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inessss [21]
3 years ago
10

A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is its acceleration?

Mathematics
2 answers:
In-s [12.5K]3 years ago
6 0
The car's acceleration can be modeled by the following formula:

a = \frac{v_f - v_i}{t}

vf represents the final velocity of the car. vi represents the initial velocity of the car. t represents the total time it took for the car to get from its initial to its final velocity.

We already have our values for our velocities and time. Plug them into the equation:

\text{Initial velocity: 28 m/s}^{2}
\text{Final velocity: 0 m/s}^2
\text{Time: 7 seconds}

\frac{0-28}{7} = -\frac{28}{7} = \boxed{-4}

The acceleration of the car is -4 m/s, or -4 meters per second.
Alika [10]3 years ago
5 0
The function that models the scenario is
v = init_v + acceleration * time

Plug the values into the equation,
0 = 28 + a * 7 => -28 = a * 7 => a = - 28/7 m/s.
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(1 point) Suppose we want a 95% confidence interval for the average amount spent on books by freshmen in their first year at col
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Answer:

a) n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40

So the answer for this case would be n=40 rounded up to the nearest integer

b) For this case if we see the formula for the margin of error

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean

\sigma=16 represent the population standard deviation

n represent the sample size  

Solution to the problem

Part a

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40

So the answer for this case would be n=40 rounded up to the nearest integer

Part b

For this case if we see the formula for the margin of error

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

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