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bekas [8.4K]
4 years ago
15

an object is thrown off a platform that is 15 ft high with an initial velocity of 8.5 ft/is what function models the height h of

the object after t seconds
Mathematics
1 answer:
NemiM [27]4 years ago
4 0

Answer:

Therefore the function that models the height of the object after t seconds is given by "H(t) = 15 + 8.5*t + 16.09*t²".

Step-by-step explanation:

Since the object has a initial velocity and it's being accelerated by gravity, than it's height is defined by:

H(t) = H(0) + V(0)*t + 0.5*g*t²

Applying the data from the problem, we have:

H(t) = 15 + 8.5*t + 0.5*32.17405*t²

H(t) = 15 + 8.5*t + 16.09*t²

Therefore the function that models the height of the object after t seconds is given by "H(t) = 15 + 8.5*t + 16.09*t²".

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a) p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

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c) (0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

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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".  

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Part a

p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

Part b

p_B represent the real population proportion for male

\hat p_B =\frac{102}{170}=0.6 represent the estimated proportion for male

n_B=170 is the sample size elected for male

z represent the critical value for the margin of error  

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p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

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(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between 0.124 \leq p_A -p_B \leq 0.316

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