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kakasveta [241]

3 years ago

9

Find the area of the triangle below

2 answers:

Aleksandr [31]3 years ago

7 0

Answer:

$A=73.5\text{ cm}^2$

Step-by-step explanation:

The formula for an area of a triangle is:

$A=\frac{1}{2}bh$

Where b is the base length and h is the vertical height.

From the triangle, we can see that the base is 21 cm while the <em>vertical</em> height is 7. Thus:

$A=\frac{1}{2}(7)(21)$

Multiply:

$A=\frac{1}{2}(147)$

Multiply:

$A=73.5\text{ cm}^2$

And that's our answer :)

ad-work [718]3 years ago

6 0

The answer is 1,102.5, to find the area of a triangle you need to multiply LxWxB/2

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3 years ago

The map shows the intersection of three roads. Malcom Way intersects Sydney Street at an angle of 162°. Park Road intersects Syd

Y_Kistochka [10]

Answer:

The angle at which malcom Way intersects Park Road = 111 °

Step-by-step explanation:

Malcom Way intersects Sydney Street at $\theta_{1}$ = 162 °

Park Road intersects Sydney Street at $\theta_{2}$ = 87 °

The angle at which Malcom Way intersects Park Road $\theta_{3}$ = ?

We know that

$\theta_{1} + \theta_{2} + \theta_{3} = 360$

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6 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

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3 years ago