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3 years ago

Item 3 Find the slope of the line that contains (-2,-2) and (-10,-9

Mathematics

1 answer:

serg [7]3 years ago

8 0

Answer:

$m=\frac{7}{8}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (-2, -2)

Point (-10, -9)

Step 2: Find slope m

Substitute [Slope Formula]: $m=\frac{-9+2}{-10+2}$
Add: $m=\frac{-7}{-8}$
Simplify: $m=\frac{7}{8}$

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4 years ago

a web music store offers two versions of a popular song. the size of the standard version is 2.9 megabytes (MB). the size of the

kenny6666 [7]

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-let x denote the number of standard version downloads.

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3 years ago

Identify the reason that justifies the statement.

Amiraneli [1.4K]

Answer:

C; Substitution property

Step-by-step explanation:

Here, we want to find the justification that justifies the written equation;

If PQ + RS = PS

and RS = XY

then PQ + XY = PS

What we simply did is to substitute RS for XY in the second equation;

The correct answer is Substitution property

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3 years ago

-The dimensions of a rectangular garden were 2 m by 4 m. Each dimension was

posledela

Answer: 4 m

Step-by-step explanation:

Let the amount each dimension was increased by be x.

Now we can set up a equation.

(2+x)(4+x) = 48

8 + 2x + 4x + x^2 = 48

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2 years ago

Read 2 more answers

The national average for the math portion of the College Board’s Scholastic Aptitude Test

Alex73 [517]

Answer:

a) 0.1587

b) 0.023

c) 0.341

d) 0.818

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 515

Standard Deviation, σ = 100

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(score greater than 615)

P(x > 615)

$P( x > 615) = P( z > \displaystyle\frac{615 - 515}{100}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 615) = 1 - 0.8413 = 0.1587 = 15.87\%$

b) b) P(score greater than 715)

$P(x > 715) = P(z > \displaystyle\frac{715-515}{100}) = P(z > 2)\\\\P( z > 2) = 1 - P(z \leq 2)$

Calculating the value from the standard normal table we have,

$1 - 0.977 = 0.023 = 2.3\%\\P( x > 715) = 2.3\%$

c) P(score between 415 and 515)

$P(415 \leq x \leq 515) = P(\displaystyle\frac{415 - 515}{100} \leq z \leq \displaystyle\frac{515-515}{100}) = P(-1 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1)\\= 0.500 - 0.159 = 0.341 = 34.1\%$

$P(415 \leq x \leq 515) = 34.1\%$

d) P(score between 315 and 615)

$P(315 \leq x \leq 615) = P(\displaystyle\frac{315 - 515}{100} \leq z \leq \displaystyle\frac{615-515}{100}) = P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.841 - 0.023 = 0.818 = 81.8\%$

$P(315 \leq x \leq 615) = 81.8\%$

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3 years ago