5 gallons x gallons
------------- = ----------------
12 hours 240 hours
using cross products
5 * 240 = 12 * x
1200 = 12 x
divide each side by 12
1200/12 = 12x/12
100 =x
You will need 100 gallons
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Researcher
Military and airline forces
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Answer:
Dimensions :
x (the longer side, only one side with fence ) = 90 ft
y ( the shorter side two sides with fence ) = 45 ft
Total fence used 45 * 2 + 90 = 180 ft
A(max) =
Step-by-step explanation: If a farmer has 180 ft of fencing to encloses a rectangular area with fence in three sides and the river on one side, the farmer surely wants to have a maximum enclosed area.
Lets call "x" one the longer side ( only one of the longer side of the rectangle will have fence, the other will be along the river and won´t need fence. "y" will be the shorter side
Then we have:
P = perimeter = 180 = 2y + x ⇒ y = ( 180 - x ) / 2 (1)
And A (r) = x * y
A(x) = x * ( 180 - x ) /2 ⇒ A(x) = (180/2) *x - x² / 2
Taking derivatives on both sides of the equation :
A´(x) = 90 - x
Then if A´(x) = 0 ⇒ 90 - x = 0 ⇒ x = 90 ft
and from : y = ( 180 - x ) / 2 ⇒ y = 90/2
y = 45 ft
And
A(max) = 90 * 45 = 4050 ft²
