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Eva8 [605]
2 years ago
10

A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is (i)a card of s

pade or an ace (ii)a red king (iii)neither a king nor a queen (iv)either a king or queen.
Thank you.
Mathematics
1 answer:
docker41 [41]2 years ago
5 0
There are 52 cards in a standard deck so that would be the whole in the part/whole equation so
i... There are 13 spade cards and then 3 other aces since there are 4 suits of cards so 16/52 or about 0.30769 or 30.8% rounded

ii....There are 2 reds of each card in a standard deck along with 2 blacks so there would be 2 red kings which would be a 2/52 chance or about 0.0384 or 3.8% rounded

iii.... There are 4 kings and 4 queens in each deck so (52-8)/52 which is a 44/52 chance, 0.8461 about, or 84.6% rounded

iv..., As I said in iii, there are 8 kings and queens so an 8/52 chance, about 0.1538, or 15.4% rounded
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3 /8 + 2/ 3 = 9/ 24 + 16/ 24 = 25/ 24 =
Alchen [17]
Convert the denominator of the first two fractions to 24 since 3 and 8 can fit into 24.

3/8=9/24

2/3=16/24

Now add them together.

9/24+16/24=25/24

Next add the second two fractions 9/24 and 16/24.

9/24+16/24=25/24


Since all fractions equal 25/24, the equation was true!

Hope this helps!

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3 years ago
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lakkis [162]

Answer:

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Step-by-step explanation:

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In 4567 which number means ones?​
Wittaler [7]

Answer:

7 means ones because it is the units

Step-by-step explanation:

4= thousand

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Which is NOT a statistical question
Aneli [31]

Answer:

How many plates are on the table as it results in only one value with no variability

Step-by-step explanation:

8 0
3 years ago
Complete the square and write in standard form. Show all work.What would be the conic section:CircleEllipseHyperbolaParabola
mote1985 [20]

ANSWER

This is an ellipse. The equation is:

\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1

EXPLANATION

We have to complete the square for each variable. To do so, we have to take the first two terms and compare them with the perfect binomial squared formula,

(a+b)^2=a^2+2ab+b^2

For x we have to take 16x² and -32x. Since the coefficient of x is not 1, first, we have to factor out the coefficient 16,

16x^2-32x=16(x^2-2x)

Now, the first term of the expanded binomial would be x and the second term -2x. Thus, the binomial is,

(x-1)^2=x^2-2x+1

To maintain the equation, we have to subtract 1,

16(x^2-2x+1-1)=16((x-1)^2-1)=16(x-1)^2-16

Now, we replace (16x² - 32x) from the given equation by this equivalent expression,

16(x-1)^2-16+9y^2+72y+16=0

The next step is to do the same for y. We have the terms 9y² + 72y. Again, since the coefficient of y² is not 1, we factor out the coefficient 9,

9y^2+72y=9(y^2+8y)

Following the same reasoning as before, we have that the perfect binomial squared is,

(y+4)^2=y^2+8y+16

Remember to subtract the independent term to maintain the equation,

9(y^2+8y)=9(y^2+8y+16-16)=9((y+4)^2-16)=9(y+4)^2-144

And now, as we did for x, replace the two terms (9y² + 72y) with this equivalent expression in the equation,

16(x-1)^2-16+9(y+4)^2-144+16=0

Add like terms,

\begin{gathered} 16(x-1)^2+9(y+4)^2+(-16-144+16)=0 \\ 16(x-1)^2+9(y+4)^2-144=0 \end{gathered}

Add 144 to both sides,

\begin{gathered} 16(x-1)^2+9(y+4)^2-144+144=0+144 \\ 16(x-1)^2+9(y+4)^2=144 \end{gathered}

As we can see, this is the equation of an ellipse. Its standard form is,

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

So the next step is to divide both sides by 144 and also write the coefficients as fractions in the denominator,

\begin{gathered} \frac{16(x-1)^2}{144}+\frac{9(y+4)^2}{144}=\frac{144}{144} \\  \\ \frac{(x-1)^2}{\frac{144}{16}}+\frac{(y+4)^2}{\frac{144}{9}}=1 \end{gathered}

Finally, we have to write the denominators as perfect squares, so we identify the values of a and b. 144 is 12², 16 is 4² and 9 is 3²,

\frac{(x-1)^2}{(\frac{12}{4})^2}+\frac{(y+4)^2}{(\frac{12}{3})^2}=1

Note that we can simplify a and b,

\frac{12}{4}=3\text{ and }\frac{12}{3}=4

Hence, the equation of the ellipse is,

\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1

3 0
1 year ago
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