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3 years ago

Brian rolls a fair dice and flips a fair coin.

Mathematics

1 answer:

Sergio [31]3 years ago

4 0

Answer:

The probability of obtaining a head is 1/2 or 50% chance.

The probability of obtaining a 2 is 1/6

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Which fraction and decimal forms match the long division problen?

sweet-ann [11.9K]

Answer:

B because it is a single digit number with added decimals

Step-by-step explanation:

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3 years ago

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Write an equation giving the following: m=-1, y-int =8

Karolina [17]

Answer:

y = -x + 8

Step-by-step explanation:

Equation of a line is:

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c = 8 (y-intercept)

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3 years ago

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Which transformation is not a rigid transformation?

Ivan

Answer:

A) Dilation

Step-by-step explanation:

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3 years ago

The sum of the digits is 11. When rounded to the nearest hundred, it's 500. Rounding to the nearest 10 makes it 530. What's the

kupik [55]

If rounded to the nearest 10 is 530, then this number must be between 525 and 534. The only one which digits sum up to 11 is 533.

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4 years ago

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Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago