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aniked
2 months ago
5

A recent survey concluded that the proportion of american teenagers who have a cell phone is 0.27. the true population proportio

n of american teenagers who have a cell phone is 0.29. for samples of size 1,000 that are selected at random from this population, what are the mean and standard deviation, respectively, for the sampling distribution of the sample proportion of american teenagers who have a cell phone?
Mathematics
1 answer:
Svet_ta [12.7K]2 months ago
4 0

Answer:

The average for the sampling distribution of the sample proportion is 0.29

The standard deviation for this sampling distribution is 0.01435

Step-by-step explanation:

The mean of the sampling distribution for the sample proportion equals the actual population proportion, which is p = 0.29 in this scenario.

The standard deviation for the sampling distribution of the sample proportion is computed as follows;

\sqrt{\frac{p(1-p)}{n} }

Utilizing the provided values;

p = 0.29

1 - p = 0.71

n = 1000

The standard deviation computes to;

\sqrt{\frac{(0.29)(1-0.29)}{1000} } \\

Thus, the standard deviation is 0.01435.

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Answer:

The illustration provided shows the graph

Step-by-step explanation:

Let

x ------> number of times Emma cuts the grass

y ------> hours Emma spends babysitting

it is known that

6x+8y\geq 48 ------> represents the inequality of the scenario

The solution is the shaded region above the solid line where both x and y are positive

The equation for the solid line is represented as 6x+8y=48

The slope of the line is negative m=-\frac{3}{4}

The y-intercept is at the point (0,6) (the y-value when x is zero)

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The illustration provided shows the graph

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2 months ago
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PIT_PIT [12445]
Past due amount = $87.50
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graph representing the function f of x equals 30 multiplied by 0.88 raised to the exponent of x

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Where a signifies the starting quantity (a=30), r is the decay rate (r=0.12) and x denotes the time intervals, measured in days for this case. By inserting the given values into the formula:

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The savings account offering which of these APRs and compounding periods offers the best APY?
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\bf \qquad \qquad \textit{Annual Yield Formula}
\\\\
~~~~~~~~~~~~\textit{4.0784\% compounded monthly}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0784\%\to \frac{4.0784}{100}\to &0.040784\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12
\end{cases}
\\\\\\
\left(1+\frac{0.040784}{12}\right)^{12}-1\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}\\\\


\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2
\end{cases}
\\\\\\
\left(1+\frac{0.040798}{2}\right)^{2}-1\\\\
-------------------------------\\\\


\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0730\%\to \frac{4.0730}{100}\to &0.040730\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{daily, thus 365}
\end{array}\to &365
\end{cases}
\\\\\\
\left(1+\frac{0.040730}{365}\right)^{365}-1
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