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Sav [38]
3 years ago
14

NFL player Gerald Sensabaugh recorded a 46 inch standing vertical jump at the 2005 NFL Combine, at that time the highest for any

NFL player in the history of the Combine. Sensabaugh weighed about 200 lb when he set the record. Part A What was his speed as he left the floor
Mathematics
1 answer:
andrew11 [14]3 years ago
7 0

Answer:

His speed as he left the floor is 4.83 m/s.

Step-by-step explanation:

Given: 46 inches = 1.1684 m and mass = 200 lb = 90.7185 Kg.

From the third equation of motion under free fall,

  V^{2} = U^{2} - 2gs

Where; V is the final velocity (0), U is the initial velocity (unknown), g is the value of gravity - 10 m/s^{2} and s is the distance = 1.1684 m.

Then;

0 = U^{2} - 2gs

U^{2} = 2gs

   = 2 × 10 × 1.1684

  = 23.368

⇒ U = \sqrt{23.368}

       = 4.8340 m/s

The initial velocity, U = 4.83 m/s.

Therefore, his speed as he left the floor is 4.83 m/s.

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Clarinex is a drug used to treat asthma. In clinical tests of this drug, 1655 patients were treated with 5- mg doses of Clarinex
bagirrra123 [75]

Answer:

z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363  

p_v =P(z>3.363)=0.00039  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

\hat p=0.021 estimated proportion of interest

p_o=0.012 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:  

Null hypothesis:p \leq 0.012  

Alternative hypothesis:p > 0.012  

When we conduct a proportion test we need to use the z statisitic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>3.363)=0.00039  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

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Answer choices:<br> 3<br> 2 sq root 3<br> 3 sq root 3<br> 6<br> 9
Vlada [557]

Answer:

third

Step-by-step explanation:

hope it helped!!!

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