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azamat
2 years ago
8

Data from a sample of 576 people in a city was used to estimate the mean monthly bill

Mathematics
1 answer:
stiv31 [10]2 years ago
3 0

Using the z-distribution, it is found that the margin of error for a 95% confidence interval is of $0.04.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • \sigma is the standard deviation for the sample.

The margin of error is given by:

M = z\frac{\sigma}{\sqrt{n}}

In this problem, the values of the parameters are:

z = 1.96, \sigma = 0.48, n = 576.

Hence, the <em>margin of error</em>, in dollars, is given by:

M = z\frac{\sigma}{\sqrt{n}}

M = 1.96\frac{0.48}{\sqrt{576}}

M = 0.04.

More can be learned about the z-distribution at brainly.com/question/25890103

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3. A rare species of aquatic insect was discovered in the Amazon rainforest. To protect the species, environmentalists declared
navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

8 0
2 years ago
Help! First one to answer I’ll mark brainliest
Blizzard [7]

Answer:

p=38

Step-by-step explanation:

hope that helps!

3 0
3 years ago
If you make $11.75 an hour and you want to buy a car for $40,000. How many hours would you have to work to pay for that car. Set
Alex787 [66]
11.75x=40000. Answer is 3,404.255319148936 or 3405
7 0
2 years ago
Read 2 more answers
Please help me with this question! Will mark brainliest :)
Tcecarenko [31]

Answer:

Y=-2x + 5

Step-by-step explanation:

The line goes through the point (0,5) and the slope is -2.

4 0
3 years ago
Read 2 more answers
What is the value of x?<br><br><br><br> Enter your answer in the box.
Lapatulllka [165]

We can see that

there are six sides

so, n=6

so, firstly we will find total angle

total angle =(n-2)*180

total angle =(6-2)*180

total angle =720

so, sum of all angles must be 720

so, we get

x+120+100+128+133+112=720

now, we can solve for x

x+593=720

x=127.............Answer



5 0
3 years ago
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