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djyliett [7]
2 years ago
11

Any help with this question kinda brained

Mathematics
1 answer:
kodGreya [7K]2 years ago
4 0
I believe the answer is is two
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What is 60 - 56 divided by 8•2 will give brainliest
Paladinen [302]

Answer:

1/4

Step-by-step explanation:

if you subtract the top, and multiply the bottom, the outcome is 4/16 or 1/4

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Marcus monthly salary is $1,500 plus 10% of sales. How much could Marcus earn per month with salary and 9% sales? Please explain
sergejj [24]

If Marcus monthly salary is $1,500 plus 10% of sale, then he could earn 1500 + 0.09x per month with salary and 9% sales

Monthly salary of Marcus = $1500

The percentage incentives for sales = 10%

If the percentage of incentive for sales is 9 %

Consider the amount of sales is $x

Therefore the monthly salary of Marcus = Monthly salary of Marcus + (The amount of sales × 9/100)

Substitute the values in the equation

The monthly salary of Marcus = 1500 + (x×9/100)

Divide the numbers in the equation

= 1500 + (x × 0.09)

Multiply the numbers the numbers

= 1500 + 0.09x

Therefore, the monthly salary of Marcus will be 1500 + 0.09x

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8 0
1 year ago
A store has apple on sale for $20.00 for 8 pounds. If an apple is approximately 5 ounces, how manyapples can you buy for $120.00
lubasha [3.4K]

Answer: 24 apples

Step-by-step explanation: $20.00 = 8 pounds and 16 ounces = 1 pound So you can do $120.00 divided by 5 = 24 apples

4 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
14.
vladimir2022 [97]

Answer:

`15

Step-by-step explanation:

5 0
2 years ago
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