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2 years ago

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Question 4 Are there any outliers in the data sets? If there are outliers, do they lie at the lower end or at the higher end of

the data sets?
(data sets are provided below!)

Walden’s data set: 2, 2, 2, 3, 3, 3, 3, 4, 4, 4
Drake’s data set: 3, 3, 4, 4, 4, 4, 5, 6, 7, 9

*PLS HELP, THIS IS PASS DUE*

2 answers:

Komok [63]2 years ago

3 0

Answer:

The center of Walden’s data, 3, is less than the center of Drake’s data set, 6.

Step-by-step explanation:

nevsk [136]2 years ago

3 0

Answer:

Walden’s data set doesn’t have any outliers. Drake’s data has an outlier with the value 9, which is at the higher end of the distribution.

Step-by-step explanation:

sample answer

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8. The length of the support is 7. 9 m

9. The length of the conveyer is 12m

3. a = 59m

A = 44°

B = 52°

4. c = 88. 6 mm

b = 49. 1 m

B = 34°

<h3>How to solve the trigonometry</h3>

8. We have the angle to be 20 degrees

Opposite side = x

hypotenuse = 23m

Using the sine ratio

sin θ = opposite/ hypotenuse

$sin 20 = \frac{x}{23}$

Cross multiply

$sin 20$ × $23 = x$

$x = 0. 3420$ × $23$

x = 7. 9 m

The length of the support is 7. 9 m

9. The angle of elevation is 37. 3 degrees

Hypotenuse = 19 . 0m

Opposite = x

Using the sine ratio

sin θ = opposite/ hypotenuse

$sin 37. 3 = \frac{x}{19}$

cross multiply

$x = 0. 6059$ × $19$

x = 11.5

x = 12 m in 2 significant figures

The length of the conveyer is 12m

3. To determine the sides and angles, we use the sine rule;

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

For side a, we use the Pythagorean theorem

$c^2 = a^2 + b^2$

$85^2 = a^2 + 67^2$

$a = \sqrt{89^2-67^2}$

$a = \sqrt{3432}$

a = 58. 58, a = 59m

To find angle A and B, use the sine rule

$\frac{59}{sin A } = \frac{85}{sin 90}$

cross multiply

$sin A$ × $85$ = $sin 90$ × $59$

make sin A subject of formula

$sin A = \frac{59}{85}$

$sin A = 0. 6941$

A = $sin^-^1(0. 6941)$

A = 44°

$\frac{67}{sin B} = \frac{85}{sin 90}$

cross multiply

$sin B$ × $85$ = $sin 90$ × $67$

make sin b subject of formula

$sin B = \frac{67}{85}$

$sin B = 0. 7882$

B = $sin^-^1( 0. 7882)$

B = 52°

4. To find the sides, we use the sine rule;

$\frac{74. 0}{sin 56. 6} = \frac{c}{sin 90}$

Cross multiply

$sin 56. 6$ × $c = sin 90$ × $74$

make 'c' subject of formula

$c = \frac{74}{0. 8348}$

c = 88. 6 mm

To find length b, we use the Pythagorean theorem

$c^2 = a^2 + b^2$

$b^2 = c^2 - a^2$

$b^2 = 88. 8^2 - 74^2$

$b = \sqrt{7885. 44 - 5476}\\\\ b = \sqrt{2409. 44}$

b = 49. 1 m

$\frac{74. 0}{sin 56. 6} = \frac{49. 1}{sin B}$

cross multiply

$sin B = \frac{40. 99}{74. 0}$

B = $sin^-^1(0. 5539)$

B = 34°

Learn more about trigonometric identity here:

brainly.com/question/7331447

#SPJ1

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What is the equation of the circle graphed below? Use the caret ( ^ ) to enter any exponents; for example, write x2 as x^2.

kvv77 [185]

Equation of a circle in standard form:
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(h,k) is the center.

In this case:
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r=4

Therefore:
(x-0)²+(y-0)²=4²
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