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Lelu [443]
3 years ago
6

Lindsey is using a map to find the distance between her house and Juanita’s house. On the map, the distance is 2.5 in. If the ma

p scale is 1/8 in. : 1.5 mi, how far from Juanita does Lindsey live?
Mathematics
1 answer:
telo118 [61]3 years ago
7 0
I like using decimals instead of fractions.
1/8: 0.125
So the scale is:
0.125 in: 1.5 mi
What you do to one side you have to do to the other side. We have the map distance, which is 2.5. Let's see what it was multiplied by to find the map distance of their houses.
2.5÷ 0.125= 20
Let's multiply the other side by 20.
20×1.5
30
So, Juanita and Lindsey live 30 minutes away.
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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

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