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2 years ago

If a fair die is rolled 3 times, what is the probability, to the nearest thousandth, of getting exactly 2 fives?

Mathematics

1 answer:

Alexxandr [17]2 years ago

5 0

Answer:

The answer is .578 you just do 5/6 due to the fact that there are 6 sides on a die and 1 side is a 2 so if you don’t get a two there is 5 sides left so you do 5/6 to the power of 3 leaving you with .578 and some

Step-by-step explanation:

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Please Help me solve this (Who ever answers correctly will get brainlest)

nlexa [21]

Answer:

15 2/3

Step-by-step explanation:

11 feet + 4 2/3 feet = 15 2/3

11+4= 15 + 2/3 = 15 2/3

Its actually very simple.

7 0

2 years ago

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Mamont248 [21]

Answer:

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2 years ago

The number of customers waiting for gift-wrap service at a department store is an rv X with possible values 0, 1, 2, 3, 4 and co

julia-pushkina [17]

Answer:

P[X=3,Y=3] = 0.0416

Step-by-step explanation:

Solution:

- X is the RV denoting the no. of customers in line.

- Y is the sum of Customers C.

- Where no. of Customers C's to be summed is equal to the X value.

- Since both events are independent we have:

P[X=3,Y=3] = P[X=3]*P[Y=3/X=3]

P[X=3].P[Y=3/X=3] = P[X=3]*P[C1+C2+C3=3/X=3]

P[X=3]*P[C1+C2+C3=3/X=3] = P[X=3]*P[C1=1,C2=1,C3=1]

P[X=3]*P[C1=1,C2=1,C3=1] = P[X=3]*(P[C=1]^3)

- Thus, we have:

P[X=3,Y=3] = P[X=3]*(P[C=1]^3) = 0.25*(0.55)^3

P[X=3,Y=3] = 0.0416

6 0

2 years ago

3. A balancing balloon toy is in the shape of a hemisphere (half-sphere) attached to the base of a cone. If the toy is 4ft tall

Katen [24]

Answer:

The volume of the toy is $V=5.23\ ft^3$

Step-by-step explanation:

step 1

Find the volume of the hemisphere

The volume of the hemisphere is given by the formula

$V=\frac{2}{3}\pi r^{3}$

In this problem, the wide of the toy is equal to the diameter of the hemisphere

$D=2\ ft$

$r=2/2=1\ ft$ ----> the radius is half the diameter

substitute

$V=\frac{2}{3} \pi (1)^{3}=\frac{2}{3} \pi\ ft^3$

step 2

Find the volume of the cone

The volume of the cone is given by

$V=\frac{1}{3}\pi r^{2}h$

we know that

The radius of the cone is the same that the radius of the hemisphere

$r=1\ ft$

The height of the cone is equal to subtract the radius of the hemisphere from the height of the toy

$h=4-1=3\ ft$

substitute the given values

$V=\frac{1}{3}\pi (1)^{2}(3)=\pi\ ft^3$

step 3

Find the volume of the toy

we know that

The volume of the toy, is equal to the volume of the cone plus the volume of the hemisphere.

$V=(\frac{2}{3} \pi+\pi)\ ft^3$

$V=(\frac{5}{3}\pi)\ ft^3$

assume

$\pi=3.14$

$V=\frac{5}{3}(3.14)=5.23\ ft^3$

5 0

3 years ago

The topic is volume & surface area, can someone help?

marishachu [46]

This pic should help

3 0

3 years ago