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2 years ago

At a high school 7 student volunteer for a committtee how many different 5 person committees can be chosen

1 answer:

5 0

Total number of volunteer students for a committee = 7
Number of committee = 5

Number of ways of selecting the committee from the 7 volunteers = 7C5 = 7!/(5!*2!) = 21 ways.

There are 21 ways in which the 5-committee member can be selected from the 7 volunteers.

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a family has two cars. the first car has a fuel efficiency of 20 miles per gallon of gas and the second has a fuel efficiency of

Kryger [21]

Let the number of gallons consumed be the first car be x, and that of the second car, y. Then:
x + y = 50 . . . . . . . (1)
20x + 15y = 850. . . (2)

(1) x 20 => 20x + 20y = 1000 . . . (3)
(2) - (3) => -5y = -150 => y = -150/-5 = 30

From (1), x + 30 = 50 => x = 50 - 30 = 20

Therefore, first car consumed 20 gallons and the second consumed 30 gallons.

3 0

3 years ago

Write 84 as a product of primes.<br>Use index notation where appropriate

elena-s [515]

Answer:

2 x 2 x 3 x 7

Step-by-step explanation:

2x2=4

4x3=12

12x7=84

3 0

2 years ago

The table below shows the scoring averages for the players on the Supersonics intramural basketball team last season.

lozanna [386]

Answer:

C. Graph A, Set A

Step-by-step explanation:

(I took the test)

Your X axis is the age of players and the Y axis is the scoring average.

8 0

2 years ago

Read 2 more answers

8. The probability that Ava gets promoted is 7/10. Find the odds in favor of Ava getting promoted.

agasfer [191]

Very likely to be promoted but you didnt put the actually answers

6 0

1 year ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago