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2 years ago

Sketch the region bounded by the curves y=3x3, x+y=4, and y=0. Find the coordinates of the centroid.

Mathematics

1 answer:

Scilla [17]2 years ago

5 0

The graph of the region is attached. The coordinates of the centroid are (5/3, 1).

The coordinates of the centroid are found by averaging the coordinates of the region;

Oₓ = (Aₓ+Bₓ+Cₓ)/3 = (0+1+4)/3 = 5/3
O(y) = (A(y) + B(y) + C(y)) = (0+3+0)/3=3/3=1

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The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years. Based on th

elena-14-01-66 [18.8K]

Answer: 0.9013

Step-by-step explanation:

Given mean, u = 10, standard deviation =8

P(X) =P(Z= X - u /S)

We are to find P(X> or =12)

P(X> or = 12) = P(Z> 12-10/8)

P(Z>=2/8) = P(Z >=0.25)

P(Z) = 1 - P(Z<= 0.25)

We read off Z= 0.25 from the normal distribution table

P(Z) = 1 - 0.0987 = 0.9013

Therefore P(X> or=12) = 0.9013

Note the question was given as an incomplete question the correct and complete question had to be searched online via Google. So the data used are those gotten from the online the Googled question.

3 0

1 year ago

1) The cost of constructing a path 5 m broad inside the boundary of a square lawn at Rs 36.25 per sq. metre is Rs 90625. What is

Svetach [21]

Let , side lawn is a.

Area of boundary is :

$A=4\times5\times a\\\\A=20a\ m^2$

Now, total cost is given by :

$T=20a\times 36.25\\\\90625=725a\\\\a=125\ m$

Area left is :

$A'=a^2-20a\\\\A'=125^2-20\times 125\ m^2\\\\A'=13125\ m^2$

Price to empty space with turfs is :

$P=A'\times 20\\\\P=13125\times 20\\\\P=262500$

Therefore, the cost of covering the empty space with turfs at the rate of Rs 20 per sq. metre is Rs 262500.

Hence, this is the required solution.

3 0

2 years ago

Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well

laiz [17]

Answer:

The value is $P(A) = 0.133617$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 7.5$

The standard deviation is $\sigma = 0.2$

The safest water level is between 7.2 and 7.8

Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

$P(A) = 1 - P(7.2 \le X \le 7.8 )$

Here

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } < \frac{X - \mu }{ \sigma }$

Generally $\frac{X - \mu }{ \sigma } = Z (The \ standardized \ value \ of X )$

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - 7.5 }{0.2 } < Z$

$P(7.2 < X < 7.8 ) = P(-1.5 < Z$

=> $P(7.2 < X < 7.8 ) = P(Z < 1.5) - P( Z < - 1.5)$

From the z-table the probability of (Z < -1.5) and ( Z <1.5) are

$P(Z < 1.5) = 0.93319$

and

$P(Z < -1.5) = 0.066807$

$P(7.2 < X < 7.8 ) =0.93319 - 0.066807$

$P(7.2 < X < 7.8 ) =0.0866383$

$P(A) = 1 - 0.0866383$

=> $P(A) = 0.133617$

5 0

1 year ago

Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.

Tanya [424]

Answer:

The larger cross section is 24 meters away from the apex.

Step-by-step explanation:

The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.

The length of the side of the hexagon is equal to the radius of the circle that inscribes it.

The area is

$A=\frac{3\sqrt{3} }{2} r^2$

Where $r$ is the radius of the inscribing circle (or the length of side of the hexagon).

Now we are given the areas of the two cross sections of the right hexagonal pyramid: $A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2$

From these areas we find the radius of the hexagons:

$r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}$

$r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}$

Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that $r_1$ $r_2$ form similar triangles with length $H$