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3 years ago

5

En el acuario donde trabaja Alexandra, han requerido 4627 peces para distribuirlos en 21 peceras ¿cuantos peces se deben colocar

en casa una de las peceras? ¿ se hará necesario agregar una pecera más? ¿Cuantos peces colocarán en esa pecera?

1 answer:

brilliants [131]3 years ago

7 0

Answer: Se colocará 220 peces en cada pecera pero necesitaremos 1 pecera más para colocar los 7 peces restantes.

Step-by-step explanation:

4627/21=220.33....

21*220=4620

4627-4620=7 peces restantes en la última pecera

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Complete the statement. 469.1 mg = g

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A medical clinic is reducing the number of incoming patients by giving vaccines before flu season. During week 5 of flu season,

8_murik_8 [283]

Answer:

$f(x) = -6x +120$

Step-by-step explanation:

Let's call y the number of patients treated each week

Let's call x the week number.

If the reduction in the number of patients each week is linear then the equation that models this situation will have the following form:

$y = mx + b$

Where m is the slope of the equation and b is the intercept with the x-axis.

If we know two points on the line then we can find the values of m and b.

We know that During week 5 of flu season, the clinic saw 90 patients, then we have the point:

(5, 90)

We know that In week 10 of flu season, the clinic saw 60 patients, then we have the point:

(10, 60).

Then we can find m and b using the followings formulas:

$m=\frac{y_2-y_1}{x_2-x_1}$ and $b=y_1-mx_1$

In this case: $(x_1, y_1) = (5, 90)$ and $(x_2, y_2) = (10, 60)$

Then:

$m=\frac{60-90}{10-5}$

$m=-6$

And

$b=90-(-6)(5)$

$b=120$

Finally the function that shows the number of patients seen each week at the clinic is:

$f(x) = -6x +120$

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3 years ago

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According to government data, 20% of employed women have never been married. If 10 employed women are selected at random, what i

Ierofanga [76]

Answer:

a) $P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

b) $P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

c) For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

And replacing we got:

$P(X\geq 8)=0.0000779$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Bin (n=10 ,p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Solution to the problem

Let X the random variable "number of women that have never been married" , on this case we now that the distribution of the random variable is:

$X \sim Binom(n=10, p=0.2)$

Part a

We want to find this probability:

$P(X=2)$

And using the probability mass function we got:

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

Part b

For this case we want this probability:

$P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

We can find the individual probabilities and we got:

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

Part c

For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

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