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3 years ago

How many combinations of 3 students can a teacher choose from 32 students? A. 96 B. 4960 C. 29,760 D. 32,768

Mathematics

2 answers:

rusak2 [61]3 years ago

5 0

${32\choose3}=\frac{32!}{3!*(32-3)!}=\frac{30*31*32}{6}=4960$

Allisa [31]3 years ago

3 0

$\frac{32*31*30}{3*2*1}$

The answer is: $\frac{29760}{6}$ or $4960$

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Answer:

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3. 2nd option

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Use formula for compounded interest

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In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.$

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

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<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

$A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.$

The best will be 2nd option, because $8032.58>$7936.39

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