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3 years ago

How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?

Mathematics

1 answer:

FromTheMoon [43]3 years ago

5 0

This is essentially asking how many different ways to partition 6 into 3 segments.

I am assuming "no ball in a box" is not allowed.

6 can be partitioned as

(4,1,1), (3,2,1), and (2,2,2)

So, calculate each partition, we get

(6 choose 4) + (6 choose 3)*(3 choose 2) + (6 choose 2) * (4 choose 2)

= 15 + 20*3 + 15*6

=165

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5, 3, 7,0-2,-1, -7 1-5 least to greatest<br>

ANTONII [103]

-7, -2, -1, 0, 3, 5, 7

6 0

2 years ago

Read 2 more answers

The volume of a right circular cone with radius r and height h is V = pir^2h/3.

Scorpion4ik [409]

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

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

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9 and dr = 6.8 - 5.9 = 0.9

h = 4 and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b). The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47 and dr = 6.45 - 6.47 = -0.02

h = 10 and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

8 0

2 years ago

Two students compared the scores on their math tests. Their results on 9 tests are shown in the box plots below.

Pani-rosa [81]

The statements which are true regarding the students whose results on 9 tests are as shown are; The median of Nadia's data is equal to the median of Ben's data.

Nadia had the highest score on a test.

<h3>Which statements are true regarding the scores on the tests?</h3>

According to the task content, it follows that the results of the students in discuss are indicated by means of a box plot as in the attached image.

Consequently, it follows from observation that the median of Nadia and Ben as indicated in the attached box plot is 92 in both cases as indicated by the vertical line in both boxes.

Additionally, the highest score by Nadia is 100 while that for Ben is; 99.

Hence, Nadia had the highest score on a test.

Read more on median in box plots;

brainly.com/question/14277132

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