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sergeinik [125]
2 years ago
14

Determine whether the random variable is discrete or continuous. In each​ case, state the possible values of the random variable

. ​(a) The number of points scored during a basketball game. ​(b) The amount of rain in City Upper B during April.
Mathematics
1 answer:
Alex_Xolod [135]2 years ago
7 0

Answer:

a) Discrete, because the number of point scored during basket ball is countable.

For instance, the amount of point scored in a basketball could be 75, 103, 63 etc. The numbers are countable

b) Continuous, because the amounts of rainfall is a random variable that is uncountable.

For instance, the amount of rainfall in City Upper B during April could be 0.10 inches of rain per hour, 0.30 inches of rain per hour. This numbers are not countable, they are rather approximated or rounded off.

Step-by-step explanation:

A random variable is considered discrete if its possible values are countable while a random variable is considered to be continuous if it's possible values are not countable.

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Ben has a part time job at the fun station. Suppose he worked 13.5 hours last week and $81. How much does Ben earn per hour
scoundrel [369]

To get the answer, you will just use this operation: Division.
Let's divide it. 81 ÷ 13.5 = 6.

So, the answer is
Ben earns $6 per hour at the fun station.

8 0
3 years ago
Read 2 more answers
For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f
Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

1.

\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0

We want to find f such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)

\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C

\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}

so \vec F is conservative.

2.

\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)

\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)

\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C

\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}

so \vec F is conservative.

3.

\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0

so \vec F is not conservative.

4.

\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)

\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)

\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C

\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}

so \vec F is conservative.

4 0
3 years ago
which has greater cenetic energy a car traveling 30.0 km/hr or one twice as heavy traveling at 15 km/hr?​
iVinArrow [24]

Answer:

30 km/h car

Step-by-step explanation:

From analysis the   car traveling at 30 km/h has greater kinetic energy

we can deduce it from the expression of kinetic energy which is

KE=\frac{1}{2} mv^2

Assuming the mass m= 1 kg

 

For the 30 km/h

KE=\frac{1}{2}*1*30^2 \\\\KE=\frac{1}{2}*1*900\\\\\KE=450 J

   

For the 15 km/h

KE=\frac{1}{2}*2*15^2 \\\\ KE=\frac{1}{2}*2*225 \\\\\ KE=\frac{1}{2}*450 J\\\\\ KE=225 J

Though the kinetic energy is a function of mass and velocity, but from our analysis the faster moving object has more KE

8 0
3 years ago
P and Q are two points on the line x - y + 1 = 0 and are at distance 5 from the origin. Find the area of the triangle OPQ. ​
Alecsey [184]

Answer:

P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.

Step-by-step explanation:

P and Q are the intersection points of

x-y+1 = 0 and the circle x^2 + y^2 = 25

sub y = x+1 into the circle

x^2 + (x+1)^2 = 25

x^2 + x^2 + 2x + 1 - 25 = 0

x^2 + x - 12 = 0

(x+4)(x-3) = 0

x = 3 or x = -4

y = 4 or y = -3

so P(3,4) and Q(-4,3) are our two points

Height of triangle.

h = |0 - 0 + 1|/√2 = 1/√2

PQ = √( (-7)^2 + 1^2) = √50 = 5√2

area POQ = (1/2)(1/√2)(5√2) = 5/2 square units

hope this helped

7 0
2 years ago
Express sin UU as a fraction in simplest terms.
alukav5142 [94]

The expression of sinU as a fraction is 12/13

Find the diagram attached.

To get the fraction represented by sinU, we will use the SOH CA TOA identity.

From the diagram;

  • Hypotenuse = SU = 13
  • Opposite = ST = 12 (angle opposite to m<U)

Since sin theta = opp/hyp

SinU = ST/SU

SinU = 12/13

Hence the expression of sinU as a fraction is 12/13

Learn  more on SOH CAH TOA here: brainly.com/question/20734777

5 0
2 years ago
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