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8090 [49]
3 years ago
6

A ball is thrown strait up from the top of a 64 foot tall building with an initial speed of 48 feet per second. The height of th

e ball as a function of time can be modeled by the function h(t)=-16t^2+48t+64. At what other time will the ball be at a height of 64 feet? How long will it take for f ball to reach its maximum height? What is he maximum height reached by the ball? How long willlit take for the ball to hit the ground?
Mathematics
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

see below

Step-by-step explanation:

h(t)=-16t^2+48t+64

Find when h(t) = 64

64 = -16t^2+48t+64

Subtract 64 from each side

0 =  -16t^2+48t

Factor

0 = -16t( t-3)

Using the zero product property

-16t =0    t-3 =0

t=0 and t=3

Other than at t=0,  so at t=3 seconds

How long will it take for f ball to reach its maximum height?

Find the vertex

t = -b/2a = -48 / ( 2 * -16) = -48/ -32 = 1.5 seconds

It will reach the maximum height at 1.5 seconds

What is the maximum height reached by the ball?

The height will be

h(1.5) = -16*(1.5)^2+48(1.5)+64

       = -36+72+64

       = 100

The maximum height is 100 ft

How long will it take for the ball to hit the ground?

h(t) =0

0=-16t^2+48t+64

Factor

0 = -16( t^2 - 3t -4)

0=- 16( t-4) (t+1)

Using the zero product property

t-4=0     t+1=0

t =4     t = -1

Since time cannot be negative

t=4 seconds

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To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of t
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Answer:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

p_v =P(t_{(18)}

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_{1}=19.1 represent the mean for the sample mean for 1

\bar X_{2}=23.3 represent the mean for the sample mean for 2

s_{1}=4.818 represent the sample standard deviation for the sample 1

s_{2}=5.559 represent the sample standard deviation for the sample 2

n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

Null hypothesis:\mu_{1} \geq \mu_{2}

Alternative hypothesis:\mu_{1} < \mu_{2}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=10+10-2=18

Since is a one sided test the p value would be:

p_v =P(t_{(18)}

Conclusion

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

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