Solve for X A) -6 C) 12 E) 9 B) 8 D) 3

1.609/5280 = 3.2/x....1.609 km to 5380 ft = 3.2 km to x ft
cross multiply because this is a proportion
(1.609)(x) = (5280)(3.2)
1.609x = 16896
x = 16896/1.609
x = 10,500 ft <===

5 0

3 years ago

Read 2 more answers

The circumference of a circular garden is 131.88 feet. What is the radius of the garden?

soldi70 [24.7K]

Answer:

21 ft

Step-by-step explanation:

C = 2πr

Filling in the given values, we have ...

131.88 ft = 2·3.14·r . . . . we can find r by dividing by its coefficient

131.88 ft/6.28 = r = 21 ft

3 0

2 years ago

If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with you reduce stress? To e

Pepsi [2]

The outlier of a dataset is a data element that is relatively far from the remaining data elements

99 is an outlier of pet group
See attachment for the parallel box plots

(a) Prove that 99 is an outlier for Pet

We have:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

$n = 15$

The quartiles positions are:

$Q_1 = \frac{n + 1}{4}$

$Q_1 = \frac{15 + 1}{4}$

$Q_1 = \frac{16}{4}$

$Q_1 = 4th$

$Q_3 = Q_1 \times 3$

$Q_3 = 4th \times 3$

$Q_3 = 12th$

So, we have:

$Q_1 = 4th$

$Q_3 = 12th$

From the pet group:

The data elements at the 4th and 12th positions are 68 and 79

So, we have:

$Q_1 = 68$

$Q_3= 79$

The lower and upper limits of the outlier are:

$L = Q_1 - 1.5 \times (Q_3 - Q_1)$

$U = Q_3 + 1.5 \times (Q_3 - Q_1)$

So, we have:

$L = 68 - 1.5 \times (79 - 68)$

$L = 51.5$

$U = 79+ 1.5 \times (79 - 68)$

$U = 95.5$

This means that data below 51.5 or above 95.5 are outliers.

Hence, 99 is an outlier because 99 is greater than 95.5

(b) The parallel box plot

The three groups are:

Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99

Erlento: 88 80 80 81 92 87 88 81 82 80 87 92 87 80 82

Alone: 62 70 73 75 77 80 84 84 84 87 87 87 90 91 99

See attachment for the parallel box plots

Read more about box plots and outliers at:

brainly.com/question/14940764

5 0

2 years ago