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Anna [14]
2 years ago
13

If a single card is drawn from a standard 52-card deck, what is the probability that it is an ace or a spade?

Mathematics
2 answers:
klasskru [66]2 years ago
7 0

A 52-card deck is made up of an equal number of diamonds, hearts, spades, and clubs. Because there are 4 suits, there is a 1/4 chance to draw one of them, in our case, spades.

There are 4 aces in a 52-card deck, so the chance of drawing one is 4/52, or 1/13.

The question asks for the probability of drawing an ace or a spade. Because it uses the word "or," we add the probabilities together. This is because there is a chance of drawing either of the cards; it doesn't have to meet both requirements to satisfy the statement.

However, if the question were to say "and," we would multiply the two probabilities.

Let's add 1/4 and 1/13. First, we can find a common denominator. We can use 52 because both fractions can multiply into it (since the ratio came from a deck of 52 cards as well).

\frac{1}{4} -- > \frac{13}{52}

\frac{1}{13} -- > \frac{4}{52}

Now we can add them together.

\frac{13}{52} + \frac{4}{52} = \frac{13+4}{52} = \frac{17}{52}

This cannot be simplified further, so the probability is 17 in 52, or 33%.

hope this helps!

Likurg_2 [28]2 years ago
5 0

Answer:

Step-by-step explanation:

4/13

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Answer:

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Step-by-step explanation:

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total area = 12^2*pi = 144pi

144pi/8 = 18pi

part b:

total circumference = 12*2*pi = 24pi

24pi/8 = 3pi

3 0
3 years ago
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7 0
3 years ago
PLEASE HELP ASAP WILL MEDAL BRAINLIEST
Naddik [55]
Given
Present investment, P = 22000
APR, r = 0.0525
compounding time = 10 years
Future amount, A 

A. compounded annually
n=10*1=10
i=r=0.0525
A=P(1+i)^n
=22000(1+0.0525)^10
=36698.11

B. compounded quarterly
n=10*4=40
i=r/4=0.0525/4
A=P(1+i)^n
=22000*(1+0.0525/4)^40
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Therefore, by compounding quarterly, she will get, at the end of 10 years investment, an additional amount of
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5 0
3 years ago
The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
Anit [1.1K]

Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

a_n = ar^{n-1}

a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}

the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

4 0
3 years ago
What is factor ? someone explain pls !!
Lubov Fominskaja [6]

Step-by-step explanation:

Factor, in mathmatechis, a number or algebric expression that divides another number or expression evenly.

For example, 3 and 6 are factors of 12 because 12 divided by 3 equals 4 exactly and 12 divided by 6 equals 2 exactly. The other factors of 12 are 1, 2, 4, and 12.

3 0
3 years ago
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