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2 years ago

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Two drivers, Alison and Kevin, are participating in a drag race. Beginning from a standing start, they each proceed with a const

ant acceleration. Alison covers the last 1/4 of the distance in 3 seconds, whereas Kevin overs the last 1/3 of the distance in 4 seconds. Who wins and by how much time?

1 answer:

Vilka [71]2 years ago

4 0

Answer:

Alison wins against Kevin by 0.93 s

Step-by-step explanation:

Alison covers the last 1/4 of the distance in 3 seconds, at a constant acceleration $a_a$ , we have the following equation of motion

$s/4 = a_at_a^2/2$

where s (m) is the total distance, ta = 3 s is the time

$s = 4a_a3^2/2 = 18a_a$

$a_a = s/18$

Similarly, Kevin overs the last 1/3 of the distance in 4 seconds, at a constant acceleration $a_k$ , we have the following equation of motion:

$s/3 = a_kt_k^2/2$

tk = 4 s is the time

$s = 3a_k4^2/2 = 24a_k$

$a_k = s/24$

Since $a_a = s/18$ we can conclude that $a_a > a_k$ , so Alison would win.

The time it takes for Alison to cover the entire track

$s = a_aT_a^2/2$

$T_a^2 = 2s/a_a = 2s/(s/18) = 36$

$T_a = \sqrt{36} = 6 s$

The time it takes for Kevin to cover the entire track

$s = a_kT_k^2/2$

$T_k^2 = 2s/a_k = 2s/(s/24) = 48$

$T_a = \sqrt{48} = 6.93 s$

So Alison wins against Kevin by 6.93 - 6 = 0.93 s

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Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45

vovikov84 [41]

Answer:

a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544

b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390

ii. There are no statistical significant evidence that the prep course helped

d. i. The 95% confidence interval for the change in average test scores is 3.47467 < μ₁ - μ₂ < 14.52533

ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

Step-by-step explanation:

The mean of the standardized test = 1,000

The number of students test to which the test is administered = 453 students

The mean score of the sample of students, $\bar{x}$ = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

$CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}$

At 95% confidence level, z = 1.96, therefore, we have;

$CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}$

Therefore, we have;

1,022.94559 < μ < 1,003.0544

b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. The parameters of the students taking the test are;

The number of students, n = 503

The number of hours preparation the students are given, t = 3 hours

The average test score of the student, $\bar{x}$ = 1019

The number of test scores of the student, s = 95

At 95% confidence level, z = 1.96, therefore, we have;

The confidence interval, C.I., for the difference in mean is given as follows;

$C.I. = \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

Therefore, we have;

$C.I. = \left (1013- 1019 \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}$

Which gives;

-18.955390 < μ₁ - μ₂ < 6.955390

ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped

d. The given parameters are;

The number of students taking the test = The original 453 students

The average change in the test scores, $\bar{x}_{1}- \bar{x}_{2}$ = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = $\bar{x}_{1}- \bar{x}_{2}$ + 1.96 × Δs/√n

∴ C.I. = 9 ± 1.96 × 60/√(453)

i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533

ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

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True or false: when comparing two populations, the larger the standard deviation, the more dispersion the distribution has,

Nataliya [291]

<span>It would be...
C) true, because the standard deviation describes how far, on average, each observation is from the typical value. a larger standard deviation means that observations are more distant from the typical value, and therefore, more dispersed.
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An American car travels 32 miles on one gallon of gas. A European car travels 12.7 kilometers on one liter of gas. Which car get

BabaBlast [244]

32 miles is the same as 19.8839 kilometers

There are 3.78541 liters in a gallon

Now we divide the two.

19.8839 / 3.78541 ≈ 5.25

The American car gets 5.25 km per liter. So the European car is better on gas economy.

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2 years ago

Which are the solutions of x2 = 19x + 1?

Finger [1]

Answer:

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

Step-by-step explanation:

we have

$x^2=19x+1$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$-x^{2}-19x-1=0$

so

$a=1\\b=-19\\c=-1$

substitute in the formula

$x=\frac{-(-19)\pm\sqrt{-19^{2}-4(1)(-1)}} {2(1)}$

$x=\frac{19\pm\sqrt{365}} {2}$

$x=\frac{19+\sqrt{365}} {2}$

$x=\frac{19-\sqrt{365}} {2}$

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

therefore

StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction

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Read 2 more answers

The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and

Alexus [3.1K]

Answer:

11.70% probability someone will spend no more than 30 minutes reading online national news reports.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 49, \sigma = 16$

What is the probability someone will spend no more than 30 minutes reading online national news reports?

This is at most 30 minutes, so 30 or less minutes. This is the pvalue of Z when X = 30.

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30 - 49}{16}$

$Z = -1.19$

$Z = -1.19$ has a pvalue of 0.1170.

So there is a 11.70% probability someone will spend no more than 30 minutes reading online national news reports.

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