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iris [78.8K]
3 years ago
7

A card is selected from a standard deck of 52 cards . what are the odds of selecting a red 9 ?

Mathematics
2 answers:
krok68 [10]3 years ago
8 0
B. 1:13 is the answer hope it helps
morpeh [17]3 years ago
7 0
A. 1:26
Because there are 2 red nines in a card deck of 52 cards. So 2:52 or simplified to 1:26.
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50 points. Please quick.
Allisa [31]

Answer:

Condition A.  

A rectangle with four right angles  

There can be many quadrilaterals satisfying this condition.

Condition B.

A square with one side measuring 5 inches

There can be only one quadrilateral satisfying this condition.

Condition C.

A rhombus with one angle measuring 43°

There can be many quadrilaterals satisfying this condition.

Condition D.

A parallelogram with one angle measuring 32°

There can be many quadrilaterals satisfying this condition.

Condition E.

A parallelogram with one angle measuring 48° and adjacent sides measuring 6 inches and 8 inches.

There can be only one quadrilateral satisfying this condition.

Condition F.

A rectangle with adjacent sides measuring 4 inches and 3 inches.

There can be only one quadrilateral satisfying this condition

Step-by-step explanation:

3 0
3 years ago
What is each of the expression written using each base only once? 10^10 x 10^10
aleksklad [387]
The answer is 10 thousand
4 0
3 years ago
the area of a rectangle is 95 square yard. if the perimeter is 48 yards find the length and width of rectangle
Vladimir [108]
Try this option:
according to the condition w+l=48 and w*l=95.
Using these two equation:
\left \{ {{l+w=48} \atop {lw=95}} \right. \ =\ \textgreater \  \   \left[\begin{array}{ccc}l=24- \sqrt{481};w=24+ \sqrt{481}\\ \\l=24+ \sqrt{481};w=24- \sqrt{481}\end{array}
4 0
3 years ago
Calculate the volume of the triangular pyramid below.
eduard

Answer:

90

Step-by-step explanation:

1/2x10x3.6=18

18x5=90

8 0
3 years ago
in a standard casino dice game the roller wins on the first roll if he rolls a sum of 7 or 11. what is the probability of winnin
Alborosie

<u>Answer-</u>

<em>The probability of winning on the first roll is </em><em>0.22</em>

<u>Solution-</u>

As in the game of casino, two dice are rolled simultaneously.

So the sample space would be,

|S|=6^2=36

Let E be the event such that the sum of two numbers are 7, so

E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

|E|=56

\therefore P(E)=\dfrac{|E|}{|S|}=\dfrac{6}{36}

Let F be the event such that the sum of two numbers are 11, so

F = {(6,5), (5,6)}

|F|=2

\therefore P(F)=\dfrac{|F|}{|S|}=\dfrac{2}{36}

Now,

P(\text{sum is 7 or 11)}=P(E\ \cup\ F)=P(E)+P(F)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}=0.22

8 0
4 years ago
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