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irga5000 [103]
3 years ago
14

If you roll the cube 120 times, how factors of 6?

Mathematics
2 answers:
Olegator [25]3 years ago
5 0

Answer:

120/6=20

Step-by-step explanation:

eduard3 years ago
4 0

Answer:

20

Step-by-step explanation:

You would get a statistical distribution. As given in another answer, the expected value is 20. By the central limit theorem, it would approach a normal distribution with expected value of 20 (nP) and variance of 16.67 (nP(1-P)) or standard deviation 4.08.

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I need help on this one
Dominik [7]

Answer:

5.9625

Step-by-step explanation:


6 0
3 years ago
Which one of the quotients is correct? A) 876 ÷ 43 = 20r17B) 876 ÷ 43 = 20r16C) 876 ÷ 43 = 20r7D) 876 ÷ 43 = 20r6
Olegator [25]

To check which of the quotients is correct, multiply 43 times 20 and add the remainder. The result must be equal to 876.

First, notice that 20 times 43 equals 860.

A)

The remainder is 17. 860 + 17 = 877, which is not equal to 876.

B)

The remainder is 16. 860 + 16 = 876, which is equal to 876.

C)

The remander is 7. 860 + 7 = 867, which is not equal to 876.

D)

The remainder is 6. 860 + 6 = 866, which is not equal to 876.

Since 20*43 + 16 = 876, then the correct quotient is shown in option B:

876 ÷ 43 = 20 r 16

4 0
1 year ago
3. The three angles of a triangle measure x°, y° and 46°. Which of the following statements must be true?
Brrunno [24]

Answer:es ist die d

Step-by-step explanation:weil

3 0
3 years ago
Read 2 more answers
Given g(x) = x2-8x-20, which statement is true?
sveticcg [70]

Answer:C

Step-by-step explanation:

3 0
2 years ago
3-2i/1+4i include each step necessary in simplifying
Likurg_2 [28]

Answer:

\frac{3-2i}{1+4i}=\frac{11}{17}-\frac{10}{17}i

Step-by-step explanation:

\frac{3-2i}{1+4i} <-- Given

\frac{3-2i}{1+4i}*\frac{1-4i}{1-4i} <-- Multiply by the conjugate of the denominator as a factor of 1

\frac{(3-2i)(1-4i)}{(1+4i)(1-4i)}

\frac{(3)(1)+3(-4i)-2i(1)-2i(4i)}{(1)(1)+1(-4i)+4i(1)+4i(-4i)} <-- Use the Distributive Property and FOIL

\frac{3-12i-2i-8i^2}{1-4i+4i-16i^2}

\frac{3-10i-8i^2}{1-16i^2}

\frac{3-10i-8(-1)}{1-16(-1)} <-- Rewrite i^2 as -1

\frac{3-10i+8}{1+16}

\frac{11-10i}{17}

\frac{11}{17}-\frac{10}{17}i <-- Rewrite in a\pm bi form

8 0
3 years ago
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