Please help...

For the function {(0,1), (1,-3), (2,-4), (-4,1)}, write the domain and range.

shepuryov [24]

Answer:

pretty sure it's B

Step-by-step explanation:

domain is x numbers and that has all the x numbers in it

4 0

3 years ago

Read 2 more answers

Historical data for a local manufacturing company show that the average number of defects per product produced is 2. In addition

charle [14.2K]

Answer:

The probability that there will be a total of 7 defects on four units is 0.14.

Step-by-step explanation:

A Poisson distribution describes the probability distribution of number of success in a specified time interval.

The probability distribution function for a Poisson distribution is:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...$

Let X = number of defects in a unit produced.

It is provided that there are, on average, 2 defects per unit produced.

Then in 4 units the number of defects is, $(2\times4)=8$ .

Compute the probability of exactly 7 defects in 4 units as follows:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14$

Thus, the probability of exactly 7 defects in 4 units is 0.14.

8 0

3 years ago

En la división sintética sólo son tomados en cuenta los coeficientes del divisor?

Airida [17]

La división sintetica sirve para el cociente entre polinomios, la cual es particularmente util para cuando tenemos un numerador de grado alto y un denominador de grado 1.

Veremos que si, en la división sintética solo son tomados en cuenta los coeficientes del divisor.

Para explicarla, usaremos un ejemplo.

x^3 - 2x^2 + 5x + 3 será el numerador

x - 2 será el denominador.

Para poder aplicar la división sintetica, el maximo exponente en el denominador debe ser uno.

Así, nuestro cociente es:

$\frac{x^3 - 2x^2 + 5x + 3}{x-2}$

Procedemos a igualar el denominador a cero:

x - 2 = 0

x = 2

Ahora veremos los coeficientes del numerador (el numerador es el divisor).

Lo que debemos hacer es tomar cada coeficiente (comenzando por los que multiplican a los terminos con mayor exponente) y multiplicarlos por este número.

En este caso, el primer coeficiente es 1.

2*1 = 2

Ahora le sumamos esto al proximo coeficiente, que es -2:

-2 + 2 = 0

Multiplicamos esto por el valor de x, que es 2.

2*0 = 0

Ahora sumamos esto al proximo coeficiente, que es 5

5 + 0 = 5

Multiplicamos esto por el valor de x, que es 2:

5*2 = 10

Sumamos esto al ultimo coeficiente, que es 3.

10 + 3 = 13

Este ultimo valor es el residuo, y los primeros 3 números que obtuvimos son los coeficientes del cociente, entonces el cociente será:

1*x^2 + 0*x + 10 = x^2 + 10

Como vimos, solo usamos los coeficientes del numerador/divisor para realizar esta operación, por lo que si, solo los coeficientes del divisor son tomados en cuenta.

Si quieres aprender más, puedes leer:

brainly.com/question/10466403

3 0

3 years ago

Angle TVZ makes a linear pair with angle ZVW so the measure of angle TVZ plus the measure of angle ZVW=180 degrees. Angle ZVW is

ss7ja [257]

Answer:

Step-by-step explanation:

2x +15 +2x +5 = 180

4x = 180 - 20

x = 160/4

x = 40

TVZ + 2x +5 =180

TVX +2(40) = 180 -5

TVX = 175 - 80

TVX = 95

6 0

3 years ago

A sample of 900900 computer chips revealed that 66f% of the chips fail in the first 10001000 hours of their use. The company's p

Alik [6]

Answer:

No, there is not enough evidence at the 0.02 level to support the manager's claim.

Step-by-step explanation:

We are given that the company's promotional literature states that 68% of the chips fail in the first 1000 hours of their use. Also, a sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use.

And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;

Null Hypothesis, $H_0$ : p = 0.68 {means that the actual percentage that fail is same as the stated percentage}

Alternate Hypothesis, $H_1$ : p 0.68 {means that the actual percentage that fail is different from the stated percentage}

The test statistics we will use here is;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, p = actual percentage of chip fail = 0.68

$\hat p$ = percentage of chip failed in a sample of 900 chips = 0.66

n = sample size = 900

So, Test statistics = $\frac{0.66 -0.68}{\sqrt{\frac{0.66(1- 0.66)}{900} } }$

= -1.267

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have sufficient evidence to accept null hypothesis.

Therefore, we conclude that the actual percentage that fail is same as the stated percentage and the manager's claim is not supported.

5 0

4 years ago