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Dmitry [639]
3 years ago
5

A race car is driven around a circular track at a constant speed of 180mph. If the diameter is 1/2 mile, what is the angular spe

ed of the car? Express your answer in revolutions per hour. The formula for angular speed is w=theta (in radians)/time
Mathematics
1 answer:
Bogdan [553]3 years ago
4 0
If the diameter is 1/2 mile, that is 0.5 mile, then the radius is 0.25 -half of it!

the angular speed is theta/time, which means

 w=theta/time

but another formula for it is Velocity*radians

w=velocity*radians

velocity is 180 mph and radians are (revolution/2\pi radius), that is:

w=180*(1/2\pi*0.25)
and if we calculate this, we have: w=114.59, which is the correct answer!
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Answer:

1. Combine the constant terms by adding 7 and –2.

4. Combine the like variable terms by adding the coefficients

5. 5 k and negative 3 k are like variable terms

6.The constants 7 and –2 are like terms.

7. The equivalent expression is 5 + 2 k + n.

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23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r
Nostrana [21]

Answer:

a) p_v =P(Z          

b) Since p_v we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) P(\bar X >90)=1-P(Z

1-P(Z

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}

And if we solve for n we got:

n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3

Step-by-step explanation:

Previous concepts  and data given  

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".  

We can calculate the sample mean and deviation with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

\bar X=88.96 represent the sample mean    

s=1.011 represent the sample standard deviation  

n=5 represent the sample selected  

\alpha significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:    

Null hypothesis:\mu \geq 90    

Alternative hypothesis:\mu < 90    

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907      

Part a

P-value  

First we need to calculate the degrees of freedom given by:

df=n-1=5-1 = 4

Then since is a left tailed test the p value would be:    

p_v =P(Z    

Part b

Since p_v we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >90)=1-P(Z

1-P(Z

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}

And if we solve for n we got:

n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3

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