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almond37 [142]
3 years ago
15

What is the area of this ^2

Mathematics
1 answer:
Ad libitum [116K]3 years ago
8 0
<h2>Question:- Area of the given diagram</h2>

<h2>Answer :- </h2>

Divide the diagram as shown in attachment.

Now first calculate the area of the rectangle having side 1 and 12

So area = length ×breadth

Area = 1× 12 = 12 unit²

Now find the Area of the remaining triangle with sides 12,13,5 units

So

Area =  \frac{1}{2}  \times base  \times height \:

Now put the value as -> base= 5 and height= 12

Area =  \frac{1}{2}  \times base  \times height \:  \\ Area =  \frac{1}{2}  \times 5 \times 12 \\ Area =  \frac{1}{ \cancel2}  \times 5 \times  \cancel{12}^{ \:  \: 6}  \\ Area = 5 \times 6 \\ Area = 30 \: units^{2}

Now add both areas for your answer

Area{ \tiny{1}} = 12 {units}^{2}  \\ Area{ \tiny2} = 30 {units}^{2}  \\ Area{ \tiny{total}} = Area{ \tiny{1}} + Area{ \tiny{2}} \\ Area{ \tiny{total}} = 12 + 30 = 42 {units}^{2}

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he Nassau Bahamas Community College (NBC) claims that students who take statistics spend on average 19 hours a week working. The
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Using the t-distribution, it is found that since the <u>test statistic is greater than the critical value</u> for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.

At the null hypothesis, it is <u>tested if they spend on average 19 hours a week working</u>, that is:

H_0: \mu = 19

At the alternative hypothesis, it is <u>tested if they spend more than 19 hours a week working</u>, that is:

H_1: \mu > 19

We have the <u>standard deviation for the sample</u>, hence, the <em>t-distribution</em> is used.

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

Searching the problem on the internet, it is found that the values of the <em>parameters </em>are:

\overline{x} = 24.2, \mu = 19, s = 12.59, n = 25

Hence, the value of the <em>test statistic</em> is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{24.2 - 19}{\frac{12.59}{\sqrt{25}}}

t = 2.07

The critical value for a <u>right-tailed test</u>, as we are testing if the mean is greater than a value, with a <u>significance level of 0.05</u> and 25 - 1 = <u>24 df</u> is of t^{\ast} = 1.71

Since the <u>test statistic is greater than the critical value</u> for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

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