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Serggg [28]
2 years ago
7

Chuck-A-Luck: To win any money, at least two of your dice must show the same number. What's the likelihood of winning anything a

fter one roll?
Mathematics
1 answer:
kondaur [170]2 years ago
5 0

Answer:

0.2778

Step-by-step explanation:

Each sides in a fair die has a probability of = 1/6

The probability of getting the same number on both dice is 6 from 36 outcomes, that is;

= 6/36 = 1/6

the probability of get same number on both dice after one roll = 1/6

therefore = 1/6 x 1/6 = 1/36 = 0.2778

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Move the decimal point in 250,000 to The left as many places as necessary to. Find a number that is greater than or equal to 1 a
kotykmax [81]
Your answer is 2.50 (And the rest of the 0s, but they aren't needed...) Since that is greater then 1, but less than 10. 
8 0
3 years ago
Read 2 more answers
Please help!!!!!!!!!!
defon

Answer: 90 + 4, third choice   and    90 - -4, last choice

Her score is obviously higher when the mistake is fixed, adding the 4 points gives 94.

3 0
2 years ago
Find of the areas of the squares PQRSto that of ABCD where PQ=9cm and AB=5cm​
Ket [755]

Answer: Area of square PQRS is 81cm^2 and of ABCD is 25cm^2

You know its a square. Which means all the sides are equal and same length. If you dont know what's a square then here is the definition:

Square is a plane figure with four equal sides and four right (90°) angles.

So now we know both of are squares, in case of the first square, PQRS, one side length is already given which is PQ=9cm. Now one side is 9cm which means all the sides are 9cm's (PQ=9cm, QR=9cm, RS=9cm,SP=9cm). So to find area just multiply them:

9cm * 9cm = 81cm^2

Now you found the area of the first square. So for the second square, ABCD, one side length is 5 cm so all the rest three sides are also 5 cm. Again, multiply them in order to find the area:

5cm * 5cm = 25cm^2

So square PQRS is 81cm^2 to that of square ABCD is 25cm^2

CAUTIONS: IT IS NOT THAT HARD YA KNOW

By the way, hope it helps! ^^

7 0
2 years ago
Find the solution of the following equation whose argument is strictly between 270^\circ270 ∘ 270, degree and 360^\circ360 ∘ 360
Natasha2012 [34]

\rightarrow z^4=-625\\\\\rightarrow z=(-625+0i)^{\frac{1}{4}}\\\\\rightarrow x+iy=(-625+0i)^{\frac{1}{4}}\\\\ x=r \cos A\\\\y=r \sin A\\\\r \cos A=-625\\\\ r \sin A=0\\\\x^2+y^2=625^{2}\\\\r^2=625^{2}\\\\|r|=625\\\\ \tan A=\frac{0}{-625}\\\\ \tan A=0\\\\ A=\pi\\\\\rightarrow z= [625(\cos (2k \pi+pi) +i \sin (2k\pi+ \pi)]^{\frac{1}{4}}\\\\k=0,1,2,3,4,....\\\\\rightarrow z=(625)^{\frac{1}{4}}[\cos \frac{(2k \pi+pi)}{4} +i \sin \frac{(2k\pi+ \pi)}{4}]

\rightarrow z_{0}=(625)^{\frac{1}{4}}[\cos \frac{pi}{4} +i \sin \frac{\pi)}{4}]\\\\\rightarrow z_{1}=(625)^{\frac{1}{4}}[\cos \frac{3\pi}{4} +i \sin \frac{3\pi}{4}]\\\\ \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]\\\\ \rightarrow z_{3}=(625)^{\frac{1}{4}}[\cos \frac{7\pi}{4} +i \sin \frac{7\pi}{4}]

Argument of Complex number

Z=x+iy , is given by

If, x>0, y>0, Angle lies in first Quadrant.

If, x<0, y>0, Angle lies in Second Quadrant.

If, x<0, y<0, Angle lies in third Quadrant.

If, x>0, y<0, Angle lies in fourth Quadrant.

We have to find those roots among four roots whose argument is between 270° and 360°.So, that root is

   \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]

5 0
3 years ago
[6x4(15/5)[+[2^2+(1x-5)]
Vikki [24]

Answer:

71

Step-by-step explanation:

[6x4(15/5)]+[2^2+(1x-5)]

BIDMAS (aka PEDMAS etc.)

Brackets

Indices

Division/Multiplication

Addition/Subtraction

[6 x 4(15/5)] + [2^2 + (1x-5)]

[6 x 4(3)] + [2^2 + (-5)]

[6 x 4(3)] + [4 + -5]

[6 x 12] + [4 + -5]

[72] + [4 + -5]

72 + [-1]

72 + - 1

72 - 1

71

6 0
3 years ago
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