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Citrus2011 [14]
2 years ago
8

Statistical models predict that the price (in dollars) of a 500 GB hard drive will change according to the function p(t) = 600 −

2t2, where t is the month. Which expression gives the number of months t (passed since January 1) in terms of the price p?
Mathematics
2 answers:
slega [8]2 years ago
7 0

Answer:

t(p)=\sqrt{\frac{1}{2}(600-p)}

Step-by-step explanation:

Here, the expression that gives p in term of t is,

p(t)=600-2t^2 -----(1)

Where, p represents the price and t represents the number of months t (passed since January 1).

From equation (1),

p=600-2t^2

2t^2=600-p

t^2=\frac{1}{2}(600-p)

t=\pm \sqrt{\frac{1}{2}(600-p)}

But, number of months can not be negative,

\implies t=\sqrt{\frac{1}{2}(600-p)}

Since, in this expression t is in the term of p,

Hence, the required expression that gives the number of months t in terms of the price p is,

t(p)=\sqrt{\frac{1}{2}(600-p)}

Leni [432]2 years ago
4 0

Let p be a fixed value of price. The expression for t given a value of p is as follows:

p=600-2t^2\\2t^2 = 600-p\\t^2 = \frac{600-p}{2}\\|t| = \sqrt{\frac{600-p}{2}}\\t=\sqrt{\frac{600-p}{2}},\,\,\,\,\,\,t\geq 0,\,\,p\leq 600

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Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p
posledela

Answer:

<em>P=0.0000037</em>

<em>P=0.00037%</em>

Step-by-step explanation:

<u>Probability</u>

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

\displaystyle \frac{4}{52}=\frac{1}{13}

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

\displaystyle \frac{3}{51}=\frac{1}{17}

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

\displaystyle \frac{2}{50}=\frac{1}{25}

Finally, the probability of getting the remaining ace out of the 49 cards is:

\displaystyle \frac{1}{49}

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}

\displaystyle P= \frac{1}{270,725}

P=0.0000037

P=0.00037%

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