The circumference of the inner circle is 66 ft. The distance between the inner circle and the outer circle is 44 ft. By how many
1 answer:
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Answer:
2.67 inches.
Step-by-step explanation:
Assuming that we represent the size of the squares with the letter y, such that after the squares are being cut from each corner, the rectangular length of the box that is formed can now be ( 23 - 2y), the width to be (13 - 2y) and the height be (x).
The formula for a rectangular box = L × B × W
= (23 -2y)(13-2y) (y)
= (299 - 46y - 26y + 4y²)y
= 299y - 72y² + 4y³
Now for the maximum volume:
dV/dy = 0
This implies that:
299y - 72y² + 4y³ = 299 - 144y + 12y² = 0
By using the quadratic formula; we have :
where;
a = 12; b = -144 and c = 299
Since the width is 13 inches., it can't be possible for the size of the square to be cut to be 9.33
Thus, the size of the square to be cut out from each corner to obtain the maximum volume is 2.67 inches.
Answer:
D. The center remains constant, and the area in the tails of the distribution decreases.
Step-by-step explanation:
Hello!
Be it two independent random variables, X~N(μ;σ²) and U~Xₙ², the variable t is determined by the quotient between a random variable N(0;1) and the square root of a Chi-Square variable divided by its degrees of freedom:
As a consequence of this, the structure of the distribution depends on the parameter n (degrees of freedom), it is centered in zero and has a bell-shape similar to the normal distribution.
It has a mean and variance:
E(t)= 0 for n > 1
V(t)= n > 2
As you can see the variance of the distribution is directly affected by its degrees of freedom, which means that when the degrees of freedom change, the variance of the distribution change and so does its shape.
When ↑n ⇒ ↑V(t) ⇒ The area under the tails increases.
When ↓n ⇒ ↓V(t) ⇒ The area under the tails decreases.
In this example, the degrees of freedom of the distribution decreased from 40 to 20, then the variance of the distribution decreases and it "flattens", i.e. the area under the tails gets lowered.
The E(t) isn't affected by the modification of n, so the distribution remains centered in zero.
I hope this helps!