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Airida [17]
3 years ago
7

What's the value of given expression :

sqrt{9} " alt=" \sqrt{9} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
sdas [7]3 years ago
7 0

\sqrt{9}  = 3

\:

{\boxed{\sf{3 \times 3 = 9}}}

shusha [124]3 years ago
4 0

Answer:

\sqrt{9}  =  \sqrt{3 \times 3}  = 3 \\ thank \: you

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Find the value of x if B is the midpoint of AC, AB = 2x + 9 and BC = 37.
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Step-by-step explanation:

B is the midpoint of AC, in other words it is the halfway point.

So A to B should be equal to B to C

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Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
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The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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3 years ago
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The symmetric property of equality.

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