All categories

3 years ago

How many ternary strings of length 2 n are there in which the zeroes appear only in odd-numbered positions?

1 answer:

8 0

In a string of length $2n$ , there are exactly $n$ odd-numbered posiitons. In the remaning $n$ even-numbered positions, we want to choose from only two possible digits, either a 1 or a 2. This means there are $2^n$ such ternary strings.

You might be interested in

Which of the following is not an essential feature of scientific explanations?

sineoko [7]

Answer:

D I think I might be wrong its been a while scense I did something like that

4 0

3 years ago

On a highway, a car is driven 80. kilometers

nydimaria [60]

The average speed of the car for the entire trip can be calculate by using:

$v=\frac{S}{t}$

where S is the total distance covered by the car, and t is the total time taken.

The total distance travelled by the car is:

$S=80 km+50 km+40 km=170 km$

while the total time taken is:

$t=1.00 h+0.50 h+0.50 h=2.00 h$

so, the average speed of the car is:

$v=\frac{S}{t}=\frac{170 km}{2.00 h}=85 km/h$

so, the correct answer is (3) 85 km/h.

7 0

3 years ago

Read 2 more answers

A graph of angular position v. time has the following equation:

Harman [31]

Answer:

you can simply answer by derivative = 3.5x^2+25x+250-y=0 you can derivate this eqn 7x +25-1=0 7x=24 yo u can divide you get it

5 0

3 years ago

A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?

Montano1993 [528]

Net Force = (mass) x (acceleration) (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

<em>Net Force = 300 Newtons down</em>

6 0

2 years ago

A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance

Minchanka [31]

Answer:

$F=43570.9N$

Explanation:

We can calculate the acceleration experimented by the passenger using the formula $v_f^2=v_i^2+2ad$ , taking the initial direction of movement as the positive direction and considering it comes to a rest:

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}$

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

$F=ma=\frac{-mv_i^2}{2d}$

Which for our values is:

$F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N$

6 0

3 years ago