answer:To determine the percentage composition of nitrogen in a fertilizer sample using elemental analysis, you will need to follow these steps: 1. Obtain a representative sample of the fertilizer. 2. Perform an elemental analysis on the sample using an appropriate method, such as the Kjeldahl method, Dumas method, or CHN analyzer. 3. Calculate the percentage composition of nitrogen in the sample. For example, let's say you performed an elemental analysis on a fertilizer sample and found that it contains 5 grams of nitrogen in a 20-gram sample. To calculate the percentage composition of nitrogen, use the following formula: Percentage composition of nitrogen = (mass of nitrogen / total mass of sample) x 100 Percentage composition of nitrogen = (5 grams / 20 grams) x 100 = 25% So, the percentage composition of nitrogen in the fertilizer sample is 25%.

answer:First, apply the rule that a number and the sum of its digits are congruent modulo any integer. In this case, we need to find the sum of the digits of 7835 and then find the remainder when that sum is divided by 11. Step 1: Calculate the sum of the digits of 7835. [ 7 + 8 + 3 + 5 = 23 ] Step 2: Determine the remainder when 23 is divided by 11. [ 23 div 11 = 2 text{ remainder } 1 ] Thus, the remainder when 7835 is divided by 11 is boxed{1}.

answer:To determine if the sample mean is significantly different from the known population mean, we'll perform a one-sample t-test. First, let's state the null and alternative hypotheses: - The null hypothesis (H0) is that the sample mean is equal to the population mean (80.3 = 78.7). - The alternative hypothesis (H1) is that the sample mean is not equal to the population mean (80.3 ≠ 78.7). Using the given data: - Sample mean (x̄) = 80.3 - Sample standard deviation (s) = 4.8 - Sample size (n) = 15 - Population mean (μ) = 78.7 The t-statistic is calculated as: t = (x̄ - μ) / (s / sqrt(n)) t = (80.3 - 78.7) / (4.8 / sqrt(15)) t ≈ 1.6 / 1.24 t ≈ 1.29 The degrees of freedom for a one-sample t-test is n - 1, which is 14. Using a t-distribution table or calculator, we find the critical t-value for a two-tailed test at a 0.05 significance level with 14 degrees of freedom is approximately 2.145. Since our calculated t-statistic (1.29) is less than the critical t-value (2.145), we fail to reject the null hypothesis. Therefore, at a significance level of 0.05, we cannot conclude that the sample mean is significantly different from the known population mean. The difference between the sample mean and the population mean may be due to random chance.

answer:In the calculation of the t-statistic, I used the formula: t = (x̄ - μ) / (s / sqrt(n)). The reason for using sqrt(n) is due to the concept of standard error. The standard error (SE) is a measure of the amount of variation or dispersion of a sample statistic, such as the sample mean, from the true population parameter. The standard error of the mean (SEM) is calculated as the sample standard deviation (s) divided by the square root of the sample size (n): SEM = s / sqrt(n). The square root of n is used because the standard error of the mean decreases as the sample size increases. This is known as the law of large numbers, which states that the average of a large sample will be closer to the true population mean than the average of a small sample. When calculating the t-statistic, we're essentially asking how many standard errors away from the population mean our sample mean is. By dividing the difference between the sample mean and the population mean (x̄ - μ) by the standard error (s / sqrt(n)), we're standardizing the difference and allowing for a meaningful comparison. Using sqrt(n) in the denominator helps to account for the effect of sample size on the standard error, which is essential for making accurate inferences about the population mean.