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maks197457 [2]
3 years ago
12

If someone offers you a lottery ticket that has a 25 percent chance of winning $100, a 50 percent chance of winning $200, and a

25 percent chance of winning $1,000, what is the expected value of that lottery ticket?
Mathematics
1 answer:
g100num [7]3 years ago
3 0

Answer: $375

Step-by-step explanation:

Given : The probability of winning $100 : P(100)= 0.25

The probability of winning $200 : P(200)= 0.50

The probability of winning $1000 : P(1000)= 0.25

Now, the the expected value of that lottery ticket is given by :-

P(100)\times 100+P(200)\times200+P(1000)\times1000\\\\=0.25\times100+0.50\times200+0.25\times1000\\\\=375

Hence, the the expected value of that lottery ticket =$375

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How would I solve <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20x-%5Cfrac%7B5%7D%7B3%7D%20%3D-%5Cfrac%7B1%7D%7B2%7D%
sdas [7]
Multiply both sides of the equation by 12. Move the variables to the left-hand side and change its sign. Move the constant to the left-hand side and change its sign. Collect like terms. Add the numbers. Divide both sides of the equation by 12.

1/2 x - 5= - 1/2 x + 19/4
6x -20= - 6x + 57
6x + 6x= 57 + 20
12x= 57 + 20
12x= 77
ANSWER
x = 77/12
Alternative Form .
x = 6 5/12, x= 6.416
7 0
3 years ago
Read 2 more answers
Find the minimum sample size needed to be 99% confident that the sample's variance is within 30% of the population's variance.
liq [111]

The Minimum sample size table is attached below

Answer:

X=173

Step-by-step explanation:

From the question we are told that:

Confidence Interval CI=99\%

Variance \sigma^2=30\%

Generally going through the table the

Minimum sample size is

X=173

7 0
2 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
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11Alexandr11 [23.1K]

Answer:

777 rice balls

Step-by-step explanation:

700700700 divided by 100100100 equals 7. 7 times 111 equals 777.

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3 years ago
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sineoko [7]
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