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ra1l [238]
4 years ago
10

The boundaries of the shaded region are the y-axis, the line y = 15, and the curve y = 15 4 x . Find the area of this region by

writing x as a function of y and integrating with respect to y.

Mathematics
1 answer:
Feliz [49]4 years ago
8 0

Answer:

Area = 3

Step-by-step explanation:

In given figure curve is given as:

                               y=15\sqrt[4]{x}

                               \frac{y}{15}=\sqrt[4]{x}

                               (\frac{y}{15})^{4}=x

                               x=\frac{y^{4}}{15^{4}} ---(1)

Area of bounded region is found by integrating above equation (1) between limits 0 to 15.

                 A=\frac{1}{15^{4}}\int\limits^{15}_{0} {y^{4}} \, dy

                 A=\frac{1}{15^{4}}[\frac{y^{5}}{5}]^{15}_{0}

                 A=\frac{1}{15^{4}}[\frac{15^{5}}{5}-0]

                 A=\frac{15}{5}

                 A=3

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The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard
mezya [45]

Answer:

0.6563 or 65.63% of brook trout caught will be between 12 and 18 inches

Step-by-step explanation:

Mean trout length (μ) = 14 inches

Standard deviation (σ) = 3 inches

The z-score for any given trout length 'X' is defined as:  

z=\frac{X-\mu}{\sigma}  e interval

For a length of X =12 inches:

z=\frac{12-14}{3}\\z=-0.6667

According to a z-score table, a score of -0.6667 is equivalent to the 25.25th percentile of the distribution.

For a length of X =18 inches:

z=\frac{18-14}{3}\\z=1.333

According to a z-score table, a score of 1.333 is equivalent to the 90.88th percentile of the distribution.

The proportion of trout caught between 12 and 18 inches, assuming a normal distribution, is the interval between the equivalent percentile of each length:

P(12\leq X\leq 18) = 90.88\% - 25.25\%\\P(12\leq X\leq 18) = 65.63\%

8 0
3 years ago
If 1200 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the b
Colt1911 [192]

Answer:

The largest possible volume V is ;

V = l^2 × h

V = 20^2 × 10 = 4000cm^3

Step-by-step explanation:

Given

Volume of a box = length × breadth × height= l×b×h

In this case the box have a square base. i.e l=b

Volume V = l^2 × h

The surface area of a square box

S = 2(lb+lh+bh)

S = 2(l^2 + lh + lh) since l=b

S = 2(l^2 + 2lh)

Given that the box is open top.

S = l^2 + 4lh

And Surface Area of the box is 1200cm^2

1200 = l^2 + 4lh ....1

Making h the subject of formula

h = (1200 - l^2)/4l .....2

Volume is given as

V = l^2 × h

V = l^2 ×(1200 - l^2)/4l

V = (1200l - l^3)/4

the maximum point is at dV/dl = 0

dV/dl = (1200 - 3l^2)/4

dV/dl = (1200 - 3l^2)/4 = 0

3l^2= 1200

l^2 = 1200/3 = 400

l = √400

I = 20cm

Since,

h = (1200 - l^2)/4l

h = (1200 - 20^2)/4×20

h = (800)/80

h = 10cm

The largest possible volume V is ;

V = l^2 × h

V = 20^2 × 10 = 4000cm^3

4 0
3 years ago
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