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Tatiana [17]
3 years ago
10

Help PLZ HAVENT HAD HELP IN TWO HOUR JUST ONE QUESTION One card is to be drawn from a deck of 52 cards.The deck includes four ty

pes of cards.Red heart, red stars , black triangles , black circles .Each type has 13 cards numbered 1-13.What is the probability of drawing a red one a, black three , or a six of hearts.Give your answer in the simplify form .
Mathematics
1 answer:
Vitek1552 [10]3 years ago
7 0

Answer:

FOR red one = 1/26

FOR black three = 1/26

FOR six of hearts = 1/52

Step-by-step explanation:

TECHNICALLY EVERY CARD WOULD BE 1/52 BUT SINCE THERE R TWO DIFFERENT TYPES OF BLACKS AND REDS THE FIRST TWO ARE 1/26

HOPE I HELPED

PLS MARK BRAINLIEST  

DESPERATELY TRYING TO LEVEL UP

✌ -ZYLYNN JADE ARDENNE

JUST A RANDOM GIRL WANTING TO HELP PEOPLE!

                     PEACE!

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The  total cost for the group of ten people after discount is 10 t - 0.5 t.

<h3>What is Algebra?</h3>

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In order to determine the values, these symbols are also subjected to various addition, subtraction, multiplication, and division arithmetic operations.

Given:

Discount = 5%

Total tickets= 10

We have t represents the cost of one ticket.

So, For 10 person the cost of the ticket = 10t

Now, after discount the price of tickets

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Hence, the total cost for the group of ten people after discount is 10 t - 0.5 t.

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1 year ago
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vazorg [7]

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5 0
2 years ago
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Please help me!!!!!​
denpristay [2]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π               → A = π - (B + C)

                                               → B = π - (A + C)

                                               → C = π - (A + B)

Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]

Use the following Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → RHS:</u>

LHS:                        sin A - sin B + sin C

                             = (sin A - sin B) + sin C

\text{Sum to Product:}\quad 2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Given:}\qquad 2\cos \bigg(\dfrac{\pi -(B+C)}{2}+\dfrac{B}{2}}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad =2\cos \bigg(\dfrac{\pi -C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

.\qquad \qquad =2\cos \bigg(\dfrac{\pi}{2} -\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

\text{Cofunction:} \qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)

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\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]

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\text{LHS = RHS:}\quad 4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)=4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\quad \checkmark

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3 years ago
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disa [49]
What’s the question referring to here.
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Factor using a GCF:<br> ​ 16xy2+28xy+8y
Lorico [155]

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Step-by-step explanation:

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8y = 2*2*2*y

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2*2*y( 2*2*x*y + 7 *x+2)

4y( 4xy + 7x +2)

7 0
3 years ago
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