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3 years ago

Un profesor realiza una prueba con 70 preguntas, cada una con igual valor. Un

Mathematics

1 answer:

Valentin [98]3 years ago

3 0

Tenemos un problema sobre porcentajes. Veremos que la calificación del alumno es un 80%.

En este caso sabemos que la prueba tiene 70 preguntas, por lo que la cantidad de referencia (el 100%) será 70.

Y sabemos que el estudiante responde correctamente 56 preguntas.

Lo que queremos hacer es ver que porcentaje representa este segundo número.

Para ver esto planteamos dos ecuaciones:

56 = x

70 = 100%

Para resolver esto, debemos ver el cociente entre esas ecuaciones:

56/70 = x/100%

Ahora resolvemos para x:

(56/70)*100% = x = 80%

Es decir, el alumno obtiene una clasificación del 80%.

Si quieres aprender más puedes leer:

brainly.com/question/18546712

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there are 350 red flowers in a room. There are 70 more yellow flowers than red flowers. There are 140 fewer white flowers than y

Leno4ka [110]

Red=350
yellow= 350+70=420
white=420-140=280
ratio of red to yellow to white= 350/420/280 or 35/42/28
hope this helps :)

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3 years ago

Determine whether the second polynomial is a factor of the first.

arsen [322]

Answer:

No, because the polynomial 6x² + 5x + 54 evaluated at 4 is equal to 170.

Step-by-step explanation:

A polynomial should have a factor (let's say a) that would make the polynomial equal to zero when we substitute that factor (a) into the polynomial.

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3 years ago

Please help me with translating words into equations<br> "four more than 3 times a number is 25"

schepotkina [342]

Answer:

3x + 4 = 25

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Faith is a 95% free throw shooter. At practice, each player shoots 20 free throws. Let X = the number of free throws Faith makes

nasty-shy [4]

Using the binomial distribution, it is found that:

The mean of X is of 19.

The standard deviation of X is of 0.975.

P(X = 19) = 0.3774.

There is a 0.9246 = 92.46% probability that practice ends early.

<h3>Binomial probability distribution</h3>

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

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

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.

The mean of X is:

$E(X) = np$

The standard deviation of X is:

$\sqrt{V(X)} = \sqrt{np(1 - p)}$

In this problem:

Faith is a 95% free throw shooter, hence $p = 0.95$ .

Each player shoots 20 free throws, hence $n = 20$ .

The <em>mean </em>and the <em>standard deviation</em> are:

$E(X) = np = 20(0.95) = 19$

$\sqrt{V(X)} = \sqrt{20(0.95)(0.05)} = 0.975$

P(X = 19) is given by:

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

$P(X = 19) = C_{20,19}.(0.95)^{19}.(0.05)^{1} = 0.3774$

The probability that <u>practice ends early</u> is:

$P(X \geq 18) = P(X = 18) + P(X = 19) + P(X = 20)$

In which:

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

$P(X = 18) = C_{20,18}.(0.95)^{18}.(0.05)^{2} = 0.1887$

$P(X = 19) = C_{20,19}.(0.95)^{19}.(0.05)^{1} = 0.3774$

$P(X = 20) = C_{20,20}.(0.95)^{20}.(0.05)^{0} = 0.3585$

Then:

$P(X \geq 18) = P(X = 18) + P(X = 19) + P(X = 20) = 0.1887 + 0.3774 + 0.3585 = 0.9246$

There is a 0.9246 = 92.46% probability that practice ends early.

You can learn more about the binomial distribution at brainly.com/question/24863377

4 0

3 years ago

What is the slope intercept form of the equation passing through the points (0,0) and (0,10)?

Naily [24]

This is it! remember that it is undefined because you can't divide by 0.

8 0

4 years ago