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3 years ago

Find the number of all the 2-digit numbers satisfying the

Mathematics

1 answer:

Alona [7]3 years ago

5 0

Use the Chinese remainder theorem.

Start with x = 3×5 + 2×7 = 15 + 14 = 29. Now,

• 29 ≡ 15 ≡ 1 (mod7)

• 29 ≡ 14 ≡ 4 (mod5)

Adjust for this by multiplying the first term in x by 3, and the second term by 3 (because 4×3 ≡ 12 ≡ 2 (mod5)).

So now x = 3×5×3 + 2×7×3 = 45 + 42 = 87, and

• 87 ≡ 45 ≡ 3 (mod7)

• 87 ≡ 42 ≡ 2 (mod5)

The CRT then says that x ≡ 87 (mod(7×5)) ≡ 87 (mod35), which is to say any number x = 87 + 35n satisfies both congruences (where n is any integer).

So there are 3 possible 2-digit numbers that work: {17, 52, 87}.

To confirm:

• 17 ≡ 15 + 2 ≡ 2 (mod5) and 17 ≡ 14 + 3 ≡ 3 (mod7)

• 52 ≡ 50 + 2 ≡ 2 (mod5) and 52 ≡ 49 + 3 ≡ 3 (mod7)

• 87 ≡ 85 + 2 ≡ 2 (mod5) and 87 ≡ 84 + 3 ≡ 3 (mod7)

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A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300. The

zubka84 [21]

Answer:

the variance of the refund payment to the couple = 9463.394

Step-by-step explanation:

Given that :

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300.

It is possible that the couple won't be able to refund up 100 per day or more than 100 per day.

SO; let assume that the refund payment happens to be 0, 100,200, 300

Let X be the total refund payment on the seven day cruise.

We can say X = 0, if there is no rain on all 7 days.

$P(X = 0) = _nC_x * P^x * (1 - P)n-x$

$P(X = 0) = _7C_o * 0.2^0 * (1-0.2)^{7-0$

$P(X = 0) =1 * 1* (1-0.2)^{7$

$P(X = 0) =(0.8)^{7$

$P(X = 0) =0.2097152$

If it rains on any one day; then X = 100

$P(X = 100) = _nC_x * P^x * (1 - P)n-x$

$P(X = 100) = _7C_1 * 0.2^1 * (1-0.2)^{7-1$

$P(X = 0) =7 * 0.2* (1-0.2)^{6$

$P(X = 100) =7* 0.2* (0.8)^{6$

$P(X = 100) =0.3670016$

if it rains on any two day ; then X = 200

$P(X = 200) = _nC_x * P^x * (1 - P)n-x$

$P(X = 200) = _7C_2 * 0.2^2 * (1-0.2)^{7-2$

$P(X = 200) = 21 * 0.2^2 * (0.8)^{5$

$P(X = 200) = 0.2752512$

if it rains on any three day or more than that ; then X = 300

$P(X \ge 300) = 1 - P(X < 300) \\ \\ P(X \ge 300) = 1 - [P(X = 0) + P(X = 100) + P(X = 200)] \\ \\ P(X \ge 300) = 1 - [0.2097152 + 0.3670016 + 0.2752512] \\ \\ P(X \ge 300) = 0.148032$

Now; we have our probability distribution function as:

P(X = 0) = 0.2097152

P(X = 100) = 0.3670016

P(X = 200) = 0.2752512

P(X = 300) = 0.148032

In order to determine the variance of the refund payment to the couple; we use the formula:

variance of the refund payment to the couple $[Var X] =E [X^2] - (E [X])^2$

where;

$E[X^2] = \sum x^2 \times p \\ \\ E[X^2] = 0^2 * 0.2097152 + 100^2 * 0.3670016 + 200^2 * 0.2752512 + 300^2 * 0.148032 \\ \\ E[X^2] = 0 + 3670.016 + 11010.048+ 13322.88 \\ \\ E[X^2] =28002.944$

$(E [X]) = \sum x * p\\ \\ (E [X]) = 0 * 0.2097152 + 100 * 0.3670016 + 200 * 0.2752512 + 300 * 0.148032 \\ \\ (E [X]) = 0 + 36.70016 + 55.05024 + 44.4096\\ \\ (E [X]) = 136.16 \\ \\ (E [X])^2 = 136.16^2 \\ \\ (E [X])^2 = 18539.55$