answer:To solve the given definite integrals using Python and SymPy, we'll follow these steps:1. Define the integrals in SymPy.2. Calculate the definite integrals for the given limits.3. Print the results.Let's break down and solve each integral step-by-step. Integral (1): (int_{-1}^{1}x^{3}dx)We'll calculate the definite integral of (x^3) from -1 to 1. Integral (2): (int_{2}^{e+1} frac{1}{x-1}dx)We'll calculate the definite integral of (frac{1}{x-1}) from 2 to (e+1).Here is the Python code to compute these integrals using SymPy:```pythonimport sympy as sp# Define the variablex = sp.symbols('x')# Define the integralsintegral_1 = sp.integrate(x3, (x, -1, 1))integral_2 = sp.integrate(1/(x-1), (x, 2, sp.E + 1))# Print the resultsprint(integral_1)print(integral_2)``````output01```The results of the definite integrals are:1. (int_{-1}^{1} x^3 , dx = 0)2. (int_{2}^{e+1} frac{1}{x-1} , dx = 1)Thus, the results can be boxed as:[ boxed{0} ][ boxed{1} ]These results are accurate.

answer:To solve the problem, we need to determine the probability of drawing two balls from the bag such that one is white and one is red. We'll break down the problem into manageable steps and implement the solution in Python using SymPy. Steps to Solve the Problem:1. Understand the Problem: - We have a total of 15 balls in the bag. - 10 balls are white. - 5 balls are red.2. Define the Probability Formula: The probability of an event is defined as the number of favorable outcomes divided by the total number of possible outcomes.3. Calculate Total Possible Outcomes: The total number of ways to draw 2 balls out of 15 is given by the combination formula ( binom{15}{2} ).4. Calculate Favorable Outcomes: The number of ways to draw one white ball and one red ball can be calculated by multiplying the number of ways to choose one white ball ( binom{10}{1} ) by the number of ways to choose one red ball ( binom{5}{1} ).5. Calculate the Probability: The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes.Let's implement this in Python using SymPy:```pythonimport sympy as sp# Define the number of white and red ballstotal_balls = 15white_balls = 10red_balls = 5# Calculate the total number of ways to draw 2 balls out of 15total_outcomes = sp.binomial(total_balls, 2)# Calculate the number of ways to draw 1 white and 1 red ballfavorable_outcomes = sp.binomial(white_balls, 1) * sp.binomial(red_balls, 1)# Calculate the probabilityprobability = favorable_outcomes / total_outcomes# Display the resultprint(probability)``````output10/21```The probability of drawing two balls from the bag such that one is white and one is red is (boxed{frac{10}{21}}).

answer:To solve the problem, let's break it down step-by-step:1. Determine the number of bugs eaten by the gecko.2. Calculate the number of bugs eaten by the lizard, which is half the number of bugs eaten by the gecko.3. Calculate the number of bugs eaten by the frog, which is three times the number of bugs eaten by the lizard.4. Calculate the number of bugs eaten by the toad, which is 50% more than the number of bugs eaten by the frog.5. Sum the number of bugs eaten by the gecko, lizard, frog, and toad to get the total number of bugs eaten.We can use Python code to calculate the number of bugs eaten by each animal and then find the total.Let's start by writing the code:```python# Number of bugs eaten by the geckogecko_bugs = 12# Number of bugs eaten by the lizard (half as many as the gecko)lizard_bugs = gecko_bugs / 2# Number of bugs eaten by the frog (three times as many as the lizard)frog_bugs = lizard_bugs * 3# Number of bugs eaten by the toad (50% more than the frog)toad_bugs = frog_bugs * 1.5# Total number of bugs eatentotal_bugs = gecko_bugs + lizard_bugs + frog_bugs + toad_bugsprint(total_bugs)``````output63.0```The total number of bugs eaten by the gecko, lizard, frog, and toad is (boxed{63}).

answer:To solve the problem of finding how many ways (47) can be written as the sum of two primes, we'll follow these steps:1. List Primes Less Than 47: Identify all prime numbers less than 47.2. Check Sum Combinations: For each prime number ( p ), check if ( 47 - p ) is also a prime number.3. Count Valid Combinations: Count the number of valid prime pairs ((p, 47 - p)).We can implement this step-by-step approach in Python using the sympy library for prime number verification.```pythonimport sympy as sp# List all primes less than 47primes = list(sp.primerange(1, 47))# Initialize counter for the number of wayscount_ways = 0# Check each prime and see if its complement (47 - p) is also a primefor p in primes: if sp.isprime(47 - p) and (47 - p) >= p: count_ways += 1print(count_ways)``````output0```It appears that the output is 0, meaning there are no ways to write 47 as the sum of two primes. Let's verify the process and see if the output is indeed correct. Step-by-step Verification:1. Prime Numbers Less Than 47: The primes are: [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 ]2. Check Pair Sums: We need to check each prime ( p ) to see if ( 47 - p ) is also a prime: - ( 47 - 2 = 45 ) (45 is not prime) - ( 47 - 3 = 44 ) (44 is not prime) - ( 47 - 5 = 42 ) (42 is not prime) - ( 47 - 7 = 40 ) (40 is not prime) - ( 47 - 11 = 36 ) (36 is not prime) - ( 47 - 13 = 34 ) (34 is not prime) - ( 47 - 17 = 30 ) (30 is not prime) - ( 47 - 19 = 28 ) (28 is not prime) - ( 47 - 23 = 24 ) (24 is not prime) - ( 47 - 29 = 18 ) (18 is not prime) - ( 47 - 31 = 16 ) (16 is not prime) - ( 47 - 37 = 10 ) (10 is not prime) - ( 47 - 41 = 6 ) (6 is not prime) - ( 47 - 43 = 4 ) (4 is not prime)As we can see, none of the prime pairs satisfy the condition. Thus, our Python code was correct.The number of ways to write (47) as the sum of two primes is indeed (boxed{0}).

answer:To find the length of the repeating sequence for the fraction (frac{1}{221}), we can use the following steps:1. Compute the decimal expansion of the fraction (frac{1}{221}).2. Identify and extract the repeating part of the decimal.3. Determine the length of the repeating sequence.Let's use Python to perform these calculations.```pythonfrom decimal import Decimal, getcontext# Set the precision high enough to detect the repeating sequencegetcontext().prec = 1000 # Calculate the decimal expansion of 1/221fraction = Decimal(1) / Decimal(221)decimal_str = str(fraction)[2:] # Remove the "0."# Function to find the length of the repeating sequencedef find_repeating_sequence(s): for i in range(1, len(s)): for j in range(i + 1, len(s)): if s[i:j] == s[j:2*j-i]: return j - i return None# Find and print the repeating sequence lengthrepeating_length = find_repeating_sequence(decimal_str)print(repeating_length)``````output48```The length of the repeating sequence for the fraction (frac{1}{221}) is (boxed{48}).

answer:Fishing in a river with a constant current can be challenging, but it's definitely possible. Here are a few tips: 1) Position yourself downstream so your bait moves with the current, mimicking the natural movement of prey. 2) Use a heavier sinker to keep your bait steady. 3) Cast upstream and allow your bait to float down with the current. Remember, safety is paramount. Always be aware of your surroundings and follow local fishing regulations.