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2 years ago

6

An empty propane tank for a grill weighs 17 pounds. A full propane tank weighs 37 pounds. If a propane tank weighs 35 pounds, wh

at percentage of propane remains in the tank?

1 answer:

Vinil7 [7]2 years ago

7 0

Answer:

not 100

Step-by-step explanation:

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I need help with this question please

Bess [88]

Answer:

no understand I live in myanmar .I donot know mathamatic

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1 year ago

Which point would map onto itself after a reflection across the line y= -x

lutik1710 [3]

I need help on that too

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2 years ago

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Ganezh [65]

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2 years ago

Doug, a student in this class, knows how to write programs in JAVA. Everyone who knows how to write programs in JAVA can get a h

In-s [12.5K]

Missing Part of Question

Explain which rules of inference are used in the above

Answer:

Universal instantiation, Modus ponens and Existential generalization

Step-by-step explanation:

Splitting the Statement into two, we have

"Doug knows how to write program is Java"

and

"Doug can get a high paying job"

Represent Doug with x. Then the statements is rewritten as

P(x) = "x knows how to write program is Java"

and

Q(x) = "x can get a high paying job"

In logic, A premise is a statement in an argument that provides reason or support for the conclusion.

So, Statement 1 can be written as

1. P(x) ---- Premise

Then, we have

2. Vx(P(x) --> Q(x)) -- Premise. This mean that P(x) for all values in the domain.

The universal instantiation of (2) leads to

3. P(Doug) --> Q(Doug)

The modus ponens of the above gives

4. Q(Doug)

The existential generalisation from above gives

5. ƎxQ(x)

Which means someone in this class can get a high paying job.

PS:

Universal Instantiation is a valid rule of inference from a truth about each member of a class of individuals to the truth about a particular individual of that class; it is represented by VxP(x).

Existential Generalisation is a valid rule of inference that allows one to move from a specific statement, or one instance, to a quantified generalized statement, or existential proposition.

It is represented as ƎxP(x)

4 0

2 years ago

A $16,000 robot depreciates linearly to zero in 10 years. (a) find a formula for its value as a function of time, t, in years.

SVETLANKA909090 [29]

We are given, cost of the robot for 0 number of year = $16,000.

0 represents initial time of the robot.

After 10 year cost of the robot is = $0

The problem is about the number of the years and cost of the robot over different number of years.

So, we could take x coordinate by number of hours and y coordinate for y number of hours.

So, from the problem, we could make two coordinates for the given situation.

(x1,y1) = (0, 16000) and (x2,y2) = (10, 0).

In order to find the function of time, we need to find the rate at which robot rate depreciates each year.

Slope is the rate of change.

So, we need to find the slope of the two coordinates we wrote above.

We know, slope formula

$Slope (m) = \frac{y2-y1}{x2-x1}$

Plugging values in formula, we get

$m=\frac{0-16000}{10-0} = \frac{-16000}{-10} = -1600.$

Brecasue of depreciation we got a negative number for slope or rate of change.

Therefore, rate of depreciation is $1600 per year.

We already given inital cost, that is $16,000.

So, we can setup an a function

f(x) = -1600x + 16000.

But the problem is asked to take the variable t for time.

Replacing x by t, we get

f(t) = -1600t + 16000.

8 0

2 years ago