Ask question

Ask question

All categories

2 years ago

8

An engineering student answered 81 questons correctly on a 90 question test. What percent of the question did she answer correct

ly?

1 answer:

anyanavicka [17]2 years ago

6 0

Answer:

90% of the questions she answered correctly

Step-by-step explanation:

%age = 81/90 * 100= 90 % correctness

You might be interested in

Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

natima [27]

Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ . where $\phi$ is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4 $\sqrt[]{3}$ if we take the positive square root of 48. So, let us calculate the angle between the vector $a=(0,4\sqrt[]{3},4)$ and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Therefore, over the circle, $\phi = \frac{\pi}{3}$ . Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since $z = \rho \cos \phi$ , then over the plane we have that $\rho = \frac{4}{\cos \phi}$ Then, the following is the desired integral

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ where the new factor is the jacobian for the spherical coordinates.

d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution $u = 64-r^2$ then $\frac{-du}{2} = r dr$ , then

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

3 0

2 years ago

Susan and ronald dugan agreed upon the price of 256,000 for their new home. They plan to make 30 percent down payment and financ

Eduardwww [97]

Answer:

The total amount to be paid for new home is $838,014.72

Step-by-step explanation:

Given as :

The price of the new house = $256,000

The down payment amount = 30% of house price

So, The down payment price = 30% of $256,000

i.e The down payment price = $\dfrac{30}{100}$ × 256000

Or, The down payment price = $76,800

Now, rest amount is finance

So, The finance Amount = p = $256000 - $76800 = $179,200

The rate of interest applied = r = 7.5%

The time period of finance = t = 20 years

Let The Amount after 20 years of finance = $A

Let The total amount to be paid for new home = $B

<u>Now, From Compound Interest </u>

Amount = Principal × $(1+\dfrac{\textrm rate}{100})^{\textrm time}$

Or, A = p × $(1+\dfrac{\textrm r}{100})^{\textrm t}$

Or, A = $179,200 × $(1+\dfrac{\textrm 7.5}{100})^{\textrm 20}$

Or, A = $179,200 × $(1.075)^{\textrm 20}$

Or, A = $179,200 × 4.24785

∴ A = $761,214.72

So,The Amount after 20 years of finance = A = $761,214.72

<u>Now, Again</u>

The total amount to be paid for new home = Down payment amount + The Amount after 20 years of finance

Or, B = $76,800 + A

Or, B = $76,800 + $761,214.72

Or, B = $838,014.72

So, The total amount to be paid for new home = B = $838,014.72

Hence, The total amount to be paid for new home is $838,014.72 Answer

7 0

2 years ago

If AD and BD are bisectors of CAB and CBA respectively. Find the sum of angles x and y.

olchik [2.2K]

The picture in the attached figure

step 1
we know that
It is given that AD and BD are bisectors of ∠CAB and ∠CBA respectively.
Therefore,
x = ∠CAB/2 -----> equation 1
y = ∠CBA/2 -----> equation 2

step 2
In triangle ABC,
∠CAB + ∠CBA + ∠ACB = 180° ----> [The sum of all three angles of a triangle is 180°]
∠CAB + ∠CBA + 110° = 180°
∠CAB + ∠CBA = 180° - 110°
∠CAB + ∠CBA = 70° ------> divide by 2 both sides
∠CAB/2 + ∠CBA/2 = 70/2 -------> equation 3
substitute equation 1 and equation 2 in equation 3
x+y=35

hence

the answer is
x+y =35°
⇒ x + y = 35° ...[From equation (1) and (2)]

5 0

2 years ago

The equation 4x2 – 24x + 4y2 + 72y = 76 is equivalent to

DerKrebs [107]

Answer:

Option 4 is correct.

The equation $4x^2 -24x + 4y^2 + 72y = 76$ is equivalent to $4(x-3)^2 + 4(y+9)^2 =436$

Step-by-step explanation:'

Given equation: $4x^2 -24x + 4y^2 + 72y = 76$

First group the terms with x and those with y;

$(4x^2-24x)+(4y^2+72y) = 76$

Next, we complete the squares.

We can do this by adding a third term such that the x terms and the y terms are perfect squares.

For this we must either add the same value on the other side of the equation or subtract the same value on the same side so that the equality is maintained.

⇒ $4(x^2-6x) +4(y+18y) = 76$

or

$4(x^2 -6x +3^2 -3^2) + 4(y^2 +18y +9^2 -9^2) = 76$

$4(x^2-6x + 3^2) - 36 + 4(y^2+18y +9^2) - 324 = 76$

$4(x-3)^2 + 4(y+9)^2 - 360 =76$

Add 360 on both sides we get;

$4(x-3)^2 + 4(y+9)^2 =360 +76$

Simplify:

$4(x-3)^2 + 4(y+9)^2 =436$

Therefore, the given equation is equivalent to $4(x-3)^2 + 4(y+9)^2 =436$

5 0

2 years ago

A blimp is 1100 meters high in the air and measures the angles of depression to two stadiums to the west of the blimp. If those

const2013 [10]

Answer:

The two stadiums are approximately 3115.1 meters away from each other

Step-by-step explanation:

Since we can construct two right angle triangles between the blimp and the two stadiums as shown in the attached image, then the distance "x" between the two can be find as the difference between the right triangle legs that extend on the ground.

In order to find the size of such legs, one can use the tangent function of the given depression angles as shown below:

$tan(75.2^o)=\frac{1100}{a} \\a=\frac{1100}{tan(75.2^o)}\\a\approx 290.6\,\,meters$

and for the other one:

$tan(17.9^o)=\frac{1100}{b} \\b=\frac{1100}{tan(17.9^o)}\\b\approx 3405.7\,\,meters$

The the distance between the stadiums is the difference:

b - a = 3405.7 - 290.6 meters = 3115.1 meters

8 0

2 years ago