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USPshnik [31]
2 years ago
10

Type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar. chris is playing a

card game with his friends. the game uses only the four aces from a standard deck of cards. for each move, a player draws one card, replaces it, and then draws the second card. the probability that a player draws two cards of the same color is . the probability that a red ace is drawn first and then a black ace is
Mathematics
1 answer:
Paraphin [41]2 years ago
4 0

The probability that a red ace is drawn first and then a black ace is mathematically given as

x=1/4

<h3>What is the probability that a red ace is drawn first and then a black ace?</h3>

Where

Sample mean=(x,y)\, x, y \in \{h,d,c,s\}

If we want a player to draw two cards of the same color from their deck, the following combinations of cards are appropriate choices:

(h,h),\ (h,d),\ (d,h),\ (d,d),\ (c,c),\ (c,s),\ (s,c),\ (s,s)

Generally, 8 possible pairings above 16. This indicates that the likelihood of a player drawing two cards of the same color is 8/16, which is equivalent to a probability of half.

In conclusion, In a similar vein, the following pairs of cards illustrate the likelihood of obtaining a black ace after having drawn a red ace in the previous round:

(h,c),\ (h,s),\ (d,c),\ (d,s)

Giving

x=4/16

x=1/4

Read more about probability

brainly.com/question/795909

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3 years ago
What is the 60th term of the sequence 2, -3, -8, -13?
aleksandrvk [35]

Answer:

  -293

Step-by-step explanation:

The first term of the arithmetic sequence is a1 = 2. The common difference is d = -3-2 = -5. The general term is ...

  an = a1 +d(n -1)

  an = 2 -5(n -1)

The 60th term is ...

  a60 = 2 -5(60 -1) = -293

4 0
3 years ago
Solve 5 tan x = 5√(3) for 0° ≤ x ≤ 180°.<br><br> a. 60°<br> b. 150°<br> c. 30°<br> d. 120°
ladessa [460]

Answer:

\boxed{a.\:\:\:60\degree}

Step-by-step explanation:


The given trigonometric equation is


5\tan(x)=5\sqrt{3} for 0\degree \le x\le 180\degree.


We divide through by 5 to get;


\tan(x)=\sqrt{3}


We take the inverse tangent of both sides to obtain;


x=tan^{-1}(\sqrt{3})


x=60\degree


Since the tangent ratio is positive, it means x could also be in the 3rd quadrant. But this is outside the given range.


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6 0
3 years ago
Jayce looks around at an assembly. He notices his younger sister to his right and his older brother 48 feet ahead of him. If Jay
olganol [36]

Answer:

I think it's 50

Step-by-step explanation:

48

49

50

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50 8s inbetween the two

6 0
3 years ago
9. A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?
SSSSS [86.1K]

Answer:

Part 4) r=84\ units

Part 9) sin(\theta)=-\frac{\sqrt{5}}{3}

Part 10) sin(\theta)=-\frac{9\sqrt{202}}{202}

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

C=2\pi r

using proportion

\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}

simplify

\frac{r}{180^o}=\frac{56}{120^o}

solve for r

r=\frac{56}{120^o}(180^o)

r=84\ units

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

cos^2(\theta)+sin^2(\theta)=1

we have

cos(\theta)=-\frac{2}{3}

substitute the given value

(-\frac{2}{3})^2+sin^2(\theta)=1

\frac{4}{9}+sin^2(\theta)=1

sin^2(\theta)=1-\frac{4}{9}

sin^2(\theta)=\frac{5}{9}

square root both sides

sin(\theta)=\pm\frac{\sqrt{5}}{3}

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

sin(\theta)=-\frac{\sqrt{5}}{3}

Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?    

see the attached figure to better understand the problem

In the right triangle ABC of the figure

sin(\theta)=\frac{BC}{AC}

Find the length side AC applying the Pythagorean Theorem

AC^2=AB^2+BC^2

substitute the given values

AC^2=11^2+9^2

AC^2=202

AC=\sqrt{202}\ units

so

sin(\theta)=\frac{9}{\sqrt{202}}

simplify

sin(\theta)=\frac{9\sqrt{202}}{202}

Remember that      

The point (11,-9) lies in Quadrant IV

then      

The value of sin(∅) is negative

therefore

sin(\theta)=-\frac{9\sqrt{202}}{202}

5 0
3 years ago
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