GCF 32 24 40
^ ^ ^
2 16 2 12 2 20
^ ^ ^
2 8 2 6 2 10
^ ^ ^
2 4 2 3 2 5
^
2 2
(2) (2 & 3) (2 & 5)
Answer:
Step-by-step explanation:
Hello!
X: the lifespan of a new computer monitor of Glotech.
The average life is μ= 85 months and the variance δ²= 64
And a sample of 122 monitors was taken.
You need to calculate the probability that the sample mean is greater than 86.6 months.
Assuming that the variable has a normal distribution X~N(μ;δ²), then the distribution of the sample mean is X[bar]~N(μ;δ²/n)
To calculate this probability you have to work using the sampling distribution and the following formula Z= (X[bar]-μ)/δ/√n ~N(0;1)
P(X[bar]>86.6)= 1 - P(X[bar]≤86.6)
1 - P(Z≤(86.6-85)/(8/√122))= 1 - P(Z≤2.21)= 1 - 0.98645= 0.013355
The probability of the sample mean is greater than 0.013355
I hope this helps!
Answer: 3sqrt6
Step-by-step explanation:
Answer:
15.81
Step-by-step explanation:
Please mark brainliest if this has helped. Have a good day and goodluck with your work!
Answer:
9abc(ac² - 3)
Step-by-step explanation:
<u>Given polynomial:</u>
<u>Factors of the terms:</u>
- 9a²bc³ = 3*3*a*a*b*c*c*c
- 27abc = 3*3*3*a*b*c
<u>Common factors of the above: </u>
<u>So the answer:</u>
- 9a²bc³ -27abc = 9abc(ac² - 3)
