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Sergeu [11.5K]
4 years ago
14

Two kids at a summer camp, Tina and Winston, are competing in a potato sack race. Tina is younger, so she is given a head start

of 33 meters. When the race starts, Tina hops at a rate of 1 meter per second, and Winston hops 4 meters per second. Eventually, Winston will overtake Tina. How long will that take? How far will Winston have to hop? It will take ___ seconds for Winston to hop ___ meters and catch up to Tina.
Mathematics
1 answer:
agasfer [191]4 years ago
7 0

Answer:

It will take Winston 11 seconds to hop 44 meters and catch up to Tina

Step-by-step explanation:

Set up an equation where x represents the number of seconds:

Tina will be represented by s + 33 since she goes 1 m/sec and she has a 33 meter head start, and Winston will be 4s. We need to find when Winston will catch up, so we need an equal sign.

s + 33 = 4s

Solve for s:

s + 33 = 4s

33 = 3s

11 = s

= 11 seconds

Then, plug in 11 as s to find how many meters he will go:

s + 33

11 + 33 = 44

= 44 meters

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Select the postulate of equality or inequality that is illustrated.
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Answer: Transitive Postulate of Inequality (last option)

Explanation:

Let's say we have a vacation where we go from City A, to City B, then to City C. We can use the notation A \to B \to C. If all we cared about was the first and last cities, then we basically say A \to C taking a shortcut so to speak.

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It might help to draw out a number line and pick values for a and b like a = 0 and b = 1. That way you can see how a < b, b < c and a < c all tie together.

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3 years ago
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  

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

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

\hat p=\frac{115}{180}=0.639 estimated proportion of adults with the medication was effective

p_o=0.65 is the value that we want to test

\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.

Calculate the statistic  

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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