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sergeinik [125]
3 years ago
8

En una multiplicación, si un factor es un número natural y el otro es un número entero negativo, el producto es siempre menor qu

e cada uno de los factores. Es verdadero o falso
Mathematics
1 answer:
forsale [732]3 years ago
3 0

Answer:

Falso.

Step-by-step explanation:

Acá tenemos la proposición:

"En una multiplicación, si un factor es un número natural y el otro es un número entero negativo, el producto es siempre menor que cada uno de los factores."

Primero tratemos de demostrar que esto es falso, para ello debemos encontrar un solo ejemplo en el que la proposición sea falsa.

Elijamos al número 1 como el número natural,

Elijamos -10 como el número entero negativo.

El producto es:

1*-10 = -10

Ahora veamos si el producto es menor que cada uno de los factores.

-10 < 1 ?

Si, -10 es menor que 1.

Ahora veamos con el otro factor:

-10 < - 10?

No, un número no puede ser menor que si mismo.

Entonces el producto no siempre es menor que cada uno de los factores.

Entonces la proposición es falsa.

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Answer:

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Step-by-step explanation:

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Since all of them have the denominator as 100,

the total no. of sweets must be 100. So,

No of strawberries= 100-15-40-35

= 10

p(taking a strawberry sweet)= 10/100

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3 years ago
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A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o
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Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is  p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑_{x=0 b( x, n, p )

= ³∑_{x=0 b( x, 10, 0.6 )

= ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.6)^x( 1 - 0.6 )^{10-x

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.3 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.4 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.5 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

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Answer:

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Option C is correct answer.

Step-by-step explanation:

We need to find Which expression belongs in the box?

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The property applied here is associative property of multiplication.

Associative property of multiplication states that:

(a \times b)\times c = a \times (b \times c)

So, if we apply the property to our question we have:

a = 3

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So, the expression will become: (3 \times 8)  \times 4= 3 \times (8 \times 4)

So, the expression that belongs in the box is 8 x 4

Option C is correct answer.

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