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question:What is the value of dbinom{8}{2}?

answer:The expression binom{8}{2} represents the number of ways to choose 2 items from a set of 8 distinct items. This can be calculated using the formula binom{n}{k} = frac{n!}{k!(n-k)!}, where n! represents the factorial of n. Plugging in n=8 and k=2 into the formula, we get binom{8}{2} = frac{8!}{2!(8-2)!}. Simplifying, we have binom{8}{2} = frac{8!}{2!6!}. Calculating the factorials, we have binom{8}{2} = frac{8 cdot 7 cdot 6!}{2! cdot 6!}. Canceling out the common factors, we are left with binom{8}{2} = frac{8 cdot 7}{2 cdot 1}. Evaluating the expression, we find binom{8}{2} = frac{56}{2} = boxed{28}. The answer is: 28

question:Let x and y be positive real numbers such that 4x + 9y = 60. Find the maximum value of xy.

answer:By AM-GM, we have [60 = 4x + 9y ge 2 sqrt{(4x)(9y)} = 12 sqrt{xy},]so sqrt{xy} le 5. Squaring both sides, we get xy le 25. Equality occurs when 4x = 9y and 4x + 9y = 60, which leads to x = 15/2 and y = 10/3, so the maximum value is boxed{25}. The answer is: 25

question:What is the area, in square units, of a trapezoid bounded by the lines y = x, y = 10, y = 5 and the y-axis? Express your answer as a decimal to the nearest tenth.

answer:The trapezoid is bounded by the lines y = x, y = 10, y = 5, and the y-axis. To find the area, we need to find the lengths of the bases and the height. The bases of the trapezoid are the vertical distances between the lines y = x and y = 5, and between the lines y = 10 and the y-axis. The height of the trapezoid is the horizontal distance between the lines y = 10 and y = 5. The lines y = x and y = 5 intersect at the point (5, 5), so the length of the upper base is 5 units. The lines y = 10 and the y-axis intersect at the point (0, 10), so the length of the lower base is 10 units. The lines y = 10 and y = 5 are parallel and have the same x-intercept, so the height of the trapezoid is 5 units. The area of a trapezoid is given by the formula A = frac{1}{2}(b_1 + b_2)h, where b_1 and b_2 are the lengths of the bases and h is the height. Plugging in the values, we have A = frac{1}{2}(5 + 10)(5) = frac{1}{2}(15)(5) = 37.5. Rounded to the nearest tenth, the area of the trapezoid is boxed{37.5} square units. The answer is: 37.5

question:According to the chart shown, what was the average daily high temperature in Addington from September 15th, 2008 through September 19th, 2008, inclusive? Express your answer as a decimal to the nearest tenth. [asy] size(300); defaultpen(linewidth(.7pt)+fontsize(6pt)); int[] highs={49,62,58,57,46}; int[] lows={40,47,45,41,39}; path unithump=(0,0)--(0,1)--(1,1)--(1,0); path unitcorner=(0,1)--(1,1)--(1,0); for(int i = 0; i < 5; ++i) { transform high=shift(5+20*i,0)*scale(5,highs[i]); transform low=shift(10+20*i,0)*scale(5,lows[i]); fill(high*unitsquare,gray); fill(low*unitsquare,lightgray); draw(high*unithump); draw(low*unitcorner); label(Label(string(highs[i])), (7.5+20*i, highs[i]), N); label(Label(string(lows[i])), (12.5+20*i, lows[i]), N); label(Label(format("Sept %i", i+15)), (10+20*i, 0), S); } xaxis(xmin=0, xmax=100, above=true); yaxis(ymin=0, ymax=77, Ticks(Step=10, begin=false, beginlabel=false, step=10, Size=3), above=true); label(rotate(90)*"Temperature (^{circ}F)", (-15,40), fontsize(8pt)); label("Daily High/Low Temperature in Addington", (50,90), fontsize(10pt)); label("(September 15, 2008 -- September 19, 2008)", (50, 85), fontsize(8pt)); label("underline{KEY}", (105, 65), E, fontsize(8pt)); label("High Temp.", (113, 59), E, fontsize(8pt)); label("Low Temp.", (113, 53), E, fontsize(8pt)); filldraw(shift(109,58)*scale(3)*unitsquare, gray, black); filldraw(shift(109,52)*scale(3)*unitsquare, lightgray, black); [/asy]

answer:To find the average daily high temperature, we need to find the sum of the daily high temperatures and divide by the number of days. From the chart, we can see that the daily high temperatures for September 15th to September 19th are 49, 62, 58, 57, and 46. The sum of these temperatures is 49 + 62 + 58 + 57 + 46 = 272. Since there are 5 days, the average daily high temperature is frac{272}{5} = 54.4. Rounded to the nearest tenth, the average daily high temperature is boxed{54.4}.The answer is: 54.4