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nirvana33 [79]
3 years ago
7

A person with a mass of 92kg is running. They accelerate 1.2 m/s2 as they are getting up to speed. What is their force?

Mathematics
1 answer:
Dafna11 [192]3 years ago
4 0

Answer:

The answer to your question is  F = 110.4 N

Step-by-step explanation:

Data

mass = 92 kg

acceleration = 1.2 m/s²

Force = ?

Process

To solve this problem use Newton's second law which states that the acceleration of an object is dependent upon two variables, the net force acting upon the body and the mass of the object.

Formula

              F = ma

-Substitution

              F = (92)(1.2)

-Result

              F = 110.4 N

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A manufacturer of a new medication on the market for Alzheimer's disease makes a claim that the medication is effective in 65% o
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Answer:

z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309  

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So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

Step-by-step explanation:

Data given and notation

n=180 represent the random sample taken

X=115 represent the adults with the medication was effective

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\alpha=0.05 represent the significance level

Confidence=95% or 0.95

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 proportion is less than 0.65.:  

Null hypothesis:p \geq 0.65  

Alternative hypothesis:p < 0.65  

When we conduct a proportion test we need to use the z statisitc, 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.

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

z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309  

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=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

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Answer:

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