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Answer:
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³
Step-by-step explanation:
From the given information;
Let a be the cost of the box
Let b be one side of the square base ; &
h to be the height of the box
We know that the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
The base = (0.65)a²
The top = (0.2)a²
The side = (0.2) a × 25/a²
= 5/a
For the four sides of the box now ;
= (0.2) 4a × 25/a²
= 0.8 × 25/a
= 20 /a
The total cost of the box is:
b = 0.65a² + 0.2a² + 20 /a
b = 0.85 a² + 20 /a
Taking differential of b with respect to a ;we have:
db/da = 1.7a - 1/a²(20) = 0
1.7 a³ - 20 = 0
1.7 a³ = 20
a³ = 20/1.7
a³ = 11.77
a =
a = 2.274 ft
Thus; the cost for the base of the box = (0.65)a²
the cost for the base of the box =(0.65) × ( 2.274)²
the cost for the base of the box = 3.362
The top of the box = (0.2)a²
The top of the box = (0.2)× ( 2.274)²
The top of the box = 1.034
The four sides of the box = 20 /a
The four sides of the box = 20/2.274
The four sides of the box = 8.795
the total cost = b = 0.85 a² + 20 /a
the total cost = 0.85 (2.274)² + 20 /2.274
the total cost = 4.395 + 8.795
the total cost = 13.19
Recall that:
the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
h = 10/ 2.274²
h 1.934
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³
Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:
What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So
has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79