%5C%20dx%3D0.75%5C%5C%5C%5C%5CRightarrow%5C%20%5Cint%20%5E%7Ba%7D_%7B10.82%7D%5Cdfrac%7B100%7D%7B1435%7D%5C%20dx%3D0.75%5C%5C%5C%5C%5CRightarrow%5Cdfrac%7B100%7D%7B1435%7D%5Bx%5D%5E%7Ba%7D_%7B10.82%7D%3D0.75%5C%5C%5C%5C%5CRightarrow%20%5Cdfrac%7B100%7D%7B1435%7D%28a-10.82%29%3D0.75%5C%5C%5C%5C%5CRightarrow%20a-10.82%3D0.75%5Ctimes%5Cdfrac%7B1435%7D%7B100%7D%5C%5C%5C%5C%5CRightarrow%20a-10.82%3D10.7625%5C%5C%5C%5C%5CRightarrow%20a%3D%2010.7625%2B10.82%3D%09%24%2421.5825%5Capprox%09%24%2421.58)
Hence, 75% of all days the stock is below $21.58 or Third quartile =$21.58 .
The answer is 3(2x-5) (x+12)