Which of the following statements accurately describes a pedigree chart?

1. A pedigree chart is used to predict the genotype of offspring.
2. A pedigree chart tracks inherited traits from parents to offspring.
3. A pedigree chart predicts the percentage of offspring that will have a specific trait.
4. A pedigree chart identifies the phenotype of the organism.

Answer:

o.75

Explanation:

A train moving at a velocity of 15 m/s is accelerated to 24m/s over a 12 second period, time traveled by the train is 12 seconds.

The initial velocity u of the train is 15 m/s, the final velocity v is 24 m/s, and the time t is 12 seconds. The train is undergoing uniform acceleration.

We can use the kinematic equation to relate the initial velocity, final velocity, acceleration, and time:

v = u + at

Solving for t:

[tex]\[ t = \dfrac{v - u}{a} \][/tex]

Given that a is the acceleration and is calculated as:

[tex]\[ a = \dfrac{v - u}{t} \][/tex]

Substitute the given values:

[tex]\[ a = \dfrac{24 \, \text{m/s} - 15 \, \text{m/s}}{12 \, \text{s}} \][/tex]

[tex]\[ a = \dfrac{9 \, \text{m/s}}{12 \, \text{s}} \\\\= 0.75 \, \text{m/s}^2 \][/tex]

Now substitute the calculated acceleration back into the equation for \( t \):

[tex]\[ t = \dfrac{v - u}{a} \\\\= \dfrac{24 \, \text{m/s} - 15 \, \text{m/s}}{0.75 \, \text{m/s}^2} \][/tex]

[tex]\[ t = \dfrac{9 \, \text{m/s}}{0.75 \, \text{m/s}^2} \\\\= 12 \, \text{s} \][/tex]

Thus, the time traveled by the train is 12 seconds.

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Your question seems incomplete, the probable complete question is:

A train moving at a velocity of 15 m/s is accelerated to 24m/s over a 12 second period. Determine the time travelled​