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Mariana [72]
3 years ago
6

Which of the following statements is true? When rolling a fair number cube - ____________ A) the theoretical probability is alwa

ys greater than the experimental probability. B) the theoretical probability is always less than the experimental probability. C) the more times the cube is rolled the closer the experimental probability gets to theoretical probability. D) the number 7 will appear more than the
Mathematics
2 answers:
tiny-mole [99]3 years ago
7 0
I believe that the answer is C because lets say you decide that the cube will roll on 1 six times. if you roll the cube 6 times, then there is a 1/6 probability of the cube landing on 1, but the more times you roll the cube, the more chances there will be of the number landing that many times as you theorized 
Softa [21]3 years ago
3 0
The answer is c hope that helps
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just olya [345]
The empty or null set.
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A fair, twenty-faced die has $19$ of its faces numbered from $1$ through $19$ and has one blank face. another fair, twenty-faced
lawyer [7]
Represent 24 as a sum of two numbers, first number from first die and second number from second die.
24=19+5=18+6=17+7=16+8=14+10=13+11=12+12=11+13=10+14=9+15=8+16=7+17=6+18=5+19=4+20 (the sums 15+9 and 20+4 are absent, because there aren't numbers: 20 on the first die and number 9 on the second die). Totally, you receive 15 different representations of 24.
<span>The probability that the sum of the two numbers facing up will be 24 is
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</span><span>P= \frac{15}{20\cdot 20} = \frac{3}{80} (here 20\cdot 20 means that you have 20 possibilities to roll first number or blank face on the first die and 20 possibilities to roll number or blank face on the second die).
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8 0
3 years ago
Solve the equation. (find x)​
Irina-Kira [14]

Answer:

Step-by-step explanation:

(x^2-3x+1)^{(2x^2+x-6)}=1\ we \ take \ the \ logarithm\\\\(2x^2+x-6)*ln(x^2-3x+1)=0\\\\(2x^2+x-6)=0\ or\ x^2-3x+1=1\\\\1)\\2x^2+x-6=0\\2x^2+4x-3x-6=0\\2x(x+2)-3(x+2)=0\\(x+2)(2x-3)\\x=-2\ or\ x=\dfrac{3}{2} \\\\2)\\x^2-3x=0\\x(x-3)=0\\x=0\ or\ x=3\\\\sol=\{-2,0,\dfrac{3}{2},3\}\\

8 0
3 years ago
Rewrite in simplest radical form square root of x times the fourth root of x. Show each step of your process.
Ipatiy [6.2K]

Answer:

\sqrt[4] {x^3}

Step-by-step explanation:

At this point, we can transform the square root into a fourth root by squaring the argument, and bring into the other root:

\sqrt x \cdot \sqrt[4] x =\sqrt [4] {x^2} \cdot \sqrt[4] x = \sqrt[4]{x^2\cdot x} = \sqrt[4] {x^3}

Alternatively, if you're allowed to use rational exponents, we can convert everything:

\sqrt x \cdot \sqrt[4] x = x^{\frac12} \cdot x^\frac14 = x^{\frac12 +\frac14}= x^{\frac24 +\frac14}= x^\frac34 = \sqrt[4] {x^3}

5 0
2 years ago
Find the product please i need it ?
Ipatiy [6.2K]

Answer:a2=5 , a1=1 , a0=3, a5=2 , a4=5 , a3=2 , a6=0 , a10=5 , a9=3 , a8=8 , a7=2

Step-by-step explanation:

63 x 45=2835

5 0
3 years ago
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