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3 months ago

Using the extended Euclidean algorithm, find the multiplicative inverse of a. 1234 mod 4321 b. 24140 mod 40902

Mathematics

1 answer:

AnnZ [12.3K]3 months ago

8 0

(a) The multiplicative inverse of 1234 (mod 4321) is x so that 1234*x ≡ 1 (mod 4321). We can apply Euclid's algorithm:

4321 = 1234 * 3 + 619

1234 = 619 * 1 + 615

619 = 615 * 1 + 4

615 = 4 * 153 + 3

4 = 3 * 1 + 1

Now we will express 1 as a linear combination of 4321 and 1234:

1 = 4 - 3

1 = 4 - (615 - 4 * 153) = 4 * 154 - 615

1 = 619 * 154 - 155 * (1234 - 619) = 619 * 309 - 155 * 1234

1 = (4321 - 1234 * 3) * 309 - 155 * 1234 = 4321 * 309 - 1082 * 1234

This reduces to

1 ≡ -1082 * 1234 (mod 4321)

Thus, the inverse is

-1082 ≡ 3239 (mod 4321)

(b) Since both 24140 and 40902 are even, their GCD cannot equal 1, indicating no inverse exists.

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Respuesta, explicación paso a paso:

A. Con el ejercicio anterior podemos deducir que se trata de un número de ventas en una tienda de comestibles, la frecuencia relativa para la cantidad de unidades vendidas se muestra a continuación:

unidades vendidas. frecuencia relativa. frecuencia acumulada. intervalo de números aleatorios

30. 0.16. 0.16. 0.00 <0.16

40. 0.24. 0.4. 0.16 <0.4

50. 0.3. 0.7. 0.4 <0.7

60. 0.2. 0.9. 0.7<0.9

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B. Para el siguiente punto, nos dan algunos números aleatorios y se comparan con la simulación de 10 días de ventas:

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número vendido

0.12. 30

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los dos listados se comparan de modo que cada uno tiene como resultado de la simulación

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2 months ago

One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples

babunello [11817]

Answer:

Step-by-step explanation:

Hello!

To determine whether boys excel in math classes compared to girls, two random samples were collected:

Sample 1

X₁: score achieved by a boy in calculus

n₁= 15

X[bar]₁= 82.3%

S₁= 5.6%

Sample 2

X₂: score obtained by a girl in calculus

n₂= 12

X[bar]₂= 81.2%

S₂= 6.7%

To estimate a confidence interval for the difference between the average percentages of boys and girls in calculus, it's essential that both variables come from normally distributed populations.

For utilizing a pooled variance t-test, it is also required that the population variances, though unknown, are assumed to be equal.

The confidence interval can then be calculated with:

[(X[bar]_1 - X[bar]₂) ± $t_{n_1+n_2-2;1-\alpha /2}$ * $Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ ]

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{14*(5.6^2)+11*(6.7^2)}{15+12-2} }= 6.108= 6.11$

$t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708$

[(82.3 - 81.2) ± 1.708 * (6.11 * $\sqrt{\frac{1}{15}+\frac{1}{12} }$ ]

[-2.94; 5.14]

Using a 90% confidence level, the interval [-2.94; 5.14] is expected to encompass the true difference between the average percentages achieved by boys and girls in calculus.

I hope this is of assistance!

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3 months ago

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Answer:

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3 months ago

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AnnZ [12381]

First, we determine the percentage increase for the tax.

Initial amount = 25×9 = 225 ⇒ The 9 signifies 9 lots of thousands representing the house's value.
New amount = 28×9 = 252

Tax increase = 252 - 225 = 27
Percentage increase = (27÷225) ×100 = 12%

Therefore, the yearly rent should rise by 12%

Monthly rent is $60
Yearly rent = 60×12 = $720
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Monthly rent becomes 806.4÷12 = $67.20, resulting in an additional $7.20 each month.

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2 months ago

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Answer:

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