Determine the 1-unit growth factor for f f. [We encourage you to write the expression that represents the value of the growth fa
ctor rather than a rounded decimal. If you choose to use rounded answers, round to five decimal places.]
1 answer:
Answer: The complete question is found in the attachment
Step-by-step explanation:
a) To determine the 1-unit growth factor for <em>f</em>
= 2.1/1.75 = 1.2
b) To determine the 1-unit percent change for <em>f</em>
<em>=100 x 1-1.2/1 = 0.2 x 100= 20%</em>
<em>c) to determine the initial value of f</em>
<em>f(x)= abˣ , b =1.2 a=initial value</em>
<em>at (-1,2.1)</em>
<em>2.1= (1.2)⁻¹ x a</em>
<em>a = 2.1/(1.2)⁻¹ </em>
<em>a = 2.1 x 1.2 = 2.52</em>
<em />
<em>d) to determine the function formula for f</em>
<em>f(x)= abˣ = 2.52(1.2)ˣ</em>
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Answer:
a) 20.61%
b) 21.82%
c) 42.36%
d) 4 withdrawals
Step-by-step explanation:
This situation can be modeled with a binomial distribution, where p = probability of “success” (completing the course) equals 80% = 0.8 and the probability of “failure” (withdrawing) equals 0.2.
So, the probability of exactly k withdrawals in 20 cases is given by
a)
We are looking for
P(0;20)+P(0;1)+P(0;2) =
0.0115292150460685 + 0.0576460752303424 + 0.136909428672063 = 0.206084718948474≅ 0.2061 or 20.61%
b)
Here we want P(20;4)
c)
Here we need
But we already have P(0;20)+P(0;1)+P(0;2) =0.2061 and
d)
For a binomial distribution the <em>expectance </em>of “succeses” in n trials is np where p is the probability of “succes”, and the expectance of “failures” is nq, so the expectance for withdrawals in 20 students is 20*0.2 = <em>4 withdrawals.</em>