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2 years ago

Write the expanded form of the expression. 3(4x - 6) =

Mathematics

1 answer:

Bad White [126]2 years ago

7 0

Answer:

12x - 18

Step-by-step explanation:

3 x 4x = 12x

3 x -6 = -18 (Negative and positive make a negative.)

12x - 18

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A man can dig a ditch in 3 hours and his daughter can dig he same size ditch in 5 hours. Working together, how long will it take

Marrrta [24]

I believe that this would be 4 hours

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3 years ago

3.Escribe un polinomio que cumpla con las condiciones dadas

Free_Kalibri [48]

Answer: 4x^6

Step-by-step explanation:

Para responder esta pregunta debemos conocer que es un monomio, su grado y el coeficiente.

Un monomio es un polinomio que posee 1 solo termino, el grado se refiera a la potencia a la que esta elevada la variable (en este caso usaremos la letra x)

El coeficiente es el número que está a la izquierda de una variable:

Entonces:

Monomio de grado 6 y coeficiente 4

4x^6

7 0

3 years ago

A chef boiled four eggs and put them in a basket with eight eggs that were not boiled. All of the eggs look the same. Randy sele

SashulF [63]

4/12 times 3/11 = .091 so a 9.1% chance

8 0

3 years ago

Select the correct answer.

Leokris [45]

Answer:

(x + 2) (x +4) (x^2 + 25)

Step-by-step explanation:

Just did this problem

( + 2 ) ( ^3 + 4 ^2 + 2 5 + 1 0 0 )

( + 2 ) ( + 4 ) ( ^2 + 2 5 )

5 0

3 years ago

Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal diseas

zysi [14]

Answer:

(A) 0.297

(B) 0.595

Step-by-step explanation:

Let,

H = a person who suffered from a heart attack

G = a person has the periodontal disease.

Given:

P (G|H) = 0.79, P(G|H') = 0.33 and P (H) = 0.15

Compute the probability that a person has the periodontal disease as follows:

$P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.15)+(0.33\times(1-0.15))\\=0.399$

(A)

The probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is:

$P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.15}{0.399} \\=0.29699\\\approx0.297$

Thus, the probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is 0.297.

(B)

Now if the probability of a person having a heart attack is, P (H) = 0.38.

Compute the probability that a person has the periodontal disease as follows:

$P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.38)+(0.33\times(1-0.38))\\=0.5048$

Compute the probability of a person having a heart attack given that he or she has the disease:

$P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.38}{0.5048}\\ =0.59469\\\approx0.595$

The probability of a person having a heart attack given that he or she has the disease is 0.595.

4 0

3 years ago