Please solve 6^x=45 for x

El costo de producir 10 juegos de video al día es de 350 dólares , mientras que producir 30 juegos del mismo tipo al día cuestan 850 dólares . Si suponemos un modelo lineal como el descrito anteriormente , determine:

¿Cuál es el costo de producir un juego de vídeo?

(Explanation will be held in Spanish)

El modelo lineal es un polinomio de primer orden de la forma:

$y = a\cdot x + b$

Donde:

$x$ - Variable independiente (eje x - Cantidad de juegos de vídeo producidos - Adimensional)

$y$ - Variable dependiente (eje y - Coste de producción de juegos de vídeo - Dólares)

$a$ - Pendiente de la función.

$b$ - Intercepto en el eje y.

Dados los dos costes para dos cuotas distintas de producción, se construye el siguiente sistema de ecuaciones lineales:

$10\cdot a + b = 350$

$30 \cdot a + b = 850$

Se resuelve el sistema con algo de manipulación algebraica:

$(30\cdot a + b) - (10\cdot a + b) = 850 - 350$

$20 \cdot a = 500$

$a = 25$

Luego:

$b = 850 - 30\cdot a$

$b = 100$

La ecuación lineal es:

$y = 25\cdot x +100$

Finalmente, el coste de producir un juego de vídeo es:

$y = 25 \cdot (1) + 100$

$y = 125$

El coste de producir un juego de vídeo es de 125 dólares.

7 0

4 years ago

An insurance company estimates 45 percent of its claims have errors. The insurance company wants to estimate with 99 percent con

nydimaria [60]

Answer:

We need a sample size of at least 657.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

45 percent of its claims have errors.

So $\pi = 0.45$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

What sample size is needed if they wish to be within 5 percent of the actual

This is a sample size of at least n, in which n is found when M = 0.05.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.05 = 2.575\sqrt{\frac{0.45*0.55}{n}}$

$0.05\sqrt{n} = 1.28$

$\sqrt{n} = \frac{1.28}{0.05}$

$\sqrt{n} = 25.62$

$\sqrt{n}^{2} = (25.62)^{2}$

$n = 656.4$

We need a sample size of at least 657.

6 0

3 years ago

The output of a chemical process is continually monitored to ensure that the concentration remains within acceptable limits. Whe

marissa [1.9K]

Answer:

a) 0.31 = 31%

b) 0.03 = 3%

c) 0.36 = 36%

d) 2 times

Step-by-step explanation:

If $F_X(x)$ is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

$P(X \leq x) = F_X(x)$

If additionally the random variable is discrete (only has non-negative integers as outcomes as is the case in this problem) then

$P(X=a)=F_X(a)-\lim_{x \to a^-}F_X(x)$

a)

We are looking for P(X<2)

$P(X < 2) = P(X\leq 2)-P(X=2)=F_X(2)-\lim_{x \to 2^-}F_X(x)=0.84-0.53=0.31$

b)

In this case we want P(X>3)

$P(X >3) = 1-P(X\leq 3)=1-F_X(3)=1-0.97=0.03$

c)

Now, we are interested in P(X=1)

$P(X =1) =F_X(1)-\lim_{x \to 1^-}F_X(x)=0.53-0.17=0.36$

d)

The expected number of times that the process is recalibrated during the week is the expected value of the probability distribution:

P(X=1)+2P(X=2)+...+nP(X=n)+...

But it is easy to see that P(X=n) = 0 if n is an integer >4

So, the expected value is

P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)

We already have P(X=1) and P(X=2). Let's compute the rest

$P(X =3) =F_X(3)-\lim_{x \to 3^-}F_X(x)=0.97-0.84=0.13$

$P(X =4) =F_X(4)-\lim_{x \to 4^-}F_X(x)=1-0.97=0.03$

and the expected value is

0.36 + 2*0.53+3*0.13+4*0.03= 1.93 = 2 times rounding to the nearest integer.

8 0

4 years ago