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Andrej [43]
2 years ago
7

The formula for the volume of a sphere is mc012-1.jpg. What is the formula solved for r? mc012-2.jpg mc012-3.jpg mc012-4.jpg mc0

12-5.jpg
Mathematics
2 answers:
Tju [1.3M]2 years ago
8 0

we know that

The volume of the sphere is equal to

V=\frac{4}{3} \pi r^{3}

where

r is the radius of the sphere

Solve for r

V=\frac{4}{3} \pi r^{3}\\\\3V=4\pi r^{3}\\ \\r^{3}=\frac{3V}{4\pi}\\ \\r=\sqrt[3]{\frac{3V}{4\pi}}

therefore

<u>the answer is</u>

r=\sqrt[3]{\frac{3V}{4\pi}}

NikAS [45]2 years ago
3 0
Cannot see your image, but the formula for the volume of a sphere is
V=(4/3)πr³
to solve for r: r³=v÷(4/3)π=v*3/(4π)=3v/(4π) (three v out of 4 pi)
r=∛(3v/4π)
r equals the cubic root of (three v over 4π)
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The position of a particle moving to the right on the x-axis is given by x(t), where x(t) is measured in inches and t is measure
wlad13 [49]

Answer:

No options

Step-by-step explanation:

Knowing that V=in/min = x(t)/t and according to the graph,

V=20in/20min=in/min, so

No options since speed is not inches per minute, spedd (V) = inches over minutes

8 0
2 years ago
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During a catered lunch, an average of 4 cups of tea are poured per minute. The lunch will last 2 hours. How many gallons of tea
allsm [11]
The answer is 30.
There are 16 cups in a gallon. And 120 minutes in 2 hours. so multiple 
16*120 which is 480. Then divide 480/120 and get 30 as your answer
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2 years ago
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11. A basketball team of 16 players consists of 6 guards, 7 forwards, and 3 centers. The coach decides to randomly select 5 play
Citrus2011 [14]

Answer:

0.2163

Step-by-step explanation:

Firstly, we need to evaluate the total number of possible outcomes. Since there are 16 players, and we are selecting just 5, the total number of possibilities is 16C5= 4,368

Now, we know we need 2 guards from 6 , 2 forwards from 7 and 1 center from 3 to start the game. Since we are selecting, it is a combination problem. These can be done in the following number of ways:

6C2 * 7C2* 3C1

The probability is thus (6C2 * 7C2* 3C1 )/16C5 = 945/4368 = 0.2163

8 0
2 years ago
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390

<u>For X=8</u>

This happens when everybody is late

P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147

8 0
2 years ago
Mr. Donley drove 504 km in each direction on a vacation
mestny [16]

Answer:

Data that we know:

He drove 504km in each direction.

Going, the average speed was 14km/h more than returning.

The total trip lasted 21 hours.

Ok, to solve this first:

Let's define the variables:

Sg = Average speed when going.

Sr = Average speed when returning.

Then we can write one of our relationships as:

Sg = Sr + 14km/h.

Now, you can recall the relation:

Time = Distance/speed.

Then we can write the equation that represents the total time of the travel as:

504km/Sg + 504km/Sr = 21h

Now we can replace the Sg by Sr + 14km/h, and get:

504km/(Sr + 14km/h) + 504km/Sr = 21h.

Now we must multiplacate by each denominator, in order to remove them:

504km*Sr + 504km*(Sr + 14km/h) = 21h*Sr*(Sr + 14km/h)

Sr*1008km + 7056km^2/h = 21h*Sr^2 + 294km*Sr

Now we can write a cuadratic equation, where i will ignore the units so it is easier to read, as:

21*Sr^2 -714*Sr - 7056 = 0

The solutions are given by the Bhaskara formula:

Sr = \frac{714 +- \sqrt{714^2 - 4*21*(-7056)} }{2*21}  = \frac{714 +- 1050}{42}

We have two solutions, one negative and one positive, and because Sr represents an average speed, it must be a positive number, then we choose the positive solution:

Sr = (714 + 1050)/42  km/h = 42km/h

Now we have the average speed for the returning, with this we can find Sg.

Sg = Sr + 14km/h = 42km/h + 14km/h = 56km/h.

6 0
2 years ago
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