A teacher is choosing 4 students from a class of 30 to represent the class at a science fair. In how many ways can the teacher c

hoose the students?

Mathematics

1 answer:

marin [14]3 years ago

4 0

${30 \choose 4}=\frac{30!}{4!26!}=\frac{27\cdot28\cdot29\cdot30}{2\cdot 3\cdot4}=27\cdot7\cdot29\cdot5=27405$

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Country days scholarship rounds receive a gift of $135000. The money is invested in stock, bonds, and CDs. CDs pay 2.75% interes

Vsevolod [243]

Answer:

CDs — $10,000
bonds — $80,000
stocks — $45,000

Step-by-step explanation:

Let the variables c, b, s represent the dollar amounts invested in CDs, stocks, and bonds, respectively. Then the problem statement gives us 3 relations between these 3 variables:

c + b + s = 135000 . . . . . . . . . . . . . . . . . total invested

0.0275c +.045b +0.104s = 8555 . . . . . total income earned

-c + b = 70000 . . . . . . . . . . . . . . . . . . . . . 70,000 more was in bonds than CDs

Using the third equation to write an expression for b, we can substitute into the other two equations.

b = 70000 +c . . . . . . . . . . . . . . . . expression we can substitute for b

c + (70000 +c) +s = 135000 . . . . substitute for b in the first equation

2c +s = 65000 . . . . . . . . . . . . . . . . [eq4] simplify

.0275c +.045(70000 +c) +.104s = 8555 . . . . . substitute for b in 2nd eqn

.0725c +.104s = 5405 . . . . . . . . . . [eq5] simplify

Using [eq4], we can write an expression for s that can be substituted into [eq5].

s = 65000 -2c . . . . . . . expression we can substitute for s

0.0725c +0.104(65000 -2c) = 5405

-0.1355c = -1355 . . . . . . . . . . . . . . . . . . . . subtract 6760, simplify

c = 1355/.1355 = 10,000

s = 65000 -2×10000 = 45,000

b = 70000 +10000 = 80,000

The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

_____

Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

$\left[\begin{array}{ccc|c}1&1&1&135000\\0.0275&0.045&0.104&8555\\-1&1&0&70000\end{array}\right]$