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grigory [225]
3 years ago
7

A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65m/s in 5 seconds what

is the acceleration?
Mathematics
2 answers:
Lerok [7]3 years ago
5 0

Answer:

6m/s2

Step-by-step explanation:

artcher [175]3 years ago
3 0

The acceleration is defined as the ratio between the change in velocity and the time elapsed to perform such a change.

These "changes" are indicated with the capital greek letter delta, \Delta, and when you write \Delta x you mean the difference between the finial and the inital values of the variable x:

\Delta x = x_{\text{fin}} - x_{\text{init}}

So, the acceleration is defined as

a = \dfrac{\Delta v}{\Delta t} = \dfrac{v_{\text{fin}} - v_{\text{init}}}{t_{\text{fin}} - t_{\text{init}}}

In this case, the initial velocity is 35, the final velocity is 65. Assuming we start the clock at the beginning of the observation, the inital time is 0 and the final time is 5. So, we have

a = \dfrac{65-35}{5-0} = \dfrac{30}{5} = 6m/s^2

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The scores on a standardized exam are normally distributed with a mean of 400 and a standard deviation of 50.
S_A_V [24]

Using the normal distribution, it is found that approximately 40% of the scores are greater than 413.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, we have that the mean and the standard deviation of the scores are given by:

\mu = 400, \sigma = 50

Approximately 40% of the scores are greater than the 60th percentile, which is <u>X when Z = 0.253</u>.

Then:

Z = \frac{X - \mu}{\sigma}

0.253 = \frac{X - 400}{50}

X - 400 = 50(0.253)

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Rounding up, approximately 40% of the scores are greater than 413.

More can be learned about the normal distribution at brainly.com/question/24663213

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