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stiv31 [10]
3 years ago
14

Omar ordered his sister a birthday card from a company that randomly selects a card from their inventory. The company has 21 tot

al cards in inventory. 14 of those cards are birthday cards. What is P(not a birthday card)
Mathematics
2 answers:
Lera25 [3.4K]3 years ago
4 0
P(not a birthday card) is \frac{1}{3}    ( \frac{7}{21} ) because if 14 of the 21 are birthday cards then 21-14=7 are not.
Evgen [1.6K]3 years ago
3 0

Answer:  \frac{1}{3} or 33.33%

Step-by-step explanation:

Given: The total number of cards in inventory = 21

The number of birthday cards = 14

Then the number of cards which are not a birthday card = 21-14=7

If company that randomly selects a card from their inventory, then the probability of getting a card which is not a birthday card will be :-

\text{P(not a birthday card)}=\frac{\text{not a birthday cards}}{\text{total cards}}\\\\\Rightarrow\ \text{P(not a birthday card)}=\frac{7}{21}\\\\\Rightarrow\ \text{P(not a birthday card)}=\frac{1}{3}

In percent, \text{P(not a birthday card)}=\frac{1}{3}\times100=33.33\%

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3 years ago
You are designing a rectangular garden. the garden will be enclosed by fencing on three sides and by a house on the fourth side.
RUDIKE [14]
Suppose the dimensions of the rectangle is x by y and let the side enclosed by a house be one of the sides measuring x, then the sides that is to be enclosed are two sides measuring y and one side measuring x.

Thus, the length of fencing needed is given by

P = x + 2y

The area of the rectangle is given by xy,

i.e.

xy = 288  \\  \\  \\  \\ \Rightarrow y= \frac{288}{x}

Substituting for y into the equation for the length of fencing needed, we have

P=x+2\left( \frac{288}{x} \right)=x+ \frac{576}{x}

For the amount of fencing to be minimum, then

\frac{dP}{dx} =0 \\  \\ \Rightarrow1- \frac{576}{x^2} =0 \\  \\ \Rightarrow \frac{576}{x^2} =1 \\  \\ \Rightarrow x^2=576 \\  \\ \Rightarrow x=\sqrt{576}=24

Now, recall that

y= \frac{288}{x} = \frac{288}{24} =12

Thus, the length of fencing needed is given by

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3 years ago
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Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select
Blizzard [7]

Answer:

Probability is:   $ \frac{\textbf{13}}{\textbf{51}} $

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $ \frac{number \hspace{1mm} of  \hspace{1mm} black  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm} of  \hspace{1mm} cards} $  $ = \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}} $

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $ \frac{number  \hspace{1mm} of  \hspace{1mm} red  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm}of  \hspace{1mm} cards} $  $ = \frac{26}{51}  $

Now, the probability of both black and red card = $ \frac{1}{2} \times \frac{26}{51} $

$ = \frac{\textbf{13}}{\textbf{51}} $

Hence, the answer.

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