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3 years ago

Solve each equation by substitution

Mathematics

2 answers:

Artemon [7]3 years ago

5 0

Answer:

(1, 7)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Algebra I

Solving systems of equations using substitution/elimination

Step-by-step explanation:

Step 1: Define systems

y = 3x + 4

y = 6x + 1

Step 2: Solve for x

Substitute in y: 3x + 4 = 6x + 1
Subtract 3x on both sides: 4 = 3x + 1
Subtract 1 on both sides: 3 = 3x
Divide 3 on both sides: 1 = x
Rewrite: x = 1

Step 3: Solve for y

Define original equation: y = 6x + 1
Substitute in x: y = 6(1) + 1
Multiply: y = 6 + 1
Add: y = 7

Aneli [31]3 years ago

4 0

Answer:

(1, 7)

Step-by-step explanation:

y = 3x + 4

y = 6x + 1

6x + 1 = 3x + 4 ⇒ x = 1

y = 3(1) + 4 = 7

(1, 7)

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

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See attachment for the parallel box plots

(a) Prove that 99 is an outlier for Pet

We have:

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$n = 15$

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

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