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olchik [2.2K]
3 years ago
13

a stone is thrown downward with an intitial velocity of 29.5 m/sec will travel a distance of a meters, where s(t) = 4.9t^2 + 29.

4 and t is in seconds. if a stone is thrown downward at 29.4m/sec from a height of 132.3 m, how long will it take the stone to hit the ground?
Mathematics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

t = 9 s

Step-by-step explanation:

A stone is thrown downward with an intitial velocity of 29.5 m/sec will travel a distance of a meters, where

s(t) = -4.9t^2 + 29.4+132.3

We need to find when the stone will hit the ground. It means when s(t) = 0. So,

-4.9t^2 + 29.4t+132.3=0\\\\t=-3\ and\ t=9

So, the stone will take 9 seconds to hit the ground. Hence, this is the required solution.

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Answer:

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Step-by-step explanation:

To evaluate, substitute the values of y and solve.

So, when y =-2

0.67y + 4 = -1.34 + 4 = 2.66

And when y = 2

0.67y + 4 = 1.34 + 4 = 4.34

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Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

    So  

         Z_{\frac{\alpha }{2} } =  1.96

The  margin of error is mathematically represented as

      E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       \r p  =  \frac{210}{500}

       \r p  = 0.42

Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

The critical value of \frac{\alpha }{2} from the normal distribution table is  

      Z_{\frac{\alpha }{2} }  =  2.58

  Generally the margin of error is mathematically represented as

       E =  Z_{\frac{\alpha }{2} }*  \sqrt{ \frac{\r p (1- \r p )}{n} }

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

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     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

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