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Pepsi [2]
4 years ago
12

There are 4 jacks and 13 clubs in a standard, 52-card deck of playing cards. What is the probability that a card picked at rando

m from a standard deck of playing cards is a club or a jack?
Mathematics
1 answer:
ale4655 [162]4 years ago
3 0

Answer:

16/52, or 4/13.

Step-by-step explanation:

First, since we know that the question is asking for the probability of a club <u>or</u> a jack, we know that we have to add the two probabilities. The first probability is that of picking a club, which is 13/52. The probability of picking a jack (be sure not to overlap; don't double count the jack of clubs) is 3/52. Adding these two gives us 13/52+3/52=16/52, which simplifies to 4/13.

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I dont understand this
AURORKA [14]

Answer:

you have to add them all together to get the right answer.

Step-by-step explanation:

6 0
4 years ago
What is negative 3/8 plus 7/8
Luba_88 [7]

Hi there!

To add fractions there are three simple steps:

1. Make sure the bottom numbers (denominators) are the same. → In your case you're all good, both your denominators are 8.

2. Add the top numbers (numerators) and put that answer over the denominator :

-3 + 7 = 4 → \frac{4}{8}

3. Simplify the fraction (if needed). → In your case, you do need to simplify your fraction :

Divide both your numerator and denominator by 4 to get your simplified fraction :

4 ÷ 4 = 1

8 ÷ 4 = 2

Your simplified fraction is : \frac{1}{2}


There you go! I really hope this helped, if there's anything just let me know! :)

8 0
3 years ago
Read 2 more answers
a basketball dropped from a height of 76 in on Bounce one it rebounds to a height of 38 in on bounce to it rebounds to a height
adelina 88 [10]
Ok so 38 is half of 76 and 19 is half of 38. If you follow this pattern then theoretically you should get 9.5
3 0
3 years ago
Can someone help me please
JulsSmile [24]

Answer:

yes

Step-by-step explanation:

8 0
3 years ago
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WILL MARK BRAINLIEST! Which of the following is the expansion of (3c+d^2)^6? USE BINOMIAL THEROEM
svetlana [45]

Answer:

(3c+d^2)6=729c^6+1458c^5d^2+1215c^4d^4+540c^3d^6+135c^2d^8+18cd^10+d^12

Step-by-step explanation:

The expansion is given by the following formula: (a+b)n=∑k=0n(nk)an−kbk,

where (nk)=n!(n−k)!k! and n!=1⋅2⋅3...n.  

We have that a=3c, b=d2, n=6.

Therefore, (3c+d2)6=∑k=06(6k)(3c)6−k(d2)k

Now, calculate the product for every value of k from 0 to 6.

k=0: (60)(3c)6−0(d2)0=6!(6−0)!0!(3c)6(d2)0=729c6

k=1: (61)(3c)6−1(d2)1=6!(6−1)!1!(3c)5(d2)1=1458c5d2

k=2: (62)(3c)6−2(d2)2=6!(6−2)!2!(3c)4(d2)2=1215c4d4

k=3: (63)(3c)6−3(d2)3=6!(6−3)!3!(3c)3(d2)3=540c3d6

k=4: (64)(3c)6−4(d2)4=6!(6−4)!4!(3c)2(d2)4=135c2d8

k=5: (65)(3c)6−5(d2)5=6!(6−5)!5!(3c)1(d2)5=18cd10

k=6: (66)(3c)6−6(d2)6=6!(6−6)!6!(3c)0(d2)6=d12

Finally, (3c+d2)6=∑k=06(6k)(3c)6−k(d2)k=(60)(3c)6−0(d2)0+(61)(3c)6−1(d2)1+(62)(3c)6−2(d2)2+(63)(3c)6−3(d2)3+(64)(3c)6−4(d2)4+(65)(3c)6−5(d2)5+(66)(3c)6−6(d2)6=729c6+1458c5d2+1215c4d4+540c3d6+135c2d8+18cd10+d12

Answer is above :)

3 0
3 years ago
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