10 7/8 meters long when the two sections are put together.
Answer:
c
Step-by-step explanation:
the final speed in m/s of the 10.0 kg is 2.53 m/s .
<u>Step-by-step explanation:</u>
Here we have , A 10.0 kg and a 2.0 kg cart approach each other on a horizontal friction less air track. Their total kinetic energy before collision is 96 ). Assume their collision is elastic. We need to find What is the final speed in m/s of the 10.0 kg mass if that of the 2.0 kg mass is 8.0 m/s . Let's find out:
We know that in an elastic collision :
⇒ Total kinetic energy before collision = Total kinetic energy after collision
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Therefore , the final speed in m/s of the 10.0 kg is 2.53 m/s .
Answer
7
Step-by-step explanation:
13x=91
91/13=x
x=7
Answer:
29.2
Step-by-step explanation:
Mean = 21.4
Standard deviation = 5.9%
The minimum score required for the scholarship which is the scores of the top 9% is calculated using the Z - Score Formula.
The Z- score formula is given as:
z = x - μ /σ
Z score ( z) is determined by checking the z score percentile of the normal distribution
In the question we are told that it is the students who scores are in the top 9%
The top 9% is determined by finding the z score of the 91st percentile on the normal distribution
z score of the 91st percentile = 1.341
Using the formula
z = x - μ /σ
Where
z = z score of the 91st percentile = 1.341
μ = mean = 21.4
σ = Standard deviation = 5.9
1.341= x - 21.4 / 5.9
Cross multiply
1.341 × 5.9 = x - 21.4
7.7526 = x -21.4
x = 7.7526 + 21.4
x = 29.1526
The 91st percentile is at the score of 29.1526.
We were asked in the question to round up to the nearest tenth.
Approximately, = 29.2
Step-by-step explanation: