Anyone knows the answer

What two numbers add up to -20 and multiply to -1125

Oliga [24]

I don't have my calculator handy just now. But just looking at that problem, why don't you try +20 and -45 and see if they do the job.

8 0

2 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

2 years ago

Problema de capacidades Una empresa aceitera ha envasado 3000 litros de aceite en 1200 botellas de dos y de cinco litros. ¿Cuánt

mariarad [96]

Answer:

Hay 200 botellas de 5 litros y 1000 botellas de 2 litros.

Step-by-step explanation:

Un sistema de ecuaciones lineales es un conjunto de dos o más ecuaciones de primer grado, en el cual se relacionan dos o más incógnitas.

Resolver un sistema de ecuaciones consiste en encontrar el valor de cada incógnita para que se cumplan todas las ecuaciones del sistema.

En este caso, las variables a calcular son:

x= cantidad de botellas de 2 litros.
y= cantidad de botellas de 5 litros.

Una empresa aceitera ha envasado 3000 litros de aceite en 1200 botellas de dos y de cinco litros. Entonces es posible plantear el siguiente sistema de ecuaciones:

$\left \{ {{2*x+5*y=3000} \atop {x+y=1200}} \right.$

Existen varios métodos para resolver un sistema de ecuaciones. Resolviendo por el método de sustitución, que consiste en despejar o aislar una de las incógnitas y sustituir su expresión en la otra ecuación, despejas x de la segunda ecuación:

x= 1200 - y

Sustituyendo la expresión en la primer ecuación:

2*(1200 - y) + 5*y=3000

Resolviendo se obtiene:

2*1200 - 2*y + 5*y= 3000

2400 +3*y= 3000

3*y= 3000 - 2400

3*y= 600

y= 600÷3

y= 200

Reemplazando en la expresión x= 1200 - y:

x= 1200 - y

x=1200 -200

x= 1000

<u><em>Hay 200 botellas de 5 litros y 1000 botellas de 2 litros.</em></u>

7 0

2 years ago

Starting with x=4.62 repeating express x as a fraction in simplest form

arsen [322]

$x=4.62=4+0.62=4+\dfrac{62}{100}=4+\dfrac{62:2}{100:2}=4+\dfrac{31}{50}\\\\\text{as a mixed number}\ 4\dfrac{31}{50}\\\\\text{as an improper fraction}\ \dfrac{4\cdot50+31}{50}=\dfrac{231}{50}$

5 0

2 years ago

A jar contains 4 brown marbles and 9 yellow marbles.

denis23 [38]

Ratio brown to yellow after drawing=3:8

8 0

3 years ago

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