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3 years ago

3x - 5y = 15.

Mathematics

1 answer:

antoniya [11.8K]3 years ago

4 0

Answer:

$(0,-3)$

$(5,0)$

$(10,3)$

$(20,9)$

Step-by-step explanation:

Given

$3x - 5y = 15$

Required

Form 4 pairs of (x,y) values

This question will be answered based on assumptions of x and y values;

Haven said that:

For point 1

Take x = 0

Substitute 0 for x in $3x - 5y = 15$

$3(0) -5y = 15$

$-5y = 15$

Solve for y

$y = 15/-5$

$y = -3$

So; the first pair is:

$(0,-3)$

For point 2:

Take y = 0

Substitute 0 for y in $3x - 5y = 15$

$3x - 5(0) = 15$

$3x = 15$

Solve for x

$x = 15/3$

$x = 5$

So, the pair is

$(5,0)$

For point 3:

Take x = 10

Substitute 10 for x in $3x - 5y = 15$

$3(10) - 5y = 15$

$30 - 5y = 15$

Collect Like Terms

$-5y = 15 - 30$

$-5y = -15$

Solve for y

$y = -15/-5$

$y = 3$

So, the pair is:

$(10,3)$

For point 4:

Take y = 9

Substitute 9 for y in $3x - 5y = 15$

$3x - 5(9) = 15$

$3x - 45 = 15$

Collect Like Terms

$3x = 45 + 15$

$3x = 60$

Solve for x

$x = 60/3$

$x = 20$

So, the pair is:

$(20,9)$

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I'll use text forms of the logical symbols because copying/pasting on mobile is annoying.

⊕ = XOR

∨ = OR

∧ = AND

~ = NOT

↔ = IFF (meaning "if and only if")

→ = IMP (short for "implies")

and for logical equivalence I'll use a regular equal sign.

Then

p XOR q = (p OR q) AND NOT (p AND q)

(a) By definition of XOR, we have

p XOR p = (p OR p) AND NOT (p AND p)

= p AND NOT p

and this is tautologically FALSE.

(b) To rewrite XOR, first distribute the NOT over the AND using DeMorgan's law:

NOT (p AND q) = NOT p OR NOT q

Then we can interchange the AND and ORs using their respective distributive properties. That is,

p XOR q = (p OR q) AND (NOT p OR NOT q)

= (p AND (NOT p OR NOT q)) OR (q AND (NOT p OR NOT q))

= ((p AND NOT p) OR (p AND NOT q)) OR ((q AND NOT p) OR (q AND NOT q))

Both (p AND NOT p) and (q AND NOT q) are tautologically FALSE, so the truth value of either OR depends only on the other statements, so we have

p XOR q = (p AND NOT q) OR (q AND NOT p)

Recall the definition of IFF:

p IFF q = (p IMP q) AND (q IMP p)

Also recall that IMP is logically equivalent to

p IMP q = p OR NOT q

Then we have

p IFF q = (p OR NOT q) AND (q OR NOT p)

so that its negation is

NOT (p IFF q) = NOT ((p OR NOT q) AND (q OR NOT p))

= NOT (p OR NOT q) OR NOT (q OR NOT p)

= (NOT p AND q) OR (NOT q AND p)

The statements in bold match, so p XOR q is indeed equivalent to NOT (p IFF q).

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2 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

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Answer:

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(b) 0.657

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

$p = \frac{70}{100} = 0.7$

Let P(a vehicle not passing the test) = q

q = 1 - p

q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

= 0.7 × 0.7 × 0.7

= 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

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= 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

= 0.657

= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

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(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

= 0.063 + 0.063 + 0.063 + 0.027

= 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

= (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

= 0.973

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

$= \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}$

$= \frac{0.343}{0.973}$

= 0.353

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