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ehidna [41]
2 years ago
5

The letters in the word JANUARY were put in a bag and drawn randomly. What is the probability of picking an a?

Mathematics
2 answers:
sesenic [268]2 years ago
7 0

Answer: 2/7

Step-by-step explanation:

There are 7 letters in January, and 2 "a"s in order to find the probability, simply take 2/7. You can either leave it in the fraction form, or use the calculator, showing you have a 0.2857% chance of picking an "a"

Alecsey [184]2 years ago
5 0

2/7

there are 7 letters in jAnuAry and as you can see there are 2 A's so there is a 2/7 chance which is higher of a chance than the rest of the letters which are 1/7 each so it is most likely you would draw A although I can't fully say that because there is a 5/7 chance of drawing the other letters. Another way to say it is there is a 0.28571428571% chance of drawing A but there is a 0.71428571428% chance of drawing any other letter.

                Hope this helps :3

                              Have a good day!

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Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

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Step-by-step explanation:

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We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2

Equating it to zero we get,

x^2 + 36x - 6400 = 0

We use the quadratic formula to find the values of x:

x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}, where a, b and c are coefficients of x^2, x^1 , x^0 respectively.

Putting these value we get x = -100, 64

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\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x

At x = 64,  \displaystyle\frac{d^2(P(x))}{dx^2} < 0

Hence, maxima occurs at x = 64.

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