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7nadin3 [17]
3 years ago
13

An investment website can tell what devices are used to access their site. The site managers wonder whether they should enhance

the facilities for trading via smartphones so they want to estimate the proportion of users who access the site that way. They draw a random sample of
131


investors from their customers. Suppose that the true proportion of smartphone users is


26


​%.


Complete parts​ a) through​ c) below.


​a) What would the managers expect the shape of the sampling distribution for the sample proportion to​ be?


Skewed to the left


Unimodal and symmetric


Skewed to the right


Nonsymmetric


​b) What would be the mean of this sampling​ distribution?


(Round to two decimal places as​ needed.)


​c) What would be the standard deviation of the sampling​distribution?


(Round to three decimal places as​ need
Mathematics
1 answer:
lbvjy [14]3 years ago
6 0

Answer: a)  Unimodal and symmetric

b) 0.26

c) 0.038

Step-by-step explanation:

Given: Sample size of investors (n)= 131

True proportion of smartphone users(p)  =26%

a) Since sampling distribution for the sample proportion is approximately normal when n is larger.

Normal distribution is Unimodal and symmetric.

So correct option : Unimodal and symmetric

b)  mean of this sampling​ distribution = p = 0.26

c)  standard deviation of the sampling​distribution = \sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26\times (1-0.26)}{131}}=\sqrt{0.00146870229008}

=0.0383236518364\approx0.038

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Answer:

<em>(a) x=2, y=-1</em>

<em>(b)  x=2, y=2</em>

<em>(c)</em> \displaystyle x=\frac{5}{2}, y=\frac{5}{4}

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.

We call the determinant of the system

\Delta=\begin{vmatrix}a &b \\c  &d \end{vmatrix}

We also define:

\Delta_x=\begin{vmatrix}p &b \\q  &d \end{vmatrix}

And

\Delta_y=\begin{vmatrix}a &p \\c  &q \end{vmatrix}

The solution for x and y is

\displaystyle x=\frac{\Delta_x}{\Delta}

\displaystyle y=\frac{\Delta_y}{\Delta}

(a) The system to solve is

\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.

Calculating:

\Delta=\begin{vmatrix}1 &1 \\1  &-2 \end{vmatrix}=-2-1=-3

\Delta_x=\begin{vmatrix}1 &1 \\4  &-2 \end{vmatrix}=-2-4=-6

\Delta_y=\begin{vmatrix}1 &1 \\1  &4 \end{vmatrix}=4-3=3

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1

The solution is x=2, y=-1

(b) The system to solve is

\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.

Calculating:

\Delta=\begin{vmatrix}4 &-1 \\1  &-1 \end{vmatrix}=-4+1=-3

\Delta_x=\begin{vmatrix}6 &-1 \\0  &-1 \end{vmatrix}=-6-0=-6

\Delta_y=\begin{vmatrix}4 &6 \\1  &0 \end{vmatrix}=0-6=-6

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2

The solution is x=2, y=2

(c) The system to solve is

\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.

Calculating:

\Delta=\begin{vmatrix}-1 &2 \\1  &2 \end{vmatrix}=-2-2=-4

\Delta_x=\begin{vmatrix}0 &2 \\5  &2 \end{vmatrix}=0-10=-10

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\displaystyle x=\frac{5}{2}, y=\frac{5}{4}

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\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.

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\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7

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4 0
3 years ago
Which value of p makes the equation 18+ 2 (3p – 8) = –37 true?
Volgvan

Answer:

p = -6.5

Step-by-step explanation:

<u>Given:</u>

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<u>Solving for p:</u>

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Option 2 is correct in the list

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