All categories

2 years ago

Describe the relationship between the two quantities. Distance traveled by car; Amount of gas in the car.

Mathematics

2 answers:

Novosadov [1.4K]2 years ago

6 0

Amount of gas in the car helps the car go and reach its destinations.

icang [17]2 years ago

6 0

Answer:

Inverse relation between distance traveled and amount of gas in the car

Step-by-step explanation:

Because amount of gas in the decreases as distance traveled increases.

Suppose you have 2 liter fuel in the car and as you start the car for travelling so distance increases and fuel decreases.

You might be interested in

The value of the given digits the 4s in 244

Digiron [165]

244

The 4's are in the ONE's and TEN's places

4 0

2 years ago

13+y =40<br> explain how the equal sign is use to balance this equation

seropon [69]

Answer:

Y = 27.

Step-by-step explanation:

40 - 13 = 27. If we check our math 13 +27 does equal 40 making the equation correct.

5 0

2 years ago

What is 56% written as a ratio A)56 to 1

stellarik [79]

It c) because it 56 out of 100

5 0

2 years ago

The topic is volume & surface area, can someone help?

marishachu [46]

This pic should help

3 0

2 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

2 years ago