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3 years ago

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A water balloon is thrown out of a window 96 feet above the ground. The balloon is traveling downward at 16 feet per second init

ially. If h is the balloon’s height in feet after t seconds, then h = -16t 2 -16t + 96, It will take the balloon how many seconds to hit the ground.

2 answers:

nignag [31]3 years ago

4 0

Answer:

The balloon is thrown out a window that is 96 feet. We don't know if it is thrown at an angle we just know that it is moving at 16 feet per second so we would divide 16 from 96 to get 6.

torisob [31]3 years ago

3 0

It hits the ground at h=0

0=-16t^2-16t+96

0=-16(t^2+t-6)

0=-16(t-2)(t+3)

0= 2 and -3

Since you can't have negative time that excludes the -3. So 2 is the answer.

I've seen other people answering 96/16=6. The question says the balloon drops at 16feet/sec INITIALLY. That's why there is a specific formula for this. If your answer ends in something like "I think" then please don't answer.

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In a large midwestern university (the class of entering freshmen is 6000 or more students), an SRS of 100 entering freshmen in 1

Serga [27]

Answer:

The p-value of the test is 0.0228, which is less than the standard significance level of 0.05, which means that there is evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

1999:

20 out of 100 in the bottom third, so:

$p_1 = \frac{20}{100} = 0.2$

$s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

2001:

10 out of 100 in the bottom third, so:

$p_2 = \frac{10}{100} = 0.1$

$s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

Test if proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

At the null hypothesis, we test if the proportion is still the same, that is, the subtraction of the proportions in 1999 and 2001 is 0, so:

$H_0: p_1 - p_2 = 0$

At the alternative hypothesis, we test if the proportion has been reduced, that is, the subtraction of the proportion in 1999 by the proportion in 2001 is positive. So:

$H_1: p_1 - p_2 > 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the two samples:

$X = p_1 - p_2 = 0.2 - 0.1 = 0.1$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05$

Value of the test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{0.1 - 0}{0.05}$

$z = 2$

P-value of the test and decision:

The p-value of the test is the probability of finding a difference of at least 0.1, which is the p-value of z = 2.

Looking at the z-table, the p-value of z = 2 is 0.9772.

1 - 0.9772 = 0.0228.

The p-value of the test is 0.0228, which is less than the standard significance level of 0.05, which means that there is evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

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Answer:

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Answer:

Ground Morgans are 52% the price of Floating Jumps.

Step-by-step explanation:

A pair of Ground Morgans $GM$ is 20% less expensive than a pair of Elite Butterflies $EB$ , this means

$GM= 0.8 EB$ <em>(GM is 80% the price of EB's)</em>

Elite Butterflies $EB$ are 30% more expensive than Cloud Trainers $CT$ 's, this means

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Now, if we want the price of Ground Morgans in terms of Floating Jumps, then, since EB = 1.3 CT,

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and since CT = 0.5F, we have

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alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

So, no option is correct. And the answer is

$T^{-1}(x,y)=(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

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