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2 years ago

1. There are 125 juniors at the high school.How many ways can they elect a

Mathematics

1 answer:

valentinak56 [21]2 years ago

7 0

Answer: 1. 1,906,500

2. 840

Step-by-step explanation:

1. Given: There are 125 juniors at the high school.

Total positions = 3 [president,vice president, and secretary]

Then, Number of ways to elect a president,vice president, and secretary( which is in order) out of 125 juniors = $^{125}P_{3}=\dfrac{125!}{(125-3)!}$ [Using permutations]

$=\dfrac{125!}{122!}\\\\=\dfrac{125\times124\times123\times122!!}{122!}\\\\=125\times124\times123\\\\=1906500$

Hence, Required ways =1906500

(2) Given word : "RHOMBUS"

Total letters= 7

By permutation, the umber of 4-letter arrangements = $^7P_4=\dfrac{7!}{(7-4)!}=\dfrac{7!}{3!}$

$=\dfrac{7\times6\times5\times4\times3!}{3!}\\\\= 7\times6\times5\times4\\\\=840$

Hence, required ways = 840

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You just measured a sugar cube and obtained the following information: mass = 3.48 g height = length = width = 1.3 cm Determine

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Answer: Volume of sugar cube=2.20 cm³

Density of sugar cube = 1.60 g/cm³.

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Step-by-step explanation:

Given : Mass of cube = 3.48 g

Also, dimension of sugar cube :

height = length = width = 1.3 cm

Since , the volume of cube = (side)³

Volume of sugar cube=(1.3) ³= 2.197 ≈ 2.20 cm³

Also, Density = $\dfrac{\text{Mass}}{\text{Volume}}$

⇒ Density of sugar cube = $\dfrac{3.48}{2.2}=1.5818\approx1.60\ g/cm^3$

∴ Density of sugar cube = 1.60 g/cm³.

It is known that the density of water is 1 g/cm³.

Here , density of sugar cube > density of water

⇒ The sugar cube will sink .

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Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

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Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

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