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anzhelika [568]
3 years ago
13

Using Point Slope Formula, write an equation that passes through (-2, 6) and ( 8, 9)

Mathematics
1 answer:
Ulleksa [173]3 years ago
5 0

Answer:

y - 9 = 3/10(x - 8)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

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

Point-Slope Form: y - y₁ = m(x - x₁)  

  • x₁ - x coordinate
  • y₁ - y coordinate
  • m - slope

Step-by-step explanation:

<u>Step 1: Define</u>

Point (-2, 6)

Point (8, 9)

<u>Step 2: Find slope </u><em><u>m</u></em>

Simply plug in the 2 coordinates into the slope formula to find slope <em>m</em>

  1. Substitute [SF]:                    \displaystyle m=\frac{9-6}{8+2}
  2. Subtract/Add:                      \displaystyle m=\frac{3}{10}

<u>Step 3: Write Function</u>

<em>Plug in variables into general form.</em>

y - 9 = 3/10(x - 8)

y - 6 = 3/10(x + 2)

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It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha
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Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.54% of the calls receive a busy signal, hence  p = 0.0054.
  • A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

P(X \geq 5) = 1 - P(X < 5)

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009

P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062

P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218

P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513

P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

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