Flagstad Inc. presented the following data : Net income $2,790,000 Preferred stock: 57,300 shares outstanding, $100 par, 9% cumulative,

1. United States : 2013 real GDP 15,779.54, 2014 : 16,152.72. Canada : 2013 real GDP 1,738.66, 2014 : 1,781.963. Japan : 2013 real GDP 464,321.4, 2014 : 495,576.94. Italy: 2013 real GDP 1,549.08, 2014 : 1,539.33 5. Australia : 2013 real GDP : 1,473.74, 2014: 1,512.096. United Kingdom : 2013 real GDP : 1,642.37, 2014: 1,690.09Inflation rate:1. United States: 1.7%2. Canada: 2.6% 3. Japan: 1.6%4. italy: 1.0%5. Australia: 0.28 %6. United Kingdom: 1.62 %... Show More

as many as you're willing to give 🙂Step-by-step explanation:... Show More

it is not equal but if you round it, it is 96it is 23.0857142857... Show More

like 1.5kStep-by-step explanation:... Show More

Term 1 = (0.616 × 10⁻⁵)Term 2 = (7.24 × 10⁻⁵)Term 3 = (174 × 10⁻⁵)Term 4 = (317 × 10⁻⁵)(σ ₑ/ₘ) / (e/m) = (499 × 10⁻⁵) to the appropriate significant figures.Explanation:(σ ₑ/ₘ) / (e/m) = (σᵥ /V)² + (2 σᵢ/ɪ)² + (2 σʀ /R)² + (2 σᵣ /r)²mean measurementsVoltage, V = (403 ± 1) V, σᵥ = 1 V, V = 403 VCurrent, I = (2.35 ± 0.01) A σᵢ = 0.01 A, I = 2.35 ACoils radius, R = (14.4 ± 0.3) cmσʀ = 0.3 cm, R = 14.4 cmCurvature of the electron trajectory, r = (7.1 ± 0.2) cm. σᵣ = 0.2 cm, r = 7.1 cmTerm 1 = (σᵥ /V)² = (1/403)² = 0.0000061573 = (0.616 × 10⁻⁵)Term 2 = (2 σᵢ/ɪ)² = (2×0.01/2.35)² = 0.000072431 = (7.24 × 10⁻⁵)Term 3 = (2 σʀ /R)² = (2×0.3/14.4)² = 0.0017361111 = (174 × 10⁻⁵)Term 4 = (2 σᵣ /r)² = (2×0.2/7.1)² = 0.0031739734 = (317 × 10⁻⁵)The relative value of the e/m ratio is a sum of all the calculated terms.(σ ₑ/ₘ) / (e/m)= (0.616 + 7.24 + 174 + 317) × 10⁻⁵= (498.856 × 10⁻⁵)= (499 × 10⁻⁵) to the appropriate significant figures.Hope this Helps!!!... Show More

thx for the pointsExplanation:... Show More

To make the essay more convincing and important. Explanation: Using specific details in an essay is important. Specific details or supporting evidence makes an essay more intriguing and convincing. It keeps the readers to keep on reading as an essay with specific details does not speak in air. When a writer uses supporting evidence in the essay, the readers are able to know the facts also. Essays that are not supported by the evidences are usually unconvincing and boring and vague. In the given passage, the writer has made the use of specific details to make the readers aware of the exactness and the facts about the Great Barrier reef. These supporting details will enable the readers convinced about the essay and will know the importance of it.... Show More

We conclude that the true average percentage of organic matter in such soil is different from 3%.Step-by-step explanation:We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively. Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.Let = true average percentage of organic matter in such soilSO, Null Hypothesis, : = 3%   {means that the true average percentage of organic matter in such soil is equal to 3%}Alternate Hypothesis, : 3%   {means that the true average percentage of organic matter in such soil is different than 3%}The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;                            T.S.  =  ~ where,   = sample mean amount of organic matter = 2.481%               s = sample standard deviation = 1.616%               n = sample of soil specimens = 30So, test statistics  =    ~                                =  -1.759Now, P-value of the test statistics is given by;        P-value = P( > -1.759) = 0.046 or 4.6%If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.... Show More

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.Step-by-step explanation:We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.Let = true average percentage of organic matterSo, Null Hypothesis, : = 3%      {means that the true average percentage of organic matter in such soil is 3%}Alternate Hypothesis, : 3%      {means that the true average percentage of organic matter in such soil is something other than 3%}The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;                          T.S.  =    ~ where, = sample mean percentage of organic matter = 2.481%              s = sample standard deviation = 1.616%             n = sample of soil specimens = 30So, the test statistics =    ~                                      =  -1.76The value of t-test statistics is -1.76.(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.... Show More