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Murrr4er [49]
3 years ago
15

A set of cards contains the numbers 1 through 20. Mathieu chooses a card at random, records the number of the card, and then ret

urns the card to the set. He conducts 200 trials of this event. Based on the theoretical probability, how many times can Mathieu expect to choose a multiple of 5?​
Mathematics
2 answers:
Svetach [21]3 years ago
7 0
There are 4 multiples of 5 between 1-20 5.10.15.20 which are 4/20 aka 1/5 so 1/5 of 200 is 20 times Yaga yeet
JulsSmile [24]3 years ago
3 0

Answer:

40

Step-by-step explanation:

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6 0
3 years ago
The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives an
Mashcka [7]

Answer:

Step-by-step explanation:

To calculate ;

1) the expected value of the job satisfaction score for senior executives ;

expected value = Summation (Px)

= 1 x 0.05 + 2 x 0.09 + 3 x 0.03 + 4 x 0.42 + 5 x 0.41

= 4.05

2) the expected value of the job satisfaction score for middle managers;

= 1 x 0.04 + 2 x 0.10 + 3 x 0.12 + 4 x 0.46 + 5 x 0.28

= 3.84

c) the variance of job satisfaction scores for executives and middle managers (to 2 decimals).

Executives ; Variance = Summation(PX^2 - Summation(PX)^2

i) For Executive Managers = 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2 = 1.246 = 1.25

ii) for middle managers ; 1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 = 1.134 = 1.13

d) the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals). Executives, Middle managers;

For Executives = square root [ 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2] = 1.12

For Middle Managers ; Square root [1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 ] = 1.06

e) from the values gotten for the variance of both executive and middle managers, the variance of the former is more than that of the latter as such higher satisfaction with the executive managers.

5 0
3 years ago
Juarez Distribution Company conducted a time study of Martha Smith’s job as an installer. Using the observation chart, find her
Alex777 [14]
Average time on task 1 = \frac{3.4+4.1+6.5+8.1}{4}=5.5 seconds
Average time on task 2 = \frac{3.5+4.3+6.8+8.3}{4}=5.7 seconds
Average time on task 3 = \frac{4.3+4.1+6.5+7.9}{4}=5.7 seconds
Average time on task 4 = \frac{2.9+4.6+6.6+7.1}{4} =5.3 seconds
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4 0
3 years ago
Janet is designing a frame for a client. She wants to prove to her client that m∠AGE ≅ m∠CHE in her sketch. What is the missing
Amiraneli [1.4K]

Answer:

I believe the answer is transative property. I'm currently taking the test so I'm not 100% sure if I'm right yet.

Step-by-step explanation:

In step one, AGE and HGB are verticle angles since they are opposite of each other. then step two is alternate interior angles because they are interior angles opposite of each other, self-explanatory. the reason I choose the transitive property over the corresponding angles theorem is because the transitive property states: among A B and C, if A=B and B=c then A=C and in this case since AGE=HGB andHGB=CHE then AGE=CHE. I hope this helps and if you know the right answer just put it in the comments, if I'm wrong ofc.  

6 0
3 years ago
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
3 years ago
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