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dolphi86 [110]
3 years ago
13

Solve the inequality

Mathematics
1 answer:
GenaCL600 [577]3 years ago
3 0
-1 < c + 2 < 3....subtract 2 from all sections
-1 - 2 < c + 2 - 2 < 3 - 2...simplify
-3 < c < 1
==============================
32 > 16 - 4g > 12....subtract 16 from all sections
32 - 16 > 16 - 16 - 4g > 12 - 16....simplify
16 > -4g > -4 ...now divide all sections by -4, and change inequality signs
16/-4 < (-4/-4)g < -4/-4...simplify
-4 < g < 1
============================
6y + 1 > = 10
6y > = 10 - 1
6y > = 9
y > = 9/6 which reduces to 3/2 or 1 1/2

-3/2y > = 9 ....multiply both sides by -2/3, cancelling the -3/2 on the left...and change the inequality sign
y < = 9 * -2/3
y < = -18/3 which reduces to - 6

so y > = 9/6(or 1 1/2) or y < = -6


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3 years ago
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
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Answer:

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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Answer:

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Step-by-step explanation:

Blue angle: 34°: 180 (sum of angles in a triangle) - 92 - 54 = 34

x = 180 (flat angle) - 34 = 146°

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For the standard normal probability distribution, the area to the right of the mean is:
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