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3 years ago

Which of the statements best compares the pairs of perpendicular and parallel lines?

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

8 0

Wich statements. Let me know

quester [9]3 years ago

7 0

Answer:

perpendicular lines intersect, and parallel lines never meet.

Step-by-step explanation:

this is pretty obvious in itself, not a whole lot of explainaton needed. parallel lines are exactly the same distance from each other and no matter how far they extend, they will never touch each other. perpendicular lines have 2 lines that intersect at a 90 degree angle, both vertically and horizontally. please mark brainliest.

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The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0

2 years ago

What is the solution set of the equation 30/x^2-9 +1=5/x-3

gizmo_the_mogwai [7]

$\frac{30}{x^{2} - 9 + 1} = \frac{5}{x - 3} \\\frac{30}{x^{2} - 8} = \frac{5}{x - 3} \\5(x^{2} - 8) = 30(x - 3) \\5(x^{2}) - 5(8) = 30(x) - 30(3) \\5x^{2} - 40 = 30x - 90 \\5x^{2} + 50 = 30x \\5x^{2} - 30x + 50 = 0 \\5(x^{2}) - 5(6x) + 5(10) = 0 \\5(x^{2} - 6x + 10) = 0 \\\frac{5(x^{2} - 6x + 10)}{5} = \frac{0}{5} \\x^{2} - 6x + 10 = 0 \\x = \frac{-(-6) \± \sqrt{(-6)^{2} - 4(1)(10)}}{2(1)} \\x = \frac{6 \± \sqrt{36 - 40}}{2} \\x = \frac{6 \± \sqrt{-4}}{2} \\x = \frac{6 \± 2i}{2} \\x = 3 \± i \\x = 3 + i\ or\ x = 3 - i$

The answer is D.

8 0

3 years ago

Read 2 more answers

PLEASE HELP FAST NO WRONG ANSWERS

liberstina [14]

Answer:

Option D.

Step-by-step explanation:

Segment VW is intersected by planes XUV and SRV

4 0

2 years ago

Read 2 more answers

Which of the following is equal to the expression below?

erik [133]

(4^-5)^3 =
4^(-5*3) =
4^-15 =
1/(4^15).....answer C

8 0

3 years ago

Solve the following pair of equations simultaneously.

Alchen [17]

Answer:

No solutions.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Algebra I

Solving systems of equations using substitution/elimination
Solving systems of equations by graphing
Expanding
Finding roots of a quadratic
Standard Form: ax² + bx + c = 0
Quadratic Formula: $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Step-by-step explanation:

Step 1: Define systems

2x - y = 9

4x² + 3y² - 2x + y = 16

Step 2: Rewrite systems

2x - y = 9

Subtract 2x on both sides: -y = 9 - 2x
Divide -1 on both sides: y = 2x - 9

Step 3: Redefine systems

y = 2x - 9

4x² + 3y² - 2x + y = 16

Step 4: Solve for x

Substitution

Substitute in y: 4x² + 3(2x - 9)² - 2x + (2x - 9) = 16
Expand: 4x² + 3(4x² - 36x + 81) - 2x + (2x - 9) = 16
Distribute 3: 4x² + 12x² - 108x + 243 - 2x + 2x - 9 = 16
Combine like terms: 16x² - 108x + 234 = 16
Factor GCF: 2(8x² - 54x + 117) = 16
Divide 2 on both sides: 8x² - 54x + 117 = 8
Subtract 8 on both sides: 8x² - 54x + 109 = 0
Define variables: a = 8, b = -54, c = 109
Resubstitute: $x=\frac{54\pm\sqrt{(-54)^2-4(8)(109)} }{2(8)}$
Exponents: $x=\frac{54\pm\sqrt{2916-4(8)(109)} }{2(8)}$
Multiply: $x=\frac{54\pm\sqrt{2916-3488} }{16}$
Subtract: $x=\frac{54\pm\sqrt{-572} }{16}$

Here we see that we start to delve into imaginary roots. Since on a real number plane, we do not have imaginary roots, there would be no solution to the systems of equations.

Step 5: Graph systems

We can verify our results.

8 0

2 years ago