Help with these questions

Suppose that the members of a student governance committee will be selected from the 40 members of the student senate. There are

Len [333]

Answer:

The total number of ways to form a student governance committee is 1,211,760.

Step-by-step explanation:

The students senate consists of a total of 40 students.

The students are either Sophomores or Juniors or Seniors.

The number of students in each of these categories are as follows:

Sophomores = 18

Juniors = 12

Seniors = 10

A governance committee have to be selected from the students senate.

The committee have to made up of 2 sophomores, 2 juniors and 3 seniors.

Combinations can be used to select 2 sophomores from 18, 2 juniors from 12 and 3 seniors from 10.

Combinations is a mathematical technique used to determine the number of ways to select <em>k</em> items from <em>n</em> distinct items.

The formula is:

${n\choose k}=\frac{n!}{k!(n-k)!}$

(1)

Compute the number of ways to select 2 sophomores from 18 as follows:

${n\choose k}=\frac{n!}{k!(n-k)!}$

${18\choose 2}=\frac{18!}{2!(18-2)!}=\frac{18\times 17\times 16!}{2\times 16!}=153$

Thus, there are 153 ways to select 2 sophomores from 18.

(2)

Compute the number of ways to select 2 juniors from 12 as follows:

${n\choose k}=\frac{n!}{k!(n-k)!}$

${12\choose 2}=\frac{12!}{2!(12-2)!}=\frac{12\times 11\times 10!}{2\times 10!}=66$

Thus, there are 66 ways to select 2 juniors from 12.

(3)

Compute the number of ways to select 3 seniors from 10 as follows:

${n\choose k}=\frac{n!}{k!(n-k)!}$

${10\choose 3}=\frac{10!}{3!(10-3)!}=\frac{10\times 9\times 8\times 7!}{2\times 3\times 7!}=120$

Thus, there are 120 ways to select 3 seniors from 10.

The total number of ways to form a student governance committee that must have 2 sophomores, 2 juniors and 3 seniors is:

Total number of ways = ${18\choose 2}\times {12\choose 2}\times {10\choose 3}$

$=153\times 66\times 120\\=1211760$

Thus, the total number of ways to form a student governance committee is 1,211,760.

7 0

3 years ago

Given trapezoid abcd with bases ab and cd, draw diagonals ac and bd. let e be the midpoint of ac and f the midpoint of bd. prove

mylen [45]

Consider the picture.

Let MN be the midsegment of the trapezoid.

That is M is the midpoint of AD, N is the midpoint of BC.

Being the midsegment of the trapezoid, MN is parallel to the bases.

Let O and K be the intersections of the diagonals with the midsegment.

MN//AB, so MO//AB, and since M is the midpoint of DA, O must be the midpoint of DB,

Similarly we prove that K is the midpoint of CA.

Thus O is F and K is E.

O and K lie on the midsegment MN, so F and E lie on the midsegment.

MO is a midsegment of triangle ABD so |MO|=1/2 |AB|=1/2 * 10=5

MK is a midsegment of triangle ADC, so |MK|=1/2 * |DC|=1/2 * 22=11

|OK|=|MK|-|MO|=11-5=6 (units)

7 0

4 years ago

Five times a number is the same as 30 more than 8 times the number.find the number

Goshia [24]

Answer:

hey here's your answer!☺️

since number = n

so, 5 times of number = 5×n = 5n

and , 30 more than 8 times the number = 8n + 30

so, According to the question,

equation formed :-

" 5n = 8n + 30 "

hope it is helpful ✌️!

6 0

3 years ago

HELPPPP IT DOWN BELLOW

Alborosie

Answer:

paul put it at 3,8 and its supposed to be at 0,6

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Write in word form and expanded form: 23,876.901

kondor19780726 [428]

Word form:

Twenty-three thousand, eight hundred seventy-six and nine hundred one thousandths.

Expanded form:

20, 000 + 3, 000 + 800 + 70 + 6 + 0.9 + 0.001

6 0

3 years ago