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2 years ago

May someone please help me, please? This is about exponents by the way. About 5 questions I need help with.

Mathematics

2 answers:

Rina8888 [55]2 years ago

8 0

Anything that is raised to 0 ALWAYS equals to 1.

Leno4ka [110]2 years ago

5 0

Answer:

1. 1

2. 1

3. 16/81

4. 3

5. 3

Step-by-step explanation:

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Need help. If u answer fast or if u are first i will give u brainliest

aksik [14]

Answer:

It is correct.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Solve the equation for y. Identify the slope and y-intercept then graph the equation.

drek231 [11]

Answer:

y = 1 m = -3 b = 1

Step-by-step explanation:

y = mx + b y = -3x + 1

y = 1

m = -3

b = 1

5 0

3 years ago

For a certain online store, the distribution of number per hour is approximately normal with mean 1200 purchases and standard de

Advocard [28]

Answer: 16%

Step-by-step explanation:

When we have a normal distribution, the 100% of our sample will be placed in a "gaussian bell".

Where the middle of this bell coincides with the mean.

The mean, in this case, is 1200, and the standard deviation is 200.

In the image below you can see that between the points.

Mean + standard deviation and the end of the graph, we have:

13.5% + 2.35% + 0.15% = 16%

And particularly, if we want to know the percentage that exceed 1400, will be the same percentage above the mean plus the standard deviation.

mean = 1200

SD = 200

mean + standard deviation = 1200 + 200 = 1400.

Then the percentage that exceeds 1400 is equal to the percentage that we found above, then the correct option is C: 16%

5 0

3 years ago

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

Amiraneli [1.4K]

Answer:

Step-by-step explanation:

bet

7 0

3 years ago

Read 2 more answers