Answer:
Mean track length for this rock specimen is between 10.463 and 13.537
Step-by-step explanation:
99% confidence interval for the mean track length for rock specimen can be calculated using the formula:
M±
where
- M is the average track length (12 μm) in the report
- t is the two tailed t-score in 99% confidence interval (2.977)
- s is the standard deviation of track lengths in the report (2 μm)
- N is the total number of tracks (15)
putting these numbers in the formula, we get confidence interval in 99% confidence as:
12±
=12±1.537
Therefore, mean track length for this rock specimen is between 10.463 and 13.537
X^2+x-30 -- find two numbers that add to one and multiply to -30, which would be 6 and -5. so now you have x^2+6x-5x-30. factor out x and 5 to get x(x+6)-5(x+6). so you get (x+6)*(x-5)
Answer:
Value of x =20
Step-by-step explanation:
Given: EF = 9x+14 units , FG = 56 units and EG = 250 units.
Segment Addition Postulates states the following for 3 points that are collinear.
i.e, Let three points A, B and C are collinear and B is between A and C.
i,e AC = AB + BC
By Segment addition postulates; solve for x;
EG = EF + FG
Substitute the given values we get;
250 = 9x + 14 + 56
or
250 = 9x + 70
Subtract 70 on both sides we get;
250 -70 = 9x + 70 -70
Simplify:
180 = 9x
Divide both sides by 9 we get;

Simplify:
x = 20
Therefore, the value of x =20
45000 x 4 = 180,000
180,000 * 5 = 900,000 ml in 5 days
1 liter = 1000 ml
900,000 / 1000 = 900 liters of water