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shtirl [24]
3 years ago
10

An investment analyst has tracked a certain bluechip stock for the past six months and found that on any given day, it either go

es up a point or goes down a point. Furthermore, it went up on 25% of the days and down on 75%. What is the probability that at the close of trading four days from now, the price of the stock will be the same as it is today? Assume that the daily fluctuations are independent event
Mathematics
1 answer:
anastassius [24]3 years ago
7 0

Answer:

0.2109 or 21.09%

Step-by-step explanation:

In order to maintain the same price after two days, the stock must go up (U) on two days and go down (D) on two days, the sample space for this event is:

S={UUDD, UDUD, UDDU, DDUU, DUDU, DUUD}

There are 6 equally likely possible outcomes. The probability that the price of the stock will be the same as it is today is:

P =6* (0.25*0.25*0.75*0.75)\\P=0.2109=21.09\%

The probability is 0.2109 or 21.09%.

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Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/
SVEN [57.7K]

Given p,q>0 and \log_9p=\log_{12}q=\log_{16}(p+q)=x (say).

Then,

p=9^x\\ q=12^x\\ p+q=16^x

From the above 3 equations,

\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\

From the equations, we get

\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}

Since p,q>0, the negative value is rejected.

\frac{q}{p}=\frac{1 + \sqrt{5}}{2}

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3 years ago
Can someone help me with this?
Vladimir79 [104]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
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A ladder is leaning against a tall wall with the foot of the ladder placed at 7 feet from the base of the wall and the angle of
djverab [1.8K]

Complete question is;

A 21 ft ladder is leaning against a tall wall with the foot of the ladder placed at 7 feet from the base of the wall and the angle of elevation is?

Answer:

θ = 70.5°

Step-by-step explanation:

The angle of elevation simply means the angle that the ladder makes with the ground. Let's call this angle θ.

I've attached a diagram showing the triangle made by this ladder and the wall.

From the attached diagram, we can see the triangle formed by the ladder and the wall.

We can find the angle of elevation θ from trigonometric ratios.

Thus;

7/21 = cos θ

cos θ = 0.3333

θ = cos^(-1) 0.3333

θ = 70.5°

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