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question:Evaluate (x a + 3)^2 if the value of x is (a + 9).
answer:Substituting x = a + 9 into the expression (x - a + 3)^2: 1. Replace x with (a + 9) in the expression: [ x - a + 3 = (a + 9) - a + 3 ] 2. Simplify the expression: [ (a + 9) - a + 3 = 9 + 3 = 12 ] 3. Square the result: [ (12)^2 = 144 ] Therefore, (x - a + 3)^2 evaluates to boxed{144} when x = a + 9.
question:a certain sum of money for 3 yrs at 8 percent / annum is 1 / 2 the ci on rs . 4000 for 2 yrs at 10 percent / annum . the sum placed on si is ?
answer:To find the sum placed on simple interest (SI), we first need to calculate the compound interest (CI) on Rs. 4000 for 2 years at 10% per annum. The formula for compound interest is: CI = P [(1 + r/n)^(nt) - 1] Where: P = principal amount (initial investment) r = annual interest rate (in decimal) n = number of times interest is compounded per year t = number of years In this case, since the interest is compounded annually (n = 1), the formula simplifies to: CI = P [(1 + r)^t - 1] Given: P = Rs. 4000 r = 10% per annum = 0.10 t = 2 years Let's calculate the compound interest: CI = 4000 [(1 + 0.10)^2 - 1] CI = 4000 [(1.10)^2 - 1] CI = 4000 [1.21 - 1] CI = 4000 [0.21] CI = Rs. 840 Now, we know that the simple interest on a certain sum of money for 3 years at 8% per annum is half of this compound interest. The formula for simple interest is: SI = P * r * t Where: SI = simple interest P = principal amount (the sum we need to find) r = annual interest rate (in decimal) t = number of years Given: SI = 1/2 * CI = 1/2 * 840 = Rs. 420 r = 8% per annum = 0.08 t = 3 years Let's rearrange the formula to solve for P: P = SI / (r * t) P = 420 / (0.08 * 3) P = 420 / 0.24 P = Rs. 1750 Therefore, the sum placed on simple interest is Rs. boxed{1750} .
question:order to test students' mastery of high school mathematics knowledge, two opaque boxes, Box A and Box B, are prepared. Box A contains 2 conceptual description questions and 2 calculation questions; Box B contains 2 conceptual description questions and 3 calculation questions (all questions are different). Two students, A and B, come to draw questions to answer; each student randomly draws two questions from Box A or Box B one by one. Each student first draws one question to answer, does not put it back after answering, then draws another question to answer (not answering on the question paper). After answering the two questions, the two questions are put back into the original box. (1) If student A draws two questions from Box A, what is the probability that the second question drawn is a conceptual description question? (2) If student A draws two questions from Box A, answers them, and mistakenly puts the questions into Box B. Student B then continues to draw questions to answer from Box B. If he draws two questions from Box B, what is the probability that the first question drawn is a conceptual description question?
answer:# Step-by-Step Solution Part 1 Let's denote A_{i} as "the i-th question drawn from Box A is a conceptual description question", where i=1,2. - Initially, the probability of drawing a conceptual description question as the first question is P(A_{1})=frac{2}{4}=frac{1}{2}, since there are 2 conceptual and 2 calculation questions in Box A. - If the first question drawn is a conceptual description question (A_{1}), then one conceptual question is removed, leaving 1 conceptual and 2 calculation questions. Thus, P(A_{2}|A_{1})=frac{1}{3}. - Conversely, if the first question drawn is not a conceptual description question (overline{A_{1}}), then there are still 2 conceptual and 1 calculation question left. Hence, P(A_{2}|overline{A_{1}})=frac{2}{3}. Therefore, the probability that the second question drawn is a conceptual description question is calculated as follows: [ P(A_{2})=P(A_{1})times P(A_{2}|A_{1})+P(overline{A_{1}})times P(A_{2}|overline{A_{1}})=frac{1}{2}timesfrac{1}{3}+frac{1}{2}timesfrac{2}{3}=frac{1}{2} ] So, the final answer for part (1) is boxed{frac{1}{2}}. Part 2 Let's define the events for student A drawing from Box A: - B_{1}: Drawing two conceptual description questions. - B_{2}: Drawing two calculation questions. - B_{3}: Drawing one conceptual description question and one calculation question. And let C denote the event that student B draws from Box B and the first question is a conceptual description question. Calculating the probabilities for each event: - P(B_{1})=frac{{C}_{2}^{2}}{{C}_{4}^{2}}=frac{1}{6}, since there are 2 ways to choose 2 conceptual questions out of 4 total ways to choose any 2 questions. - P(B_{2})=frac{{C}_{2}^{2}}{{C}_{4}^{2}}=frac{1}{6}, similarly for choosing 2 calculation questions. - P(B_{3})=frac{{C}_{2}^{1}{C}_{2}^{1}}{{C}_{4}^{2}}=frac{2}{3}, for choosing 1 conceptual and 1 calculation question. Given these events, the probabilities that student B draws a conceptual description question first from Box B are: - P(C|B_{1})=frac{{C}_{4}^{1}{C}_{3}^{1}+{A}_{4}^{2}}{{A}_{7}^{2}}=frac{4}{7}, considering the new composition of Box B after B_{1}. - P(C|B_{2})=frac{{C}_{2}^{1}{C}_{5}^{1}+{A}_{2}^{2}}{{A}_{7}^{2}}=frac{2}{7}, after B_{2}. - P(C|B_{3})=frac{{C}_{3}^{1}{C}_{4}^{1}+{A}_{3}^{2}}{{A}_{7}^{2}}=frac{3}{7}, after B_{3}. Thus, the overall probability that student B draws a conceptual description question first from Box B is: [ P(C)=P(B_{1})P(C|B_{1})+P(B_{2})P(C|B_{2})+P(B_{3})P(C|B_{3})=frac{1}{6}timesfrac{4}{7}+frac{1}{6}timesfrac{2}{7}+frac{4}{6}timesfrac{3}{7}=frac{3}{7} ] Therefore, the final answer for part (2) is boxed{frac{3}{7}}.
question:There are 7 eggs in a box and Daniel puts some more eggs inside. Now, there are 11 eggs in the box. How many eggs did Daniel put inside?
answer:Daniel put 11 - 7 = boxed{4} eggs inside the box.