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domjuan1456
12.07.2019 •
Mathematics
Aresearcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for suvs equipped with the tires. the mean braking distance for suvs equipped with tires made with compound 1 is 7676 feet, with a population standard deviation of 6.56.5. the mean braking distance for suvs equipped with tires made with compound 2 is 8484 feet, with a population standard deviation of 12.912.9. suppose that a sample of 4141 braking tests are performed for each compound. using these results, test the claim that the braking distance for suvs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. let μ1μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. use the 0.050.05 level of significance.
step 2 of 4 :
compute the value of the test statistic. round your answer to two decimal places.
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Ответ:
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Form- 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.
Forma totalment