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Anarel [89]
3 years ago
6

Can someone plz help me with this one problem plzzzzz (I’M MARKING BRAINLIEST)

Mathematics
2 answers:
Blizzard [7]3 years ago
7 0
1: 23
2: 46
3:69
6: 138
Juli2301 [7.4K]3 years ago
3 0
23s+75=c
S c
1(23)+75=98
2(23)+75=121
3(23)+75=144
6(23)+75=213
Hope this helps
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The following probability distributions of job satisfaction scores for a sample of information
kodGreya [7K]

Answer:

a. 4.05 b. 3.84 c. 1.2475 and 1.1344 d. 1.1169 and 1.0651  e. We can say that the overall job satistaction of senior executives and middle managers is about 4; however, there is more variability in the job satisfaction for senior executives than in the job satisfaction for middle managers.

Step-by-step explanation:

a. (1)(0.05)+(2)(0.09)+(3)(0.03)+(4)(0.42)+(5)(0.41) = 4.05

b. (1)(0.04)+(2)(0.1)+(3)(0.12)+(4)(0.46)+(5)(0.28) = 3.84

c. We compute the variances as follow: [(1)^2(0.05)+(2)^2(0.09)+(3)^2(0.03)+(4)^2(0.42)+(5)^2(0.41)] - 4.05^2 = 1.2475 and [(1)^2(0.04)+(2)^2(0.1)+(3)^2(0.12)+(4)^2(0.46)+(5)^2(0.28)]-3.84^2 = 1.1344

d. The standard deviation is the squared root of the variance, therefore, we have \sqrt{1.2475} = 1.1169 and \sqrt{1.1344} = 1.0651

e. The expected value of the job satisfaction score for senior executives is very similar to the job satisfaction score for middle managers. We can say that the overall job satistaction of senior executives and middle managers is about 4; however, there is more variability in the job satisfaction for senior executives than in the job satisfaction for middle managers.

4 0
3 years ago
the indian ocean is 2/10 if the area of the world's oceans.What fraction represents the area of the remaining oceans that make u
aleksklad [387]
8/10=4/5, your answer is my pleasure!

6 0
3 years ago
-18 subtracted from - 7
castortr0y [4]
The answer is 11 my friend :)
5 0
2 years ago
Read 2 more answers
A box contains 16 ​transistors, 5 of which are defective. if 5 are selected at​ random, find the probability that
ExtremeBDS [4]
Part (a): All are defective
Only one way of selecting the 5 defective transistors:

Number of ways of selections available = 6C5 = 16!/[5!*(15-5)!] = 4368

Probability they are all defective = Number of ways of selecting 5 defectives/Total number of ways possible = 1/4368 ≈ 0.000229

Part (b): None are defective
Total number of non defectives = 16 -5 = 11
Number of ways of selecting 5 non defective = 11C5 = 462 ways
Total number of ways possible = 16C5 = 4368
Probability of selecting 5 non defectives = 462/4368 = 11/104 ≈ 0.1058


5 0
4 years ago
Read 2 more answers
89.51 68.168 11.047 in order least to greatest
Goryan [66]

Answer:

11.047---68.168---89.51

Step-by-step explanation:

just switch 11 and 89

4 0
3 years ago
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