Using Chebyshev's Theorem, considering a standard deviation of 40, we have that at least 75% of the students scored between 360 and 520.
<h3>What does Chebyshev’s Theorem state?</h3>When we have no information about the population distribution, Chebyshev's Theorem is used. It states that:
In this problem, considering a standard deviation of 40, we have that:
440 - 2 x 40 = 360.
440 + 2 x 30 = 520.
Within 2 standard deviations of the mean, no information about the distribution, hence, at least 75% of the students scored between 360 and 520.
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a = amount invested at 7%
b = amount invested at 9%
we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000. We can also say that
since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.
The maximum volume of the box is 40√(10/27) cu in.
Here we see that volume is to be maximized
The surface area of the box is 40 sq in
Since the top lid is open, the surface area will be
lb + 2lh + 2bh = 40
Now, the length is equal to the breadth.
Let them be x in
Hence,
x² + 2xh + 2xh = 40
or, 4xh = 40 - x²
or, h = 10/x - x/4
Let f(x) = volume of the box
= lbh
Hence,
f(x) = x²(10/x - x/4)
= 10x - x³/4
differentiating with respect to x and equating it to 0 gives us
f'(x) = 10 - 3x²/4 = 0
or, 3x²/4 = 10
or, x² = 40/3
Hence x will be equal to 2√(10/3)
Now to check whether this value of x will give us the max volume, we will find
f"(2√(10/3))
f"(x) = -3x/2
hence,
f"(2√(10/3)) = -3√(10/3)
Since the above value is negative, volume is maximum for x = 2√(10/3)
Hence volume
= 10 X 2√(10/3) - [2√(10/3)]³/4
= 2√(10/3) [10 - 10/3]
= 2√(10/3) X 20/3
= 40√(10/27) cu in
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