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BlackZzzverrR [31]
3 years ago
9

Why would a conservative investor invest in more short term bonds than long term bonds?

Mathematics
2 answers:
castortr0y [4]3 years ago
7 0
<span>A conservative investor invests in more short term bonds than long term bonds because they believe long term bonds are not efficient.
</span>Some investors think that bonds are “safe,” , stocks are “risky”. 

<span>Short-term bonds on the other hand, seems to be attractive to many investors, specifically the conservative ones because they don't require you to tie up your money for long.  </span>
faust18 [17]3 years ago
6 0

Answer and  explanation:

Conservative investor invest more in short term bonds than long term bonds because the longer the term of bond is , the higher the risk it possess.

Risk associated with the long term bonds is that if there is increase in inflation, it could reduce the value of payments.

And higher  interest rate reduce the price of bonds.

Due to that reasons, conservative investor prefers short term bonds to long term bonds.

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54,36,24 whats the next term in the geometric sequence
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16

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Solve the system of elimination -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5
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x = 1 , y = 1 and z = 0

Step-by-step explanation:

-2x+2y+3z=0    ------(1)

-2x-y+z=-3     --------(2)

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<u>To find the values of x,y and z</u>

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

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

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3 years ago
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