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Alisiya [41]
3 years ago
14

In a survey of 23312331 ​adults, 730730 say they believe in UFOs. Construct a 99 %99% confidence interval for the population pro

portion of adults who believe in UFOs. A 9999​% confidence interval for the population proportion is ​(nothing​,nothing​). ​(Round to three decimal places as​ needed.) Interpret your results. Choose the correct answer below. A. With 9999​% ​probability, the population proportion of adults who do not believe in UFOs is between the endpoints of the given confidence interval. B. The endpoints of the given confidence interval shows that 9999​% of adults believe in UFOs. C. With 9999​% ​confidence, it can be said that the sample proportion of adults who believe in UFOs is between the endpoints of the given confidence interval. D. With 9999​% ​confidence, it can be said that the population proportion of adults who believe in UFOs is between the endpoints of the given confidence interval.

Mathematics
1 answer:
serg [7]3 years ago
8 0

Answer:

0.288 < p < 0.338

Step-by-step explanation:

We have 2331 adults polled and 730 said they believe in UFOs.  We are to construct a 99% confidence interval for the population proportion.  We are told to round to 3 decimal places as needed.  

We need to find p-hat and q-hat

p-hat = 730/2331 = 0.313

q-hat = 1 - p-hat = 1 - 0.313 = 0.687

Since the sample size is greater than 30, we use a z-value for 99% confidence, which is 2.575

See attached photo for the construction of the confidence interval

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2. A 12-ounce bottle of fruit juice costs
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Answer:

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7 0
3 years ago
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The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t,
VashaNatasha [74]

Functions can be used to model real life scenarios

  • The reasonable domain is \mathbf{[0,\infty)}.
  • The average rate of change from t = 0 to 2 is 20 persons per week
  • The instantaneous rate of change is \mathbf{V'(t) = -3t^2 + 2t + 12}.
  • The slope of the tangent line at point (2,V(20) is 10
  • The rate of infection at the maximum point is 8.79 people per week

The function is given as:

\mathbf{V(t) = -t^3 + t^2 + 12t}

<u>(a) Sketch V(t)</u>

See attachment for the graph of \mathbf{V(t) = -t^3 + t^2 + 12t}

<u />

<u>(b) The reasonable domain</u>

t represents the number of weeks.

This means that: <em>t cannot be negative.</em>

So, the reasonable domain is: \mathbf{[0,\infty)}

<u />

<u>(c) Average rate of change from t = 0 to 2</u>

This is calculated as:

\mathbf{m = \frac{V(a) - V(b)}{a - b}}

So, we have:

\mathbf{m = \frac{V(2) - V(0)}{2 - 0}}

\mathbf{m = \frac{V(2) - V(0)}{2}}

Calculate <em>V(2) and V(0)</em>

\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}

\mathbf{V(0) = (0)^3 + (0)^2 + 12 \times 0 = 0}

So, we have:

\mathbf{m = \frac{20 - 0}{2}}

\mathbf{m = \frac{20}{2}}

\mathbf{m = 10}

Hence, the average rate of change from t = 0 to 2 is 20

<u>(d) The instantaneous rate of change using limits</u>

\mathbf{V(t) = -t^3 + t^2 + 12t}

The instantaneous rate of change is calculated as:

\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}

So, we have:

\mathbf{V(t + h) = (-(t + h))^3 + (t + h)^2 + 12(t + h)}

\mathbf{V(t + h) = (-t - h)^3 + (t + h)^2 + 12(t + h)}

Expand

\mathbf{V(t + h) = (-t)^3 +3(-t)^2(-h) +3(-t)(-h)^2 + (-h)^3 + t^2 + 2th+ h^2 + 12t + 12h}\mathbf{V(t + h) = -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h}

Subtract V(t) from both sides

\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h - V(t)}

Substitute \mathbf{V(t) = -t^3 + t^2 + 12t}

\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h +t^3 - t^2 - 12t}

Cancel out common terms

\mathbf{V(t + h) - V(t)= -3t^2h -3th^2 - h^3  + 2th+ h^2  + 12h}

\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}} becomes

\mathbf{V'(t) = \lim_{h \to \infty} \frac{ -3t^2h -3th^2 - h^3  + 2th+ h^2  + 12h}{h}}

\mathbf{V'(t) = \lim_{h \to \infty} -3t^2 -3th - h^2  + 2t+ h  + 12}

Limit h to 0

\mathbf{V'(t) = -3t^2 -3t\times 0 - 0^2  + 2t+ 0  + 12}

\mathbf{V'(t) = -3t^2 + 2t + 12}

<u>(e) V(2) and V'(2)</u>

Substitute 2 for t in V(t) and V'(t)

So, we have:

\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}

\mathbf{V'(2) = -3 \times 2^2 + 2 \times 2 + 12 = 4}

<em>Interpretation</em>

V(2) means that, 20 people were infected after 2 weeks of the virus spread

V'(2) means that, the rate of infection of the virus after 2 weeks is 4 people per week

<u>(f) Sketch the tangent line at (2,V(2))</u>

See attachment for the tangent line

The slope of this line is:

\mathbf{m = \frac{V(2)}{2}}

\mathbf{m = \frac{20}{2}}

\mathbf{m = 10}

The slope of the tangent line is 10

<u>(g) Estimate V(2.1)</u>

The <em>value of 2.1 </em>is

\mathbf{V(2.1) = (-2.1)^3 + (2.1)^2 + 12 \times 2.1}

\mathbf{V(2.1) = 20.35}

<u />

<u>(h) The maximum number of people infected at the same time</u>

Using the graph, the maximum point on the graph is:

\mathbf{(t,V(t) = (2.361,20.745)}

This means that:

The maximum number of people infected at the same time is approximately 21.

The rate of infection at this point is:

\mathbf{m = \frac{V(t)}{t}}

\mathbf{m = \frac{20.745}{2.361}}

\mathbf{m = 8.79}

The rate of infection is <em>8.79 people per week</em>

Read more about graphs and functions at:

brainly.com/question/18806107

6 0
3 years ago
Round 4806 to the nearest 1000?
grandymaker [24]
If you want to round to the nearest thousand, you cannot look at the thousands place to round.  You will have to look at the number in the hundreds place.  The rules of rounding are 
<em>5 and above-round up.  Example-2743.  7 is greater than 5, so 2743 rounded to the nearest thousand is 3000.
</em>
<em>4 and below-round down.  Example-4390.  Three is less than 5, so 4390 rounded to the nearest thousand is 4000.  
</em>
Apply these rules to your problem.  8 is greater than 5.  So we will round up.  You get 5000.<em>

</em>
7 0
3 years ago
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