Victoria had the following car expenses last year:

Could you help me and answer a couple questions for me?

kari74 [83]

Answer:

I think no. <em>D</em> is the answer

5 0

3 years ago

Explain the steps in calculating the mean absolute deviation of a set of data.

Anni [7]

Answer:

Step-by-step explanation:

Step 1: Calculate the mean.

Step 2: Calculate how far away each data point is from the mean using positive distances. These are called absolute deviations.

Step 3: Add those deviations together.

Step 4: Divide the sum by the number of data points.

7 0

3 years ago

Read 2 more answers

Light travels at 186,000 miles per second. How many FEET does it travel in a microsecond? (A microsecond is a millionth of a sec

Maslowich

1 mile is 5,280 ft, and 1 second has 1,000,000 microseconds.
So 186,000 miles per second multiplied by 5,280 feet per mile gives 982,080,000 feet per second.
Dividing this value by 1,000,000 microseconds per second gives the final answer of 982.08 feet per microsecond.

7 0

3 years ago

Which side in the image corresponds to Side length T V in the pre-image? Side length U T Side length V U Side length T S Side le

marshall27 [118]

Answer:

Answer: The side in the image that corresponds to Side length TV in the pre-image is TV

Step-by-step explanation:

4 0

3 years ago

The percent frequency distributions of job satisfaction scores for a sample of information systems (IS) senior executives and mi

tiny-mole [99]

Corrected Question

a. Develop a probability distribution for the job satisfaction score of a senior executive.

b. Develop a probability distribution for the job satisfaction score of a middle manager.

c. What is the probability a senior executive will report a job satisfaction score of 4 or 5?

d. What is the probability a middle manager is very satisfied?

Answer:

(c)0.83

(d)0.28

Step-by-step explanation:

The percent frequency distributions of job satisfaction scores os give below:

$\left|\begin{array}{c|c|c}$Job Satisfaction Score&$IS Senior Executives(\%)&$IS Middle Managers(\%)\\1&5&4\\2&9&10\\3&3&12\\4&42&46\\5&41&28\end{array}\right|$

(a)Probability distribution for the job satisfaction score of a senior executive.

$\left|\begin{array}{c|c}$Job Satisfaction Score&$Relative Frequency (f_1(x))\\------&------\\1&0.05\\2&0.09\\3&0.03\\4&0.42\\5&0.41\end{array}\right|$

(b)Probability distribution for the job satisfaction score of a middle manager.

$\left|\begin{array}{c|c}$Job Satisfaction Score&$Relative Frequency (f_2(x))\\1&0.04\\2&0.10\\3&0.12\\4&0.46\\5&0.28\end{array}\right|$

(c)Probability a senior executive will report a job satisfaction score of 4 or 5

P(a senior executive will report a job satisfaction score of 4 or 5)

$=f_1(4)+f_1(5)\\=0.42+0.41\\=0.83$

(d)Probability a middle manager is very satisfied

The probability a middle manager is very satisfied

$=f_2(5)\\=0.28$

7 0

3 years ago