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Marizza181 [45]
3 years ago
8

A locker combination has 2 nonzero digits. Digits in a combination are not repeated and range from 2 through 9. Event A = choosi

ng an odd digit for the first number Event B = choosing an odd digit for the second number If all of the digits are equally likely to be chosen, what is P(B|A) expressed in simplest form?.
Mathematics
1 answer:
Anna [14]3 years ago
6 0
It must be B.

There are two possibilities; in the first, the first digit is below 5, in the second, it is above. The probability of the first possibility is 3/8, because there are 3 possible digits below 5 and 8 total digits. In this scenario, the chance of the second digit being below 5 is 2/7, because one of the digits is taken. In the other possibility, which has a chance of 5/8, the probability of choosing a number below 5 is 3/7, since all of them are still available. Doing the arithmetic, you find that the total probability is 2/7. Hope this helps!
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From a thin piece of cardboard 50 in. by 50 in., square corners are cut out so that the sides can be folded up to make a box. Wh
mixer [17]

Answer:

When dimension of box is 33.33 inches × 33.33 inches ×8.33  then its volume is maximum and is 9259.26 cubic inches.

Step-by-step explanation:

Let h be the length (in inches) of the square corners that has been cut out from the cardboard and that would be the height of the cardboard box.

Since the squares have been cut from cardboard, both sides of the cardboard would reduce by 2h.

Thus, The dimension of box is  (50 – 2h) × (50 – 2h) × h in dimensions.

The volume V of rectangular box = (Length × Breadth × Height) cubic inches.

V=(50-2h) \times (50-2h) \times h

V=(50-2h)^2 \times h  ..............(1)

Using (a-b)^2=a^2+b^2-2ab

V=h(2500+4h^2-200h)

V=2500h+4h^3-200h^2

For obtaining a box of maximum volume, maximize V as a function of h.


Differentiate both sides with respect to h,

\frac{dV}{dh}=2500+12h^2-400h

\frac{dV}{dh}=4(625+3h^2-100h)

Solving quadratic equation,625+3h^2-100h

\frac{dV}{dh}=4(3h^2-25h-75h+625)

\frac{dV}{dh}=4(h(3h-25)-25(3h-25))

\frac{dV}{dh}=4((h-25)(3h-25))

For maximum, \frac{dV}{dh}=0  

thus,4((h-25)(3h-25))=0

⇒ h= 25 or h=\frac{25}{3}

Now check (1) for h= 25 and h=\frac{25}{3}.

h= 25 is not possible as when h is 25 inches then length and breadth becomes 0.

When h=\frac{25}{3}.

(1) ⇒ V=(50-2(\frac{25}{3}))^2 \times\frac{25}{3}=9259.2592593  

This is the maximum volume the box can assume.

Thus, when dimension of box is 33.3 inches × 33.3 inches ×8.3  then its volume is maximum and is 9259.26 cubic inches.

6 0
3 years ago
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LekaFEV [45]
Let x = the number


6x + 12 = 8x


Subtract 6x on both sides of equation.

12 = 2x


Divide both sides of equation by 2.

6 = x


The number is 6.
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Answer:

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