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djyliett [7]
3 years ago
11

Running from the top of a flagpole to a hook in the ground there is a rope that is 13 meters long. If the hook is 5 meters from

the flagpole, how tall is the flagpole?
Mathematics
2 answers:
pickupchik [31]3 years ago
7 0

Answer:

The flagpole  12 meters tall

Step-by-step explanation:

Running from the top of a flagpole to a hook in the ground there is a rope that is 13 meters long

∵ The hook is 5 meters from the flag pole

∵ The flagpole and the ground ⊥ to each other

By using Pythagoras theorem:

hypothenus side is 13m

base is 5m

how tall is the flagpole is x

13^2=x^2+5^2\\\\169=x^2+25\\\\x^2=169-25\\\\x^2=144\\\\x=12m

The flagpole  12 meters tall

bagirrra123 [75]3 years ago
6 0

Answer:

12 meters

Step-by-step explanation:

Looking at the problem we can see that it typifies a right angled triangle. The rope running from the top of the flagpole to the hook on the ground is the hypotenuse of the triangle. Let us call this hypotenuse c. Let the distance between the hook and the foot of the flagpole be b. Let the height of the flagpole be a.

From Pythagoras theorem;

c^2 = a^2 + b^2

a^2= c^2 - b^2

a= √c^2-b^2

From the question

c= 13 metres

b= 5 metres

a= the unknown

a= √c^2-b^2

a= √(13)^2 - (5)^2

a= √169 - 25

a= √144

a= 12 meters

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Step-by-step explanation:

Coefficient of variation is a measure of dispersion, showing the variability of data in relation to the mean.

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Coefficient of variation = 50 / 5

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3 years ago
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Step-by-step explanation:

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The following results come from two independent random samples taken of two populations.
photoshop1234 [79]

Answer:

(a)\ \bar x_1 - \bar x_2 = 2.0

(b)\ CI =(1.0542,2.9458)

(c)\ CI = (0.8730,2.1270)

Step-by-step explanation:

Given

n_1 = 60     n_2 = 35      

\bar x_1 = 13.6    \bar x_2 = 11.6    

\sigma_1 = 2.1     \sigma_2 = 3

Solving (a): Point estimate of difference of mean

This is calculated as: \bar x_1 - \bar x_2

\bar x_1 - \bar x_2 = 13.6 - 11.6

\bar x_1 - \bar x_2 = 2.0

Solving (b): 90% confidence interval

We have:

c = 90\%

c = 0.90

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.90

\alpha = 0.10

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.10/2}

z_{\alpha/2} = z_{0.05}

The z score is:

z_{\alpha/2} = z_{0.05} =1.645

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.645 * \sqrt{0.0735+0.2571}

2.0 \± 1.645 * \sqrt{0.3306}

2.0 \± 0.9458

Split

(2.0 - 0.9458) \to (2.0 + 0.9458)

(1.0542) \to (2.9458)

Hence, the 90% confidence interval is:

CI =(1.0542,2.9458)

Solving (c): 95% confidence interval

We have:

c = 95\%

c = 0.95

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.95

\alpha = 0.05

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.05/2}

z_{\alpha/2} = z_{0.025}

The z score is:

z_{\alpha/2} = z_{0.025} =1.96

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.96 * \sqrt{0.0735+0.2571}

2.0 \± 1.96* \sqrt{0.3306}

2.0 \± 1.1270

Split

(2.0 - 1.1270) \to (2.0 + 1.1270)

(0.8730) \to (2.1270)

Hence, the 95% confidence interval is:

CI = (0.8730,2.1270)

8 0
3 years ago
Help please 25 pts. I will mark. One month Justin rented 2 movies and 5 video games for a total of $32. The next month he rented
umka21 [38]

Answer:

The movie costs $3.50 and the video games cost $5.

Step-by-step explanation:

a = 1 movie

b = 1 video game

2a + 5b = $32

8a + 3b = $43

2a x -4 = -8a

5b x -4 = -20b

32 x -4 = -128

   8a + 3b = $43

-   8a + 20b = $128

--------------------------------

-17b = -85

17b = 85

/17     /17

b = 5

2a + 5(5) = 32

2a + 25 = 32

-       25 = -25

----------------------

2a = 7

/2 = /2

a = 3.5

a = $3.50

b = $5

6 0
3 years ago
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