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3 years ago

Neil works for 3 days a week and earn a salary of$728.56 per week how much does he earn per month

1 answer:

5 0

728.56 x3 (because that is the days he is working) = $2185.68
2185.68 x4 (because that’s how many week are in a month) = $8742.72 per month

Neil will be earning $8742.72 per month

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The_________states that the sum measures of the interior angles of a triangle is 180

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The Triangle sum theorem states that the sum measures of the interior angles of a triangle is 180 degree

Solution:

We know that,

The Triangle Sum Theorem states that if you add all three interior angles, those are the angles inside the triangle, they would always add up to 180 degrees.

It is easy to remember that we add the three angle measurements to get 180 degrees because of the word sum in the name of the theorem. To make it short and sweet:

Consider a triangle with interior angles a, b, and c. Then we can say,

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2 years ago

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

Amanda [17]

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}

n = 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 t confidence interval.

The critical value of t 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 t-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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$\bf \begin{array}{ccll}
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