the number of hours,h, it takes to wash c numbers of cars at a school fundraiser is inversely proportional to the number of stud

ents,s, needed to wash the cars. Write a equation to represent his relationship

Mathematics

1 answer:

Alex787 [66]2 years ago

3 0

$\bf \qquad \qquad \textit{inverse proportional variation}
\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\stackrel{\textit{\underline{h}ours to wash a car, is inversely proportional to \underline{s}tudents working on it}}{h=\cfrac{k}{s}}$

You might be interested in

Find an equation of a line that comes close to the points listed in the table.

sergey [27]

$y=x+2\\x=6\to y=6+2=8\\x=10\to y=10+2=12\\x=12\to y=12+2=14\\x=16\to y=16+2=18\\x=18\to y=18+2=20\\-----------------------\\y=x+3\\x=6\to y=6+3=9\\x=10\to y=10+3=13\\x=12\to y=12+3=15\\x=16\to y=16+3=19\\x=18\to y=18+3=21\\-----------------------\\Answer:y=x+1.5$

3 0

2 years ago

Which function has a greater rate of change that the function represented in the table

Galina-37 [17]

I don't see a table

3 0

2 years ago

Graph the line that represents this equation:<br> y = -5.1 +2

Mrac [35]

That's for the equation I got from the picture

4 0

2 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$