MWizard2k04

52 Cards

Number of red cards = \(13 \times 2 \\ = 26\)

Number of jacks = 4

⁵²P₂ = 2652 ways to get two cards.

Number of ways to get 2 red cards:

²⁶P₂ = 650 ways to get two red cards

Number of ways to get 2 jacks:

⁴P₂ = 12 ways

As the Jack of Hearts and Jack of Diamonds are considered two red cards, subtract one way.

We get 650 + 12 - 1 = 661 ways.

P(2 red card or 2 jacks) = \(661 \over 2652\)

                                      = 0.24925 (correct to 5 significant figures)

                                      = 24.925%

If \(7 ∣ r\), then \(π = \frac{22}{7}\).

As \(r = 6\), \(6 ∤ 7\). 

\(\therefore π = 3.14\)

\(C = 2πr \\ = 2 \times 3.14 \times 6 \\ = 6.28 \times 6 \\ = 37.68\)

TheSmartOne2.0's answers:

a) 30cm

b)90

c)1363

d)2033

e)I'm just that good

x + 2 = y + 3              | - 2

x = y + 1                    | - 1

y = x - 1

A randomiser of n bits (base 2 unsigned)

When n ≥ 2, we have 2ⁿ possible ways to store numbers.

Number of consecutive Heads:

10 heads + 90 random (1st ~ 10th try, 2nd ~ 11th try ... 91st ~ 100th try)

91 tries × 2⁹⁰ = 112 652 543 574 969 605 015 820 304 384

11 heads + 89 random (1st ~ 11th, 2nd ~ 12th ... 90th ~ 100th try)

90 tries × 2⁸⁹ = 55 707 301 767 842 112 370 460 590 080

12 heads + 88 random

89 tries × 2⁸⁸ = 27 544 165 874 099 711 116 505 513 984

...

The information above is an assumption; we do it by working backwards and finding a pattern.

97 heads + 3 random (97H, T, T, T; 97H, T, T, H; 97H, T, H, T; 97H, T, H, H  )

4 tries

98 heads + 2 random (98 heads, 1 tail, 1 head or 98 heads, 2 tails)

2 tries

99 heads + 1 tails

1 try

100 heads

1 try

                       100H   99H         98H            97H        96H            95H                  11H                  10H

Thus, we have 1 + (2¹ - 2⁰) + (2² - 2¹) + (2³ - 2²) + (2⁴ - 2³) + (2⁵ - 2⁴) + ... + (2⁸⁹ - 2⁸⁸) + (2⁹⁰ - 2⁸⁹) = 2⁹⁰ = 

                        ||

                        2⁰

1 237 940 039 285 380 274 899 124 224 ways.

Do some cancellation. Remove the brackets as there is only + outside of brackets, and a + (b - c) = a + b - c.

Thus, the probability of getting at least 10 consecutive heads is:

Percentage : \(8.0779356694631608 \times 10^{-28} %\)

Fraction : \(1 \over 1 237 940 039 285 380 274 899 124 224\)         

My Calculator

Brand: Casio

Model: fx-96SG PLUS

Type: Natural-V.P.A.M.

Digit Display: 10 digits

Calculation Input: 15 digits

Calculation Range: \(1 \times {10}^{-99} ≤ |x| < 1 \times {10}^{100} \, or \, 0\)

Let the number of meals ordered and the number of orders = M

a) $2 M + $9.50 M

= $11.50M

b) $3 M + $9.50 M 

= $12.50M

c) $3 M + $2(M-1) + $9.50 M

= $(2(M-1) + 12.50) M

d) $2 M + $9.50 M

= $11.50M 

Number of lollipops at first = \(10 \times 20 = 200\)

Number of lollipops eaten  = \(200 \times \frac{1}{3} \\ ≈ 67\)

This answer is rounded to the nearest whole number.

Let the number of games Luke bought = x.

Let the number of games Bella bought = y.

How much Luke paid = \($99.95 + $19.99 \times x = $99.95+$19.99x\)

How much Bella paid = \($125.79 + $15.99 \times y = $125.79 + $15.99y\)

If how much Luke paid is equal or lesser than how much Bella paid, then

99.95 + 19.99x ≤ 125.79 + 15.99y                | - 99.95

             19.99x ≤ 25.84 + 15.99y

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