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solong [7]
3 years ago
5

What is the distance between the points (2, 1) and(6, 7)?

Mathematics
1 answer:
icang [17]3 years ago
3 0

Answer:

\displaystyle d = 2\sqrt{13}

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> </u>

<u>Algebra I</u>

  • Coordinates (x, y)

<u>Algebra II</u>

  • Distance Formula: \displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (2, 1)

Point (6, 7)

<u>Step 2: Find distance </u><em><u>d</u></em>

Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>

  1. Substitute in points [Distance Formula]:                                                         \displaystyle d = \sqrt{(6 - 2)^2 + (7 - 1)^2}
  2. [√Radical] (Parenthesis) Subtract:                                                                   \displaystyle d = \sqrt{4^2 + 6^2}
  3. [√Radical] Evaluate exponents:                                                                       \displaystyle d = \sqrt{16 + 36}
  4. [√Radical] Add:                                                                                                 \displaystyle d = \sqrt{52}
  5. [√Radical] Simplify:                                                                                           \displaystyle d = 2\sqrt{13}
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Ghella [55]

Answer:

Step-by-step explanation:

Given the first two numbers of a sequence as 2, 6...

If it is an arithmetic difference, the common difference will be d = 6-2 = 4

Formula for calculating nth term of an ARITHMETIC sequence Tn = a+(n-1)d

a is the first term = 2

d is the common difference = 4

n is the number if terms

Substituting the given values in the formula.

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2) If the sequence us a geometric sequence

Nth term of the sequence Tn = ar^(n-1)

r is the common ratio

r is gotten by the ratio of the terms I.e

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r = 6/2

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3 0
3 years ago
An office manager has received a report from a consultant that includes a section on equipment replacement. the report indicates
wolverine [178]

Answer:

a) 22.663%

b) 44%

c) 38.3%

Explanation:

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates that scanners have a service life that is normally distributed with a mean of 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 41 months

σ is the population standard deviation = 4 months

a. Before 38 months of service

Before in z score score means less than 38 months

Hence,

z = 38 - 41/4

z = -0.75

Probability value from Z-Table:

P(x<38) = 0.22663

Converting to percentage = 0.22663 × 100

= 22.663%

b. Between 40 and 45 months of service

For x = 40 months

z = 40 - 41/4

z = -0.2

Probaility value from Z-Table:

P(x =40) = 0.40129

For x = 45

z = 45 - 41/4

z = 1

Probability value from Z-Table:

P(x = 45) = 0.84134

Between 40 and 45 months of service

= 0.84134 - 0.40129

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Converting to Percentage

= 0.44005 × 100

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c. Wihin ± 2 months of the mean life

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For x = 43

z = 43 - 41 /4

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P-value from Z-Table:

P(x = 43) = 0.69146

For x = 39

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Probability value from Z-Table:

P(x = 39) = 0.30854

Within ± 2 months of the mean life

= 0.69146 - 0.30854

= 0.38292

= 38.3%

Learn more about z-score:

brainly.com/question/17436641

#SPJ4

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2 years ago
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Leto [7]
$12.80



Explanation:


3=$2.40 3\2.40 =0.80 0.80 • 16 = $12.80
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