Your answer would 72.6 inches. The way you solved the answer is by adding all the numbers up, then dividing by how many ever numbers you have. Then, you’ll have your answer.
Answer:
Brother's money = $6
Money I have = $18
Step-by-step explanation:
Hi, to answer this question we have to write an equation with the information given:
<em>I have 3 times as much money as my brother:
</em>
Money I have = 3x
Where x is the money that my brother has:
Brother's money = x
<em>Our total is $24
</em>
So, 3x+x =24
Solving for x
3x+x=24
4x =24
x =24/4
x = 6
Replacing with the value of x.
Brother’s money = x = $6
Money i have = 3x =3(6) = $18
1. 7 : 07
2. 9 : 45
3. 6
4. 2 : 25
5. 8
6. 17
Assuming that your question is (x-1)(x+2) = [5(x-1)]/x-1
On the right side, the x-1's will cancel out, leaving you with (x-1)(x+2) = 5
expand the left side, giving you x^2 + x -2 = 5
which goes to x^2 +x -7 = 0
the possible values for x are 2.93 and -3.93. I don't think this was your question, so I'll do the other possible question that you might have been asking.
(x-1)(x+2) = 5(x-1)
divide by x-1 on both sides, leaving you with x+2=5
x+2=5
x=5-3
x=3
La opinión correcta es la de Camila que argumenta que la frecuencia relativa de la cara con el #6 será un valor cercano a 0,166.
<h3>Qué es la frecuencia relativa?</h3>
La frecuencia relativa es un término matemático que se utiliza en la estadística para referirse al número de veces que un evento se repite durante un experimento.
Por otra parte, es necesario aclara que la frecuencia relativa no se modifica en gran medida si se aumenta el número de veces de una prueba, debido a que la probabilidad de ocurrencia de este evento va a ser la misma.
De acuerdo a lo anterior, Camila tiene razón debido a que considera que el valor de la frecuencia relativa no se modifica en gran medida entre 300 o 600 lanzamientos de los dados.
Nota: Esta pregunta está incompleta porque no está la tabla. No obstante, la puedo responder basado en mi conocimiento previo.
Aprenda más sobre probabilidad en: brainly.com/question/16019390
