What is the distance between the points (-3,4) and (4,6)
1 answer:
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Answer:
maximum height reached = 35 feet
and
Step-by-step explanation:
writing linear motion equations
where s is the total displacement, u the initial velocity, t the time travelled, and a is the acceleration.
given u = 48 ft/s, and a = acceleration due to gravity g = -9.8
1 m = 3.28 feet therefore g becomes -9.8×3.28
here negative sigh comes as acceleration due to gravity is in opposite direction of initial velocity.
therefore f(t) becomes
to find max height we should find differentiation of f(t) and equate it to 0
therefore we get 48 = 32.144t
t = 1.49 s
therefore max height f(1.49) = 71.67-36.67 = 35 feet
Answer:
hello your question is incomplete below is the complete question
Case-Control Calculation Enter only numeric values into the answer line. Place a leading 0 before the decimal and remember to round to two decimal places. Note your answers for the next short essay question. Swine Influenza (SI) is a growing concern at the local Broncoville Health Department in Lawrence, Kansas. You are working for a pharmaceutical company which is developing a human vaccine against SI. You are in charge of testing the vaccine’s effectiveness. You need to determine if one or two doses of vaccine is required to provide immunity. You field test and collect the following information about the vaccine status of persons who did and did not contract SI after getting either one or two doses of vaccine. Develop a 2 x 2 table with the following information: Of the 727 cases of SI, 203 received two doses of vaccine. Of 1,239 controls, 771 received two doses of vaccine for SI. What is the likelihood of contracting SI with only one dose of vaccine?
1. Calculate the incidence of persons with one dose of vaccine for SI
2. Calculate the incidence of persons with two doses of vaccine for SI
3. Calculate the odds ratio of developing SI among those with one dose of vaccine for SI
answer : 1) ≈ 0.53 , (2) ≈ 0.21, (3) ≈ 1.12
Step-by-step explanation:
cases of SI = 727
number of SI cases that received two doses of vaccine = 203
number of SI cases that received one dose of vaccine = 727 -203 = 524
controls = 1239
number of controls that received two doses of vaccine = 771
number of controls that received one dose of vaccine = 1239 - 771 = 468
I) The incidence of persons with one dose of vaccine for SI
= 524 / 992 = 0.528 ≈ 0.53
2) The incidence of persons with two doses of vaccine for SI = 203/974 = 0.208 ≈ 0.21
3) The Odds of SI in one dose group
= 524/468 = 1.119 ≈ 1.12
