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larisa86 [58]
1 year ago
5

What is the value of x? Enter your answer as a decimal to the nearest tenth in the box.

Mathematics
2 answers:
gogolik [260]1 year ago
7 0

ANSWER

23.6 feet

EXPLANATION

The given triangle is a right triangle.

The hypotenuse is 27 ft.

The given angle I s 29°

The unknown side x, is adjacent to the given angle.

We use the cosine ratio to get,

\cos(29 \degree)  =  \frac{adjacent}{hypotenuse}

\cos(29 \degree)  =  \frac{x}{27}

We multiply both sides by 27 to get;

x = 27\cos(29 \degree)

x = 23.6ft

to the nearest tenth.

k0ka [10]1 year ago
3 0

Answer:

x = 23.6

Step-by-step explanation:

<u>Points to remember</u>

<u>Trigonometric ratios </u>

Sin θ  = Opposite side/Hypotenuse

Cos θ = Adjacent side/Hypotenuse

Tan θ = Opposite side/Adjacent side

<u>To find the value of x</u>

From the figure we can see a right angled triangle, ABC

Cos 29 = Adjacent side/Hypotenuse

 = AB/AC

 = x/27

x = 27 * Cos 29

 = 27 * 0.87

 = 23.6

Therefore x = 23.6

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Answer:

1) The linear regression model is y = -0.0348·x + 13.989

2) The correlation coefficient is -0.0725

3) The strength of the model is strong - association

Step-by-step explanation:

1)

                         X            Y          XY       X²

                         27            13           351          729

                         65             12          780         4225

                         83              11          913         6889

                         109            10          1090      11881

                         142            9            1278     20164

                         175              8           1400      30625

                ∑      601              63 5812 74513

From y = ax + b, we have

a = \frac{n\sum xy - \sum x\sum y }{n\sum x^{2}-\left (\sum x  \right )^{2}} = \frac{6 \times 5812  - 601 \times 63}{6 \times 74513-601^{2}} = - 0.0348

b = 1/n(∑y -a∑x) = 1/6(63 - (0.0348)×601) = 13.989

Therefore, the linear regression model is y = -0.0348·x + 13.989

2)

r = \frac{n\sum xy - \sum x\sum y }{\sqrt{[n\sum x^{2}-\left (\sum x  \right )^{2}] [n\sum y^{2}-\left (\sum y  \right )^{2}]}}  = \frac{6 \times 5812  - 601 \times 63}{\sqrt{[6 \times 74513-601^{2}] [6  \times 3969 - 63^2]} } = - 0.0725

3) The strength is - association.

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