All categories

2 years ago

Do you go to the movies at least twice a week?

Mathematics

1 answer:

egoroff_w [7]2 years ago

5 0

Answer:

We are given the below table of information:

Yes No Total

Male 35 45 80

Female 67 28 95

Total 102 73 175

We have to find the probability that a person who does not go to the movies at least twice a week is female.

We know that there are 73 students who does not go to the movies at least twice a week

Also we know that there are 28 female's who does not go to the movies at least twice a week.

Therefore, the probability that a person who does not go to the movies at least twice a week is female is:

$\bold{\frac{28}{73}=0.384}$

You might be interested in

a port and a radar station are 2 mi apart on a straight shore running east and west. a ship leaves the port at noon traveling no

Rom4ik [11]

Answer:

The rate of change of the tracking angle is 0.05599 rad/sec

Step-by-step explanation:

Here the ship is traveling at 15 mi/hr north east and

Port to Radar station = 2 miles

Distance traveled by the ship in 30 minutes = 0.5 * 15 = 7.5 miles

Therefore the ship, port and radar makes a triangle with sides

2, 7.5 and x

The value of x is gotten from cosine rule as follows

x² = 2² + 7.5² - 2*2*7.5*cos(45) = 39.04

x = 6.25 miles

By sine rule we have

$\frac{sin A}{a} = \frac{sin B}{b}$

Therefore,

$\frac{sin 45}{6.25} = \frac{sin \alpha }{7.5}$

α = Angle between radar and ship α

∴ α = 58.052

Where we put

$\frac{sin 45}{6.25} = \frac{sin \alpha }{x}$ to get

$\frac{x}{6.25} = \frac{sin \alpha }{sin 45}$ and differentiate to get

$\frac{\frac{dx}{dt} }{6.25} = cos\alpha\frac{\frac{d\alpha }{dt} }{sin 45}$

$\frac{15sin45 }{6.25cos\alpha } =\frac{d\alpha }{dt} }$ = 3.208 degrees/second = 0.05599 rad/sec.

6 0

3 years ago

Kars

8090 [49]

vaka laka vakanlaka vaka laka

6 0

3 years ago

If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?

zavuch27 [327]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2866883

_______________

   dy
Find —— for an implicit function:
    dx

x²y – 3x = y³ – 3

First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

$\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\
\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}$

Applying the product rule for the first term at the left-hand side:

$\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\
\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}$

   dy
Now, isolate —— in the equation above:
   dx

$\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\
\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\
\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\
\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}$

Compute the derivative value at the point (– 1, 2):

x = – 1 and y = 2

$\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>

6 0

2 years ago

If f(x)=x^2 then evaluate (2+3i)

Virty [35]

Answer:

-5 + 12 i

Step-by-step explanation:

(2 + 3i)^2 = 4 + 6i + 6i + 9i^2

= 4 + 12 i + 9(-1)

= - 5 + 12 i

3 0

1 year ago

How to determine proofs

wariber [46]

Answer:

check footprints and then check finger print the finger print pour powder on it and Trace

6 0

3 years ago