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kari74 [83]
3 years ago
10

A sum lent out at simple interest becomes rs4480 in 3 years and rs 4800 in 5years.find the rate of interest​

Mathematics
1 answer:
jek_recluse [69]3 years ago
8 0

Answer: rate = 4%

Step-by-step explanation:

SI = 4800 - 4480 = 320 for two years

For 1 year, it is 320/2 = 160

For 5 years, it is 160 * 5 = 800

Principal = Amount -SI

P = 4800 - 800 = 4000

SI = prt / 100

800 = (4000 * r * 5) / 1000

r = 800 * 100 / 4000 * 5

r = 4%

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The probability that you would choose lemon-lime and then orange is 3/11 =.273.  

Step-by-step explanation:

These are 'dependent events', which mean that your the event is affected by previous events.  So, because you have eleven total bottles (five lemon-lime and six orange) and you do not replace the first bottle, that would only leave you with ten bottles remaining.  The probability that you will pick the lemon-lime on the first choice is 5/11 because all of the bottles are there.  However, your second choice will only include ten total bottles since you already took one.  The probability that you would choose orange would be 6/10.  When you multiply these two fractions and reduce to simplest form, you get 3/11.  


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3 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]

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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.

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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\\\\

P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021

8 0
3 years ago
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uysha [10]
P=2 then 2^2+4=8
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