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3 years ago

What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?

Mathematics

2 answers:

Likurg_2 [28]3 years ago

8 0

Force = mass x exceleration
1100 x 0.5 = 550

AlexFokin [52]3 years ago

4 0

Answer: 550 N

Identify the given information.
m = 1100 kg
a = 0.5 m/s²

Choose which formula you should use. In this case, you are finding the force.
f = ma, where m is the mass and a is the acceleration.

Substitute and solve.
f = (1100 kg)(0.5 m/s²)
f = 550 N

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Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

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The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

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So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

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The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292$

The 95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

A similar problem is given at brainly.com/question/16807970

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