Answer:
0.9963 is the probabilities that mean number of daily activity of 6 lean people exceed 410 minutes.
Step-by-step explanation:
Solution:
6 Mildly obese people exceed 410 minutes.
Given:
Mean = µ = 376
Standard deviation = ∂ = 70
n = 6
p (x> 410) = p (z > x - µ/ ∂ / √n)
= p (z > 410 – 376 / 70 /√6 )
= p (z > 1.185)
= 1 – p (z < 1.185)
= 1 – 0.38199
= 0.618
0.618 is the probability that the mean number of minutes of daily activity of 6 mildly obese people exceed 410 minutes.
6 lean people exceed 410 minutes:
Given:
Mean = µ = 526
Standard deviation = ∂ = 106
n = 6
=p(x > 410)
= p (z > x - µ/ ∂ /√n)
=p (z > (410 – 526 ) / 106 / √ 6 )
= p (z > - 2.67)
= p (z <2.68)
= 0.9963
0.9963 is the probabilities that mean number of daily activity of 6 lean people exceed 410 minutes.
The correct answer is A. 17
Hope this helps
Answer: There is no quetion this is not helpful i can't help if i don't no the problem
Step-by-step explanation:
We can to make the method of substitution.
Look:
3x + 15y = 35
Isolating 3x
15y = 35 - 3x
Dividing both the sides by 15
y = 35/15 - 3x/15
Simpliting:
y = 7/3 - x/5
Now let's equate this expression with the other
7/3 - x/5 = - x/5 + 7/3
0 = 0
Oh noOo!
There is solution infinatly to this question!
Because, the two equation are the same.
Hope this helps!
