Answer:
the answer to this is 7 over 13
Answer
\((y + 3) (y^{2} - 3 y + 9) V)
Explanation
Explanation
Detail ste
Apply \(a^3\pm b^3=(a\pm b)(a^2\ mp ab+b^2)V : \((y + 3) (y^[2} - y \times 3 + 3^{2}) V
Multiply the monomials: \((y + 3) (y^{2} -3y + 3^{23) V
Calculate the power: \((y + 3) (y^{2} - 3 y + 9) V
<span>Members lifted weights: W=32%
</span><span>Members preferred aerobics: A=41%
</span>Members preferred both: B=6%
Members lifted weights only: Wo=W-B=32%-6%→Wo=26%
Members preferred aerobics only: Ao=A-B=41%-6%→Ao=35%
F<span>ind the probability that the member prefers another method neither weights nor aerobics to exercise: N=?
N=100%-Wo-Ao-B
N=100%-26%-35%-6%
N=100%-67%
N=33%
Answer: T</span>he probability that the member prefers another method neither weights nor aerobics to exercise is 33%
10 images per day. Since it can receive 3 mb per second for 11 hours a day, that’s up to 118,800 megabits it can receive in one day. By multiplying the amount of gigabits in a typical picture (11.2) by the amount of megabits in a gigabit (1024) you get that there’s 11,468.8 megabits in each picture. Lastly, divide the number of megs that the station receives in one day by the amount of megs in a picture, and you get 10 and some change, therefore it can receive up to ten FULL pictures in a day
