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2 years ago

A tractor travels at an average speed of 36km/hr for 10 sec.calculate the distance the tractor covers in 10 sec

Mathematics

2 answers:

Katen [24]2 years ago

8 0

1 hour = 3600 seconds

10 seconds is 10/3600 = 1/360 of an hour

36 km per hour x 1/360 of an hour = 0.1 km (100 meters)

DIA [1.3K]2 years ago

5 0

Answer:

.1km

Step-by-step explanation:

The tractor can move .01 km in a second so 10 times .01 is .1

so the tractor can move .1km every 10 seconds

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An object is dropped from a bridge and allowed to freefall to the ground. The height of the object over time can be modeled usin

kkurt [141]

Answer:

<em>t = 3 seconds</em>

Step-by-step explanation:

We have the equation, as a function of time, that describes the height of the object that is dropped from the bridge.

The equation is:

$h (t) = 144 - 16t ^ 2$

Where t is the time in seconds and h is the height in feet.

To know how long it takes the object to fall to the ground we do h (t) = 0 and solve for t.

So:

$0 = 144 - 16t ^ 2\\\\16t ^ 2 = 144\\\\t ^ 2 = \frac{144}{16}\\\\t ^ 2 = 9\\\\t_1 = 3\\t_2 = -3$

We take the positive solution.

Therefore the object takes 3 seconds to reach the ground

4 0

2 years ago

Read 2 more answers

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$