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2 years ago

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For a drag race car that weighs 1600 kilograms, the velocity v (in kilometers per hour) reached by the end of a drag race can be

modeled by the function v=23.8\sqrt[3]{p} , where p is the car's power (in horsepower). Estimate the power of a 1600-kilogram car that reaches a velocity of 220 kilometers per hour. Round your answer to the nearest whole number.

1 answer:

vovangra [49]2 years ago

8 0

Answer:

790

Step-by-step explanation: i just got it from someone else

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The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA ba

PSYCHO15rus [73]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA basketball tournament (not all private schools report their coach's salaries). Consider these 62 salaries to be a sample from the population of salaries of all 346 NCAA Division I basketball coaches.

Question 1. Use the 62 salaries from the TOTAL PAY column to construct a 95% confidence interval for the mean salary of all basketball coaches in NCAA Division I.

xbar = $1,465,752

SD = $1,346,046.2

lower bound of confidence interval ________

upper bound of confidence interval _______

Question 2. Coach Mike Krzyzewski's high salary is an outlier and could be significantly affecting the confidence interval results. Remove Coach Krzyzewski's salary from the data and recalculate the 95% confidence interval using the remaining 61 salaries.

xbar = $1,371,191

SD = $1,130,666.5

lower bound of confidence interval _________

upper bound of confidence interval. ________

Answer:

Question 1:

lower bound of confidence interval = $1,124,027

upper bound of confidence interval = $1,807,477

Question 2:

lower bound of confidence interval = $1,081,512

upper bound of confidence interval = $1,660,870

Step-by-step explanation:

Question 1:

The sample mean salary of 62 couches is

$\bar{x} = 1,465,752$

The standard deviation of mean salary is

$s = 1,346,046.2$

The confidence interval for the mean salary of all basketball coaches is given by

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } )$

Where $\bar{x}$ is the sample mean, n is the sample size, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 62 - 1 = 61

From the t-table at α = 0.025 and DoF = 61

t-score = 1.999

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (\frac{1,346,046.2}{\sqrt{62} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (170948.04 ) \\\\CI = 1,465,752 \pm 341,725 \\\\LCI = 1,465,752 - 341,725 = 1,124,027 \\\\UCI = 1,465,752 + 341,725 = 1,807,477\\\\$

Question 2:

After removing the Coach Krzyzewski's salary from the data

The sample mean salary of 61 couches is

$\bar{x} = 1,371,191$

The standard deviation of the mean salary is

$s = 1,130,666.5$

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 61 - 1 = 60

From the t-table at α = 0.025 and DoF = 60

t-score = 2.001

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (\frac{1,130,666.5}{\sqrt{61} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (144767 ) \\\\CI = 1,371,191 \pm 289,678.8 \\\\LCI = 1,371,191 - 289,678.8 = 1,081,512 \\\\UCI = 1,371,191 + 289,678.8 = 1,660,870\\\\$

6 0

3 years ago

Câu 5 (3.0 điểm) Cho đường tròn (O) bán kính R ngoại tiếp tam giác ABC có ba góc nhọn.

Luda [366]

Answer:

almost nobody here understands your language sorry

3 0

3 years ago

Please answer this question for brainliest!

kkurt [141]

The original side length is given as 3.00

Multiply the original length by the scale factor:

3.00 x 0.75 = 2.25

The answer is B. 2.25 cm

3 0

3 years ago

A pair of jeans at a store in Seattle costs $220. Since the jeans are from last season, they are 15% off. Because of a Thanksgiv

Lostsunrise [7]

The total percent discount on the jeans is 35% amounting to $77

<h3>Discount</h3>

Cost of a pair of jean = $220

Discount from last season = 15%
Discount from Thanksgiving day = 20%

Total percentage discount = 15% + 20%

= 35%

Amount of discount = Total percentage discount × Cost of a pair of jean

= 35% × $220

= 0.35 × 220

= $77

Therefore, the total percent discount on the jeans is 35% amounting to $77

Learn more about discount:

brainly.com/question/7459025

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6 0

2 years ago

HELP ME DO THIS PLZ

cestrela7 [59]

Answer:

It landed at a height of 14 feet

Step-by-step explanation:

Given

$y = -\frac{1}{8}x^2 +4x$ --- Path of a t-shirt

$3y = 2x - 14$ --- height of bleachers

Required

Height when t-shirts land in bleachers

$3y = 2x - 14$

Make y the subject

$y = \frac{2}{3}x - \frac{14}{3}$

Substitute $y = \frac{2}{3}x - \frac{14}{3}$ in $y = -\frac{1}{8}x^2 +4x$

$\frac{2}{3}x - \frac{14}{3} = -\frac{1}{8}x^2 +4x$

Multiply through by 24

$16x- 112 = -3x^2 +96x$

Express as:

$3x^2 + 16x -96x- 112 = 0$

$3x^2 -80x- 112 = 0$

Using a calculator:

$x = 28$ or $x = -\frac{4}{3}$

x can not be negative:

So:

$x = 28$

Substitute $x = 28$ in $y = -\frac{1}{8}x^2 +4x$ to calculate the height it landed

$y = -\frac{1}{8} * 28^2 + 4 * 28$

$y = 14$

3 0

3 years ago