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Klio2033 [76]
3 years ago
6

Mayor's claim?

Mathematics
1 answer:
Georgia [21]3 years ago
7 0

Answer:

Previously you had said obscenities about someone's name this reflects the help you get- currently you have over 10 questions unanswered doesn't this mean by being funny doesn't help you - facts are I cannot report you as I would lose brainiest  on other questions helped - knowing this where people read this without such explanation given they know straight away their efforts just like mine was- would end up - with no offense intended- would end up in the bin due to your own rude obscenities

Step-by-step explanation:

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PLEASE HELP
True [87]

Answer:A

Step-by-step explanation:

I got it correct on the test

4 0
3 years ago
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
3 years ago
The distance that an object falls from rest, when air resistance is negligible, varies directly as the square of the time that i
eimsori [14]

Since the distance varies directly as the square of time, then its expression looks like this:

d\text{ = k}\cdot t^2

Where d is the distance, "k" is the proportionality constant and t is the time the object is falling. We know that after 6 seconds the stone travels 304 feet. With this information we can determine the value of "k".

\begin{gathered} 304=k\cdot(6)^2 \\ 304=k\cdot36 \\ k=\frac{304}{36} \\ k=8.44 \end{gathered}

Therefore the complete expression is:

d=8.44\cdot t^2

We want to know the distance after 7 seconds, therefore t = 7.

\begin{gathered} d=8.44\cdot(7)^2 \\ d=8.44\cdot49=413.56\text{ feet} \end{gathered}

The stone will travell approximatelly 314 feet in 7 seconds.

7 0
2 years ago
Please help i need to pass
Oduvanchick [21]
5.0000001.0000000
+
900000000.00000
5 0
3 years ago
What is the length of AB
Anettt [7]

Answer:

\boxed{\boxed{\overline{AB}=16}}

Step-by-step explanation:

In the triangle ABC, m∠ABC=90°. Hence, it is right angle triangle.

\tan \theta=\dfrac{p}{b}=\dfrac{AB}{64}  --------1

In the triangle ABD, m∠ABD=90°. Hence, it is right angle triangle.

\cot (90-\theta)=\dfrac{b}{p}=\dfrac{4}{AB}

As \cot (90-\theta)=\tan \theta, so

\tan \theta=\dfrac{b}{p}=\dfrac{4}{AB}  ---------2

From equation 1 and 2,

\Rightarrow \dfrac{AB}{64}=\dfrac{4}{AB}

\Rightarrow AB^2=64\times 4

\Rightarrow AB^2=256

\Rightarrow AB=\sqrt{256}

\Rightarrow AB=16


5 0
3 years ago
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