The linear function for the number of trimmers assembled is:
y = 7 + 4x.
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A linear function has the following format:
![y = ax + b](https://tex.z-dn.net/?f=y%20%3D%20ax%20%2B%20b)
In which:
- a is the rate of change.
- b is the fixed amount.
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- In an earlier shift, 7 trimmers had already been produced, thus 7 is the fixed amount, that is,
. - Diane assembles 4 trimmers per hour, thus the rate of change is 4, that is,
![a = 4](https://tex.z-dn.net/?f=a%20%3D%204)
- The <u>amount of trimmers y produced after x hours</u> is given by:
![y(x) = 4x + 7](https://tex.z-dn.net/?f=y%28x%29%20%3D%204x%20%2B%207)
A similar problem is given at brainly.com/question/16302622
This exponential growth/decay (in this case decay because r<1) of the form:
f=ir^t, f=final value, i=initial value, r=common ratio or "rate", t=time.
Since the population decreases by 4.5% each year the common ratio is:
r=(100-4.5)/100=0.955 so we can say
P(t)=8500(0.955^t)
....
7000=8500(0.955^t)
14/17=(955/1000)^t taking the natural log of both sides
ln(14/17)=t ln(955/1000)
t=ln(14/17)/ln(955/1000)
t≈4.22 years (to nearest hundredth of a year)
Since t is the years since 2010, the population will fall to 7000 in the year (2010+4.22=2014.22, more than four years will have elapsed) 2015.
Total = 14
jhon = 2 / 14 * 20 = 2.8
mina = 5 / 14 * 20 = 7.1
tim = 7 / 14 * 20 = 10
since sweet can't be divided in decimal places so it will be 3, 7, 10
Answer:
The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
Suppose a sample of 333 tankers is drawn. Of these ships, 257 did not have spills.
333 - 257 = 76 have spills.
This means that
80% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 - 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.199](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.228%20-%201.28%5Csqrt%7B%5Cfrac%7B0.228%2A0.772%7D%7B333%7D%7D%20%3D%200.199)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 + 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.257](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.228%20%2B%201.28%5Csqrt%7B%5Cfrac%7B0.228%2A0.772%7D%7B333%7D%7D%20%3D%200.257)
The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).