answer:First, we calculate the total number of ways to draw 4 marbles from the bag: [ text{Total ways} = binom{17}{4} = frac{17 times 16 times 15 times 14}{4 times 3 times 2 times 1} = 2380 ] Next, calculate the probability of drawing all red, all white, or all blue: - All red: ( binom{3}{4} = 0 ) (impossible to draw 4 reds as there are only 3) - All white: ( binom{6}{4} = 15 ) - All blue: ( binom{8}{4} = 70 ) Now, sum the probabilities: [ P(text{all same color}) = frac{binom{6}{4} + binom{8}{4}}{binom{17}{4}} = frac{15 + 70}{2380} = frac{85}{2380} ] Simplifying the fraction: [ frac{85}{2380} = frac{17}{476} ] Conclusion: [ boxed{frac{17}{476}} ]

answer:(1) From the given data, we have n=10, bar{x} = frac{1}{n} sum_{i=1}^{10} x_i = frac{80}{10} = 8, bar{y} = frac{1}{n} sum_{i=1}^{10} y_i = frac{20}{10} = 2. Next, we calculate the slope of the linear regression equation: hat{b} = frac{sum_{i=1}^{10} x_i y_i - nbar{x}bar{y}}{sum_{i=1}^{10} x_i^2 - nbar{x}^2} = frac{184 - 10 times 8 times 2}{720 - 10 times 8^2} = frac{24}{80} = 0.3 Then, we calculate the intercept of the linear regression equation: hat{a} = bar{y} - hat{b} bar{x} = 2 - 0.3 times 8 = -0.4 Therefore, the linear regression equation is hat{y} = 0.3x - 0.4. Since the slope hat{b} = 0.3 > 0, the variables x and y are positively correlated. (2) To predict the monthly savings of a household with a monthly income of 7 thousand yuan, we substitute x=7 into the linear regression equation: hat{y} = 0.3 x - 0.4 = 0.3 times 7 - 0.4 = 1.7 Therefore, the predicted monthly savings of a household with a monthly income of 7 thousand yuan is boxed{1.7} thousand yuan.

answer:1. **Convert Alicia's new hourly wage to cents**: Since Alicia earns 25 per hour, converting this to cents: [ 25 text{ dollars} times 100 text{ cents/dollar} = 2500 text{ cents} ] 2. **Calculate the new tax deduction in cents**: With the local tax rate now at 2%, we calculate the amount deducted for taxes from Alicia's wage in cents as follows: [ 2% text{ of } 2500 text{ cents} = 0.02 times 2500 = 50 text{ cents} ] 3. **Conclusion**: Thus, 50 cents per hour of Alicia's wages are used to pay local taxes. [ 50 text{ cents} ] The final answer is boxed{mathrm{(C) } 50 text{ cents}}

answer:Let's address each part of the problem step-by-step. Part a) For ( N = 5 ) We need to determine if Aline, who starts the game, or Bruna, who plays second, is in the winning position when there are initially 5 candies. The rules are that each player must eat between 1 and half of the candies on the table. 1. A player wins if they can force the number of candies left to be exactly 1 on their opponent's turn. 2. To force the opponent into a losing position with exactly 2 candies, you need to leave them exactly 2 candies. 3. From a starting position of 5 candies: - If Aline eats 1 candy, there will be 4 candies left. Bruna then can eat up to 2 candies, leaving between 2 and 3 candies. - If Bruna eats 1 candy, she leaves exactly 3 candies (she could eat at most 2 candies). In both scenarios where Aline eats either 1 or 2 candies, Bruna can adjust her moves to force a position where Aline's subsequently forced move will allow Bruna to win. **Conclusion:** Bruna is in the winning position. boxed{text{Bruna}} Part b) For ( N = 20 ) We need to consider how Aline could force Bruna into a losing position. 1. If there are 20 candies initially, Aline plays first. 2. From experience (or previous knowledge), we know losing positions follow numbers of the form ( 2k + 1 ) where ( k ) was a previous losing position. From part a), we established 5 as a losing position. The next losing positions would be compiled similarly: - ( k = 2 rightarrow 2 cdot 2 + 1 = 5 ) - ( k = 5 rightarrow 2 cdot 5 + 1 = 11 ) - ( k = 11 rightarrow 2 cdot 11 + 1 = 23 ) (thus surpassing 20) 3. For Aline to force Bruna into ( 11 ) candies, she must eat: - ( 9 ) candies making ( 20 - 9 = 11 ) With Bruna stuck at 11 candies, following through the sequence will then demonstrate more easily that Bruna’s forced subsequent plays will lose to Aline. **Conclusion:** Aline is in the winning position. boxed{text{Aline}} Part c) For ( 100 < N < 200 ), find ( N ) where Bruna wins We seek all values ( N ) between 100 and 200 ensuring Bruna is in the winning position: Bruna wins if the initial value is a losing position for Aline. 1. The known series follows: 2, 5, 11, 23, 47, 95, 191, ldots 2. Locate values in provided interval ( 100 < N < 200 ): - The value falling within the interval is ( 191 ) from the compiled sequence. **Conclusion:** The values where Bruna wins are constrained to singleton 191 within our bounds. boxed{191}