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dlinn [17]
3 years ago
12

From two cities that are 500 mi apart, two cars left simulaneously moving towards each other. The speed of one car was 10 mph gr

eater than the speed of the other car. In 5 hours the cars met. Find the speed of each car.
Mathematics
1 answer:
Ne4ueva [31]3 years ago
6 0

Given:

Distance between to cities = 500 mi

Two cars left simultaneously moving towards each other.

The speed of one car was 10 mph greater than the speed of the other car.

They meet in 5 hours.

To find:

The speed of each car.

Solution:

Let x mi/h be the speed of one car.

So, speed of second car = (x + 10) mi/h

Two cars left simultaneously moving towards each other.

So, their relative speed = x + (x+10) = (2x+10) mi/h

We know that,

Speed =\dfrac{Distance}{Time}

On substituting the values, we get

2x+10=\dfrac{500}{5}

2x+10=100

2x=100-10

2x=90

Divide both sides by 2.

x=\dfrac{90}{2}

x=45

Now,

Speed of one car = 45 mi/h

Speed of other car = 45+10

                                = 55 mi/h

Therefore, the speeds of two cars are 45 mi/h and 55 mi/hr.

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A <em>non</em>-repeating infinite decimal is an irrational number. √2 and π are examples of such numbers.

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Sample Size for Proportion As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of gol
Dima020 [189]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56  

And rounded up we have that n=2663

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".  

The population proportion have the following distribution

\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.025 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We can assume an estimated proportion of \hat p =0.5 since we don't have prior info provided. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56  

And rounded up we have that n=2663

6 0
3 years ago
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