answer:1. Initially, Shirley planned to buy 120 balloons for 10. This implies that the expected price per balloon is: [ text{Expected price per balloon} = frac{10 text{ dollars}}{120 text{ balloons}} = frac{1}{12} text{ dollars per balloon} ]2. The balloons were on sale for 20% less than expected. Therefore, the sale price per balloon is: [ text{Sale price per balloon} = left(1 - 0.20right) times frac{1}{12} text{ dollars per balloon} = 0.80 times frac{1}{12} text{ dollars per balloon} = frac{0.80}{12} text{ dollars per balloon} = frac{2}{30} text{ dollars per balloon} = frac{1}{15} text{ dollars per balloon} ]3. With 10, the number of balloons Shirley can buy at the sale price is: [ text{Number of balloons} = frac{10 text{ dollars}}{frac{1}{15} text{ dollars per balloon}} = 10 times 15 = 150 text{ balloons} ]Therefore, Shirley can buy 150 balloons for her 10.The final answer is boxed{150}.

answer:To solve this problem, we start by defining the common ratio of the geometric sequence {a_{n}} as q. This allows us to express terms of the sequence in terms of q and known values. Given that a_{6}=2, we can find expressions for a_{7}, a_{5}, and a_{9} as follows:- For a_{7}, since it's the next term after a_{6} in a geometric sequence, it can be expressed as a_{7}=a_{6}q=2q.- For a_{5}, being the term before a_{6}, it is found by dividing a_{6} by q, thus a_{5}=frac{a_{6}}{q}=frac{2}{q}.- For a_{9}, two terms after a_{6}, it is a_{6}q^3=2q^3.Given that a_{7}, a_{5}, and a_{9} form an arithmetic sequence, we use the property of arithmetic sequences where the sum of the first and last terms is twice the middle term, leading to the equation a_{7}+a_{9}=2a_{5}. Substituting the expressions for a_{7}, a_{5}, and a_{9}, we get:[2q + 2q^3 = frac{4}{q}]This simplifies to a quartic equation in terms of q:[q^4 + q^2 - 2 = 0]Factoring this equation, we find:[(q^2 + 2)(q^2 - 1) = 0]This gives us two possible sets of solutions for q^2, where q^2 = -2 or q^2 = 1. Since q^2 cannot be negative in this context (as we're dealing with real numbers), we conclude that q^2 = 1.Knowing q^2 = 1, we can find a_{4}, which is one term before a_{5}, thus requiring us to divide a_{6} by q^2. Therefore, a_{4} = frac{a_{6}}{q^2} = frac{2}{1} = 2.Hence, the correct answer is boxed{D}.

answer:We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are 8 coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),white); filldraw(circle((-2,0),0.35),white); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]Then, after the second (new heads in blue);[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]And after the third (new head in green);[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),green); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]So in total, 7 of the 8 coins resulted in heads. Now we have the ratio of frac{7}{8} of the total coins will end up heads. Therefore, we have frac{7}{8}cdot64=boxed{mathbf{(D)} 56}

answer:Let the common difference of the arithmetic sequence {a_n} be d. Since a_1, a_3, a_7 form a geometric sequence, we have:(a_1 + 2d)^2 = a_1(a_1 + 6d).Expanding and rearranging, we get:a_1^2 + 4a_1d + 4d^2 = a_1^2 + 6a_1d,Subtracting a_1^2 from both sides, we have:4a_1d + 4d^2 = 6a_1d,Moving terms with d to one side gives us:4d^2 = 2a_1d,Dividing both sides by 2d (since d neq 0), we find:2d = a_1.Therefore, the common ratio q of the geometric sequence can be calculated as:q = frac{a_1 + 2d}{a_1} = frac{a_1 + a_1}{a_1} = frac{2a_1}{a_1} = 2.So, the correct answer is:boxed{C} quad 2.

answer:The focus of the parabola y^2 = -8x is (-2, 0), and the equation of the directrix is x = 2.Therefore, the radius of the circle with the focus of the parabola y^2 = -8x as its center, and tangent to the directrix of this parabola is 4.Therefore, the equation of the circle with the focus of the parabola y^2 = -8x as its center, and tangent to the directrix of this parabola is (x+2)^2 + y^2 = 16.Hence, the correct answer is boxed{D}.

answer:Since the school has a total of 3200 people and a sample of 160 individuals is selected, we have:- The probability of each individual being selected is frac{160}{3200} = frac{1}{20}.Let T represent the number of teachers in the school. Then, the number of teachers selected in the sample is frac{1}{10}T (since 150 out of 160 individuals are students, the remaining 10 must be teachers).Using the same probability as above, we have:frac{1}{10}T = 10Solving for T, we find:T = 10 times 20 = 200Therefore, the number of teachers in the school is boxed{200}.To solve this problem, we first determined the probability of each individual being selected. Then, using the given information about the number of students selected, we found the number of teachers selected. Finally, we used this information to determine the total number of teachers in the school.