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1 year ago

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If a baseball player hits a baseball from 4 feet off the ground with an initial velocity of 64 feet per second, how long will it

take the baseball to hit the ground? Use the equation h = −16t2 + 4t + 4. Round your answer to the nearest hundredth.

1 answer:

topjm [15]1 year ago

5 0

Answer:

0.64 seconds

Step-by-step explanation:

In the equation provided:

h = −16t2 + 4t + 4

h is the height of the ball and t is time. Since we want to find the time when the ball touches the floor, then height is 0. This leaves us with the equation

-16 $t^{2}$ + 4t + 4 = 0

This is a quadratic equation can be solved with the following formula:

$x= \frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

where a=-16

b=4

c=4

Solving for t we will find two different results:

$t1=\frac{-4-\sqrt{272} }{2(-16)} =0.125+0.125\sqrt{17} =0.64039$ $t2=\frac{-4+\sqrt{272} }{2(-16)} =0.125-0.125\sqrt{17} =-0.39039$

Since time can't be negative, we discard t2 and choose t1.

Since it is required to answer in the nearest hundredth, we round the result to t=0.64 seconds.

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Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

$\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5$

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

$\widehat{Y}=Y_1+Y_2+...+Y_{200}$

= $200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs$

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

S= $\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486$

Then, the standard error of $\widehat{Y} is$

$\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:$ =83.785

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]$

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

$\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45$

The sample ratio is

$r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}$ =0.1 , this is also the estimate of the population ratio R= $\frac{\overline{Y}}{\overline{X}} .$

Therefore, the estimated total weight of grain in the population using ratio estimator is

$\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\,$ lbs

4) The variance of the ratio estimator is

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Hence, the standard error of the estimate of the total population is

$\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:$ =21.556

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]$

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